├── src ├── .gitignore └── Tables.hs ├── .gitignore ├── chapter 4 ├── section4-1.pdf ├── prog.hs ├── section4-1.tex ├── section4-2.tex ├── section4-3.tex └── section4-5.tex ├── chapter 3 ├── factor.hs ├── section3.2.tex └── section3.4.tex ├── README.md ├── chapter 2 ├── section2-3.md ├── section2-2.md ├── section2-1.md └── section2-3.tex ├── introduction └── chapter0.md ├── chapter 6 ├── section6-4.tex ├── section6-1.tex └── section6-2.tex ├── chapter 1 ├── section1-3.md ├── chapter1-1.md ├── section1-4.md ├── section1-5.md └── chapter1-2.md └── chapter 5 └── section5-3.tex /src/.gitignore: -------------------------------------------------------------------------------- 1 | *hi 2 | *o 3 | Table.txt 4 | Tables 5 | 6 | -------------------------------------------------------------------------------- /.gitignore: -------------------------------------------------------------------------------- 1 | *aux 2 | *dvi 3 | *log 4 | *pdf 5 | *fmt 6 | *prv 7 | _region_.tex 8 | -------------------------------------------------------------------------------- /chapter 4/section4-1.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/psibi/how-to-prove/HEAD/chapter 4/section4-1.pdf -------------------------------------------------------------------------------- /chapter 4/prog.hs: -------------------------------------------------------------------------------- 1 | import Data.List (nub) 2 | 3 | atLeastOne :: String -> String -> Bool 4 | atLeastOne x y = length (x' ++ y') /= length (nub (x' ++ y')) 5 | where x' = nub x 6 | y' = nub y 7 | 8 | allCom :: [a] -> [(a,a)] 9 | allCom xs = [(x,y) | x <- xs, y <- xs] 10 | 11 | problem2 :: [(String, String)] 12 | problem2 = filter (\(x,y) -> atLeastOne x y) r 13 | where r = allCom ["cat", "dog", "bird", "rat"] 14 | -------------------------------------------------------------------------------- /chapter 3/factor.hs: -------------------------------------------------------------------------------- 1 | findFactors :: Integral a => a -> [a] 2 | findFactors num = take 1 $ filter (\x -> num `mod` x == 0 && x /= num) [2..num] 3 | 4 | hasFactor :: Integral a => a -> Bool 5 | hasFactor x = if (null (findFactors x)) 6 | then False 7 | else True 8 | 9 | isPrime :: Integral a => a -> Bool 10 | isPrime x = not $ hasFactor x 11 | 12 | problem3 :: Num a => a -> a 13 | problem3 n = 2*n + 13 14 | 15 | primeProblem3 :: [Integer] 16 | primeProblem3 = take 1 $ filter isPrime $ map problem3 [3..] 17 | -------------------------------------------------------------------------------- /src/Tables.hs: -------------------------------------------------------------------------------- 1 | import Data.Logic.Propositional 2 | import System.IO 3 | 4 | p = Variable (Var 'p') 5 | q = Variable (Var 'q') 6 | 7 | exp1 = (Negation p) `Disjunction` q 8 | exp1T = truthTable exp1 9 | 10 | --(S ∨ G) ∧ (¬S ∨ ¬G). 11 | 12 | s = Variable (Var 's') 13 | g = Variable (Var 'g') 14 | 15 | exp2 = (s `Disjunction` g ) `Conjunction` (Negation s `Disjunction` Negation g) 16 | exp2T = truthTable exp2 17 | 18 | -- ¬[P ∧ (Q ∨ ¬P)] 19 | 20 | exp3 = Negation $ p `Conjunction` (q `Disjunction` (Negation p)) 21 | exp3T = truthTable exp3 22 | 23 | -- (P ∨ Q) ∧ (¬P ∨ R) 24 | r = Variable (Var 'r') 25 | exp4 = (p `Disjunction` q) `Conjunction` (Negation p `Disjunction` r) 26 | exp4T = truthTable exp4 27 | 28 | main = writeFile "/home/sibi/github/prove/src/Table.txt" exp4T 29 | 30 | -------------------------------------------------------------------------------- /README.md: -------------------------------------------------------------------------------- 1 | How to Prove It: A Structured Approach 2 | ------------ 3 | 4 | Contains solution for the Velleman's book. 5 | 6 | The reason I have started studying this is to ultimately study 7 | [type theory](http://en.wikipedia.org/wiki/Type_theory). 8 | 9 | Feel free to raise 10 | [issue](https://github.com/psibi/how-to-prove/issues) if you think a 11 | proof is wrong or if it needs some clarification. Pull requests and 12 | contributions are welcome. 13 | 14 | Credits 15 | -------- 16 | 17 | * [David Cantrell](https://github.com/davecan) for Section 3.3 and 18 | various other improvements. 19 | 20 | Notes: 21 | ------- 22 | 23 | From Chapter 2, I have moved from markdown format to Latex as it helps 24 | in much easier rendering of mathematical symbols. It can be compiled 25 | using `pdflatex`: 26 | 27 | `pdflatex tex_filename` 28 | 29 | Although I did solve Chapter 3 problems, I haven't uploaded all of 30 | them yet because of my laziness. 31 | -------------------------------------------------------------------------------- /chapter 2/section2-3.md: -------------------------------------------------------------------------------- 1 | More operations on Set 2 | ----------------------- 3 | 4 | Exercise 1 5 | ----------- 6 | 7 | Analyze the logical forms of the following statements. You may use the 8 | symbols ∈, ∈, 9 | / =, =, ∧, ∨, →, ↔, ∀, and ∃ in your answers, but not 10 | ⊆, ⊆, P , ∩, ∪, \, {, }, or ¬. (Thus, you must write out the definitions 11 | of some set theory notation, and you must use equivalences to get rid of 12 | any occurrences of ¬.) 13 | (a) F ⊆ P (A). 14 | (b) A ⊆ {2n + 1 | n ∈ N}. 15 | (c) {n 2 + n + 1 | n ∈ N} ⊆ {2n + 1 | n ∈ N}. 16 | (d) P (∪ i∈I A i ) ⊆ ∪ i∈I P (A i ). 17 | 18 | (a) 19 | 20 | * `F ⊆ P (A)` 21 | * `∀x (x ∈ F -> x ∈ P(A))` 22 | * `∀x (x ∈ F -> x ∈ P(A))` 23 | * `∀x (x ∈ F -> x ∈ P(A))` 24 | * `∀x (x ∈ F -> x ⊆ A)` 25 | * `∀x (x ∈ F -> ∀y(y ∈ x -> y ∈ A))` 26 | 27 | (b) 28 | 29 | * `A ⊆ {2n + 1 | n ∈ N}` 30 | * `∀x (x ∈ A -> x ∈ {2n + 1 | n ∈ N})` 31 | * `∀x (x ∈ A -> ∃n ∈ N(x = 2n + 1))` 32 | 33 | (c) 34 | 35 | * `{n^2 + n + 1 | n ∈ N} ⊆ {2n + 1 | n ∈ N}` 36 | * `∀x (x ∈ {n^2 + n + 1 | n ∈ N} -> x ∈ {2n + 1 | n ∈ N})` 37 | * `∀x (∃n ∈ N(x = n^2 + n + 1) -> ∃n ∈ N(x = 2n + 1))` 38 | 39 | (d) 40 | 41 | * `¬(P(∪i∈IAi) ⊆ ∪i∈IP(Ai))` 42 | 43 | -------------------------------------------------------------------------------- /introduction/chapter0.md: -------------------------------------------------------------------------------- 1 | Introduction 2 | ------------ 3 | ------------ 4 | 5 | Exercise 1) a) 6 | -------------- 7 | 8 | If n is not prime, then 2^n - 1 is also not a prime. 9 | 10 | 2^15 - 1 = 32,767 11 | 12 | Factors of 15 = 5 * 3 13 | In the proof of Conjecture 2, they use two numbers x & y. 14 | x = 2 ^ b - 1 15 | 16 | So, one of the factors of 32,767 is 2 ^ 5 - 1 = 31 17 | And the other factor is 32767 / 31 = 1057 18 | 19 | Exercise 1) b) 20 | -------------- 21 | 22 | Find an integer x such that 1 < x < 2^32767 - 1 and 2^32767 - 1 is 23 | divisible by x. 24 | 25 | So, 2^32767 - 1 should be divisible by x. 26 | 27 | We know that 32767 is not a prime (from Ex 1a), so 2^32767 - 1 is also 28 | not a prime and hence has factors. 29 | 30 | Factors of 32767: 1057, 31 31 | 32 | So, from the proof of Conjecture 2, one of the factors of 2^32767 - 1 33 | should be 2^b - 1 34 | 35 | So, one factor of 2^32767 - 1 is 2^1057 - 1 = a big number. 36 | Another factor of 2^32767 - 1 is 2^31 - 1 = 2147483647 37 | 38 | Exercise 2) 39 | ------------ 40 | 41 | Since they are asking for conjecture, I can frame any bullshit 42 | arguments here. But I will try to be reasonable. :P 43 | 44 | Ok, now I have written a 45 | [program](https://github.com/psibi/rwh/blob/bb424ba002c5aef0b5edd105f98d1825446d066c/misc/Factor.hs) 46 | for generating data from 1 to 10, and check whether they are prime for 47 | each of our test case. 48 | 49 | The first column indicates number, the second one indicates 3^n -1 and 50 | the third one indicates 3^n - 2^n. 51 | 52 | * (2,False,True) 53 | * (3,False,True) 54 | * (4,False,False) 55 | * (5,False,True) 56 | * (6,False,False) 57 | * (7,False,False) 58 | * (8,False,False) 59 | * (9,False,False) 60 | * (10,False,False) 61 | 62 | Well, so looking at the second column it seems I can come up with some 63 | sensible conjecture: 64 | 65 | Sibi's (Too much, huh?) conjecture: 3^n - 1 is never a prime number for 66 | any positive integer greater than 1. 67 | 68 | For third column, I can't see any pattern there. Still I can produce 69 | some idiotic conjecture: For all numbers n greater than 5, 3^n - 2^n is 70 | not a prime number. 71 | 72 | Exercise 3) 73 | ----------- 74 | 75 | The proof of Theorem 3 gives a method for finding a prime number different 76 | from any in a given list of prime numbers. 77 | (a) Use this method to find a prime different from 2, 3, 5, and 7. 78 | (b) Use this method to find a prime different from 2, 5, and 11. 79 | 80 | The way to solve this problem is to understand Theorem 3 (Euler's 81 | Theorem ?). 82 | 83 | (a) 2.3.5.7 + 1 = 211 84 | 85 | (b) 2.3.5.7.11 + 1 = 2311 86 | 87 | Or 2 + 1 = 3 88 | 89 | 2.3 + 1 = 7 90 | 91 | They are giving redundant information (5,11) in the question to 92 | confuse you! 93 | 94 | Exercise 4 95 | ----------- 96 | 97 | Find five consecutive integers that are not prime. 98 | 99 | We need to use Theorem 4 for solving this problem. 100 | 101 | Let n = 5. By Theorem 4, 5 numbers from (n+1)! + 2 will not be prime. 102 | 103 | They are: 722, 723, 724, 72, 726 104 | 105 | Exercise 5 106 | ----------- 107 | 108 | Use the table in Figure 1 and the discussion on p. 5 to find two more perfect 109 | numbers. 110 | 111 | Two perfect numbers are already given in the book: 6,28. 112 | 113 | Using Euclid's proof on p. 5 that if 2 ^ n - 1 is prime, then (2 ^ (n - 1)) (2 ^ n - 1) is perfect, we can see that n = 2 and n = 3 gives us the 6 and 28. 114 | 115 | Using the other 2 values of n for which 2 ^ n - 1 is prime gives us two more perfect numbers. 116 | 117 | For n = 5 and n = 7 : 118 | 119 | (2 ^ (5 - 1)) (2 ^ 5 - 1) = 496 120 | 121 | (2 ^ (7 - 1)) (2 ^ 7 - 1) = 8128 122 | 123 | 124 | Exercise 6 125 | ----------- 126 | 127 | The sequence 3, 5, 7 is a list of three prime numbers such that each pair of 128 | adjacent numbers in the list differ by two. Are there any more such “triplet 129 | primes”? 130 | 131 | I don't know, it seems unlikely though. Once again I will write a 132 | [program](https://github.com/psibi/rwh/blob/78a5676662b7ecc3b2a01bdeb326986bb4d496cb/misc/Factor.hs#L35) 133 | to check this stuff. I checked it from the first 2000 numbers and I 134 | wans't able to find any triplet primes, Sorry! 135 | -------------------------------------------------------------------------------- /chapter 6/section6-4.tex: -------------------------------------------------------------------------------- 1 | %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% 2 | % Author: Sibi 3 | %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% 4 | \documentclass{article} 5 | \usepackage{graphicx} 6 | \usepackage{verbatim} 7 | \usepackage{amsmath} 8 | \usepackage{amsfonts} 9 | \usepackage{amssymb} 10 | \usepackage{tabularx} 11 | \usepackage{mathtools} 12 | \newcommand{\BigO}[1]{\ensuremath{\operatorname{O}\bigl(#1\bigr)}} 13 | \setlength\parskip{\baselineskip} 14 | \begin{document} 15 | \title{Chapter 6 (Section 6.2)} 16 | \author{Sibi} 17 | \date{\today} 18 | \maketitle 19 | 20 | % See here: http://tex.stackexchange.com/a/43009/69223 21 | \DeclarePairedDelimiter\abs{\lvert}{\rvert}% 22 | \DeclarePairedDelimiter\norm{\lVert}{\rVert}% 23 | 24 | % Swap the definition of \abs* and \norm*, so that \abs 25 | % and \norm resizes the size of the brackets, and the 26 | % starred version does not. 27 | \makeatletter 28 | \let\oldabs\abs 29 | \def\abs{\@ifstar{\oldabs}{\oldabs*}} 30 | % 31 | \let\oldnorm\norm 32 | \def\norm{\@ifstar{\oldnorm}{\oldnorm*}} 33 | \makeatother 34 | \newpage 35 | 36 | \section{Solution 1} 37 | 38 | \subsection{Solution (a)} 39 | 40 | ($\Rightarrow$) Suppose $\forall n Q(n)$. Let $n$ be an arbitrary element. Since $Q(n+1)$ is true, it follows $\forall m < n + 1(P(m))$. Let $m = n$. Since $n < n + 1$, it follows that $P(n)$. Since $n$ was arbitrary it follows that $\forall n P(n)$. 41 | 42 | ($\Leftarrow$) Suppose $\forall n P(n)$. Then it follows that $\forall k < n P(k)$. So, $Q(n)$. 43 | 44 | \subsection{Solution (b)} 45 | 46 | By mathematical induction on $n$, 47 | 48 | Base case. $n = 0$. $\forall k < n P(k)$ is vacously true. 49 | Induction step. Suppose $Q(n)$. So, $\forall k < n P(k)$. Since $k < n + 1$, it follows that $\forall k < n + 1 P(k)$. Now $\forall k < n + 1 P(k)$ is equivalent to $Q(n + 1)$. 50 | 51 | \section{Solution 2} 52 | $P(q) = \forall q \in N (q > 0 \implies \neg \exists p \in N(p/q = \sqrt{2}))$ 53 | 54 | Let $q$ be an arbitrary element in $N$. Suppose $\forall k < q(P(k))$. Suppose $q > 0$. Let us prove by contradiction. There exists some $p \in N$ such that $p/q = \sqrt{2}$. So, $p^2 = 2q^2$. It follows that $p$ and $q$ are even. So, $p = 2p'$ and $q = 2q'$. Now, $p'/q' = \sqrt{2}$. Since $q' < q$, it contradicts with the Inductive hypothesis. So, $\sqrt{2}$ is irrational. 55 | 56 | \section{Solution 3} 57 | \subsection{Solution (a)} 58 | Suppose $\sqrt{6}$ is rational. Then, $\exists p \in Z^{+} \exists q \in Z^{+}(p/q = \sqrt{6})$. So the set $S = \{q \in Z^{+} \mid \exists p \in Z^{+}(p/q = \sqrt{6})\}$ is non-empty. By well ordering principle, let $q$ be the smallest element in $S$. Since $q \in S$, we can choose some $p \in Z^{+}$ such that $p/q = \sqrt{6}$. So, $p^2 = 6q^2$. We know that both $p$ and $q$ are even. So, let $p' = 2p$ and $q' = 2q$ such that $p'/q' = \sqrt{6}$. Since $q' < q$ it contradicts that $q$ is the smallest element of the set. So, $\sqrt{6}$ is irrational. 59 | 60 | \subsection{Solution (b)} 61 | Suppose $\sqrt{2} + \sqrt{3} = p/q$ where $p \in Z^{+}$ and $q \in Z^{+}$. So, $p^2 / q^2 = 5 + 2\sqrt{6}$. So, $\sqrt{6} = (p^2 - 5q^2)/q^2$. This contradicts the part (a). 62 | 63 | \section{Solution 4} 64 | To prove: $\forall n >= 12 \exists a \in \mathbb{N} \exists b \in \mathbb{N}(3a + 7b = n)$. 65 | 66 | We have to prove it like this: 67 | 68 | \begin{itemize} 69 | 70 | \item 71 | Prove for n=12,13,14 72 | 73 | \item 74 | $\forall n >= 15 \exists a \exists b. (n = 3a + 7b)$ 75 | \end{itemize} 76 | 77 | I will just prove the second item here. Let $n >= 15$ be an arbitrary element.Suppose $\forall k( 15 <= k < n)$, there exists $a$ and $b$ such that $k = 3a + 7b$. Since $n - 3 < n$, it follows that $n - 3 = 3a + 7b$ from inductive hypothesis. So, $n = 3(a + 1) + 7b$. Let $a + 1 = a'$, So, $\exists a' \exists b (n = 3a' + 7b)$. Since $n$ was arbitrary, it follows that $\forall n >= 15 \exists a \exists b. (n = 3a + 7b)$. 78 | 79 | Also see this: http://math.stackexchange.com/a/1745857/124772 80 | 81 | \section{Solution 5} 82 | 83 | Let $n$ be an arbitrary number such that $n >= 1$. Suppose $\forall k < n$, $x^k + 1/x^k$ is an integer. Let $k = n-1$, then, 84 | \begin{gather*} 85 | x^{n-1} + \frac{1}{x^{n-1}} \text{ is an integer.} 86 | \end{gather*} 87 | 88 | Then, 89 | \begin{gather*} 90 | (x^{n-1} + \frac{1}{x^{n-1}})(x + \frac{1}{x}) \\ 91 | = x^{n-1}.x + \frac{x}{x^{n-1}} + \frac{x^{n-1}}{x} + \frac{1}{x^{n-1}.x} \\ 92 | = x^n + \frac{1}{x^n} + x^{n-1} + \frac{1}{x^{n-2}} \\ 93 | \end{gather*} 94 | 95 | We know from inductive hypothesis that $x^{n-2} + \frac{1}{x^{n-2}}$ is integer. Also $x^n + \frac{1}{x^n} + x^{n-1} + \frac{1}{x^{n-2}}$ is integer. So, $x^n + \frac{1}{x^n}$ is integer. 96 | 97 | \section{Solution 6} 98 | \subsection{Solution (a)} 99 | By mathematical induction on $n$, 100 | 101 | Base case. When $n = 0$, both sides of the equation becomes $0$. 102 | Induction step. Suppose $\sum_{i=0}^{n} F_i = F_{n+2} - 1$. Then, 103 | \begin{gather*} 104 | \begin{align} 105 | \sum_{i=0}^{n+1} F_i = \sum_{i=0}^{n} F_i + F_{n+1} \\ 106 | = F_{n+2} + F_{n+1} - 1 \\ 107 | = F_{n+3} - 1 108 | \end{align} 109 | \end{gather*} 110 | So, $\sum_{i=0}^{n} F_i \implies \sum_{i=0}^{n+1} F_i$. 111 | 112 | 113 | 114 | 115 | \end{document} 116 | -------------------------------------------------------------------------------- /chapter 1/section1-3.md: -------------------------------------------------------------------------------- 1 | Variables and Sets 2 | ------------------ 3 | ------------------ 4 | 5 | Exercise 1 6 | ----------- 7 | 8 | Analyze the logical forms of the following statements: 9 | (a) 3 is a common divisor of 6, 9, and 15. (Note: You did this in exercise 10 | 2 of Section 1.1, but you should be able to give a better answer now.) 11 | (b) x is divisible by both 2 and 3 but not 4. 12 | (c) x and y are natural numbers, and exactly one of them is prime. 13 | 14 | (a) 15 | 16 | D(x,y) = x is a divisor of y 17 | 18 | D(3,6) ∧ D(3,9) ∧ D(3,15) 19 | 20 | (b) 21 | 22 | D(x,y) = x is divisible by y 23 | 24 | D(x,2) ∧ D(x,3) ∧ ¬(D(x,4)) 25 | 26 | (c) 27 | 28 | N(x) = x is a natural number. 29 | P(x) = x is a prime number 30 | 31 | N(x) ∧ N(y) ∧ (P(x) ⊕ P(y)) 32 | 33 | ⊕ is symbol for exclusive or. It will true when "only one" of its operand is true. 34 | 35 | Exercise 2 36 | ---------- 37 | 38 | Analyze the logical forms of the following statements: 39 | (a) x and y are men, and either x is taller than y or y is taller than x. 40 | (b) Either x or y has brown eyes, and either x or y has red hair. 41 | (c) Either x or y has both brown eyes and red hair. 42 | 43 | (a) 44 | 45 | M(x) = x is men. 46 | T(x,y) = x is taller than y 47 | 48 | M(x) ∧ M(y) ∧ (T(x,y) ∨ (T(y,x))) 49 | 50 | (b) 51 | 52 | B(x) = x has brown eyes. 53 | R(x) = x has red hair. 54 | 55 | (B(x) ∨ B(y)) ∧ (R(x) ∨ R(y)) 56 | 57 | (c) 58 | 59 | (B(x) ∧ R(x)) ∨ (B(y) ∧ R(y)) 60 | 61 | Exercise 3 62 | ---------- 63 | 64 | Write definitions using elementhood tests for the following sets: 65 | (a) {Mercury, Venus, Earth, Mars, Jupiter, Saturn, Uranus, Neptune, 66 | Pluto}. 67 | (b) {Brown, Columbia, Cornell, Dartmouth, Harvard, Princeton, Univer- 68 | sity of Pennsylvania, Yale}. 69 | (c) {Alabama, Alaska, Arizona, . . . , Wisconsin, Wyoming}. 70 | (d) {Alberta, British Columbia, Manitoba, New Brunswick, Newfound- 71 | land and Labrador, Northwest Territories, Nova Scotia, Nunavut, On- 72 | tario, Prince Edward Island, Quebec, Saskatchewan, Yukon}. 73 | 74 | (a) 75 | 76 | {x | x is planet in solar system} 77 | 78 | (b) 79 | 80 | {x | x is an ivy-league university } 81 | 82 | (c) 83 | 84 | {x | x is a state of US} 85 | 86 | (d) 87 | 88 | {x | x is a province in Canada} 89 | 90 | Exercise 4 91 | ---------- 92 | 93 | Write definitions using elementhood tests for the following sets: 94 | (a) {1, 4, 9, 16, 25, 36, 49, . . .}. 95 | (b) {1, 2, 4, 8, 16, 32, 64, . . .}. 96 | (c) {10, 11, 12, 13, 14, 15, 16, 17, 18, 19} 97 | 98 | (a) 99 | 100 | {x^2 | x ∈ Z+} 101 | 102 | (b) 103 | 104 | {2^x | x ∈ N} 105 | 106 | (c) 107 | 108 | {x ∈ N | 9 < x < 20} 109 | 110 | Exercise 5 111 | ----------- 112 | 113 | Simplify the following statements. Which variables are free and which are 114 | bound? If the statement has no free variables, say whether it is true or 115 | false. 116 | (a) −3 ∈ {x ∈ R | 13 − 2x > 1}. 117 | (b) 4 ∈ {x ∈ R − | 13 − 2x > 1}. 118 | (c) 5 ∈/ {x ∈ R | 13 − 2x > c}. 119 | 120 | (a) 121 | 122 | Free variables: None 123 | Bound variables: x 124 | Statement is true since 19 > 1. 125 | 126 | (b) 127 | 128 | Free variables: None 129 | Bound variables: x 130 | Statement is false since 5 doesn't belong to R^- 131 | 132 | (c) 133 | 134 | Free variables: c 135 | Bound variables: x 136 | Statement is neither true nor false since it depends upon c. (3 > 137 | c) 138 | 139 | Exercise 6 140 | ----------- 141 | 142 | Simplify the following statements. Which variables are free and which are 143 | bound? If the statement has no free variables, say whether it is true or 144 | false. 145 | (a) w ∈ {x ∈ R | 13 − 2x > c}. 146 | (b) 4 ∈ {x ∈ R | 13 − 2x ∈ {y | y is a prime number}}. (It might make 147 | this statement easier to read if we let P = {y | y is a prime number}; 148 | using this notation, we could rewrite the statement as 4 ∈ {x ∈ R | 149 | 13 − 2x ∈ P}.) 150 | (c) 4 ∈ {x ∈{y | y is a prime number} | 13 − 2x > 1}. (Using the same no- 151 | tation as in part (b), we could write this as 4 ∈ {x ∈ P | 13 − 2x > 1}.) 152 | 153 | (a) 154 | 155 | Free variables: w, c 156 | Bound variables: x 157 | 158 | (b) 159 | 160 | Thanks for the hints! 161 | 162 | let P = {y | y is a prime number} 163 | 4 ∈ {x ∈ R | 13 − 2x ∈ P} 164 | 165 | Bound variable: x, y 166 | Statement is true since 5 is a prime number. 167 | 168 | (c) 169 | 170 | let P = {y | y is a prime number} 171 | 4 ∈ {x ∈ P | 13 − 2x > 1} 172 | 173 | Bound variable: x 174 | Statement is false since 4 is not a prime number. 175 | 176 | Exercise 7 177 | ---------- 178 | 179 | What are the truth sets of the following statements? List a few elements of 180 | the truth set if you can. 181 | (a) Elizabeth Taylor was once married to x. 182 | (b) x is a logical connective studied in Section 1.1. 183 | (c) x is the author of this book. 184 | 185 | (a) 186 | 187 | {x | Elizabeth Taylor was once married to x} = {Conrad Hilton Jr., 188 | Michael Wilding, Michael Todd, Eddie Fisher, Richard Burton, John 189 | Warner, Larry Fortensky} 190 | 191 | Copied the husband names from the answers section since I didn't know them. 192 | 193 | (b) 194 | 195 | {x | x is a logical connective studied in section 1.1} = { ∧, ∨, ¬ } 196 | 197 | (c) 198 | 199 | {x | x is the author of this book} = {Daniel J Velleman} 200 | 201 | Exercise 8 202 | ----------- 203 | 204 | What are the truth sets of the following statements? List a few elements of 205 | the truth set if you can. 206 | (a) x is a real number and x 2 − 4x + 3 = 0. 207 | (b) x is a real number and x 2 − 2x + 3 = 0. 208 | (c) x is a real number and 5 ∈ {y ∈ R | x 2 + y 2 < 50}. 209 | 210 | (a) 211 | 212 | {x ∈ R | x^2 − 4x + 3 = 0} 213 | 214 | (b) 215 | 216 | {x ∈ R | x^2 − 2x + 3 = 0} 217 | 218 | (c) 219 | 220 | {x ∈ R | 5 ∈ {y ∈ R | x^2 + y^2 < 50}} 221 | {x ∈ R | x^2 < 25} 222 | -------------------------------------------------------------------------------- /chapter 1/chapter1-1.md: -------------------------------------------------------------------------------- 1 | Deductive Reasoning and Logical connectives 2 | --------------------------------------------- 3 | --------------------------------------------- 4 | 5 | Exercise 1 6 | ----------- 7 | 8 | 1. Analyze the logical forms of the following statements: 9 | 10 | (a) We’ll have either a reading assignment or homework problems, but we 11 | won’t have both homework problems and a test. 12 | 13 | (b) You won’t go skiing, or you will and there won’t be any snow. 14 | 15 | (c) √7 ≤ 2. 16 | 17 | (a) 18 | 19 | P = We'll have reading assignment. 20 | 21 | Q = We'll have homework problems. 22 | 23 | T = We'll have test. 24 | 25 | (P ∨ Q) ∧ ¬(Q ∧ T) 26 | 27 | (b) 28 | 29 | P = You go skiing 30 | 31 | Q = There will be snow 32 | 33 | ¬P ∨ (P ∧ ¬Q) 34 | 35 | (c) 36 | 37 | √7 ≤ 2 (The question is not equal to.) 38 | 39 | P = √7 < 2 40 | 41 | Q = √7 = 2 42 | 43 | ¬(P ∨ Q) 44 | 45 | Exercise 2 46 | ----------- 47 | 48 | 2. Analyze the logical forms of the following statements: 49 | (a) Either John and Bill are both telling the truth, or neither of them is. 50 | (b) I’ll have either fish or chicken, but I won’t have both fish and mashed 51 | potatoes. 52 | (c) 3 is a common divisor of 6, 9, and 15. 53 | 54 | (a) 55 | 56 | J = John is telling the truth 57 | B = Bill is telling the truth 58 | 59 | (J ∧ B) ∨ (¬J ∧ ¬B) 60 | 61 | (b) 62 | 63 | F = I'll have fish. 64 | C = I'll have chicken 65 | P = I'll have mashed potatoes 66 | 67 | (F ∨ C) ^ ¬(F ∧ P) 68 | 69 | (c) 70 | 71 | X = 3 is a common divisor of 6 72 | Y = 3 is a common divisor of 9 73 | Z = 3 is a common divisor of 15 74 | 75 | X ∧ Y ∧ Z 76 | 77 | Exercise 3 78 | ----------- 79 | 80 | Analyze the logical forms of the following statements: 81 | (a) Alice and Bob are not both in the room. 82 | (b) Alice and Bob are both not in the room. 83 | (c) Either Alice or Bob is not in the room. 84 | (d) Neither Alice nor Bob is in the room. 85 | 86 | A = Alice is in the room. 87 | B = Bob is in the room. 88 | 89 | (a) 90 | 91 | A ∧ B = Alice and Bob are both in the room 92 | 93 | ¬(A ∧ B) 94 | 95 | (b) 96 | 97 | (¬A ∧ ¬B) 98 | 99 | (c) 100 | 101 | ¬A ∨ ¬B 102 | 103 | (d) 104 | 105 | (¬A ∧ ¬B) 106 | 107 | Exercise 4 108 | ---------- 109 | 110 | Ofcourse these two doesn't make sense: 111 | 112 | (b) ¬(P, Q, ∧R). 113 | (d) (P ∧ Q)(P ∨ R). 114 | 115 | Exercise 5 116 | ---------- 117 | 118 | Let P stand for the statement “I will buy the pants” and S for the statement 119 | “I will buy the shirt.” What English sentences are represented by the fol- 120 | lowing expressions? 121 | (a) ¬(P ∧ ¬S). 122 | (b) ¬P ∧ ¬S. 123 | (c) ¬P ∨ ¬S. 124 | 125 | (a) 126 | 127 | P ∧ ¬S = I will buy the Pant but not the shirt. 128 | = I will buy the Pant without the shirt. 129 | 130 | ¬(P ∧ ¬S) = I won't buy the Pant without the shirt. 131 | 132 | (b) 133 | 134 | ¬P ∧ ¬S = I will neither buy the pant nor the shirt. 135 | 136 | (c) 137 | 138 | ¬P ∨ ¬S = Either I won't buy the pant or won't buy the shirt. 139 | 140 | Exercise 6 141 | ----------- 142 | 143 | Let S stand for the statement “Steve is happy” and G for “George is happy.” 144 | What English sentences are represented by the following expressions? 145 | (a) (S ∨ G) ∧ (¬S ∨ ¬G). 146 | (b) [S ∨ (G ∧ ¬S)] ∨ ¬G. 147 | (c) S ∨ [G ∧ (¬S ∨ ¬G)]. 148 | 149 | The best way to solve this problem are by reducing the 150 | [expressions](http://math.stackexchange.com/questions/885946/simplifying-ambiguous-statements) 151 | as stated in the link. 152 | 153 | Exercise 7 154 | ----------- 155 | 156 | Identify the premises and conclusions of the following deductive argu- 157 | ments and analyze their logical forms. Do you think the reasoning is valid? 158 | (Although you will have only your intuition to guide you in answering 159 | this last question, in the next section we will develop some techniques for 160 | determining the validity of arguments.) 161 | (a) Jane and Pete won’t both win the math prize. Pete will win either 162 | the math prize or the chemistry prize. Jane will win the math prize. 163 | Therefore, Pete will win the chemistry prize. 164 | (b) The main course will be either beef or fish. The vegetable will be either 165 | peas or corn. We will not have both fish as a main course and corn as a 166 | vegetable. Therefore, we will not have both beef as a main course and 167 | peas as a vegetable. 168 | (c) Either John or Bill is telling the truth. Either Sam or Bill is lying. 169 | Therefore, either John is telling the truth or Sam is lying. 170 | (d) Either sales will go up and the boss will be happy, or expenses will go 171 | up and the boss won’t be happy. Therefore, sales and expenses will not 172 | both go up. 173 | 174 | (a) 175 | 176 | J = Jane will win the math prize 177 | P = Pete will win the math prize 178 | C = Pete will win the chemistry prize. 179 | 180 | Premises: 181 | 182 | ¬ (J ∧ P) 183 | P ∨ C 184 | J 185 | 186 | Conclusion: 187 | C 188 | 189 | Reasoning is valid: Jane has won the math prize so Pete can't win the 190 | math prize. So he can only will the chemistry prize according to the 191 | premises. 192 | 193 | (b) 194 | 195 | B = Beef will be the main course. 196 | F = Fish will be the main course. 197 | P = Peas will be the vegetable. 198 | C = Corn will be the vegetable. 199 | 200 | Premises: 201 | B ∨ F 202 | P ∨ C 203 | ¬(F ∧ C) 204 | 205 | Conclusion: 206 | ¬(B ∧ P) 207 | 208 | Reasoning seems invalid: In a case where you cannot have both fish and 209 | corn, you will end up with Beef and Peas. 210 | 211 | (c) 212 | 213 | J = John is telling the truth. 214 | B = Bill is telling the truth. 215 | S = Sam is telling the truth. 216 | 217 | Premises: 218 | J ∨ B 219 | ¬S ∨ ¬B 220 | 221 | Conclusion: 222 | J ∨ ¬S 223 | 224 | Reasoning seems valid: J ∨ B ∨ ¬S ∨ ¬B gives J ∨ ¬S 225 | 226 | (d) Either sales will go up and the boss will be happy, or expenses will go 227 | up and the boss won’t be happy. Therefore, sales and expenses will not 228 | both go up. 229 | 230 | S = Sales will go up. 231 | B = Boss will be happy. 232 | E = Expenses will go up. 233 | 234 | Premises: 235 | (S ∧ B) ∨ (E ∧ ¬B) 236 | 237 | Conclusion: 238 | ¬(S ∧ E) 239 | 240 | Reasoning is invalid: When S = T, E = T, B = T, them premise is True 241 | and Conclusion is false. 242 | -------------------------------------------------------------------------------- /chapter 1/section1-4.md: -------------------------------------------------------------------------------- 1 | Operations on Sets 2 | ------------------- 3 | ------------------- 4 | 5 | Exercise 1 6 | ----------- 7 | 8 | Let A = {1, 3, 12, 35}, B = {3, 7, 12, 20}, and C = {x | x is a prime 9 | number}. List the elements of the following sets. Are any of the sets 10 | below disjoint from any of the others? Are any of the sets below subsets 11 | of any others? 12 | (a) A ∩ B. 13 | (b) (A ∪ B) \ C. 14 | (c) A ∪ (B \ C). 15 | 16 | (a) 17 | 18 | A ∩ B = {3,12} 19 | 20 | (b) 21 | 22 | (A ∪ B) = {1,3,7,12,20,35} 23 | (A ∪ B) \ C = {1,12,20,35} 24 | 25 | (c) 26 | 27 | (B \ C) = {12,20} 28 | A ∪ (B \ C) = {1,3,12,20,35} 29 | 30 | No, two sets seem to be disjoint. 31 | Two sets are subsets: 32 | 33 | (A ∩ B) ⊆ (A ∪ (B \ C)) 34 | ((A ∪ B) \ C) ⊆ A ∪ (B \ C) 35 | 36 | Exercise 2 37 | ---------- 38 | 39 | Let A = {United States, Germany, China, Australia}, B = {Germany, 40 | France, India, Brazil}, and C = {x | x is a country in Europe}. List the 41 | elements of the following sets. Are any of the sets below disjoint from 42 | any of the others? Are any of the sets below subsets of any others? 43 | (a) A ∪ B. 44 | (b) (A ∩ B) \ C. 45 | (c) (B ∩ C) \ A. 46 | 47 | (a) 48 | 49 | A ∪ B = {United States, Germany, China, Australia, India, Brazil, 50 | France} 51 | 52 | (b) 53 | 54 | (A ∩ B) = {Germany} 55 | (A ∩ B) \ C = { } 56 | 57 | (c) 58 | 59 | (B ∩ C) = {Germany, France} 60 | (B ∩ C) \ A = {France} 61 | 62 | Any set along with `(A ∩ B) \ C` forms a disjoint set. Also, `(B ∩ C) 63 | \ A ⊆ A ∪ B` 64 | 65 | Exercise 3 66 | ----------- 67 | 68 | 69 | 70 | Exercise 3 & 4 & 6 71 | ------------------- 72 | 73 | Probably I will learn some graphical package to display this answer. 74 | 75 | Exercise 5 76 | ----------- 77 | 78 | (a) A \ (A ∩ B) = A \ B. 79 | (b) A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C). 80 | 81 | (a) 82 | 83 | x ∈ A \ (A ∩ B) 84 | = x ∈ A ∧ ¬(x ∈ (A ∩ B)) 85 | = x ∈ A ∧ ¬(x ∈ A ∧ x ∈ B) 86 | = x ∈ A ∧ (x ∈ ¬A ∨ x ∈ ¬B) 87 | = (x ∈ A ∧ x ∈ ¬A) ∨ (x ∈ A ∧ x ∈ ¬B) 88 | = (x ∈ A ∧ x ∈ ¬B) 89 | = A \ B 90 | 91 | (b) 92 | 93 | x ∈ (A ∪ (B ∩ C)) 94 | = (x ∈ A) ∨ (x ∈ (B ∩ C)) 95 | = (x ∈ A) ∨ (x ∈ B ∧ x ∈ C) 96 | = (x ∈ A ∨ x ∈ B) ∧ (x ∈ A ∨ x ∈ C) 97 | = (A ∪ B) ∩ (A ∪ C) 98 | 99 | Exercise 7 100 | ----------- 101 | 102 | (a) (A ∪ B) \ C = (A \ C) ∪ (B \ C). 103 | (b) A ∪ (B \ C) = (A ∪ B) \ (C \ A). 104 | 105 | (a) 106 | 107 | x ∈ (A ∪ B) \ C 108 | = x ∈ (A ∪ B) ∧ x ∈ ¬C 109 | = (x ∈ A ∨ x ∈ B) ∧ x ∈ ¬C 110 | = x ∈ ¬C ∧ (x ∈ A ∨ x ∈ B) 111 | = (x ∈ ¬C ∧ x ∈ A) ∨ (x ∈ ¬C ∧ x ∈ B) 112 | = (A \ C) ∪ (B \ C) 113 | 114 | (b) 115 | 116 | x ∈ A ∪ (B \ C) 117 | = x ∈ A ∨ (x ∈ B \ C) 118 | = x ∈ A ∨ (x ∈ B ∧ x ∈ ¬C) 119 | = (x ∈ A ∨ x ∈ B) ∧ (x ∈ A ∨ x ∈ ¬C) 120 | = (x ∈ A ∨ x ∈ B) ∧ ¬(x ∈ ¬A ∧ x ∈ C) 121 | = (A ∪ B) \ (C \ A) 122 | 123 | Exercise 8 124 | ----------- 125 | 126 | For each of the following sets, write out (using logical symbols) what it 127 | means for an object x to be an element of the set. Then determine which 128 | of these sets must be equal to each other by determining which statements 129 | are equivalent. 130 | (a) (A \ B) \ C. 131 | (b) A \ (B \ C). 132 | (c) (A \ B) ∪ (A ∩ C). 133 | (d) (A \ B) ∩ (A \ C). 134 | (e) A \ (B ∪ C) 135 | 136 | (a) 137 | 138 | x ∈ (A \ B) \ C 139 | = x ∈ (A \ B) ∧ ¬(x ∈ C) 140 | = x ∈ A ∧ ¬(x ∈ B) ∧ ¬(x ∈ C) 141 | = x ∈ A ∧ x ∈ ¬B ∧ x ∈ ¬C 142 | 143 | (b) 144 | 145 | x ∈ A \ (B \ C) 146 | = x ∈ A ∧ ¬(x ∈ B \ C) 147 | = x ∈ A ∧ ¬(x ∈ B ∧ x ∈ ¬C) 148 | = x ∈ A ∧ (x ∈ ¬B ∨ x ∈ C) 149 | = (x ∈ A ∧ x ∈ ¬B) ∨ (x ∈ A ∧ x ∈ C) 150 | 151 | (c) 152 | 153 | x ∈ (A \ B) ∪ (A ∩ C) 154 | = x ∈ (A \ B) ∨ x ∈ (A ∩ C) 155 | = (x ∈ A ∧ x ∈ ¬B) ∨ (x ∈ A ∧ x ∈ C) 156 | 157 | (d) 158 | 159 | x ∈ (A \ B) ∩ (A \ C) 160 | = x ∈ (A \ B) ∧ x ∈ (A \ C) 161 | = x ∈ A ∧ x ∈ ¬B ∧ x ∈ A ∧ x ∈ ¬C 162 | = x ∈ A ∧ x ∈ ¬B ∧ x ∈ ¬C 163 | 164 | (e) 165 | 166 | x ∈ A \ (B ∪ C) 167 | = x ∈ A ∧ ¬(x ∈ (B ∪ C)) 168 | = x ∈ A ∧ ¬(x ∈ B ∨ x ∈ C) 169 | = x ∈ A ∧ x ∈ ¬B ∧ x ∈ ¬C 170 | 171 | From the above: (a), (d) & (e) are equivalent. Also, (b) & (c) are 172 | equivalent. 173 | 174 | Exercise 9 175 | ----------- 176 | 177 | It was shown in this section that for any sets A and B, (A ∪ B) \ B ⊆ A. 178 | Give an example of two sets A and B for which (A ∪ B) \ B =/ A. 179 | (not equal to.) 180 | 181 | First some logical analysis: 182 | 183 | x ∈ (A ∪ B) \ B 184 | x ∈ (A ∪ B) ∧ x ∈ ¬B 185 | (x ∈ A ∨ x ∈ B) ∧ x ∈ ¬B 186 | x ∈ ¬B ∧ (x ∈ A ∨ x ∈ B) 187 | (x ∈ ¬B ∧ x ∈ A ) ∨ (x ∈ ¬B ∧ x ∈ B) 188 | (x ∈ ¬B ∧ x ∈ A ) ∨ (False ∧ True)[ or True ∧ False] 189 | (x ∈ ¬B ∧ x ∈ A ) ∨ (False) 190 | x ∈ ¬B ∧ x ∈ A 191 | 192 | B = {2,4} 193 | A = {1,2} 194 | A ∪ B = {1,2,4} 195 | 196 | (A ∪ B) \ B = {1} 197 | 198 | So, the key thing to figure out here is that if set B contains any 199 | elements which is present in set `A`, then `(A ∪ B) \ B` is not equal 200 | to `A`. 201 | 202 | Exercise 11 203 | ----------- 204 | 205 | (b) Give an example of sets A, B, and C for which (A ∪ B) \ C ≠ A ∪ (B \ C). 206 | 207 | A = {1, 2} 208 | B = {1, 2} 209 | C = {1, 2} 210 | 211 | (A ∪ B) \ C = ({1, 2} ∪ {1, 2}) \ {1, 2} 212 | = {1, 2} \ {1, 2} 213 | = {} or ∅ 214 | 215 | A ∪ (B \ C) = {1, 2} ∪ ({1, 2} \ {1, 2}) 216 | = {1, 2} ∪ {} 217 | = {1, 2} 218 | 219 | Hence, shown that (A ∪ B) \ C ≠ A ∪ (B \ C) 220 | 221 | Exercise 13 222 | ----------- 223 | 224 | Use any method you wish to verify the following identities: 225 | (a) (A ▵ B) ∪ C = (A ∪ C) ▵ (B \ C). 226 | (b) (A ▵ B) ∩ C = (A ∩ C) ▵ (B ∩ C). 227 | (c) (A ▵ B) \ C = (A \ C) ▵ (B \ C) 228 | 229 | (a) 230 | 231 | A ▵ B = (A \ B) ∪ (B \ A) = (A ∪ B) \ (A ∩ B) 232 | 233 | (A ▵ B) ∪ C 234 | = x ∈ (A ▵ B) ∪ C 235 | = x ∈ (A \ B) ∪ (B \ A) ∪ C 236 | = (x ∈ A ∧ x ∈ ¬B) ∨ (x ∈ B ∧ x ∈ ¬A) ∨ x ∈ C 237 | = ((x ∈ A ∧ x ∈ ¬B) ∨ (x ∈ B)) ∧ ((x ∈ A ∧ x ∈ ¬B) ∨ (x ∈ ¬A)) ∨ x ∈ C 238 | = ((x ∈ A ∨ x ∈ B) ∧ (x ∈ ¬B ∨ x ∈ ¬A)) ∨ x ∈ C 239 | = (x ∈ C ∨ (x ∈ A ∨ x ∈ B)) ∧ (x ∈ C ∨ (x ∈ ¬B ∨ x ∈ ¬A)) 240 | = ((x ∈ A ∨ x ∈ C) ∨ x ∈ B) ∧ (x ∈ ¬B ∨ x ∈ C ∨ x ∈ ¬A) 241 | = (x ∈ (A ∪ C) ∨ x ∈ B) ∧ (¬(x ∈ B ∧ x ∈ ¬C) ∨ x ∈ ¬A) 242 | = (x ∈ (A ∪ C) ∨ x ∈ B) ∧ (x ∈ (B \ C) ∨ x ∈ ¬A) 243 | = (x ∈ (A ∪ C) ∧ (x ∈ (B \ C) ∨ x ∈ ¬A)) ∨ (x ∈ B ∧ (x ∈ (B \ C) ∨ x ∈ 244 | ¬A)) 245 | = (x ∈ (A ∪ C) ∧ (B \ C)) ∨ (x ∈ (A ∪ C) ∧ x ∈ ¬A)) 246 | -------------------------------------------------------------------------------- /chapter 3/section3.2.tex: -------------------------------------------------------------------------------- 1 | %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% 2 | % Author: Sibi 3 | %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% 4 | \documentclass{article} 5 | \usepackage{graphicx} 6 | \usepackage{verbatim} 7 | \usepackage{amsmath} 8 | \usepackage{amsfonts} 9 | \usepackage{amssymb} 10 | \usepackage{tabularx} 11 | \setlength\parskip{\baselineskip} 12 | \begin{document} 13 | \title{Chapter 3 (Section 3.2)} 14 | \author{Sibi} 15 | \date{\today} 16 | \maketitle 17 | \newpage 18 | 19 | \section{Problem 1} 20 | 21 | Solution (a) 22 | 23 | %% +--------------+--------------+ 24 | %% |Givens |Goals | 25 | %% +--------------+--------------+ 26 | %% |P \implies Q |P \implies R | 27 | %% | | | 28 | %% |Q \implies R | | 29 | %% +--------------+--------------+ 30 | % This LaTeX table template is generated by emacs 24.3.1 31 | \begin{tabular}{| >{$}l<{$} | >{$}l<{$} |} 32 | \hline 33 | Givens & Goals \\ 34 | \hline 35 | P \implies Q & P \implies R \\ 36 | & \\ 37 | Q \implies R & \\ 38 | \hline 39 | \end{tabular} 40 | 41 | 42 | %% +--------------+--------------+ 43 | %% |Givens |Goals | 44 | %% +--------------+--------------+ 45 | %% |P \implies Q |R | 46 | %% | | | 47 | %% |Q \implies R | | 48 | %% | | | 49 | %% |P | | 50 | %% +--------------+--------------+ 51 | % This LaTeX table template is generated by emacs 24.3.1 52 | \begin{tabular}{| >{$}l<{$} | >{$}l<{$} |} 53 | \hline 54 | Givens & Goals \\ 55 | \hline 56 | P \implies Q & R \\ 57 | & \\ 58 | Q \implies R & \\ 59 | & \\ 60 | P & \\ 61 | \hline 62 | \end{tabular} 63 | 64 | Proof Structure 65 | 66 | Suppose $P$. 67 | 68 | [Proof of R goes here] 69 | 70 | Therefore $P \implies R$. 71 | 72 | Proof. Suppose $P$. Since $P \implies Q$, using modus ponens we can 73 | conclude $Q$. Since $Q \implies R$, again from modus ponens we can 74 | conclude R. Therefore, $P \implies R$. 75 | 76 | Solution (b) 77 | 78 | %% +--------------------+--------------+ 79 | %% |Givens |Goals | 80 | %% +--------------------+--------------+ 81 | %% |\neg R \implies (P |P \implies (Q | 82 | %% |\implies \neg Q) |\implies R) | 83 | %% +--------------------+--------------+ 84 | % This LaTeX table template is generated by emacs 24.3.1 85 | \begin{tabular}{| >{$}l<{$} | >{$}l<{$} |} 86 | \hline 87 | Givens & Goals \\ 88 | \hline 89 | \neg R \implies (P & P \implies (Q \\ 90 | \implies \neg Q) & \implies R) \\ 91 | \hline 92 | \end{tabular} 93 | 94 | $ \neg R \implies (P \implies \neg Q)$ is equivalent to $(P \land Q) 95 | \implies R$ 96 | 97 | Proof. Suppose P. Suppose Q. Since $ \neg R \implies (P \implies \neg 98 | Q) $ is equivalent to $(P \land Q) \implies R$, we can conclude R. 99 | Therefore, $P \implies (Q \implies R)$. 100 | 101 | \section{Problem 2} 102 | 103 | Solution(a) 104 | 105 | Proof. Suppose P. Since $P \implies Q$, we can conclude Q. Since $R 106 | \implies \neg Q$ is equivalent to $Q \implies \neg R$, we can conclude 107 | $\neg R$ as Q is true. Therefore, $P \implies \neg R$ is true. 108 | 109 | Solution (b) 110 | 111 | Proof. Suppose Q. We will prove by contradiction. Let $Q \implies \neg 112 | P$ be true. But $\neg P$ is true but our given $P$ is also true which 113 | leads us to a contradiction. Therefore, $\neg(Q \implies \neg P)$ is 114 | true. 115 | 116 | Alt. Proof: Suppose $Q$. Then $\neg (Q \implies \neg P) \equiv Q \land P$, 117 | because $\neg (Q \implies \neg P) \equiv \neg ( \neg Q \lor \neg P )$. 118 | Since we are given P and assumed Q, then $Q \land P$ is true, 119 | therefore if $P$ and $Q$ are true then $Q \implies \neg (Q \implies \neg P)$ 120 | is also true. 121 | 122 | \section{Problem 3} 123 | 124 | Proof. Suppse $x \in A$. Since $A \subseteq C$, it follows $x \in C$. 125 | But B and C are disjoint, so $x \notin B$. 126 | 127 | Expanded Proof. Suppose $x \in A$. Since $A \subseteq C$ then it holds that 128 | $x \in A \implies x \in C$, so $x \in C$. Since $B \cap C = \emptyset$ then 129 | $x \in C \implies x \notin B$, therefore $x \notin B$. 130 | 131 | \section{Problem 4} 132 | 133 | Proof. Suppose $x \in C$. We can conclude that $x \notin A \setminus 134 | B$. From, $x \notin A \setminus B$, it follows that $x \in B$. 135 | Therefore, if $x \in C$, then $x \in B$. 136 | 137 | \section{Problem 5} 138 | 139 | Proof. (by contradiction) Suppose $a \in A \setminus B$. 140 | It follows that $a \in A$ from our assumption. Also, 141 | $a \in C$ from which we can conclude that $a \in B$ since $A \cap C 142 | \subseteq B$ But this contradicts the fact that $a \notin B$. 143 | Therefore $a \notin A \setminus B$. 144 | 145 | \section{Problem 6} 146 | 147 | Proof. (by contradiction) Suppose $a \notin C$. From $a 148 | \notin B \setminus C$, we can conclude that either $a \notin B$ or $a 149 | \in C$ is true. Since $a \in A$, we can conclude that $a \in B$ from 150 | $A \subseteq B$. So, $a \notin B$ is false. Therefore, $a \in C$ 151 | should be true. But this contradicts the fact that $a \notin C$. 152 | Therefore $a \in C$. 153 | 154 | Note: This can be proven by focusing on $a \in B$ as well. By assuming 155 | $a \notin C$ we know that $a \notin B$ by $a \notin B \setminus C$, since this 156 | means logically that $\neg (a \in B \land a \notin C)$ which simplifies 157 | to $a \notin B \lor a \in C$. By assuming $a \notin C$ we automatically 158 | must have $a \notin B$. But since we have $a \in A$ then by the subset 159 | $A \subseteq B$ we have $\forall x ( x \in A \implies x \in B)$ and 160 | therefore $a \in A$ is sufficient to force $a \in B$. But this 161 | contradicts that $a$ cannot be in $B$ since we assumed $a \notin C$. 162 | Since $a \notin C$ causes an impossible situation, $a$ must be in $C$. 163 | 164 | \section{Problem 7} 165 | 166 | Proof. (by contradiction) Suppose $y = 0$. Solving $y + 167 | x = 2y - x$ by substituting $y=0$ we get that $x=0$. But it 168 | contradicts the fact that both $x$ and $y$ cannot be $0$. Therefore, 169 | $y \neq 0$. 170 | 171 | \section{Problem 8} 172 | 173 | Proof. Suppose $a < 1/a < b < 1/b$. Multiplying the inequality by 174 | $1/a < 1/b$ by $ab$ we should get $a < b$. From this, we can conclude 175 | that $ab$ is a negative number. Since $a < b$, we can conclude that 176 | $a$ is the negative number. Multiplying inequality $a < 1/a$ by 177 | negative number we get $a^2 < 1$. From that we can conclude that $a < 178 | -1$. Therefore, if $a < 1/a < b < 1/b$, then $a < -1$. 179 | 180 | \section{Problem 9} 181 | 182 | Proof. (by contradiction) Suppose $x^2y = 2x + y$. Suppose $y \neq 0$ Suppose $x = 0$. 183 | Substituting this into the equivalent $x^2y = 2x + y$, we get $y = 0$. 184 | But this contradicts the fact that $y \neq 0$. Therefore $x \neq 0$. 185 | Thus, if $y \neq 0$, then $x \neq 0$. 186 | 187 | \section{Problem 10} 188 | 189 | Proof. Suppose $x \neq 0$. Suppose $y = 3x^2 + 2y / x^2 + 2$. Solving 190 | it we get $x^2(3 - y) = 0$. Since $x \neq 0$, we can conclude that $y 191 | = 3$. 192 | 193 | \section{Problem 11} 194 | 195 | Solution (a) 196 | 197 | If conclusion of the theorem is false, then $\neg( x \neq 3 \land y 198 | \neq 8)$. That is equivalent to $x = 3 \lor y = 8$. 199 | 200 | Solution (b) 201 | 202 | First example: $x = 3$ and $y = 7$. 203 | Second example: $x = 2$ and $y = 8$. 204 | 205 | \section{Problem 12} 206 | 207 | Solution (a) 208 | 209 | This is plain wrong: Since $x \notin B$ and $B \subseteq C$, $x \notin 210 | C$. 211 | 212 | From $x \notin B$ and $B \subseteq C$, we cannot conclude that $x 213 | \notin C$. 214 | 215 | Expansion: 216 | 217 | $A \subseteq C$ means $\forall x ( x \in A \implies x \in B )$. Likewise, 218 | $B \subseteq C$ means $\forall x ( x \in B \implies x \in C )$. We are given 219 | that $x \in A$, therefore we know $x \in C$ by the first given. Now if we 220 | claim that $x \notin C$ because $x \notin B$ then we are committing the 221 | fallacy of denying the hypothesis, or trying to accept the inverse of 222 | $x \in B \implies x \in C$. This is clearly wrong, therefore the proof 223 | is invalid. 224 | 225 | Solution (b) 226 | 227 | $ C = \{1,2,3,4,5\} $ 228 | $ A = \{1,2\}$ 229 | $ B = \{3,4,5\}$ 230 | $ x = 1 $ 231 | \end{document} 232 | -------------------------------------------------------------------------------- /chapter 4/section4-1.tex: -------------------------------------------------------------------------------- 1 | %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% 2 | % Author: Sibi 3 | %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% 4 | \documentclass{article} 5 | \usepackage{graphicx} 6 | \usepackage{verbatim} 7 | \usepackage{amsmath} 8 | \usepackage{amsfonts} 9 | \usepackage{amssymb} 10 | \usepackage{tabularx} 11 | \setlength\parskip{\baselineskip} 12 | \begin{document} 13 | \title{Chapter 4 (Section 4.1)} 14 | \author{Sibi} 15 | \date{\today} 16 | \maketitle 17 | \newpage 18 | 19 | \section{Problem 1} 20 | 21 | Solution (a) 22 | 23 | $\{(x,y) \in P \times P \mid x \text{ is a parent of } y \} = \{(Prabakaran, Sibi) .. 24 | (Prabakaran, Madhu)\}$ 25 | 26 | Solution (b) 27 | 28 | $\{(x,y) \in C \times U \mid \text{Someone lives in } x \text{ and 29 | attends } y \}$ 30 | 31 | \section{Problem 2} 32 | 33 | Solution (a) 34 | 35 | $\{(x,y) \in P \times C \mid x \text{ lives in } y \} = \{(Sibi, Chennai) .. \}$ 36 | 37 | Solution (b) 38 | 39 | $\{(x,y) \in C \times N \mid \text{Population of } x \text{ is } y \} = \{(Chennai, 3434343)\}$ 40 | 41 | \section{Problem 3} 42 | 43 | Solution (a) 44 | 45 | $\{(0,-2), (1,-2), ..\}$ 46 | 47 | Solution (b) 48 | 49 | $\{(1,0), (2,0), (3,0), ..\}$ 50 | 51 | Solution (c) 52 | 53 | $\{(0,-2), (1,-2), ..\}$ 54 | 55 | Solution (d) 56 | 57 | $\{(2,0), (0, -2), ..\}$ 58 | 59 | \section{Problem 4} 60 | 61 | \begin{align*} 62 | A = \{1,2,3\} \\ 63 | B = \{1,4\} \\ 64 | C = \{3,4\} \\ 65 | D = \{5\} 66 | \end{align*} 67 | 68 | Solution(1) 69 | 70 | \begin{align*} 71 | B \cap C = \{4\} \\ 72 | A \times (B \cap C) = \{(1,4),(2,4),(3,4)\} \\ \\ 73 | A \times B = \{(1,1),(1,4),(2,1),(2,4),(3,1),(3,4)\} \\ 74 | A \times C = \{(1,3),(1,4),(2,3),(2,4),(3,3),(3,4)\} \\ 75 | (A \times B) \cap (A \times C) = \{(1,4),(2,4),(3,4)\} 76 | \end{align*} 77 | 78 | Solution(2) 79 | 80 | \begin{align*} 81 | B \cup C = \{1,3,4\} \\ 82 | A \times (B \cup C) = 83 | \{(1,1),(1,4),(1,3),(2,1),(2,4),(2,3),(3,1),(3,4),(3,3)\} \\ 84 | (A \times B) \cup (A \times C) = 85 | \{(1,1),(1,4),(2,1),(2,4),(3,1),(3,4),(1,3),(2,3),(3,3)\} \\ 86 | \end{align*} 87 | 88 | Solution(3) 89 | 90 | \begin{align*} 91 | A \times B = \{(1,1),(1,4),(2,1),(2,4),(3,1),(3,4)\} \\ 92 | C \times D = \{(3,5),(4,5)\} \\ \\ 93 | (A \times B) \cap (C \times D) = \emptyset \\ 94 | A \cap C = \{3\} \\ 95 | B \cap D = \emptyset \\ 96 | (A \cap C) \times (B \cap D) = \emptyset 97 | \end{align*} 98 | 99 | Solution(4) 100 | 101 | \begin{align*} 102 | A \times B = \{(1,1),(1,4),(2,1),(2,4),(3,1),(3,4)\} \\ 103 | C \times D = \{(3,5),(4,5)\} \\ 104 | (A \times B) \cup (C \times D) = 105 | \{(1,1),(1,4),(2,1),(2,4),(3,1),(3,4),(3,5),(4,5)\} \\ \\ 106 | A \cup C = \{1,2,3,4\} \\ 107 | B \cup D = \{1,4,5\}\\ 108 | (A \cup C) \times (B \cup D) = \{(1,1),(1,4),(1,5),(2,1),(2,4),(2,5),(3,1),(3,4),(3,5),(4,1),(4,4),(4,5)\} 109 | \end{align*} 110 | 111 | \section{Problem 5} 112 | 113 | Solution(a) 114 | 115 | ($ \Leftarrow $) Let $p$ be an arbitrary element of $A \times (B \cup C)$. Then by the 116 | definition of cartesian product, $p = (x,y)$ for some $x \in A$ and $y 117 | \in B \cup C$. Let us consider the cases: 118 | 119 | Case 1. $y \in B$. Since $x \in A$ and $y \in B$, it follows that $p 120 | \in A \times B$. So, $p \in (A \times B) \cup (A \times C)$. 121 | 122 | Case 2. $y \in C$. Since $x \in A$, it follows that $p \in A \times 123 | C$. So, $p \in (A \times B) \cup (A \times C)$. 124 | 125 | ($ \Rightarrow $) Let $p$ be an arbitrary element of $(A \times B) 126 | \cup (A \times C)$. Let us consider the cases: 127 | 128 | Case 1. $p \in A \times B$ Then there is some element $x$ and $y$ such 129 | that $p = (x,y)$ and $x \in A$ and $y \in B$. From $y \in B$, it 130 | follows that $y \in B \cup C$. So, $p \in A \times (B \cup C)$. 131 | 132 | Case 2. $p \in A \times C$ Then by definition of cartesian product, $p 133 | = (x,y)$ for some $x \in A$ and $y \in C$. From $y \in C$, it follows 134 | that $y \in B \cup C$. So, $p \in A \times (B \cup C)$. 135 | 136 | Solution(b) 137 | 138 | ($ \Leftarrow $) Let $p$ be an arbitrary element in $(A \times B) \cap 139 | (C \times D)$. Then it follows that $p \in A \times B$ and $p \in C 140 | \times D$. By definition of cartesian product, $p = (x,y)$ for some $x 141 | \in A$ and $y \in B$. Similarly, $x \in C$ and $y \in D$. It follows 142 | that $x \in (A \cap C)$ and $y \in (B \cap D)$. So, $(x,y) \in (A \cap 143 | C) \times (B \cap D)$. Since $p$ is an arbitrary element, $(A \times 144 | B) \cap (C \times D) \subseteq (A \cap C) \times (B \cap D)$. 145 | 146 | ($ \Rightarrow $) Let $p$ be an arbitrary element in $(A \cap C) 147 | \times (B \cap D)$. By the definition of cartesian product it follows 148 | that $p = (x,y)$ for some $x \in A \cap C$ and $y \in (B \cap D)$. 149 | From $x \in A \cap C$, we can conclude that $x \in A$ and $x \in C$. 150 | Similarly, $y \in B$ and $y \in D$. So, $(x,y) \in (A \times B) \cap 151 | (C \times D)$. Since p is arbitrary, $(A \cap C) \times (B \cap D) 152 | \subseteq (A \times B) \cap (C \times D)$. 153 | 154 | \section{Problem 6} 155 | 156 | The cases are not exhaustive. What about when $x \in A$ and $y \in D$, 157 | huh ? 158 | 159 | \section{Problem 7} 160 | 161 | $ m * n$ 162 | 163 | \section{Problem 8} 164 | 165 | ($ \Leftarrow $) Let $p$ be an arbitrary element in $A \times (B 166 | \setminus C)$. From the definition of cartesian product it follows 167 | that $p = (x,y)$ for some $x \in A$ and $y \in B \setminus C$. From $y 168 | \in B \setminus C$, it follows that $y \in B$ and $y \notin C$. 169 | Therefore $(x,y) \in (A \times B)$. Similarly, $(x,y) \in A \times C$. 170 | Since $p$ is arbitrary, we can conclude that $A \times (B \setminus C) 171 | \subseteq (A \times B) \setminus (A \times C)$. 172 | 173 | ($ \Rightarrow $) Let $p$ be an arbitrary element in $(A \times B) 174 | \setminus (A \times C)$. It follows that $p \in (A \times B)$ and $p 175 | \notin (A \times C)$. From the definition of cartesian product it 176 | follows that $p = (x,y)$ for some $x \in A$ and $y \in B$. Similarly 177 | $(x,y) \notin (A \times C)$. But we know that $x \in A$, so $y \notin 178 | C$. Therefore, $y \in B \setminus C$. So, $(x,y) \in A \times (B 179 | \setminus C)$. Since $p$ is arbitrary, we can conclude that $A \times 180 | (A \setminus C) = (A \times B) \setminus (A \times C)$. 181 | 182 | \section{Problem 9} 183 | 184 | ($ \Leftarrow $) Let $p$ be an arbitrary element in $(A \times B) 185 | \setminus (C \times D)$. It follows that $p \in A \times B$ and $p 186 | \notin (C \times D)$. From the definition of cartesian product, it 187 | follows that $p = (x,y)$ for some $x \in A$ and $y \in B$. Similarly, 188 | $(x,y) \notin (C \times D)$ which means either $x \notin c$ or $y 189 | \notin D$. Let us consider the cases separately: 190 | 191 | Case 1. $y \notin D$. We already know that $x \in A$ and $y \in B$. 192 | So, $y \in B \setminus D$. Therefore $(x,y) \in (A \times (B \setminus 193 | D))$. So, $p \in (A \times (B \setminus D)) \cup ((A \setminus C) \times B)$. 194 | 195 | Case 2. $x \notin C$. From earlier, $x \in A$ and $y\in B$. It follows 196 | that $x \in A \setminus C$. Therefore, $(x,y) \in ((A \setminus c) 197 | \times B)$. So, $p \in (A \times (B \setminus D)) \cup ((A \setminus 198 | C) \times B)$. 199 | 200 | ($ \Rightarrow $) Let $p$ be an arbitrary element in $(A \times (B 201 | \setminus D)) \cup ((A \setminus C) \times B)$. Let us consider the 202 | cases separately: 203 | 204 | Case 1. $p \in A \times (B \setminus D)$. From the definition of 205 | cartesian product, it follows that $p = (x,y)$ for some $x \in A$ and 206 | $y \in B \setminus D$. Therefore $y \in B$ and $y \notin D$. So, $p 207 | \in (A \times B)$ and since $ y \notin D$, it follows that $(x,y) 208 | \notin C \times D$. So, $p \in (A \times B) \setminus (C \times D)$. 209 | 210 | Case 2. $p \in ((A \setminus C) \times B)$ From the definition of 211 | cartesian product, it follows that $p = (x,y)$ for some $x \in A 212 | \setminus C$ and $y \in B$. From $x \in A \setminus C$, it follows 213 | that $x \in A$ and $x \notin C$. So, $p \in (A \times B)$ and $p 214 | \notin C \times D$ since $x \notin C$. So, $p \in (A \times B) 215 | \setminus (C \times D)$. 216 | 217 | \section{Problem 10} 218 | 219 | Suppose $A \times B$ and $C \times D$ are disjoint. Let $p$ be an 220 | arbitrary element in $A \times B$. Since they are disjoint, $p \notin 221 | C \times D$. From the definition of cartesian product, it follows that 222 | $p = (x,y)$ for some $x \in A$ and $y \in B$. Now since $(x,y) \in C 223 | \times D$, so either $x \notin C$ or $y \notin D$. So, if $x \notin 224 | C$, then $A \cap C = \emptyset$ or if $y \notin D$ then $B \cap D = \emptyset$. 225 | 226 | \section{Problem 11} 227 | 228 | Solution(a) 229 | 230 | Let $p$ be an arbitrary element in $\cup_{i \in I} (A_i \times B_i)$. 231 | It follows that $\exists i \in I(p \in A_i \times B_i)$. So $p \in A_i 232 | \times B_i$ for some $i$. From the definition of cartesian product, it 233 | follows that $p = (x,y)$ for some $x \in A_i$ and $y \in B_i$. From $x 234 | \in A_i$, we can conclude that $x \in \cup_{i \in I} A_i$. Similarly, 235 | $y \in \cup_{i \in I} B_i$. Therefore $p \in (\cup_{i \in I}A_i \times 236 | \cup_{i \in I}B_i)$. Since $p$ is an arbitrary, we can conclude that 237 | $\cup_{i \in I}(A_i \times B_i) \subseteq \cup_{i \in I})A_i \times 238 | \cup_{i \in I} B_i$. 239 | 240 | Solution (b) 241 | 242 | ($\Rightarrow$) Let $h$ be an arbitrary element in $\cup_{p in P}C_p$. 243 | It follows that $\exists p \in P(h \in C_p)$. So, $h \in A_i \times 244 | B_j$. From the definition of cartesian product, it follows that $h = 245 | (x,y)$ for some $x \in A_i$ and $y \in B_j$. From $x \in A_i$, we can 246 | conclude that $x \in \cup_{i \in I}A_i$. Similarly, $y \in \cup_{i \in 247 | I}B_i$. Therefore $h \in (\cup_{i \in I}A_i) \times (\cup_{i \in I}B_i)$. 248 | 249 | ($\Leftarrow$) Let $h$ be an arbitrary element in $(\cup_{i \in I}A_i) 250 | \times (\cup_{i \in I}B_i)$. Then by the definition of cartesian 251 | product, it follows that $h = (x,y)$ for some $x \in \cup_{i \in 252 | I}A_i$ and $y \in \cup_{j \in I}B_j$. From $x \in \cup_{i \in 253 | I}A_i$, it follows that for some $i$ in $I$, $x \in A_i$. Similarly, 254 | $y \in B_j$. Therefore $(x,y) \in A_i \times B_j$. So, $(x,y) \in 255 | C_{(i,j)}$ for some $i$ and $j$. Therefore $h \in \cup_{p \in P}C_p$. 256 | 257 | \section{Problem 12} 258 | 259 | What about $A = \emptyset $, $B = \{1\}$, $C = \emptyset$, $D = 260 | \{2\}$, huh? 261 | 262 | \end{document} 263 | 264 | 265 | 266 | 267 | -------------------------------------------------------------------------------- /chapter 2/section2-2.md: -------------------------------------------------------------------------------- 1 | Equivalences Involving Quantifiers 2 | ----------------------------------- 3 | ----------------------------------- 4 | 5 | Notes 6 | ----- 7 | 8 | * Universal quantification distributes over conjunction: 9 | ∀x(E(x) ∧ T (x)) is equivalent to ∀x E(x) ∧ ∀x T (x) 10 | * Existential quantification doesn't distributes over conjunction. 11 | * Existential quantification distributes over disjunction. 12 | * Universal quantification doesn't distribute over disjunction. 13 | 14 | Exercise 1 15 | ----------- 16 | 17 | Negate these statements and then reexpress the results as equivalent 18 | positive statements. (See Example 2.2.1.) 19 | (a) Everyone who is majoring in math has a friend who needs help with 20 | his homework. 21 | (b) Everyone has a roommate who dislikes everyone. 22 | (c) A ∪ B ⊆ C \ D. 23 | (d) ∃x∀y[y > x → ∃z(z 2 + 5z = y)] 24 | 25 | (a) 26 | 27 | * Everyone who is majoring in math has a friend who needs help with 28 | his homework. 29 | * ∀x(if x is majoring in math, then x has a friend who needs help 30 | with his homework.) 31 | * ∀x(M(x) -> ∃y(F(y,x) ∧ H(y))) 32 | 33 | where 34 | 35 | M(x) = x is majoring in Math. 36 | F(y,x) = y is a friend of x. 37 | H(y) = y needs help with his homework. 38 | 39 | Now negate the above logical form: 40 | 41 | * ¬∀x(M(x) -> ∃y(F(y,x) ∧ H(y))) 42 | * ∃x¬(¬M(x) ∨ ∃y(F(y,x) ∧ H(y))) 43 | * ∃x(M(x) ∧ ¬(∃y(F(y,x) ∧ H(y)))) 44 | * ∃x(M(x) ∧ ∀y(¬F(y,x) ∨ ¬H(y))) 45 | * ∃x(M(x) ∧ ∀y(F(y,x) -> ¬H(y))) 46 | 47 | There is some `x` who is majoring in Maths and for all `y`, if `y` is 48 | a friend of `x` then `y` doesn't need help with this homework. 49 | 50 | (b) 51 | 52 | Everyone has a roommate who dislikes everyone. 53 | 54 | R(x,y) = x has a roommate y. 55 | L(y,x) = y likes x. 56 | 57 | * ∀x∃y(R(x,y) ∧ ∀z¬L(y,z)) 58 | 59 | Now negate the above logical form: 60 | 61 | * ¬∀x∃y(R(x,y) ∧ ∀z¬L(y,z)) 62 | * ∃x¬y(R(x,y) ∧ ∀z¬L(y,z)) 63 | * ∃x∀y¬(R(x,y) ∧ ∀z¬L(y,z)) 64 | * ∃x∀y(¬R(x,y) ∨ ¬∀z¬L(y,z)) 65 | * ∃x∀y(¬R(x,y) ∨ ∃zL(y,z)) 66 | * ∃x∀y(R(x,y) -> ∃zL(y,z)) 67 | 68 | For some `x` and for all `y, if x has a roommate y, then y 69 | likes someone. Or 70 | There is someone such that all of his roommates have somebody 71 | they like. 72 | 73 | (c) 74 | 75 | * `A ∪ B ⊆ C \ D` 76 | * `∀x(x ∈ A ∪ B -> x ∈ C \ D)` 77 | * `∀x(x ∉ A ∪ B ∨ x ∈ C \ D)` 78 | * `¬∀x(x ∉ A ∪ B ∨ x ∈ C \ D)` (Negation) 79 | * `∃x¬(x ∉ A ∪ B ∨ x ∈ C \ D)` 80 | * `∃x(x ∈ A ∪ B ∧ x ∉ C \ D)` 81 | 82 | (d) 83 | 84 | * `∃x∀y[y > x → ∃z(z^2 + 5z = y)]` 85 | * `¬∃x∀y[y > x → ∃z(z^2 + 5z = y)]` 86 | * `∀x¬∀y[y > x → ∃z(z^2 + 5z = y)]` 87 | * `∀x∃y¬[y > x → ∃z(z^2 + 5z = y)]` 88 | * `∀x∃y¬[¬(y > x) ∨ ∃z(z^2 + 5z = y)]` 89 | * `∀x∃y(y > x) ∧ ∀z(z^2 + 5z ≠ y)` 90 | 91 | Exercise 2 92 | ----------- 93 | 94 | Negate these statements and then reexpress the results as equivalent 95 | positive statements. (See Example 2.2.1.) 96 | (a) There is someone in the freshman class who doesn’t have a roommate. 97 | (b) Everyone likes someone, but no one likes everyone. 98 | (c) ∀a ∈ A∃b ∈ B(a ∈ C ↔ b ∈ C). 99 | (d) ∀y > 0∃x(ax 2 + bx + c = y). 100 | 101 | (a) 102 | 103 | F(x) = x is in the freshman class 104 | R(x,y) = x has roommate y. 105 | 106 | Original Statement: 107 | 108 | * `∃x F(x) ∧ ¬∃yR(x,y)` 109 | 110 | Negate: 111 | 112 | * `¬∃x F(x) ∧ ∃yR(x,y)` 113 | * `∀x¬(F(x) ∧ ∃yR(x,y))` 114 | * `∀x¬F(x) ∨ ∃yR(x,y)` 115 | * `∀x ¬F(x) ∨ ∃yR(x,y)` 116 | * `∀x F(x) -> ∃yR(x,y)` 117 | 118 | Everyone in the freshman class has at least one roommate. 119 | 120 | (b) 121 | 122 | L(x,y) = x likes y 123 | 124 | * `∀x∃y L(x,y) ∧ ¬(Someone likes everyone)` 125 | * `∀x∃y L(x,y) ∧ ¬∃z∀a(L(z,a))` 126 | 127 | Negate: 128 | 129 | * `¬∀x∃y L(x,y) ∧ ¬∃z∀a(L(z,a))` 130 | * `∃x∀y ¬(L(x,y) ∧ ¬∃z∀a(L(z,a)))` 131 | * `(∃x∀y ¬L(x,y)) ∨ (∃z∀a(L(z,a)))` 132 | 133 | * (There exists `x` such that for all `y`, x doesn't like y) ∨ (Someone 134 | likes everyone) 135 | * (Someone doesn't like anyone) or (Someone likes everyone) 136 | 137 | Either someone doesn't like anyone or someone likes everyone. 138 | 139 | (c) 140 | 141 | * `¬(∀a ∈ A∃b ∈ B(a ∈ C ↔ b ∈ C))` 142 | * `∃a ∈ A ∀b ∈ B ¬(a ∈ C ↔ b ∈ C)` 143 | 144 | (d) 145 | 146 | * `¬∀y > 0 ∃x(ax^2 + bx + c = y)` 147 | * `∃y > 0 ∀x¬(ax^2 + bx + c = y)` 148 | * `∃y > 0 ∀x(ax^2 + bx + c ≠ y)` 149 | 150 | Exercise 3 151 | ---------- 152 | 153 | Are these statements true or false? The universe of discourse is N. 154 | (a) ∀x(x < 7 → ∃a∃b∃c(a 2 + b 2 + c 2 = x)). 155 | (b) ∃!x((x − 4)^2 = 9). 156 | (c) ∃!x((x − 4)^2 = 25). 157 | (d) ∃x∃y((x − 4)^2 = 25 ∧ (y − 4)^2 = 25). 158 | 159 | (a) 160 | 161 | `∀x(x < 7 → ∃a∃b∃c(a^2 + b^2 + c^2 = x))` 162 | 163 | True (Substitute value for x=6,5,4...0 and check if for yourself) 164 | 165 | (b) 166 | 167 | `∃!x((x − 4)^2 = 9)` 168 | 169 | Solving the equation, I can find two `x` solutions: 7,1 both falling 170 | in the domain of N. 171 | 172 | False 173 | 174 | (c) 175 | 176 | `∃!x((x − 4)^2 = 25)` 177 | 178 | Solving the equation, I can find two `x` solutions: 9,-1. But the 179 | solution is only 9 since -1 falls out of the domain. 180 | 181 | True 182 | 183 | (d) 184 | 185 | True (from above) 186 | 187 | Exercise 4 188 | ----------- 189 | 190 | Show that the second quantifier negation law, which says that ¬∀x P(x) 191 | is equivalent to ∃x¬P(x), can be derived from the first, which says that 192 | ¬∃x P(x) is equivalent to ∀x¬P(x). (Hint: Use the double negation 193 | law.) 194 | 195 | Solution: 196 | 197 | Given: 198 | ¬∃x P(x) is equivalent to ∀x¬P(x) 199 | or 200 | ¬∃x ¬P(x) is equivalent to ∀PX(x) (Replacing P(x) by ¬P(x)) 201 | or 202 | ∃x ¬P(x) is equivalent to ¬∀xP(x) (Double negation) 203 | 204 | Exercise 5 205 | ---------- 206 | 207 | Show that ¬∃x ∈ A P(x) is equivalent to ∀x ∈ A¬P(x) 208 | 209 | * ¬∃x ∈ A P(x) 210 | * ¬∃x((x ∈ A) ∧ P(x)) 211 | * ∀x¬((x ∈ A) ∧ P(x)) 212 | * ∀x(¬(x ∈ A) ∨ ¬P(x)) 213 | * ∀x(x ∈ A -> ¬P(x)) 214 | * ∀x ∈ A ¬P(x) 215 | 216 | Exercise 6 217 | ----------- 218 | 219 | Show that the existential quantifier distributes over disjunction. In other 220 | words, show that ∃x(P(x) ∨ Q(x)) is equivalent to ∃x P(x) ∨ ∃x 221 | Q(x). 222 | 223 | * ∃x(P(x) ∨ Q(x)) 224 | * ¬¬∃x(P(x) ∨ Q(x)) 225 | * ¬∀x¬(P(x) ∨ Q(x)) 226 | * ¬∀x(¬P(x) ∧ ¬Q(x)) 227 | * ¬(∀x¬P(x) ∧ ∀x¬Q(x)) 228 | * ∃xP(x) ∨ ∃xQ(x) 229 | 230 | Exercise 7 231 | ----------- 232 | 233 | Show that ∃x(P(x) → Q(x)) is equivalent to ∀x P(x) → ∃x Q(x) 234 | 235 | * `∃x(P(x) → Q(x))` 236 | * `∃x(¬P(x) ∨ Q(x))` 237 | * `¬∀x P(x) ∨ ∃xQ(x)` 238 | * `∀x P(x) -> ∃xQ(x)` 239 | 240 | Exercise 8 241 | ----------- 242 | 243 | Show that (∀x ∈ A P(x)) ∧ (∀x ∈ B P(x)) is equivalent to ∀x ∈ 244 | (A ∪ B)P(x). (Hint: Start by writing out the meanings of the bounded 245 | quantifiers in terms of unbounded quantifiers.) 246 | 247 | * `(∀x ∈ A P(x)) ∧ (∀x ∈ B P(x))` 248 | * `(∀x((x ∈ A) -> P(x))) ∧ (∀x((x ∈ B) -> P(x)))` 249 | * `∀x((x ∈ A) -> P(x)) ∧ ((x ∈ B) -> P(x))` 250 | * `∀x((¬(x ∈ A) ∨ P(x)) ∧ (¬(x ∈ B) ∨ P(x)))` 251 | * `∀x(((¬(x ∈ A) ∨ P(x)) ∧ (¬(x ∈ B))) ∨ ((¬(x ∈ A) ∨ P(x)) ∧ P(x)))` 252 | * `∀x(((¬(x ∈ A) ∨ P(x)) ∧ (¬(x ∈ B))) ∨ P(x)))` 253 | * `∀x((¬(x ∈ A) ∧ ¬(x ∈ B)) ∨ P(x) ∨ P(x))` 254 | * `∀x((¬(x ∈ A) ∧ ¬(x ∈ B)) ∨ P(x))` 255 | * `∀x¬((x ∈ A) ∨ (x ∈ B)) ∨ P(x)` 256 | * `∀x((x ∈ A) ∨ (x ∈ B)) -> P(x)` 257 | * `∀x ∈ (A ∪ B)P(x)` 258 | 259 | Exercise 9 260 | ----------- 261 | 262 | Is ∀x(P(x) ∨ Q(x)) equivalent to ∀x P(x) ∨ ∀x Q(x)? Explain. (Hint: Try 263 | assigning meanings to P(x) and Q(x).) 264 | 265 | I found this interesting example on 266 | [stackexchange](http://math.stackexchange.com/questions/6410/distribution-of-universal-quantifiers) 267 | to argue this: 268 | 269 | P(x) = Chess coin is black. 270 | Q(x) = Chess coin is white. 271 | 272 | ∀x(P(x) ∨ Q(x)) means: For all chess coins x, x is either black or x 273 | is white. 274 | ∀x P(x) ∨ ∀x Q(x) means: All chess coins are black or All chess coins 275 | are white. 276 | 277 | False 278 | 279 | Exercise 10 280 | ------------ 281 | 282 | (a) Show that ∃x ∈ A P(x) ∨ ∃x ∈ B P(x) is equivalent to ∃x ∈ (A ∪ B) 283 | P(x). 284 | (b) Is ∃x ∈ A P(x) ∧ ∃x ∈ B P(x) equivalent to ∃x ∈ (A ∩ B) P(x)? 285 | Explain 286 | 287 | (a) 288 | 289 | * `∃x ∈ A P(x) ∨ ∃x ∈ B P(x)` 290 | * `∃x ((x ∈ A) ∧ P(x)) ∨ ∃x (x ∈ B ∧ P(x))` 291 | * `∃x ((x ∈ A) ∧ P(x)) ∨ (x ∈ B ∧ P(x))` 292 | * `∃x ((x ∈ A) ∨ (x ∈ B ∧ P(x))) ∧ (P(x) ∨ (x ∈ B ∧ P(x)))` 293 | * `∃x ((x ∈ A) ∨ (x ∈ B ∧ P(x))) ∧ P(x)` 294 | * `∃x ((x ∈ A) ∨ P(x)) ∧ (x ∈ A ∨ x ∈ B) ∧ P(x)` 295 | * `∃x P(x) ∧ (x ∈ A ∨ x ∈ B)` (I love absorption law!) 296 | * `∃x ∈ (A ∪ B) P(x)` 297 | 298 | (b) 299 | 300 | * `∃x ∈ (A ∩ B) P(x)` 301 | * `∃x x ∈ A ∧ x ∈ B ∧ P(x)` 302 | 303 | Existential quantification doesn't distribute over conjunction, so 304 | they aren't equivalent. 305 | 306 | Exercise 11 307 | ------------ 308 | 309 | Show that the statements A ⊆ B and A \ B = ∅ are equivalent by writing 310 | each in logical symbols and then showing that the resulting formulas are 311 | equivalent. 312 | 313 | 314 | * `A ⊆ B` 315 | * `∀x (x ∈ A -> x ∈ B)` 316 | 317 | * `A \ B = ∅` 318 | * `¬∃x (x ∈ A) ∧ ¬(x ∈ B)` (Negating it because no element exists) 319 | * `¬∃x ¬(¬(x ∈ A) ∨ (x ∈ B))` 320 | * `¬∃x ¬(x ∈ A -> x ∈ B)` 321 | * `∀x (x ∈ A -> x ∈ B)` 322 | 323 | Hence they are equivalent. 324 | 325 | Exercise 12 326 | ----------- 327 | 328 | Let T (x, y) mean “x is a teacher of y.” What do the following statements 329 | mean? Under what circumstances would each one be true? Are any of 330 | them equivalent to each other? 331 | (a) ∃!yT (x, y). 332 | (b) ∃x∃!yT (x, y). 333 | (c) ∃!x∃yT (x, y). 334 | (d) ∃y∃!x T (x, y). 335 | (e) ∃!x∃!yT (x, y). 336 | (f ) ∃x∃y[T (x, y) ∧ ¬∃u∃v(T (u, v) ∧ (u = x ∨ v = y)) 337 | 338 | (a) 339 | 340 | * `∃!yT (x, y)` 341 | 342 | x is a teacher of one person. 343 | 344 | (b) 345 | 346 | * `∃x∃!yT (x, y)` 347 | 348 | Literally: For some x there exists one y such that x is a teacher of y. 349 | 350 | There is a teacher who teaches exactly one student. 351 | 352 | (c) 353 | 354 | * `∃!x∃yT (x, y)` 355 | 356 | Literally: There exists only one x for some y such that x is a teacher of y. 357 | 358 | There is a student who has only one teacher. 359 | 360 | (d) 361 | 362 | * `∃y∃!x T (x, y)` 363 | 364 | Literally: For some y there exits one x such that x is a teacher of y. 365 | 366 | There is a student who has only one teacher. (equivalent to c) 367 | 368 | (e) 369 | 370 | * `∃!x∃!yT (x, y)` 371 | 372 | Literally: There exists exactly one x and y such that x is a teacher of y. 373 | 374 | There is exactly one teacher who teaches exactly one student. 375 | 376 | (f) 377 | 378 | * `∃x∃y[T (x, y) ∧ ¬∃u∃v(T (u, v) ∧ (u ≠ x ∨ v ≠ y))` 379 | * `∃x∃y[T (x, y) ∧ ∀u∀v(¬T (u, v) ∨ ¬(u ≠ x ∨ v ≠ y))]` 380 | * `∃x∃y[T (x, y) ∧ ∀u∀v(¬T (u, v) ∨ ((u = x) ∧ v = y))]` 381 | * `∃x∃y[T (x, y) ∧ ∀u∀v(T(u,v) -> ((u = x) ∧ v = y))]` 382 | * `∃!x∃!yT (x, y)` 383 | 384 | f is equivalent to e 385 | -------------------------------------------------------------------------------- /chapter 2/section2-1.md: -------------------------------------------------------------------------------- 1 | Quantifiers 2 | ------------ 3 | 4 | Interesting threads: 5 | * [Relationship between quantifiers and logic](http://philosophy.stackexchange.com/questions/4165/how-do-quantifiers-work-in-predicate-logic) 6 | 7 | Exercise 1 8 | ---------- 9 | 10 | Analyze the logical forms of the following statements. 11 | (a) Anyone who has forgiven at least one person is a saint. 12 | (b) Nobody in the calculus class is smarter than everybody in the discrete 13 | math class. 14 | (c) Everyone likes Mary, except Mary herself. 15 | (d) Jane saw a police officer, and Roger saw one too. 16 | (e) Jane saw a police officer, and Roger saw him too. 17 | 18 | (a) 19 | 20 | F(x,y) = x has forgiven y 21 | S(x) = x is a saint. 22 | 23 | ∀x(∃yF(x,y) -> S(x)) 24 | 25 | (b) 26 | 27 | * ¬(Somebody in the calculus room is smarter than everybody in the 28 | discrete math class) 29 | * There does not exist someone in the calculus room And such that for 30 | all students y, if y is enrolled in Discrete math, then x is 31 | smarter than y. 32 | 33 | C(x) = x is in calculus room. 34 | D(x) = x is in discrete math room. 35 | S(x,y) = x is smarter than y. 36 | 37 | ¬∃x(C(x) ∧ ∀y( D(y) -> S(x,y))) 38 | 39 | 40 | (c) 41 | 42 | * If someone doesn't like Mary, then that someone must be Mary 43 | * If someone isn't Mary, then someone likes Mary. 44 | 45 | L(x,y) = x likes y 46 | ∀x(¬(x = m) -> L(x,m)) 47 | 48 | (d) 49 | 50 | S(x,y) = x saw y 51 | P(x) = x is a police officer. 52 | j = Jane 53 | r = Roger 54 | 55 | ∃x (P(x) ∧ S(j,x)) ∧ ∃y (P(y) ∧ S(r,y)) 56 | 57 | (e) 58 | 59 | ∃x (P(x) ∧ S(j,x) ∧ S(r,x)) 60 | 61 | Exercise 2 62 | ----------- 63 | 64 | Analyze the logical forms of the following statements. 65 | (a) Anyone who has bought a Rolls Royce with cash must have a rich 66 | uncle. 67 | (b) If anyone in the dorm has the measles, then everyone who has a friend 68 | in the dorm will have to be quarantined. 69 | (c) If nobody failed the test, then everybody who got an A will tutor 70 | someone who got a D. 71 | (d) If anyone can do it, Jones can. 72 | (e) If Jones can do it, anyone can. 73 | 74 | (a) 75 | 76 | If x bought Rolls Royce with cash, then he must have a rich uncle. 77 | 78 | R(x) = x bought Rolls Royce with cash. 79 | U(x) = x has a rich uncle. 80 | 81 | ∀x(R(x) -> U(x)) 82 | 83 | (b) 84 | 85 | D(x) = x is in the dorm 86 | M(x) = x in the dorm has measles 87 | F(y,x) = y is a friend of x 88 | Q(y) = Quarantine y 89 | 90 | * (Someone in the dorm has measles) -> (everyone who has a friend in 91 | the dorm will have to be quarantined) 92 | * `∃x(D(x) ∧ M(x)` -> ∀y(If `y` is a friend of `z` and `z` lives in 93 | dorm then `y` has to be quarantined) 94 | * `∃x(D(x) ∧ M(x)) -> ∀y(∃z(F(y,z) ∧ D(z)) -> Q(y))` 95 | 96 | (c) 97 | 98 | F(x) = x failed the test 99 | G(x,y) = x got y grade. 100 | T(x,y) = x will tutor y 101 | 102 | 103 | * (Nobody failed the test) -> (Everybody who got an A will tutor 104 | someone who got a D) 105 | * ¬(Somebody failed the test) -> ∀y(If y got A grade, then y will 106 | tutor someone who got a D) 107 | * `¬(∃x F(x)) -> ∀y(G(y,'A') -> ∃z(T(y,z) ∧ G(z,'D')))` 108 | 109 | (d) 110 | 111 | D(x) = x can do it 112 | 113 | ∃x D(x) -> D(J) 114 | 115 | (e) 116 | 117 | D(x) = x can do it 118 | 119 | D(J) -> ∀xD(x) 120 | 121 | Exercise 3 122 | ----------- 123 | 124 | Analyze the logical forms of the following statements. The universe of 125 | discourse is R. What are the free variables in each statement? 126 | (a) Every number that is larger than x is larger than y. 127 | (b) For every number a, the equation ax^2 + 4x − 2 = 0 has at least one 128 | solution iff a ≥ −2. 129 | (c) All solutions of the inequality x^3 − 3x < 3 are smaller than 10. 130 | (d) If there is a number x such that x^2 + 5x = w and there is a number y 131 | such that 4 − y^2 = w, then w is between −10 and 10. 132 | 133 | (a) 134 | 135 | ∀z (z > x) -> (z > y) 136 | 137 | Free variables: x and y 138 | 139 | (b) 140 | 141 | ∀a∃x (ax^2 + 4x − 2 = 0) <-> (a >= -2) 142 | 143 | No free variables. 144 | 145 | (c) 146 | 147 | ∀x [x^3 − 3x < 3 -> x < 10] 148 | 149 | No free variables. 150 | 151 | (d) 152 | 153 | (∃x (x^2 + 5x = w) ∧ ∃y (4 − y^2 = w)) -> (-10 < w < 10) 154 | 155 | Free variable: w 156 | 157 | Exercise 4 158 | ----------- 159 | 160 | Translate the following statements into idiomatic English. 161 | (a) ∀x[(H (x) ∧ ¬∃y M(x, y)) → U (x)], where H (x) means “x is a man,” 162 | M(x, y) means “x is married to y,” and U (x) means “x is unhappy.” 163 | (b) ∃z(P(z, x) ∧ S(z, y) ∧ W (y)), where P(z, x) means “z is a parent of 164 | x,” S(z, y) means “z and y are siblings,” and W (y) means “y is a 165 | woman.” 166 | 167 | (a) 168 | 169 | * `∀x[(H (x) ∧ ¬∃y M(x, y)) → U (x)]` 170 | * `∀x`(if x is a man and for some y, x isn't married to y then x is 171 | unhappy) 172 | * for all x, if x is a man and for some y, x isn't married to y then 173 | x is unhappy. 174 | * Every unmarried man is unhappy. 175 | 176 | (b) 177 | 178 | * `∃z(P(z, x) ∧ S(z, y) ∧ W (y))` 179 | * `∃z`(z is a parent of x AND z and y are siblings AND y is a woman) 180 | * y is a sibling to z. 181 | 182 | Exercise 5 183 | ----------- 184 | 185 | Translate the following statements into idiomatic mathematical English. 186 | (a) ∀x[(P(x) ∧ ¬(x = 2)) → O(x)], where P(x) means “x is a prime 187 | number” and O(x) means “x is odd.” 188 | (b) ∃x[P(x) ∧ ∀y(P(y) → y ≤ x)], where P(x) means “x is a perfect 189 | number.” 190 | 191 | (a) 192 | 193 | * `∀x[(P(x) ∧ ¬(x = 2)) → O(x)]` 194 | * `∀x`(If x is a prime number AND x is not equal to 2, then x is an 195 | odd number) 196 | * Every prime number except 2 is an odd number. 197 | 198 | (b) 199 | 200 | * `∃x[P(x) ∧ ∀y(P(y) → y ≤ x)]` 201 | * ∃x(P(x) ∧ ∀y(If y is a perfect number then y is less than or equal 202 | to x.)) 203 | * ∃x(P(x) ∧ Every perfect number is less than or equal to x.) 204 | * There exists a perfect number such that all the perfect numbers are 205 | either less than or equal to it. 206 | 207 | Exercise 6 208 | ----------- 209 | 210 | Are these statements true or false? The universe of discourse is the set of 211 | all people, and P(x, y) means “x is a parent of y.” 212 | (a) ∃x∀y P(x, y). 213 | (b) ∀x∃y P(x, y). 214 | (c) ¬∃x∃y P(x, y). 215 | (d) ∃x¬∃y P(x, y). 216 | (e) ∃x∃y¬P(x, y). 217 | 218 | (a) 219 | 220 | * `∃x∀y P(x, y)` 221 | * There exists some parent x who is parent to all the peoples. 222 | 223 | This may appear true theoretically, but one cannot be a parent to 224 | himeself and hence False. 225 | 226 | False 227 | 228 | (b) 229 | 230 | * `∀x∃y P(x, y)` 231 | * Everyone is parent to someone. 232 | 233 | This is clearly `False` because if x is parent to y, then y is not a 234 | parent. 235 | 236 | (c) 237 | 238 | See also this 239 | [thread](http://math.stackexchange.com/questions/903549/concluding-truth-value-from-universe-of-discourse) 240 | for the discussion. 241 | 242 | * `¬∃x∃y P(x, y)` 243 | * `¬∃x(∃y(P(x, y)))` 244 | * There does not exist some x, there exists some y such that x is a 245 | parent of y. 246 | * There is no x such that x is a parent of someone. 247 | * There does not exist anyone who is a parent of someone. 248 | 249 | False 250 | 251 | (d) 252 | 253 | * `∃x¬∃y P(x, y)` 254 | * For some x, there exists no y such that x is a parent of y. 255 | * `x` has no children. 256 | * Some parents have no children. 257 | 258 | True 259 | 260 | (e) 261 | 262 | * `∃x∃y¬P(x, y)` 263 | * There exists some x and y such that x is not a parent to y. 264 | 265 | True 266 | 267 | Exercise 7 268 | ----------- 269 | 270 | Are these statements true or false? The universe of discourse is N. 271 | (a) ∀x∃y(2x − y = 0). 272 | (b) ∃y∀x(2x − y = 0). 273 | (c) ∀x∃y(x − 2y = 0). 274 | (d) ∀x(x < 10 → ∀y(y < x → y < 9)). 275 | (e) ∃y∃z(y + z = 100). 276 | (f) ∀x∃y(y > x ∧ ∃z(y + z = 100)). 277 | 278 | (a) 279 | 280 | * `∀x∃y(2x − y = 0)` 281 | * For all x and some y such that `2x − y = 0` 282 | 283 | True 284 | 285 | (b) 286 | 287 | * `∃y∀x(2x − y = 0)` 288 | * There exists some number y such that `2x − y = 0` for any number x. 289 | 290 | False 291 | 292 | (c) 293 | 294 | * `∀x∃y(x − 2y = 0)` 295 | * For any number x, there exists some number y such that `x − 2y = 0`. 296 | 297 | False (CounterExample: x = 3) 298 | 299 | (d) 300 | 301 | * `∀x(x < 10 → ∀y(y < x → y < 9))` 302 | * For any number x, if `x < 10` then for any number y, `y < x` implies 303 | `y < 9`. 304 | 305 | True 306 | 307 | (e) 308 | 309 | * `∃y∃z(y + z = 100)` 310 | 311 | True 312 | 313 | (f) 314 | 315 | * `∀x∃y(y > x ∧ ∃z(y + z = 100))` 316 | * `∀x∃y(x < y ∧ ∃z(y + z = 100))` 317 | * Counterexample: y = 1000, x = 999, z = 0 318 | 319 | False 320 | 321 | Exercise 8 322 | ---------- 323 | 324 | Are these statements true or false? The universe of discourse is R. 325 | (a) ∀x∃y(2x − y = 0). 326 | (b) ∃y∀x(2x − y = 0). 327 | (c) ∀x∃y(x − 2y = 0). 328 | (d) ∀x(x < 10 → ∀y(y < x → y < 9)). 329 | (e) ∃y∃z(y + z = 100). 330 | (f) ∀x∃y(y > x ∧ ∃z(y + z = 100)). 331 | 332 | (a) 333 | 334 | * `∀x∃y(2x − y = 0)` 335 | * For all x, there exist some y such that `2x - y = 0` 336 | 337 | True 338 | 339 | (b) 340 | 341 | * `∃y∀x(2x − y = 0)` 342 | * There exists some `y`, such that for any `x` it is `2x - y = 0` 343 | 344 | False 345 | 346 | (c) 347 | 348 | * `∀x∃y(x − 2y = 0)` 349 | * For all x and y `x - 2y = 0` 350 | 351 | True 352 | 353 | (d) 354 | 355 | * `∀x(x < 10 → ∀y(y < x → y < 9))` 356 | * For all x, if `x < 10` then for all y `(y < x → y < 9)` 357 | 358 | False (x = 9.9, y = 9.8) 359 | 360 | (e) 361 | 362 | * `∃y∃z(y + z = 100)` 363 | * For some y and z, `y + z = 100` 364 | 365 | True 366 | 367 | (f) 368 | 369 | * `∀x∃y(y > x ∧ ∃z(y + z = 100))` 370 | * For all `x` and some `y`, `y > x` and there exists some `z` such 371 | that `y + z = 100` 372 | 373 | True 374 | 375 | Exercise 9 376 | ----------- 377 | 378 | Are these statements true or false? The universe of discourse is Z. 379 | (a) ∀x∃y(2x − y = 0). 380 | (b) ∃y∀x(2x − y = 0). 381 | (c) ∀x∃y(x − 2y = 0). 382 | (d) ∀x(x < 10 → ∀y(y < x → y < 9)). 383 | (e) ∃y∃z(y + z = 100). 384 | (f) ∀x∃y(y > x ∧ ∃z(y + z = 100)). 385 | 386 | (a) 387 | 388 | * `∀x∃y(2x − y = 0)` 389 | * For all `x`, there exist some `y` such that `2x - y = 0` 390 | 391 | True 392 | 393 | (b) 394 | 395 | * `∃y∀x(2x − y = 0)` 396 | * There exists some `y`, such that for any `x` it is `2x - y = 0` 397 | 398 | False 399 | 400 | (c) 401 | 402 | * `∀x∃y(x − 2y = 0)` 403 | * For any `x`, there exists some `y` such that `x − 2y = 0` 404 | 405 | False (Counterexample: x = 3) 406 | 407 | (d) 408 | 409 | * `∀x(x < 10 → ∀y(y < x → y < 9))` 410 | * For all `x`, if `x < 10` then for all `y` `(y < x → y < 9)` 411 | 412 | True 413 | 414 | (e) 415 | 416 | * `∃y∃z(y + z = 100)` 417 | * For some `y` and `z`, `y + z = 100` 418 | 419 | True 420 | 421 | (f) 422 | 423 | * `∀x∃y(y > x ∧ ∃z(y + z = 100))` 424 | * For all `x` and some `y`, `y > x` and there exists some `z` such 425 | that `y + z = 100` 426 | 427 | True 428 | -------------------------------------------------------------------------------- /chapter 2/section2-3.tex: -------------------------------------------------------------------------------- 1 | %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% 2 | % Minimalists Report template 3 | % Author: Sibi 4 | %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% 5 | \documentclass{article} 6 | \usepackage{graphicx} 7 | \usepackage{verbatim} 8 | \usepackage{amsmath} 9 | \usepackage{amsfonts} 10 | \usepackage{amssymb} 11 | \begin{document} 12 | \title{Chapter 2 (Section 2.3)} 13 | \author{Sibi} 14 | \date{\today} 15 | \maketitle 16 | \newpage 17 | 18 | \section{Problem 1} 19 | 20 | \begin{center} 21 | (1.1) $F \subseteq \wp(A)$ 22 | \end{center} 23 | \begin{align*} 24 | \forall x (x \in F \implies x \in \wp(A)) \\ 25 | \forall x (x \in F \implies x \subseteq A) \\ 26 | \forall x (x \in F \implies \forall y (y \in x \implies y \in A)) 27 | \end{align*} 28 | 29 | \begin{center} 30 | (1.2) $A \subseteq \{2n + 1 | n \in \mathbb{N}\}$ \\ 31 | \end{center} 32 | \begin{align*} 33 | \forall x (x \in A \implies x \in \{2n + 1 | n \in \mathbb{N}\}) \\ 34 | \forall x (x \in A \implies \exists n \in \mathbb{N} (x=2n+1)) 35 | \end{align*} 36 | 37 | \begin{center} 38 | (1.3) $\{n^2 + n + 1 | n \in \mathbb{N} \} \subseteq \{2n + 1 | n \in \mathbb{N} \}$ \\ 39 | \end{center} 40 | \begin{align*} 41 | \forall x (x \in \left\{n^2 + n + 1 \mid n \in \mathbb{N} \right\}) \implies (x \in \left\{2n + 1 \mid n \in \mathbb{N} \right\}) \\ 42 | \forall x (\exists n \in \mathbb{N} (x = n^2 + n + 1)) \implies (\exists n \in \mathbb{N} (x=2n+1)) \\ 43 | \end{align*} 44 | 45 | \begin{center} 46 | (1.4) $\wp(\cup_{i \in I} A_i) \nsubseteq \cup_{i \in I} \wp(A_i)$ 47 | \end{center} 48 | \begin{align*} 49 | \exists x(x \in \wp(\cup_{i \in I} A_i) \land x \notin \cup_{i \in I} \wp(A_i)) \\ 50 | \exists x(x \subseteq \cup_{i \in I} A_i \land x \notin \cup_{i \in I} \wp(A_i)) \\ 51 | \exists x( \forall y (y \in x \implies y \in \cup_{i \in I} A_i) \land 52 | x \notin \cup_{i \in I} \wp(A_i)) \\ 53 | \exists x( \forall y (y \in x \implies \exists i \in I(y \in A_i)) 54 | \land x \notin \cup_{i \in I} \wp(A_i)) \\ 55 | \exists x( \forall y (y \in x \implies \exists i \in I(y \in A_i)) 56 | \land \neg (x \in \cup_{i \in I} \wp(A_i))) \\ 57 | \exists x( \forall y (y \in x \implies \exists i \in I(y \in A_i)) 58 | \land \neg (\exists i \in I(x \in \wp(A_i)))) \\ 59 | \exists x( \forall y (y \in x \implies \exists i \in I(y \in A_i)) 60 | \land \neg (\exists i \in I(x \subseteq A_i))) \\ 61 | \exists x( \forall y (y \in x \implies \exists i \in I(y \in A_i)) 62 | \land \neg (\exists i \in I(\forall z (z \in x \implies z \in A_i)))) \\ 63 | \exists x( \forall y (y \in x \implies \exists i \in I(y \in A_i)) 64 | \land (\forall i \in I(\exists z \neg(z \in x \implies z \in A_i)))) \\ 65 | \exists x( \forall y (y \in x \implies \exists i \in I(y \in A_i)) 66 | \land (\forall i \in I(\exists z (z \in x \land \neg(z \in A_i))))) \\ 67 | \end{align*} 68 | 69 | \section{Problem 2} 70 | \begin{center} 71 | (2.1) $x \in \cup F \setminus \cup G $ 72 | \end{center} 73 | \begin{align*} 74 | x \in \cup F \land x \notin \cup G \\ 75 | \exists A \in F(x \in A) \land \exists A \in G(x \notin A) 76 | \end{align*} 77 | 78 | \begin{center} 79 | (2.2) $\left\{ x \in B \mid x \notin C\right\} \in \wp(A)$ 80 | \end{center} 81 | \begin{align*} 82 | \left\{ x \in B \mid x \notin C\right\} \subseteq A \\ 83 | \forall y(y \in \left\{ x \in B \mid x \notin C\right\} \implies y \in A) \\ 84 | \forall y(y \in B \land y \notin C \implies y \in A) \\ 85 | \end{align*} 86 | 87 | \begin{center} 88 | (2.3) $ x \in \cap_{i \in I}(A_i \cup B_i)$ 89 | \end{center} 90 | \begin{align*} 91 | \forall i \in I(x \in (A_i \cup B_i)) \\ 92 | \forall i \in I(x \in A_i \lor x \in B_i) 93 | \end{align*} 94 | 95 | \begin{center} 96 | (2.4) $x \in (\cap_{i \in I}A_i) \cup (\cap_{i \in I}B_i)$ 97 | \end{center} 98 | \begin{align*} 99 | \forall i \in I(x \in A_i)) \lor (\forall i \in I(x \in B_i) 100 | \end{align*} 101 | 102 | \section{Problem 3} 103 | \begin{align*} 104 | \left\{\left\{\varnothing\right\}, \varnothing \right\} 105 | \end{align*} 106 | 107 | \section{Problem 4} 108 | \begin{align*} 109 | F = \left\{ \left\{red, green, blue \right\}, \left\{orange, red, 110 | green\right\}, \left\{purple, red, green, blue \right\} \right\} \\ 111 | \cup F = \left\{ red, green, blue, orange, purple\right\} \\ 112 | \cap F = \left\{ red, green\right\} \\ 113 | \end{align*} 114 | 115 | \section{Problem 5} 116 | \begin{align*} 117 | F = \left\{ \left\{3,7,12 \right\}, \left\{5,7,16 118 | \right\}, \left\{5,12,23 \right\} \right\} \\ 119 | \cup F = \left\{ 3,7,12,5,16,23 \right\} \\ 120 | \cap F = \varnothing \\ 121 | \end{align*} 122 | 123 | \section{Problem 6} 124 | \begin{center} 125 | $I = \left\{2,3,4,5\right\}$ \\ 126 | $A_i = \left\{i, i+1, i-1, 2*i\right\}$ \\ 127 | \end{center} 128 | \begin{align*} 129 | A_2 = \left\{2,3,1,4\right\} \\ 130 | A_3 = \left\{3,4,2,6\right\} \\ 131 | A_4 = \left\{4,5,3,8\right\} \\ 132 | A_5 = \left\{5,6,4,10\right\} \\ 133 | \cap_{i \in I}A_i = \left\{4 \right\} \\ 134 | \cup_{i \in I}A_i = \left\{1,2,3,4,5,6,8,10 \right\} \\ 135 | \end{align*} 136 | 137 | \section{Problem 7} 138 | Too lazy to figure out when they live. 139 | Or if you want a more reasonable answer: Left as an exercise to the reader :P 140 | \section{Problem 8} 141 | \begin{center} 142 | $I = \left\{2,3\right\}$ \\ 143 | $A_i = \left\{ i,2i \right\}$ \\ 144 | $B_i = \left\{ i, i + 1 \right\}$ \\ 145 | \end{center} 146 | \begin{align*} 147 | A_2 = \left\{ 2, 4 \right\} \\ 148 | A_3 = \left\{ 3, 6 \right\} \\ 149 | B_2 = \left\{ 2, 3 \right\} \\ 150 | B_3 = \left\{ 3, 4 \right\} \\ \\ 151 | \cap_{i \in I}(A_i \cup B_i) = \left\{3,4 \right\} \\ 152 | (\cap_{i \in I}A_i) \cup (\cap_{i \in I}B_i) = \left\{ 3\right\} \\ \\ 153 | No they are not equivalent. 2(c) and 2(d) are not equivalent to each other. 154 | \end{align*} 155 | 156 | \section{Problem 9} 157 | \begin{align*} 158 | Let X = \cup_{i \in I} (A_i \cap B_i) \\ 159 | Let Y = (\cup_{i \in I}A_i) \cap (\cup_{i \in I} B_i) \\ \\ 160 | x \in X is equivalent to x \in \cup_{i \in I} (A_i \cap B_i) \\ 161 | \exists i \in I(x \in (A_i \cap B_i))) \\ 162 | \exists i \in I(x \in A_i \land x \in B_i) \\ \\ 163 | x \in Y is equivalent to x \in (\cup_{i \in I}A_i) \cap (\cup_{i \in I} B_i) \\ 164 | (x \in \cup_{i \in I}A_i) \land (x \in \cup_{i \in I} B_i) \\ 165 | \exists i \in I(x \in A_i) \land \exists i \in I (x \in B_i) \\ \\ 166 | \end{align*} 167 | Now clearly they are not equivalent to each other. A trivial example of them would be: 168 | $Let I = \left\{1,2\right\}$ \\ 169 | $A_i = \left\{i\right\}$ \\ 170 | $B_i = \left\{2i\right\}$ \\ 171 | 172 | \section{Problem 10} 173 | \begin{align*} 174 | x \in \wp(A \cap B) \\ 175 | x \subseteq (A \cap B) \\ 176 | \forall y (y \in x \implies y \in (A \cap B)) \\ 177 | \forall y (y \in x \implies y \in A \land y \in B) \\ 178 | \forall y (y \notin x \lor (y \in A \land y \in B)) \\ 179 | \forall y ((y \notin x \lor y \in A) \land (y \notin x \lor y \in B)) \\ 180 | \forall y(y \in x \implies y \in A) \land \forall y(y \in x \implies y \in B) \\ \\ 181 | x \in \wp(A) \cap \wp(B) \\ 182 | x \in \wp(A) \land x \in \wp(B) \\ 183 | x \subseteq A \land x \subseteq B \\ 184 | \forall y(y \in x \implies y \in A) \land \forall y(y \in x \implies y \in B) \\ 185 | \end{align*} 186 | %% \left\{ \right\} 187 | \section{Problem 11} 188 | \begin{align*} 189 | A = \left\{1 \right\} \\ 190 | B = \left\{2 \right\} \\ 191 | \wp(A \cup B) = \left\{\left\{1,2 \right\},\left\{1 \right\}, \left\{2 \right\}, \varnothing \right\} \\ 192 | \wp(A) \cup \wp(B) = \left\{ \left\{1 \right\},\left\{2 \right\}, \varnothing \right\} 193 | \end{align*} 194 | 195 | \section{Problem 12} 196 | Problem (a) 197 | \begin{align*} 198 | \cup_{i \in I}(A_i \cup B_i) = (\cup_{i \in I}A_i) \cup (\cup_{i \in I}B_i) \\\\ 199 | x \in \cup_{i \in I}(A_i \cup B_i) \\ 200 | \exists i \in I(x \in A_i \cup B_i) \\ 201 | \exists i \in I(x \in A_i \lor x \in B_i)\\\\ 202 | x \in (\cup_{i \in I}A_i) \cup (\cup_{i \in I}B_i) \\ 203 | x \in (\cup_{i \in I}A_i) \lor x \in (\cup_{i \in I}B_i) \\ 204 | \exists i \in I(x \in A_i) \lor \exists i \in I(x \in B_i) \\ 205 | \exists i \in I(x \in A_i \lor x \in B_i) \\ 206 | \end{align*} 207 | Problem (b) 208 | \begin{align*} 209 | (\cap F) \cap (\cap G) = \cap(F \cup G)\\ 210 | x \in (\cap F) \cap (\cap G) \\ 211 | x \in (\cap F) \land x \in (\cap G) \\ 212 | \forall A \in F(x \in A) \land \forall A \in G(x \in A)\\\\ 213 | x \in \cap(F \cup G) \\ 214 | \forall A \in (F \cup G)(x \in A)\\ 215 | \forall A(A \in (F \cup G) \implies x \in A) \\ 216 | \forall A(A \in F \lor A \in G \implies x \in A)\\ 217 | \forall A((A \notin F \land A \notin G) \lor x \in A)\\ 218 | \forall A((A \notin F \lor x \in A) \land (A \notin G \lor x \in A))\\ 219 | \forall A(A \notin F \lor x \in A) \land \forall A(A \notin G \lor x \in A)\\ 220 | \forall A(A \in F \implies x \in A) \land \forall A(A \in G \implies x \in A)\\ 221 | \forall A \in F(x \in A) \land \forall A \in G(x \in A)\\ 222 | \end{align*} 223 | Problem (c) 224 | \begin{align*} 225 | \cap_{i \in I}(A_i \setminus B_i) = (\cap_{i \in I}A_i) \setminus (\cup_{i \in I}B_i)\\ 226 | x \in \cap_{i \in I}(A_i \setminus B_i) \\ 227 | \forall i \in I(x \in (A_i \setminus B_i)) \\ 228 | \forall i \in I(x \in A_i \land x \notin B_i)\\ 229 | \forall i \in I(x \in A_i) \land \forall i \in I(x \notin B_i)\\\\ 230 | x \in (\cap_{i \in I}A_i) \setminus (\cup_{i \in I}B_i)\\ 231 | x \in (\cap_{i \in I}A_i) \land x \notin (\cup_{i \in I}B_i) \\ 232 | \forall i \in I(x \in A_i) \land \neg (x \in \cup_{i \in I}B_i) \\ 233 | \forall i \in I(x \in A_i) \land \neg (\exists i \in I(x \in B_i)) \\ 234 | \forall i \in I(x \in A_i) \land \forall i \in I (x \notin B_i) 235 | \end{align*} 236 | 237 | \section{Problem 13} 238 | $$ I = \{1,2\} $$ 239 | $$ J = \{3,4\} $$ 240 | $$ A_{i,j} = \{i, j, i+j\} $$ 241 | Problem (a) 242 | \begin{align} 243 | B_j = U_{i \in I}A_{i,j} = A_{1,j} \cup A_{2,j} \\ 244 | B_3 = A_{1,3} \cup A_{2,3} \\ 245 | = \{ 1,3,4 \} \cup \{ 2,3,5 \} \\ 246 | = \{1,2,3,4,5\} \\ \\ 247 | B_4 = A_{1,4} \cup A_{2,4} \\ 248 | = \{1,4,5\} \cup \{2,4,6\} \\ 249 | = \{1,2,4,5,6\} \\ 250 | \end{align} 251 | Problem (b) 252 | \begin{align*} 253 | \cap_{j \in J}B_j \\ 254 | B_3 \cap B_4 \\ 255 | \{1,2,4,5\} \\ 256 | \end{align*} 257 | Problem (c) 258 | \begin{align*} 259 | \cup_{i \in I}(\cap_{j \in J}A_{i,j}) \\ 260 | \cup_{i \in I}(A_{i,3} \cap A_{i,4}) \\ 261 | (A_{1,3} \cap A_{1,4}) \cup (A_{2,3} \cap A_{2,4}) \\ 262 | (\{1,3,4\} \cap \{1,4,5\}) \cup (\{2,3,5\} \cap \{2,4,6\}) \\ 263 | \{1,4\} \cup \{2\} \\ 264 | \{1,2,4\} \\ 265 | \end{align*} 266 | No, they are not equivalent. \\ 267 | Problem (d) 268 | \begin{align*} 269 | x \in \cap_{j \in J}(\cup_{i \in I}A_{i,j}) \\ 270 | \forall j \in J(x \in \cup_{i \in I}A_{i,j})\\ 271 | \forall j \in J(\exists i \in I(x \in A_{i,j}))\\ \\ 272 | x \in \cup_{i \in I}(\cap_{j \in J}A_{i,j}) \\ 273 | \exists i \in I(x \in (\cap_{j \in J}A_{i,j})) \\ 274 | \exists i \in I(\forall j \in J(x \in A_{i,j}))\\ 275 | \end{align*} 276 | No, they are not equivalent. 277 | \section{Problem 14} 278 | Problem (a) 279 | \begin{align*} 280 | x \in \cup F \\ 281 | \exists A \in F(x \in A) \\ 282 | \exists A (A \in F \land x \in A) \\ 283 | \end{align*} 284 | Now if $F$ is $\emptyset$ then the statement is obviously false. There 285 | is no $A$ that will satisfy that logical form. Hence $\cup \emptyset = 286 | \emptyset$ \\ 287 | 288 | Problem (b) 289 | \begin{align*} 290 | x \in \cap F \\ 291 | \forall A \in F(x \in A) \\ 292 | \forall x (A \in F \implies x \in A) \\ 293 | \end{align*} 294 | Now if $F$ is $\emptyset$, then the antecedent is false. But the 295 | logical form is vacously true. So every element in $U$ will satisfy 296 | that. Therefore, $\cap \emptyset = U$ 297 | \end{document} 298 | 299 | -------------------------------------------------------------------------------- /chapter 1/section1-5.md: -------------------------------------------------------------------------------- 1 | The Conditional and Biconditional Connectives 2 | ----------------------------------------------- 3 | 4 | Notes from Keith Devlin's lectures: 5 | 6 | * Implication has a truth part and causality part. 7 | * Leave the causation part to the philosophers and just in mathematics 8 | we will just focus on the truth part. 9 | * Whenever we have genuine implication (with causality), the truth 10 | behaviour of the conditional is the correct one. 11 | * When we do have a genuine implication, the definition of the 12 | conditional will agree with the way implication behaves. And when we 13 | don't have genuine implication, the conditional will still be 14 | defined. 15 | 16 | The following all mean p implies q: 17 | 18 | * if p, then q 19 | * p is sufficient for q 20 | * p only if q 21 | * q if p 22 | * q whenever p 23 | * q is necessary for p 24 | 25 | For bicondtional, p <=> q these are all the same: 26 | 27 | * p is equivalent to q is itself equivalent to. 28 | * p is necessary and sufficient for q. 29 | * p if and only if q 30 | 31 | Exercise 1 32 | ----------- 33 | 34 | Analyze the logical forms of the following statements: 35 | (a) If this gas either has an unpleasant smell or is not explosive, then it 36 | isn’t hydrogen. 37 | (b) Having both a fever and a headache is a sufficient condition for George 38 | to go to the doctor. 39 | (c) Both having a fever and having a headache are sufficient conditions 40 | for George to go to the doctor. 41 | (d) If x =/ 2, then a necessary condition for x to be prime is that 42 | x be odd. 43 | 44 | (a) 45 | 46 | S = Gas has an unpleasant smell. 47 | E = Gas is explosive. 48 | H = Gas is hydrogen. 49 | 50 | S ∨ ¬E → ¬H 51 | 52 | (b) 53 | 54 | F = George is having Fever. 55 | H = George is having Headache. 56 | D = George goes to doctor. 57 | 58 | (F ∧ H) → D 59 | 60 | (c) 61 | 62 | (F ∨ H) → D 63 | ¬(F ∨ H) ∨ D 64 | (¬F ∧ ¬H) ∨ D 65 | (D ∨ ¬F) ∧ (D ∨ ¬H) 66 | (F -> D) ∧ (H -> D) 67 | 68 | (d) 69 | 70 | T = x is two. 71 | P = x is prime. 72 | O = x is odd. 73 | 74 | p -> q means Q is necessary condition for P 75 | necessary condition for x to be prime is that x be odd 76 | necessary condition for P is O 77 | O is necessary condition for P means (P -> O) 78 | 79 | ¬T -> (P -> O) 80 | 81 | Exercise 2 82 | ----------- 83 | 84 | Analyze the logical forms of the following statements: 85 | (a) Mary will sell her house only if she can get a good price and find a 86 | nice apartment. 87 | (b) Having both a good credit history and an adequate down payment is a 88 | necessary condition for getting a mortgage. 89 | (c) John will kill himself, unless someone stops him. (Hint: First try to 90 | rephrase this using the words if and then instead of unless.) 91 | (d) If x is divisible by either 4 or 6, then it isn’t prime. 92 | 93 | (a) 94 | 95 | p -> q is the same as "p only if q" 96 | 97 | H = Mary will sell her house. 98 | P = Mary will get a good price. 99 | A = Mary will find a nice apartment. 100 | 101 | H -> P ∧ A 102 | 103 | (b) 104 | 105 | p -> q is the same as "q is the necessary for p" 106 | 107 | C = Having good credit history. 108 | D = Having adequate down payment. 109 | M = Getting mortgage. 110 | 111 | M -> C ∧ D 112 | 113 | (c) 114 | 115 | S = Someone stops John 116 | K = John kills himself 117 | 118 | Not being stopped by someone is a necessary and sufficient condition 119 | for John to kill himself. If he is stopped, then he cannot kill 120 | himself and if he kills himself, then he was not stopped. 121 | 122 | ¬S <-> K 123 | 124 | (d) 125 | 126 | D(x,y) = x is divisible by y 127 | P(x) = x is prime. 128 | 129 | D(x,4) ∨ D(x,6) -> ¬P(x) 130 | 131 | Exercise 3 132 | ---------- 133 | 134 | Analyze the logical form of the following statement: 135 | (a) If it is raining, then it is windy and the sun is not shining. 136 | Now analyze the following statements. Also, for each statement determine 137 | whether the statement is equivalent to either statement (a) or its converse. 138 | (b) It is windy and not sunny only if it is raining. 139 | (c) Rain is a sufficient condition for wind with no sunshine. 140 | (d) Rain is a necessary condition for wind with no sunshine. 141 | (e) It’s not raining, if either the sun is shining or it’s not windy. 142 | (f) Wind is a necessary condition for it to be rainy, and so is a lack of 143 | sunshine. 144 | (g) Either it is windy only if it is raining, or it is not sunny only if it is 145 | raining. 146 | 147 | (a) 148 | 149 | R = It is raining. 150 | W = It is windy. 151 | S = Sun is shining. 152 | 153 | R -> W ∧ ¬S 154 | 155 | (b) 156 | 157 | p -> q is the same as p only if q 158 | 159 | * It is windy and not sunny only if it is raining. 160 | * If it is windy and not sunny, then it is raining. 161 | 162 | W ∧ ¬S -> R (Converse (a)) 163 | 164 | (c) 165 | 166 | p -> q is the same as "p is sufficient for q" 167 | 168 | R -> W ∧ ¬S (Equivalent to (a)) 169 | 170 | (d) 171 | 172 | p -> q is the same as "q is necessary for p" 173 | 174 | W ∧ ¬S -> R (Converse of (a)) 175 | 176 | (e) 177 | 178 | p -> q is the same as "q if p" 179 | 180 | S ∨ ¬W -> ¬R (Equivalent to (a)) 181 | 182 | (f) 183 | 184 | p -> q is the same as "q is necessary for p" 185 | 186 | (R -> W) ∧ (R -> ¬S) 187 | (¬R ∨ W) ∧ (¬R ∨ ¬S) 188 | (((¬R ∨ W) ∧ ¬R) ∨ ((¬R ∨ W) ∧ ¬S)) 189 | (¬R ∨ (¬R ∧ W)) ∨ ((¬R ∨ W) ∧ ¬S) 190 | ¬R ∨ ((¬R ∨ W) ∧ ¬S) 191 | ¬R ∨ ((¬R ∧ ¬S) ∨ (¬S ∧ W)) 192 | ¬R ∨ (¬S ∧ W) 193 | R -> ¬S ∧ W (Equivalent to (a)) 194 | 195 | 196 | (g) 197 | 198 | p -> q is the same as "p only if q" 199 | 200 | (W -> R) ∨ (¬S -> R) 201 | (¬W ∨ R) ∨ (S ∨ R) 202 | ¬W ∨ R ∨ S ∨ R 203 | R ∨ ¬W ∨ S 204 | ¬R -> ¬W ∨ S (Converse of (a)) 205 | 206 | Exercise 4 207 | ----------- 208 | 209 | Use truth tables to determine whether or not the following arguments are 210 | valid: 211 | (a) Either sales or expenses will go up. If sales go up, then the boss will 212 | be happy. If expenses go up, then the boss will be unhappy. Therefore, 213 | sales and expenses will not both go up. 214 | (b) If the tax rate and the unemployment rate both go up, then there will 215 | be a recession. If the GNP goes up, then there will not be a recession. 216 | The GNP and taxes are both going up. Therefore, the unemployment 217 | rate is not going up. 218 | (c) The warning light will come on if and only if the pressure is too high and 219 | the relief valve is clogged. The relief valve is not clogged. Therefore, 220 | the warning light will come on if and only if the pressure is too 221 | high. 222 | 223 | (a) 224 | 225 | S = Sales will go up. 226 | 227 | E = Expenses will go up. 228 | 229 | B = Boss will be happy. 230 | 231 | S ∨ E 232 | S -> B 233 | E -> ¬B 234 | --------- 235 | ¬(S ∧ E) 236 | 237 | Conjunction of predicate: `(S ∨ E) ∧ (S -> B) ∧ (E -> ¬B)` 238 | 239 | Conclusion: `¬(S ∧ E)`: 240 | 241 | Truth table of `((S | E) & ((S -> B) & (E -> ~B))) -> ~(S & E)` 242 | 243 | B E S | (((S | E) & ((S -> B) & (E -> ~B))) -> ~(S & E)) 244 | -------------------------------------------------------- 245 | T T T | T 246 | T T F | T 247 | T F T | T 248 | T F F | T 249 | F T T | T 250 | F T F | T 251 | F F T | T 252 | F F F | T 253 | 254 | Voila, the argument is valid. 255 | 256 | (b) 257 | 258 | T = Tax rate goes up 259 | 260 | U = Unemployment rate goes up. 261 | 262 | R = There will be recession. 263 | 264 | G = GNP goes up. 265 | 266 | (T ∧ U) -> R 267 | G -> ¬R 268 | G ∧ T 269 | ------------- 270 | ¬U 271 | 272 | Disjunction of premises: `((T ∧ U) -> R) ∧ (G -> ¬R) ∧ (G ∧ T)` 273 | Conclusion: `¬U` 274 | 275 | Truth table of `((T ∧ U) -> R) ∧ (G -> ¬R) ∧ (G ∧ T) -> ¬U`: 276 | 277 | G R T U | ((((T & U) -> R) & ((G -> ~R) & (G & T))) -> ~U) 278 | ---------------------------------------------------------- 279 | T T T T | T 280 | T T T F | T 281 | T T F T | T 282 | T T F F | T 283 | T F T T | T 284 | T F T F | T 285 | T F F T | T 286 | T F F F | T 287 | F T T T | T 288 | F T T F | T 289 | F T F T | T 290 | F T F F | T 291 | F F T T | T 292 | F F T F | T 293 | F F F T | T 294 | F F F F | T 295 | 296 | The argument is valid. 297 | 298 | (c) 299 | 300 | W = Warning light will come 301 | 302 | P = Pressure is too high 303 | 304 | R = Relief valve is clogged. 305 | 306 | W <-> (P ∧ R) 307 | ¬R 308 | ------------- 309 | W <-> P 310 | 311 | Disjunction of premises: `(W <-> (P ∧ R)) ∧ ¬R` 312 | Conclusion: `W <-> P` 313 | 314 | Truth table for `((W <-> (P ∧ R)) ∧ ¬R) -> (W <-> P)`: 315 | 316 | P R W | (((W <-> (P & R)) & ~R) -> (W <-> P)) 317 | --------------------------------------------- 318 | T T T | T 319 | T T F | T 320 | T F T | T 321 | T F F | F 322 | F T T | T 323 | F T F | T 324 | F F T | T 325 | F F F | T 326 | 327 | The truth table isn't `T` for all the cases and hence the argument is invalid. 328 | 329 | Exercise 5 330 | ----------- 331 | 332 | (a) Show that P ↔ Q is equivalent to (P ∧ Q) ∨ (¬P ∧ ¬Q). 333 | (b) Show that (P → Q) ∨ (P → R) is equivalent to P → (Q ∨ R). 334 | 335 | (a) 336 | 337 | P <-> Q 338 | (P -> Q) ∧ (Q -> P) 339 | (¬P ∨ Q) ∧ (¬Q ∨ P) 340 | ((¬P ∨ Q) ∧ ¬Q) ∨ ((¬P ∨ Q) ∧ P) 341 | (¬P ∧ ¬Q) ∨ (Q ∧ P) 342 | (P ∧ Q) ∨ (¬P ∧ ¬Q) 343 | 344 | (b) 345 | 346 | (P -> Q) ∨ (P -> R) 347 | (¬P ∨ Q) ∨ (¬P ∨ R) 348 | ¬P ∨ Q ∨ ¬P ∨ R 349 | ¬P ∨ (Q ∨ R) 350 | P -> (Q ∨ R) 351 | 352 | Exercise 6 353 | ----------- 354 | 355 | (a) Show that (P → R) ∧ (Q → R) is equivalent to (P ∨ Q) → R. 356 | (b) Formulate and verify a similar equivalence involving (P → R) ∨ 357 | (Q → R). 358 | 359 | (a) 360 | 361 | (P -> R) ∧ (Q -> R) 362 | (¬P ∨ R) ∧ (¬Q ∨ R) 363 | (¬P ∧ ¬Q) ∨ R [Distributive Law] 364 | ¬(P ∨ Q) ∨ R 365 | (P ∨ Q) -> R 366 | 367 | (b) 368 | 369 | (P -> R) ∨ (Q -> R) 370 | (¬P ∨ R) ∨ (¬Q ∨ R) 371 | ¬P ∨ ¬Q ∨ R 372 | ¬(P ∧ Q) ∨ R 373 | (P ∧ Q) -> R 374 | 375 | Exercise 7 376 | ----------- 377 | 378 | (a) Show that (P → Q) ∧ (Q → R) is equivalent to (P → R) ∧ 379 | [(P ↔ Q) ∨ (R ↔ Q)]. 380 | (b) Show that (P → Q) ∨ (Q → R) is a tautology. 381 | 382 | (a) 383 | 384 | Truth table of `(P → Q) ∧ (Q → R)`: 385 | 386 | P Q R | ((P -> Q) & (Q -> R)) 387 | ----------------------------- 388 | T T T | T 389 | T T F | F 390 | T F T | F 391 | T F F | F 392 | F T T | T 393 | F T F | F 394 | F F T | T 395 | F F F | T 396 | 397 | Truth table of `(P → R) ∧ [(P ↔ Q) ∨ (R ↔ Q)]`: 398 | 399 | P Q R | ((P -> R) & ((P <-> Q) | (R <-> Q))) 400 | -------------------------------------------- 401 | T T T | T 402 | T T F | F 403 | T F T | F 404 | T F F | F 405 | F T T | T 406 | F T F | F 407 | F F T | T 408 | F F F | T 409 | 410 | As seen in the truth table, they are equivalent. 411 | 412 | (b) 413 | 414 | Truth table of `(P → Q) ∨ (Q → R)`: 415 | 416 | P Q R | ((P -> Q) | (Q -> R)) 417 | ----------------------------- 418 | T T T | T 419 | T T F | T 420 | T F T | T 421 | T F F | T 422 | F T T | T 423 | F T F | T 424 | F F T | T 425 | F F F | T 426 | 427 | As every row in fourth column is `T`, they are tautology. 428 | 429 | Alternative Solution: 430 | 431 | (P → Q) ∨ (Q → R) 432 | (¬P ∨ Q)∨ (¬Q v R) Conditional Law 433 | (¬P ∨ R)∨ (¬Q v Q) Associative Law 434 | In the above statement (¬Q v Q) is a tautology, 435 | (¬P ∨ R)∨ (Tautology) 436 | From tautology laws, (Statement) v (tautology) is a tautology 437 | Hence the complete statement is a tautology 438 | 439 | Exercise 8 440 | ----------- 441 | 442 | Find a formula involving only the connectives ¬ and → that is equivalent 443 | to P ∧ Q. 444 | 445 | Soln: 446 | 447 | P ∧ Q 448 | ¬(¬P ∨ ¬Q) 449 | ¬(P -> ¬Q) 450 | 451 | Exercise 9 452 | ----------- 453 | 454 | Find a formula involving only the connectives ¬ and → that is equivalent 455 | to P ↔ Q. 456 | 457 | Soln: 458 | 459 | P <-> Q 460 | (P -> Q) ∧ (Q -> P) 461 | ¬(¬(P -> Q) ∨ ¬(Q -> P)) 462 | ¬((P -> Q) -> ¬(Q -> P)) 463 | 464 | Exercise 10 465 | ------------ 466 | 467 | Which of the following formulas are equivalent? 468 | (a) P → (Q → R). 469 | (b) Q → (P → R). 470 | (c) (P → Q) ∧ (P → R). 471 | (d) (P ∧ Q) → R. 472 | (e) P → (Q ∧ R). 473 | 474 | (a) 475 | 476 | P → (Q → R) 477 | ¬P ∨ ¬Q ∨ R 478 | 479 | (b) 480 | 481 | Q → (P → R) 482 | ¬Q ∨ ¬P ∨ R 483 | 484 | (c) 485 | 486 | (P → Q) ∧ (P → R) 487 | (¬P ∨ Q) ∧ (¬P ∨ R) 488 | ¬P ∨ (Q ∧ R) 489 | 490 | (d) 491 | 492 | (P ∧ Q) → R 493 | ¬P ∨ ¬Q ∨ R 494 | 495 | (e) 496 | 497 | P → (Q ∧ R) 498 | ¬P ∨ (Q ∧ R) 499 | 500 | (a), (b) & (d) are equivalent. 501 | (c) & (e) are equivalent. 502 | -------------------------------------------------------------------------------- /chapter 5/section5-3.tex: -------------------------------------------------------------------------------- 1 | %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% 2 | % Author: Sibi 3 | %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% 4 | \documentclass{article} 5 | \usepackage{graphicx} 6 | \usepackage{verbatim} 7 | \usepackage{amsmath} 8 | \usepackage{amsfonts} 9 | \usepackage{amssymb} 10 | \usepackage{tabularx} 11 | \usepackage{mathtools} 12 | \newcommand{\BigO}[1]{\ensuremath{\operatorname{O}\bigl(#1\bigr)}} 13 | \setlength\parskip{\baselineskip} 14 | \begin{document} 15 | \title{Chapter 5 (Section 5.3)} 16 | \author{Sibi} 17 | \date{\today} 18 | \maketitle 19 | 20 | % See here: http://tex.stackexchange.com/a/43009/69223 21 | \DeclarePairedDelimiter\abs{\lvert}{\rvert}% 22 | \DeclarePairedDelimiter\norm{\lVert}{\rVert}% 23 | 24 | % Swap the definition of \abs* and \norm*, so that \abs 25 | % and \norm resizes the size of the brackets, and the 26 | % starred version does not. 27 | \makeatletter 28 | \let\oldabs\abs 29 | \def\abs{\@ifstar{\oldabs}{\oldabs*}} 30 | % 31 | \let\oldnorm\norm 32 | \def\norm{\@ifstar{\oldnorm}{\oldnorm*}} 33 | \makeatother 34 | \newpage 35 | 36 | \section{Solution 1} 37 | $R^{-1}(p) = $ the person immediately right to $p.$ 38 | 39 | \section{Solution 2} 40 | $F^{-1}(X) = \text{unique } Y \text{ such that} F(Y) = X$. 41 | So, \begin{align*} 42 | F(Y) = X \\ 43 | A \setminus Y = X \\ 44 | A \setminus X = Y \\ \\ 45 | F^{-1}(X) = A \setminus X 46 | \end{align*} 47 | 48 | \section{Solution 3} 49 | We will try to find a function $g: R \to R$ such that $f \circ g = 50 | i_\mathbb{R}$ and $g \circ f = i_\mathbb{R}$. 51 | 52 | We are hoping to find $g = f^{-1}$. So, $f(x) = \text{unique }y$ such 53 | that $f^{-1}(y) = x$. So, $\frac{2x + 5}{3} = y$. Solving it, we get 54 | $x = \frac{3y-5}{2}$. So, $g(x) = \frac{3x-5}{2}$. 55 | 56 | Let's check for $f \circ g = id$. 57 | \begin{align*} 58 | f(g(x)) \\ 59 | = f(\frac{3x-5}{2}) \\ 60 | = \frac{3x - 5 + 5}{3} \\ 61 | = x \\ 62 | = id_{\mathbb{R}}(x) 63 | \end{align*} 64 | 65 | Let's check for $g \circ f = id $ 66 | \begin{align*} 67 | g(f(x)) \\ 68 | = g(\frac{2x + 5}{3}) \\ 69 | = \frac{2x + 5 - 5}{2} \\ 70 | = x \\ 71 | = id_{\mathbb{R}}(x) 72 | \end{align*} 73 | 74 | From theorem 5.3.4, we can conclude that $f$ is one-to-one and onto. 75 | 76 | \section{Solution 4} 77 | We will try to find a function $g: R \to R$ such that $f \circ g = 78 | i_\mathbb{R}$ and $g \circ f = i_\mathbb{R}$. 79 | 80 | We are hoping to find $g = f^{-1}$. So, $f(x) = \text{unique }y$ such 81 | that $f^{-1}(y) = x$. So, $2x^3 - 3 = y$. Solving it, we get 82 | $x = (\frac{3+y}{2})^{1/3}$. So, $g(x) = (\frac{3+x}{2})^{1/3}$. 83 | 84 | Let's check for $f \circ g = id$. 85 | \begin{align*} 86 | f \circ g(x) \\ 87 | = f((\frac{3+x}{2})^{1/3}) \\ 88 | = 3 + x - 3 \\ 89 | = x \\ 90 | = id_R \\ 91 | \end{align*} 92 | 93 | Similarly, $g \circ f = id_R$ 94 | 95 | From theorem 5.3.4, we can conclude that $f$ is one-to-one and onto. 96 | 97 | \section{Solution 5} 98 | We will try to find a function $g: R^{+} \to R$ such that $f \circ g = 99 | i_\mathbb{R^{+}}$ and $g \circ f = i_\mathbb{R}$. 100 | 101 | We are hoping to find $g = f^{-1}$. So, $f(x) = \text{unique }y$ such 102 | that $f^{-1}(y) = x$.So, 103 | \begin{align*} 104 | 10^{2-x} = y \\ 105 | \log_{10}10^{2-x} = \log_{10}y \\ 106 | 2 - x = log_{10}y \\ 107 | x = 2 - log y 108 | \end{align*} 109 | 110 | So, $g(x) = 2 - log(x)$. Let's verify $f \circ g = id_{\mathbb{R^{+}}}$ 111 | 112 | \begin{align*} 113 | f \circ g(x) \\ 114 | = f(2 - log x) \\ 115 | = 10^{2 + log x - 2} \\ 116 | = 10^{log x} \\ 117 | = x \\ 118 | = id_{\mathbb{R^{+}}} 119 | \end{align*} 120 | 121 | Also, let's check $g \circ f = id_R$ 122 | \begin{align*} 123 | g \circ f(x) \\ 124 | = g(10^{2 - x}) \\ 125 | = 2 - log_{10} 10^{2-x}\\ 126 | = 2 - (2 - x) \\ 127 | = x \\ 128 | \end{align*} 129 | 130 | From theorem 5.3.4, we can conclude that $f$ is one-to-one and onto. 131 | 132 | \section{Solution 6} 133 | \subsection{Solution 6(a)} 134 | We will try to find a $g: B \to A$ such that $f \circ g = i_A$ and $g 135 | \circ f = i_A$. 136 | 137 | $f(x) = $ unique $y$ such that $f^{-1}(y) = x$. 138 | 139 | So, $\frac{3x}{x-2} = y$ \\ 140 | Solving it, we get $f^{-1}(y) = \frac{2y}{y-3}$ 141 | 142 | No $g$ is undefined at $x = 3$. So, let the set $B$ be $R \setminus {3}$. 143 | 144 | Now, let's verify $f \circ g = i_B$ 145 | \begin{align*} 146 | f \circ g(b) \\ 147 | = f(g(b)) \\ 148 | = f(\frac{2b}{b-3}) \\ 149 | = \frac{6b}{b} \\ 150 | = b \\ 151 | = i_B(b) 152 | \end{align*} 153 | 154 | Also, $g \circ f = i_A$ 155 | \begin{align*} 156 | g \circ f(a) \\ 157 | = g(f(a)) \\ 158 | = g(\frac{3a}{a-2}) \\ 159 | = \frac{6a}{a} 160 | = a \\ 161 | = i_A(a) 162 | \end{align*} 163 | 164 | \subsection{Solution 6(b)} 165 | From theorem $5.3.4$, it follows that $f^{-1}(x) = \frac{2x}{x-3}$ 166 | 167 | \section{Solution 7} 168 | \subsection{Solution 7(a)} 169 | Let $a,b$ be arbitrary element in $R$. 170 | 171 | $(\Rightarrow)$ Suppose $(a,b) \in f$. It follows that 172 | $b = a + \frac{7}{5}$. So, $(a,a+\frac{7}{5}) \in f$. Applying $a$ to 173 | $f_2 \circ f_1$, we get $a + \frac{7}{5}$. So, $(a, a+\frac{7}{5}) \in 174 | f_2 \circ f_1$. So, $f \subseteq f_2 \circ f_1$. 175 | $(\Leftarrow)$ Similar to above. 176 | 177 | \subsection{Solution 7(b)} 178 | Applying $a$ to $f^{-1}$, we get $5a - 7$. Solving for inverse 179 | functions, we get: 180 | \begin{align*} 181 | f_1^{-1}(x) = x - 7 \\ 182 | f_2^{-1}(x) = 5x 183 | \end{align*} 184 | 185 | Solving them: 186 | \begin{align*} 187 | f_1^{-1}(f_2^{-1}(a)) \\ 188 | = f_1^{-1}(51) \\ 189 | = 5a - 7 190 | \end{align*} 191 | 192 | So, yes they are same! 193 | 194 | \section{Solution 8} 195 | \subsection{Solution (a)} 196 | Let $b$ be arbitrary element of $B$. Let $a = f^{-1}(b) \in A$. Then 197 | $(b,a) \in f^{-1}$. Thus, 198 | \begin{align*} 199 | f \circ f^{-a}(b) \\ 200 | = f(f^{-1}(b)) \\ 201 | = f(a) \\ 202 | = b \\ 203 | = i_B(b) 204 | \end{align*} 205 | 206 | Since $b$ was arbitrary, we can conclude that $f \circ f^{-1} = i_B$. 207 | 208 | \subsection{Solution (b)} 209 | We know that, 210 | \begin{align*} 211 | f^{-1} \circ f = i_A \\ 212 | f^{-1} \circ f \circ f^{-1} = i_A \circ f^{-1} 213 | \end{align*} 214 | 215 | Let $b$ be arbitrary element in $B$. Suppose $f(a) = b$. Then $(a,b) 216 | \in f$ and $(b,a) \in f^{-1}$. So, 217 | \begin{align*} 218 | f^{-1} \circ (f \circ f^{-1})(b) = i_A \circ f^{-1}(b) \\ 219 | f^{-1}(b) = i_A(a) \\ 220 | f^{-1}(b) = a \\ 221 | f \circ f^{-1}(b) = f(a) \tag{Apply f} \\ 222 | f \circ f^{-a} (b) = b \\ 223 | f \circ f^{-a} (b) = i_B(b) 224 | \end{align*} 225 | Since $b$ was arbitrary, we can conclude that $f \circ f^{-1} = i_B$. 226 | 227 | \section{Solution 9} 228 | Suppose $g: B \to A$ and $f \circ g = i_B$. Let $b$ be arbitrary 229 | element in $B$. Then $f \circ g(b) = i_B(b)$. Since $g$ is a function 230 | $\exists a \in A$ such that $g(b) = a$. So, $f(a) = b$. Since $b$ was 231 | arbitrary, we can conclude that $f$ is onto. 232 | 233 | \section{Solution 10} 234 | Suppose $f: A \to B, g: B \to A, g \circ f = i_A$ and $f \circ g = 235 | i_B$. Let $(b,a)$ be arbitrary element of $B \times A$ such that 236 | $(b,a) \in g$. Then $f \circ g(b) = f(a)$. From $f \circ g = i_B$, it 237 | follows that $f(a) = b$. So, $(a,b) \in f^{-1}$. So, $g \subseteq 238 | f^{-1}$. Similarly, do the other case with $g \circ f = i_A$ and show 239 | $f^{-1} \subseteq g$. 240 | 241 | \section{Solution 11} 242 | \subsection{Solution (a)} 243 | Suppose $f:A \to B$ and $g:B \to A$. Suppose $f$ is one to one and $f 244 | \circ g = i_B$. From theorem $5.3.3$, it follows that $f$ is onto. Now 245 | from $5.3.4$, it follows that $g \circ f = i_A$. Now from theorem 246 | $5.3.5$, we can conclude that $g = f^{-1}$. 247 | 248 | \subsection{Solution (b)} 249 | Suppose $f:A \to B$ and $g:B \to A$. Suppose $f$ is onto and $g \circ 250 | f = i_A$. From theorem, $5.3.3$, it follows that $f$ is one to one. 251 | From $5.3.4$, it follows that $f \circ g = i_B$. From $5.3.5$, it 252 | follows that $g = f^{-1}$ 253 | 254 | \subsection{Solution (c)} 255 | Suppose $f \circ g = i_B$ and $g \circ f \neq i_A$. 256 | 257 | \subsubsection{Solution (i)} 258 | From theorem $5.3.3$, it follows that $f$ is onto. 259 | 260 | \subsubsection{Solution (ii)} 261 | We know that $g \circ f \neq i_A$. So, $g \neq f^{-1}$. Applying 262 | contrapositive law to $(a)$, we get $g \neq f^{-1} \implies f \circ g 263 | \neq i_B \lor f \text{ is not one to one}$. We know that $f \circ g = 264 | i_B$. So $f$ is not one to one. 265 | 266 | Rest of the proofs are based on the above technique. 267 | 268 | \section{Solution 12} 269 | Suppose $f:A \to B$ and $f$ is one to one. Let $B^{-1} = Ran(f)$. Let 270 | $b$ be an arbitrary element in $B^{-1}$. 271 | 272 | Existence proof. Since $b \in Ran(f)$, $\exists a \in A$ such that 273 | $f(a) = b$. So, $(b,a) \in f^{-1}$. 274 | 275 | Uniqueness proof. Let $(b,a_1) \in f^{-1}$ and $(b,a_2) \in f^{-1}$. 276 | It follows that $(a_1, b) \in f$ and $(a_2, b) \in f$. Since $f$ is 277 | one to one, it follows that $a_1 = a_2$. 278 | 279 | \section{Solution 13} 280 | Suppose $f: A \to B$ and $f$ is onto. Let $R = \{(x,y) \in A \times A 281 | \mid f(x) = f(y) \}$ 282 | 283 | \subsection{Solution (a)} 284 | Let $X$ be an arbitrary element in $A/R$. Then it follows that there 285 | is some $x \in A$ such that $x \in X$. 286 | 287 | Existence. Since $x \in A$ and $f: A \to B$, it follows that $\exists 288 | b \in B$ such that $f(x) = b$. So, $(X,b) \in h$. 289 | 290 | Uniqueness. Suppose $(X,b_1) \in h$ and $(X, b_2) \in h$. From $(X, 291 | b_1) \in h$, it follows that $\exists x \in X$ such that $f(x)=b_1$. 292 | Similarly from $(X,b_1) \in h$, it follows that $\exists y \in X$ such 293 | that $f(y)=b_1$. We know that $xRy$. From the definition of $R$, it 294 | follows that $f(x) = f(y)$. So, $b_1 = b_2$. 295 | 296 | \subsection{Solution (b)} 297 | From $R$, it follows that $\forall x \in A \forall y \in A(xRy \iff 298 | f(x) = f(y))$. From exercise 18 of 5.1, it follows that $h$ is one to 299 | one. 300 | 301 | Onto proof. Let $b$ be arbitrary element in $B$. Since $f$ is onto, it 302 | follows that $\exists a \in A$ such that $f(a)=b$. Since $a \in A$, 303 | there is some $X \in A/R$ such that $a \in X$. We know that $h(X)=b$. 304 | Since $b$ was arbitrary, we can conclude that $h$ is onto. 305 | 306 | \subsection{Solution (c)} 307 | Let $b$ be an arbitrary element in $B$. 308 | $(\Rightarrow)$ Suppose $a \in h^{-1}(b)$. It follows that $a \in A$ 309 | and $b \in B$. All we have to prove is $f(a) = b$. Since $f$ is onto, 310 | it follows that $f(a) =b$. So, $a \in \{x \in A \mid f(x) = b\}$. 311 | 312 | $(\Leftarrow)$ Let $a \in \{x \in A \mid f(x) = b\}$. It follows that 313 | $f(a)=b$. Now, we know that $a \in A$ so there is some $X \in A/R$ 314 | such that $a \in X$. So, $h(X) = b$ or $h^{-1}(b) = X$. So, $a \in h^{-1}(b)$. 315 | 316 | \subsection{Solution (d)} 317 | The question is likely wrong. Specifically the statement $\forall b 318 | \in B(g(b) \in h(b))$. The Domain of $h$ is $A/R$ and $b \notin A/R$. 319 | 320 | \section{Solution 14} 321 | \subsection{Solution (a)} 322 | Let $x$ be an arbitrary element in $A'$. $g \circ f(x) = g(f(x))$. 323 | There is some $b \in B$ such that $g(b) = x$. So, $g(f(x)) = 324 | g(f(g(b))) = g(b) = x$. Since $x$ was arbitrary, $\forall x \in A'(g 325 | \circ f)(x) = x$. 326 | 327 | \subsection{Solution (b)} 328 | $g$ function can be written as $B \to A'$. We know that $(f \mid A' \circ 329 | g) = i_B$. From (a), it follows that $(g \circ f \mid A') = i_A$. From 330 | theorem 5.3.4, it follows that $f \mid A'$ is one to one and onto 331 | function. From theorem 5.3.5, it follows that $g=(f \mid A)^{-1}$. 332 | 333 | \section{Solution 15} 334 | We know from Example $5.3.6$ that $f \circ g = i_B$. Since $g(x)= 335 | \sqrt{2}$, it follows that $Ran(g) = R^{+} = B$. So, $g$ function can 336 | be written as $g: B \to B$. It follows that $(f \mid B \circ g) = 337 | i_B)$. From 14 (a), it follows that $(g \circ f \mid B) = i_B$. From 338 | 14 (b), it follows that $g = (f \mid B)^{-1}$. 339 | 340 | \section{Solution 16} 341 | 342 | \subsection{Solution (a)} 343 | $y \in Ran(f)$, if $y = 4x - x^2$. Forming an equation, we get 344 | $x^2 - 4x + y = 0$. The equations will be real only if 345 | $b^2 - 4ac >= 0$ on a standard equation $a + bx^2 + c$. So, 346 | \begin{align*} 347 | (-4)^2 - 4.1.y >= 0 \\ 348 | 16 - 4y >= 0 \\ 349 | 16 >= 4y \\ 350 | 4 >= y \\ 351 | B = \{x \mid 4 >= y \} 352 | \end{align*} 353 | 354 | 355 | \subsection{Solution (b)} 356 | From 14, we know that we have to find a $g: R \to R$ such that $f 357 | \circ g = i_R$. Then $A = Ran(g)$. Let's find $g$. 358 | \begin{align*} 359 | f \circ g(x) = i_R(x) \\ 360 | f(g(x)) = x \\ 361 | \end{align*} 362 | Let $g(x) = y$. Then $f(y) = x$. So, $y^2 - 4y + x = 0$. 363 | $y = 2 + \sqrt{4-x}$. So, $g(x) = 2 + \sqrt{4 - x}$. We have to 364 | confirm that $g \circ f = i_R$. 365 | \begin{align*} 366 | g \circ f(x) = i_R(x) \\ 367 | g(4x - x^2) = 2 + \sqrt{4 - 4x + x^2} \\ 368 | = 2 + \sqrt{(x - 2)^2} \\ 369 | = 2 + x - 2 \\ 370 | = x 371 | \end{align*} 372 | 373 | So, $f \mid A$ is one to one and onto function and $(f \mid A)^{-1} 374 | (x) = 2 + \sqrt{4 - x}$. 375 | \end{document} 376 | -------------------------------------------------------------------------------- /chapter 4/section4-2.tex: -------------------------------------------------------------------------------- 1 | %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% 2 | % Author: Sibi 3 | %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% 4 | \documentclass{article} 5 | \usepackage{graphicx} 6 | \usepackage{verbatim} 7 | \usepackage{amsmath} 8 | \usepackage{amsfonts} 9 | \usepackage{amssymb} 10 | \usepackage{tabularx} 11 | \setlength\parskip{\baselineskip} 12 | \begin{document} 13 | \title{Chapter 4 (Section 4.2)} 14 | \author{Sibi} 15 | \date{\today} 16 | \maketitle 17 | \newpage 18 | 19 | \section{Problem 1} 20 | Solution (a) 21 | 22 | Domain = $\{ p \mid \text{p is a living parent} \}$ 23 | Range = $\{ q \mid \text{q is a living child} \}$ 24 | 25 | Solution(b) 26 | 27 | Domain = $\mathbb{R}$ 28 | Range = $\mathbb{R^+}$ 29 | 30 | \section{Problem 2} 31 | 32 | Solution(a) 33 | 34 | Domain = $\{ p \mid \text{p is a brother to some living person }\}$ 35 | Range = $\{q \mid \text{q has some living brother.}\}$ 36 | 37 | Solution(b) 38 | 39 | Domain = $\mathbb{R}$ 40 | Range = $\{ x \mid x \in \mathbb{R} \land x > -1 \land x < 1 \}$ 41 | 42 | \section{Problem 3} 43 | 44 | \subsection{Solution (a)} 45 | \noindent 46 | 47 | $L = \{(s,r) \in S \times R \mid \text{the student s lives in the dorm room 48 | r}\}$ 49 | 50 | $L^{-1} = \{(r,s)\} \in R \times S | (s,r) \in L\}$ 51 | 52 | $L^{-1} \circ L \text{ is relation on} S$ 53 | 54 | $L^{-1} \circ L = \{(s,t) \in S \times S \mid \exists r \in R((s,r) \in L \land 55 | (r,t) \in L^{-1})\}$ 56 | 57 | $ = \{(s,t) \in S \times S \mid \exists r \in R (\text{ student s lives in 58 | dorm room r and so is student t})\}$ 59 | 60 | $ = \{(s,t) \in S \times S \mid \text{student s and t both live in some dorm 61 | room} \}$ 62 | 63 | 64 | \subsection{Solution b} 65 | \noindent 66 | 67 | $E \circ (L^{-1} \circ L)$ is relation from S to C. 68 | 69 | $E \circ (L^{-1} \circ L) = \{(t,c) \in S \times C \mid \exists s \in 70 | S((t,s) \in (S \times S) \land (s,c) \in S \times C\}$ 71 | 72 | \begin{multline} 73 | =\{(t,c) \in S \times C \mid \exists s \in S( \text{student s and t} \\ 74 | \text{both live in some dorm room and student s is enrolled in course c)}\} 75 | \end{multline} 76 | 77 | $=\{(t,c) \in S \times C \mid \text{ t is a student who shares the 78 | dorm \\ with another student who is enrolled in course c}\}$ 79 | 80 | \section{Problem 4} 81 | 82 | \subsection{Solution (a)} 83 | \noindent 84 | 85 | $S \circ R = \{(a,b) \in A \times B \mid \exists c \in B((a,c) \in R 86 | \land (c,b) \in S)$ 87 | 88 | $ = {(1,5), (1,6), (1,4), (2,4), (3,6)}$ 89 | 90 | \subsection{Solution (b)} 91 | \noindent 92 | 93 | $S \circ S^{- 1} = \{(a,b) \in B \times B \mid \exists c \in B((a,c) 94 | \in S^{-1} \land (c,b) \in S\}$ 95 | 96 | $S \circ S^{- 1} = \{(a,b) \in B \times B \mid \exists c \in B((c,a) 97 | \in S \land (c,b) \in S\}$ 98 | 99 | $ = \{(5,6),(6,5),(5,5),(6,6),(4,4)\}$ 100 | 101 | \section{Problem 5} 102 | 103 | \subsection{Solution (a)} 104 | \noindent 105 | $S^{-1} \circ R = \{(a,b) \in A \times B \mid \exists c \in C((a,c) 106 | \in R \land (c,b) \in S^{-1}\}$ 107 | 108 | $S^{-1} \circ R = \{(a,b) \in A \times B \mid \exists c \in C((a,c) 109 | \in R \land (b,c) \in S\}$ 110 | 111 | = \{(1,4), (3,4), (3,5)\} 112 | 113 | \subsection{Solution (b)} 114 | \noindent 115 | $R^{-1} \circ S = \{(b,a) \in B \times A \mid \exists c \in C((b,c) 116 | \in S \land (c,a) \in R^{-1})\}$ 117 | 118 | $= \{(b,a) \in B \times A \exists c \in C((b,c) \in S \land (a,c) \in R\}$ 119 | 120 | $={(4,1),(4,3),(5,3)}$ 121 | 122 | \section{Problem 6} 123 | 124 | \subsection{Solution (a)} 125 | 126 | Note that both $Dom(R)$ and $Ran(R^{-1})$ are subsets of $A$. Let $a$ 127 | be an arbitrary element of $Dom(R)$. 128 | 129 | \begin{align*} 130 | a \in Dom (R) \iff \exists b \in B ((a,b) \in R) \\ 131 | \iff \exists b \in B((b,a) \in R^{-1}) \\ 132 | \iff a \in Ran(R^{-1}) 133 | \end{align*} 134 | 135 | \subsection{Solution (b)} 136 | 137 | We know that $Dom(R^{-1}) = Ran(R)$. So, it follows that 138 | $Dom((R^{-1})^{-1}) = Ran (R^{-1})$. From, $(R^{-1})^{-1} = R$, we can 139 | conclude that $Dom (R) = Ran (R^{-1})$. 140 | 141 | \subsection{Solution (c)} 142 | 143 | Suppose $(a,d)$ be an arbitrary element in $(T \circ S) \circ R$. By 144 | the definition of composition, we can chose some $b \in B$ such that 145 | $(a,b) \in R$ and $(b,d) \in T \circ S$. Similarly from $T \circ S$, 146 | we can chose some $c \in C$ such that $(b,c) \in S$ and $(c,d) \in T$. 147 | From $(a,b) \in R$ and $(b,c) \in S$, it follows that $(a,c) \in S 148 | \circ R$. Also, from $(a,c) \in S \circ R$ and $(c,d) \in T$, it 149 | follows that $(a,d) \in T \circ (S \circ R)$. 150 | 151 | \subsection{Solution(d)} 152 | \noindent 153 | 154 | $(\Rightarrow)$ Let $(c,a)$ be an arbitrary element in $(S \circ 155 | R)^{-1}$. Then it follows that $(a,c) \in (S \circ R)$. From the 156 | definition of composition, it follows that there exists some $b \in B$ 157 | such that $(a,b) \in R$ and $(b,c) \in S$. It follows that $(b,a) \in 158 | R^{-1}$ and $(c,b) \in S^{-1}$. From that we can conclude that $(c,a) 159 | \in R^{-1} \circ S^{-1}$. 160 | 161 | $(\Leftarrow)$ Let $(c,a)$ be an arbitrary element in $R^{-1} \circ 162 | S^{-1}$. From the definition of composition, it follows that there 163 | exists some $b \in B$ such that $(c,b) \in S^{-1}$ and $(b,a) \in 164 | R^{-1}$. It follows that $(b,c) \in S$ and $(a,b) \in R$. Composing, 165 | them we get $(a,c) \in S \circ R$. So, $(c,a) \in (S \circ R)^{-1}$. 166 | 167 | \section{Problem 7} 168 | 169 | An enemy of one's enemy is one's friend \\ 170 | An enemy of John's enemy is John's friend 171 | 172 | Assume: John's Enemy is Josepth, so $(Joseph, John) \in E$ 173 | Joseph's Enemy is Jane, so $(Jane, Joseph) \in E$ 174 | 175 | Now, Jane and John are friend. So, $(Jane, John) \in F$ 176 | $E \circ E = \{(Jane, Joseph)\}$ 177 | 178 | $E \circ E \subseteq F$ 179 | 180 | \section{Problem 8} 181 | \subsection{Solution(a)} 182 | Let $a$ be an arbitrary element in $Dom(S \circ R)$. Thenit follows 183 | that there exists some element $c \in C$ such that $(a,c) \in S \circ 184 | R$. From the definition of composition, it follows that there exists 185 | some element $b \in B$ such that $(a,b) \in R$ and $(b,c) \in S$. So, 186 | it follows that for some $b$, $(a,b) \in R$. Then clearly, $a \in 187 | Dom(R)$. Since $a$ is arbitrary, we can conclude that $Dom(S \circ R) 188 | \subseteq Dom(R)$. 189 | 190 | \subsection{Solution (b)} 191 | 192 | Suppose $Ran (R) \subseteq Dom(S)$. 193 | 194 | $(\Rightarrow)$ Let $a$ be an arbitrary element in $Dom(S \circ R)$. 195 | Then it follows that for some $c \in C$, $(a,c) \in S \circ R$. From 196 | the definition of composition it follows that there exists some $b \in 197 | B$ such that $(a,b) \in R$ and $(b,c) \in S$. From $(a,b) \in R$, it 198 | follows that $a \in Dom(R)$. 199 | 200 | $(\Leftarrow)$ Let $a$ be an arbitrary element in $Dom(R)$. Then it 201 | follows that for some element $b \in B$ there exists $(a,b) \in R$. 202 | From $Ran(R) \subseteq Dom(S)$, it follows that $b \in Dom(S)$. So, 203 | there exists some $c \in C$ such that $(b,c) \in S$. From $(a,b) \in 204 | R$ and $(b,c) \in S$, it follows that $(a,c) \in S \circ R$. Therefore 205 | $a \in Dom(S \circ R)$. 206 | 207 | \subsection{Solution (c)} 208 | 209 | (a) $Dom(S \circ R) \subseteq Dom(R)$ \\ 210 | $ \iff Ran((S \circ R)^{-1}) \subseteq Ran(R^{-1})$ \\ 211 | $ \iff Ran(R^{-1} \circ S^{-1}) \subseteq Ran(R^{-1})$ 212 | 213 | (b) $Ran(R) \subseteq Dom(S) \implies Dom(S \circ R) = Dom(R)$ \\ 214 | $ \iff Dom(R^{-1}) \subseteq Ran(S^{-1}) \implies Ran((S \circ 215 | R)^{-1}) = Ran(R^{-1})$ \\ 216 | $ \iff Dom(R^{-1}) \subseteq Ran(S^{-1}) \implies Ran(R^{-1} \circ 217 | S^{-1}) = Ran(R^{-1})$ 218 | 219 | \section{Problem 9} 220 | 221 | \subsection{Solution (a)} 222 | 223 | Let $(a,b)$ be an arbitrary element in $R$. Then it follows that $a 224 | \in Dom(R)$ and $b \in Ran(R)$. So, clearly $(a,b) \in Dom(R) \times Ran(R)$. 225 | 226 | \subsection{Solution(b)} 227 | 228 | Suppose $R \subseteq S$. Let $(b,a)$ be an arbitrary element in 229 | $R^{-1}$. It follows that $(a,b) \in R$. From $R \subseteq S$, it 230 | follows that $(a,b) \in S$. So, $(b,a) \in S^{-1}$. Since $(b,a)$ is 231 | an arbitrary element, it follows that $R^{-1} \subseteq S^{-1}$. 232 | 233 | \subsection{Solution (c)} 234 | 235 | $(\Rightarrow)$ Let $(b,a)$ be an arbitrary element in $(R \cup 236 | S)^{-1}$. It follows that $(a,b) \in R \cup S$. Let us consider the 237 | cases separately: 238 | 239 | Case 1. $(a,b) \in R$ It follows that $(b,a) \in R^{-1}$. So $(b,a) 240 | \in R^{-1} \cup S^{-1}$. 241 | Case 2. $(a,b) \in S$ It follows that $(b,a) \in S^{-1}$. So $(b,a) 242 | \in R^{-1} \cup S^{-1}$. 243 | 244 | $(\Leftarrow)$ Let $(b,a)$ be an arbitrary element in $R^{-1} \cup 245 | S^{-1}$. Let us consider the cases separately: 246 | 247 | Case 1. $(b,a) \in R^{-1}$ It follows that $(a,b) \in R$. So $(a,b) 248 | \in R \cup S$. Therefore $(b,a) \in (R \cup S)^{-1}$. 249 | Case 2. $(b,a) \in S^{-1}$ It follows that $(a,b) \in S$. So, $(a,b) 250 | \in R \cup S$ Therefore $(b,a) \in (R \cup S)^{-1}$ 251 | 252 | \section{Problem 10} 253 | 254 | We will prove contrapositive in both directions. 255 | 256 | $(\Rightarrow)$ Suppose $(Ran(R) \cap Dom(S)) \neq \emptyset$. Then by 257 | existential instantiation, $b \in Ran(R)$ and $b \in Dom(S)$. From $b 258 | \in Ran(R)$, it follows that there exists some $a$ such that $(a,b) 259 | \in R$. Similarly, $(b,c) \in S$. From $(a,b)in B$ and $(b,c) \in S$, 260 | it follows that $(a,c) \in S \circ R$. So $S \circ R \neq \emptyset$ 261 | 262 | $(\Leftarrow)$ Suppose $S \circ R \neq \emptyset$. Then it follows 263 | that there exists some element $(a,c)$ in $S \circ R$. From the 264 | definition of composition, it follows that $(a,b) \in R$ and $(b,c) 265 | \in S$ for some $b \in B$. From $(a,b) \in R$, it follows that $b \in 266 | Ran(R)$. Similarly $b \in Dom(S)$. Therefore, $b \in Ran(R) \cap 267 | Dom(S)$. So, $Ran (R) \cap Dom(S) \neq \emptyset$ 268 | 269 | \section{Problem 11} 270 | \subsection{Solution (a)} 271 | Let $(a,c)$ be an arbitrary element in $(S \circ R) \setminus (T \circ 272 | R)$. It follows that $(a,c) \in S \circ R$ and $(a,c) \notin T \circ 273 | R$. From $(a,c) \in S \circ R$, it follows that there is some $b \in 274 | B$ such that $(a,b) \in R$ and $(b,c) \in S$. From $(a,c) \notin T 275 | \circ R$, it follows that there is some $b \in B$ such that $(a,b) 276 | \notin R \land (b,c) \notin T$. From $(b,c) \in S$ and $(b,c) \notin 277 | T$, it follows that $(b,c) \in S \setminus T$. From $(a,b) \in R$, we 278 | can conclude that $(a,c) \in (S \setminus T) \circ R$. 279 | 280 | \subsection{Solution (b)} 281 | $b$ is not arbitrary there. There can be some other $d in D$ such that 282 | $(a,d) \in R$ and $(d,c) \in T$ making $(a,c) \in T$. 283 | 284 | \subsection{Solution (c)} 285 | Trick: Trace the wrong proof to find out counterexamples! 286 | \begin{align*} 287 | R = \{(1,3),(1,4)\} \\ 288 | S = \{(3,2)\} \\ 289 | T = \{(4,2)\} \\ \\ 290 | S \setminus T = \{(3,2)\} \\ 291 | (S \setminus T) \circ R = \{(1,2)\} \\ \\ 292 | S \circ R = \{(1,2)\} \\ 293 | T \circ R = \{(1,2)\} \\ 294 | (S \circ R) \setminus (T \circ R) = \emptyset 295 | \end{align*} 296 | 297 | \section{Problem 12} 298 | 299 | \subsection{Solution (a)} 300 | 301 | Suppose $S \subseteq T$. Let $(a,c)$ be an arbitrary element in $S 302 | \circ R$. By existential instantiation, it follows that $(a,b) \in R$ 303 | and $(b,c) \in S$. From $S \subseteq T$, it follows that $(b,c) \in 304 | T$. From $(a,b) \in R$ and $(b,c) \in T$ it follows that $(a,c) \in T 305 | \circ R$. Since $(a,c)$ is an arbitrary element, we can conclude that 306 | $S \circ R \subseteq T \circ R$. Therefore $S \subseteq T \implies S 307 | \circ R \subseteq T \circ R$. 308 | 309 | \subsection{Solution (b)} 310 | 311 | Let $(a,c)$ be an arbitrary element in $(S \cap T) \circ R$. By 312 | existential instantiation, $(a,b) \in R$ and $(b,c) \in S \cap T$. 313 | From $(b,c) \in S \cap T$, we get $(b,c) \in T$. From $(b,c) \in S 314 | \cap T$, we get $(b,c) \in S$ and $(b,c) \in T$. From $(a,b) \in R$ 315 | and $(b,c) \in S$, it follows that $(a,c) \in S \circ R$. Similarly 316 | from $(a,b) \in R$ and $(b,c) \in T$, it follows that $(a,c) \in T 317 | \circ R$. So, $(a,c) \in (S \circ R) \cap (T \circ R)$. Since $(c,c)$ 318 | is an arbitrary element, we can conclude that $(S \cap T) \circ R 319 | \subseteq (S \circ R) \cap (T \circ R)$ 320 | 321 | \subsection{Solution (c)} 322 | \begin{align*} 323 | R = \{(2,1),(2,4)\} \\ 324 | S = \{(1,1)\} \\ 325 | T = \{(4,1)\} \\ \\ 326 | S \cap T = \emptyset \\ 327 | (S \cap T) \circ R = \emptyset \\ \\ 328 | S \circ R = \{(2,1)\} \\ 329 | T \circ R = \{(2,1)\} \\ 330 | (S \circ R) \cap (T \circ R) = \{(2,1)\} 331 | \end{align*} 332 | 333 | \subsection{Solution (d)} 334 | 335 | $(\Rightarrow)$ Let $(a,c)$ be an arbitrary element in $(S \cup T) 336 | \circ R$. By existential instantiation, $(a,b) \in R$ and $(b,c) \in S 337 | \cup T$. From $(b,c) \in S \cup T$, it follows that either $(b,c) \in 338 | S$ or $(b,c) \in T$. Let us consider the cases separately: 339 | 340 | Case 1. $(b,c) \in S$ From $(a,b) \in R$, it follows that $(a,c) \in S 341 | \circ R$. So $(a,c) \in (S \circ R) \cup (T \circ R)$. 342 | Case 2. $(b,c) \in T$ From $(a,b) \in R$, it follows that $(a,c) \in T 343 | \circ R$. So $(a,c) \in (S \circ R) \cup (T \circ R)$. 344 | 345 | $(\Leftarrow)$ Let $(a,c)$ be an arbitrary element in $(S \circ R) 346 | \cup (T \circ R)$. Let us consider the cases separately: 347 | 348 | Case 1. $(a,c) \in S \circ R$. By existential instantiation, it 349 | follows that $(a,b) \in R$ and $(b,c) \in S$. So, $(b,c) \in S \cup 350 | T$. From $(a,b) \in R$, it follows that $(a,c) \in (S \cup T) \circ R$. 351 | Case 2. $(a,c) in T \circ R$ By existential instantiation, it follows 352 | that $(a,b) \in R$ and $(b,c) \in T$. So, $(b,c) \in S \cup T$. From 353 | $(a,b) \in R$, it follows that $(a,c) \in (S \cup T) \circ R$. 354 | 355 | \end{document} 356 | -------------------------------------------------------------------------------- /chapter 1/chapter1-2.md: -------------------------------------------------------------------------------- 1 | Truth Tables 2 | ------------- 3 | 4 | Exercise 1 5 | ----------- 6 | 7 | a) 8 | 9 | 10 | p q | (¬p ∨ q) 11 | -------------- 12 | T T | T 13 | T F | F 14 | F T | T 15 | F F | T 16 | 17 | 18 | b) 19 | 20 | g s | ((s ∨ g) ∧ (¬s ∨ ¬g)) 21 | --------------------------- 22 | T T | F 23 | T F | T 24 | F T | T 25 | F F | F 26 | 27 | Exercise 2 28 | ----------- 29 | 30 | a) 31 | 32 | p q | ¬(p ∧ (q ∨ ¬p)) 33 | --------------------- 34 | T T | F 35 | T F | T 36 | F T | T 37 | F F | T 38 | 39 | b) 40 | 41 | p q r | ((p ∨ q) ∧ (¬p ∨ r)) 42 | ---------------------------- 43 | T T T | T 44 | T T F | F 45 | T F T | T 46 | T F F | F 47 | F T T | T 48 | F T F | T 49 | F F T | F 50 | F F F | F 51 | 52 | Exercise 3 53 | ----------- 54 | 55 | a) 56 | 57 | p q | p + q 58 | T T | F 59 | T F | T 60 | F T | T 61 | F F | F 62 | 63 | b) 64 | 65 | Find a formula using only the connectives ∧, ∨, and ¬ that is equiv- 66 | alent to P + Q. Justify your answer with a truth table. 67 | 68 | Now in the above truth table (3 (a)) see for which rows you are 69 | getting True and write down their logical connective: 70 | 71 | (p ∧ ¬q) ∨ (¬p ∧ q) 72 | 73 | p q | ((p & ~q) | (~p & q)) 74 | --------------------------- 75 | T T | F 76 | T F | T 77 | F T | T 78 | F F | F 79 | 80 | Exercise 4 81 | ---------- 82 | 83 | Find a formula using only the connectives ∧ and ¬ that is equivalent to 84 | P ∨ Q. Justify your answer with a truth table. 85 | 86 | The truth table for `P v Q` follows like this: 87 | 88 | p q | (p v q) 89 | ------------- 90 | T T | T 91 | T F | T 92 | F T | T 93 | F F | F 94 | 95 | From De-morgan's theorem: 96 | 97 | ¬(P v Q) ≡ ¬P ∧ ¬Q 98 | (P v Q) ≡ ¬(¬P ∧ ¬Q) 99 | 100 | Exercise 5 101 | ---------- 102 | 103 | (a) 104 | 105 | p q | p ↓ q 106 | T T | F 107 | T F | F 108 | F T | F 109 | F F | T 110 | 111 | (b) 112 | 113 | ¬p ∧ ¬q 114 | 115 | (c) 116 | 117 | Find formulas using only the connective ↓ that are equivalent to ¬P, 118 | P ∨ Q, and P ∧ Q. 119 | 120 | p ↓ p 121 | ¬(p ↓ q) ≡ (p ↓ q) ↓ (p ↓ q) 122 | ¬p ↓ ¬q ≡ (p ↓ p) ↓ (q ↓ q) 123 | 124 | Exercise 6 125 | ---------- 126 | 127 | Some mathematicians write P | Q to mean “P and Q are not both true.” 128 | (This connective is called nand, and is used in the study of circuits in 129 | computer science.) 130 | (a) Make a truth table for P | Q. 131 | (b) Find a formula using only the connectives ∧, ∨, and ¬ that is equiv- 132 | alent to P | Q. 133 | (c) Find formulas using only the connective | that are equivalent to ¬P, 134 | P ∨ Q, and P ∧ Q. 135 | 136 | (a) 137 | 138 | p q | p | q 139 | T T | F 140 | T F | T 141 | F T | T 142 | F F | T 143 | 144 | (b) 145 | 146 | One of these: 147 | 148 | ¬(p ∧ q) (p | q ≡ ¬(p ∧ q)) 149 | ¬p ∨ ¬q (p | q ≡ ¬p ∨ ¬q) 150 | 151 | (c) 152 | 153 | ¬p ≡ (p | p) 154 | p ∨ q ≡ (¬p | ¬q) ≡ ((p | p) | (q | q)) 155 | p ∧ q ≡ ¬(p | q) ≡ (p | q) | (p | q) 156 | 157 | 158 | Exercise 7 159 | ----------- 160 | 161 | Use truth tables to determine whether or not the arguments in exercise 7 162 | of Section 1.1 are valid. 163 | 164 | Take the conjunction of all the premises `¬(J ∧ P) ∧ (P ∨ C) ∧ J` and 165 | draw it's truth table. 166 | 167 | C J P | (~(J & P) & ((P | C) & J)) 168 | ---------------------------------- 169 | T T T | F 170 | T T F | T 171 | T F T | F 172 | T F F | F 173 | F T T | F 174 | F T F | F 175 | F F T | F 176 | F F F | F 177 | 178 | where 179 | 180 | J = Jane will win the math prize 181 | P = Pete will win the math prize 182 | C = Pete will win the chemistry prize. 183 | 184 | Now, the result is True when `C` and `J` are true. Hence the 185 | conclusion that Pete will win the chemistry prize is true. 186 | 187 | (b) 188 | 189 | Some symbols: 190 | 191 | B = Beef will be the main course. 192 | F = Fish will be the main course. 193 | P = Peas will be the vegetable. 194 | C = Corn will be the vegetable. 195 | 196 | Take the conjunction of all the premises `(B ∨ F) ∧ (P ∨ C) ∧ ¬(F ∧ 197 | C)` and draw it's truth table: 198 | 199 | B C F P | ((B | F) & ((P | C) & ~(F & C))) 200 | ------------------------------------------ 201 | T T T T | F 202 | T T T F | F 203 | T T F T | T 204 | T T F F | T 205 | T F T T | T 206 | T F T F | F 207 | T F F T | T 208 | T F F F | F 209 | F T T T | F 210 | F T T F | F 211 | F T F T | F 212 | F T F F | F 213 | F F T T | T 214 | F F T F | F 215 | F F F T | F 216 | F F F F | F 217 | 218 | Conclusion is `¬(B ∧ P)`: 219 | 220 | B P | ~(B & P) 221 | -------------- 222 | T T | F 223 | T F | T 224 | F T | T 225 | F F | T 226 | 227 | The argument is invalid because `~(B & P)` is `F` when `B = T` and `P 228 | = T`. But in third row when `B = T` and `P = T` in conjunction of 229 | premise, it is `T`. To interpret it literally, they can have Beef as 230 | main course and Peas as the vegetable, but the conclusion of the 231 | statement says otherwise. 232 | 233 | (c) 234 | 235 | Symbols: 236 | 237 | J = John is telling the truth. 238 | B = Bill is telling the truth. 239 | S = Sam is telling the truth. 240 | 241 | Premises: 242 | J ∨ B 243 | ¬S ∨ ¬B 244 | 245 | Conclusion: 246 | J ∨ ¬S 247 | 248 | Conjunction of Premises: `(J ∨ B) ∧ (¬S ∨ ¬B)` 249 | 250 | B J S | ((J | B) & (~S | ~B)) 251 | ----------------------------- 252 | T T T | F 253 | T T F | T 254 | T F T | F 255 | T F F | T 256 | F T T | T 257 | F T F | T 258 | F F T | F 259 | F F F | F 260 | 261 | Conclusion truth table: 262 | 263 | J S | (J | ~S) 264 | -------------- 265 | T T | T 266 | T F | T 267 | F T | F 268 | F F | T 269 | 270 | Whenever the conclusion truth table is `T`, for those parameters of 271 | `J` and `S`, the premises truth table is also true which implies that 272 | the conlusion is valid. 273 | 274 | (d) 275 | 276 | Symbols: 277 | 278 | S = Sales will go up. 279 | B = Boss will be happy. 280 | E = Expenses will go up. 281 | 282 | Premises: 283 | (S ∧ B) ∨ (E ∧ ¬B) 284 | 285 | Premise truth table: 286 | 287 | B E S | ((S & B) | (E & ~B)) 288 | ---------------------------- 289 | T T T | T 290 | T T F | F 291 | T F T | T 292 | T F F | F 293 | F T T | T 294 | F T F | T 295 | F F T | F 296 | F F F | F 297 | 298 | Conclusion: 299 | ¬(S ∧ E) 300 | 301 | Conclusion truth table: 302 | 303 | E S | ~(S & E) 304 | -------------- 305 | T T | F 306 | T F | T 307 | F T | T 308 | F F | T 309 | 310 | Argument is invalid because the according to the conclusion when `E = 311 | F` and `S = F`, the premise should be valid which is not the case. 312 | 313 | 314 | Exercise 8 315 | ---------- 316 | 317 | Use truth tables to determine which of the following formulas are equiv- 318 | alent to each other: 319 | (a) (P ∧ Q) ∨ (¬P ∧ ¬Q). 320 | (b) ¬P ∨ Q. 321 | (c) (P ∨ ¬Q) ∧ (Q ∨ ¬P). 322 | (d) ¬(P ∨ Q). 323 | (e) (Q ∧ P) ∨ ¬P 324 | 325 | P Q | ((P & Q) | (~P & ~Q)) 326 | --------------------------- 327 | T T | T 328 | T F | F 329 | F T | F 330 | F F | T 331 | 332 | P Q | (~P | Q) 333 | -------------- 334 | T T | T 335 | T F | F 336 | F T | T 337 | F F | T 338 | 339 | P Q | ((P | ~Q) & (Q | ~P)) 340 | --------------------------- 341 | T T | T 342 | T F | F 343 | F T | F 344 | F F | T 345 | 346 | P Q | ~(P | Q) 347 | -------------- 348 | T T | F 349 | T F | F 350 | F T | F 351 | F F | T 352 | 353 | P Q | ((Q & P) | ~P) 354 | -------------------- 355 | T T | T 356 | T F | F 357 | F T | T 358 | F F | T 359 | 360 | From the truth table `(a)` and `(c)` are equivalent to each other. 361 | Another thing that is equivalent is `(b)` and `(e)`. 362 | 363 | Exercise 9 364 | ---------- 365 | 366 | Use truth tables to determine which of these statements are tautologies, 367 | which are contradictions, and which are neither: 368 | (a) (P ∨ Q) ∧ (¬P ∨ ¬Q). 369 | (b) (P ∨ Q) ∧ (¬P ∧ ¬Q). 370 | (c) (P ∨ Q) ∨ (¬P ∨ ¬Q). 371 | (d) [P ∧ (Q ∨ ¬R)] ∨ (¬P ∨ R). 372 | 373 | Truth tables: 374 | 375 | P Q | ((P | Q) & (~P | ~Q)) 376 | --------------------------- 377 | T T | F 378 | T F | T 379 | F T | T 380 | F F | F 381 | 382 | P Q | ((P | Q) & (~P & ~Q)) 383 | --------------------------- 384 | T T | F 385 | T F | F 386 | F T | F 387 | F F | F 388 | 389 | P Q | ((P | Q) | (~P | ~Q)) 390 | --------------------------- 391 | T T | T 392 | T F | T 393 | F T | T 394 | F F | T 395 | 396 | P Q R | ((P & (Q | ~R)) | (~P | R)) 397 | ----------------------------------- 398 | T T T | T 399 | T T F | T 400 | T F T | T 401 | T F F | T 402 | F T T | T 403 | F T F | T 404 | F F T | T 405 | F F F | T 406 | 407 | From the truth tables, `(b)` is contradiction. `(c)` and `(d)` are 408 | both tautologies. `(a)` is neither of them. 409 | 410 | Exercise 10 411 | ----------- 412 | 413 | The first was checked in text, so I will skip it. 414 | Distributive law: `P ∧ (Q ∨ R)` is equivalent to `(P ∧ Q) ∨ (P ∧ R)` 415 | 416 | P Q R | (P & (Q | R)) 417 | --------------------- 418 | T T T | T 419 | T T F | T 420 | T F T | T 421 | T F F | F 422 | F T T | F 423 | F T F | F 424 | F F T | F 425 | F F F | F 426 | 427 | P Q R | ((P & Q) | (P & R)) 428 | --------------------------- 429 | T T T | T 430 | T T F | T 431 | T F T | T 432 | T F F | F 433 | F T T | F 434 | F T F | F 435 | F F T | F 436 | F F F | F 437 | 438 | Duh, they are equivalent. 439 | 440 | Exercise 11 441 | ------------ 442 | 443 | Use the laws stated in the text to find simpler formulas equivalent to these 444 | formulas. (See Examples 1.2.5 and 1.2.7.) 445 | (a) ¬(¬P ∧ ¬Q). 446 | (b) (P ∧ Q) ∨ (P ∧ ¬Q). 447 | (c) ¬(P ∧ ¬Q) ∨ (¬P ∧ Q). 448 | 449 | (a) 450 | 451 | ¬(¬P ∧ ¬Q) 452 | => P ∨ Q (Demorgan's law) 453 | 454 | (b) 455 | 456 | (P ∧ Q) ∨ (P ∧ ¬Q) 457 | => ((P ∧ Q) ∨ P) ∧ ((P ∧ Q) ∨ ¬Q) [Distributive law] 458 | => (P ∨ (P ∧ Q)) ∧ (¬Q ∨ (P ∧ Q)) [Associative law] 459 | => P ∧ ((¬Q ∨ P) ∧ (¬Q ∨ Q)) [Absorption law & Distributive law] 460 | => P ∧ (¬Q ∨ P) 461 | => P [Absorption law] 462 | 463 | (c) 464 | 465 | ¬(P ∧ ¬Q) ∨ (¬P ∧ Q) 466 | => (¬P ∨ Q) ∨ (¬P ∧ Q) [Demorgan's law] 467 | => (¬P ∨ Q ∨ ¬P) ∧ (¬P ∨ Q ∨ Q) [Distributive law] 468 | => (¬P ∨ Q) ∧ (¬P ∨ Q) [Idempotent law] 469 | => (¬P ∨ Q) [Idempotent law] 470 | 471 | Exercise 12 472 | ------------ 473 | 474 | Use the laws stated in the text to find simpler formulas equivalent to these 475 | formulas. (See Examples 1.2.5 and 1.2.7.) 476 | (a) ¬(¬P ∨ Q) ∨ (P ∧ ¬R). 477 | (b) ¬(¬P ∧ Q) ∨ (P ∧ ¬R). 478 | (c) (P ∧ R) ∨ [¬R ∧ (P ∨ Q)] 479 | 480 | (a) 481 | 482 | ¬(¬P ∨ Q) ∨ (P ∧ ¬R) 483 | => (¬¬P ∧ ¬Q) ∨ (P ∧ ¬R) [Demorgan's law] 484 | => (P ∧ ¬Q) ∨ (P ∧ ¬R) [Double Negation law] 485 | => P ∧ (¬Q ∨ ¬R) [Distributive law] 486 | 487 | (b) 488 | 489 | ¬(¬P ∧ Q) ∨ (P ∧ ¬R) 490 | => (P ∨ ¬Q) ∨ (P ∧ ¬R) [Demorgan's law] 491 | => (P ∨ ¬Q ∨ P) ∧ (P ∨ ¬Q ∨ ¬R) [Distributive law] 492 | => (P v ¬Q) ∧ (P ∨ ¬Q ∨ ¬R) [Idempotent law] 493 | => (P v ¬Q) [Absorption law] 494 | 495 | (c) 496 | 497 | (P ∧ R) ∨ [¬R ∧ (P ∨ Q)] 498 | => (P ∧ R) ∨ (¬R ∧ P) ∨ (¬R ∧ Q) [Distributive law] 499 | => (P ∧ (R ∨ ¬R)) ∨ (¬R ∧ Q) [Distributive law] 500 | => P ∨ (¬R ∧ Q) 501 | 502 | Exercise 13 503 | ------------ 504 | 505 | Use the first DeMorgan’s law and the double negation law to derive the 506 | second DeMorgan’s law. 507 | 508 | First Demorgan's law: `¬(P ∧ Q) is equivalent to ¬P ∨ ¬Q` 509 | Second Demorgan's law: `¬(P ∨ Q) is equivalent to ¬P ∧ ¬Q` 510 | 511 | ¬(P ∨ Q) [LHS of second demorgan's law] 512 | => ¬(¬¬P ∨ ¬¬Q) [Double negation] 513 | => ¬(¬(¬P ∧ ¬Q)) [Demorgan's first law] 514 | => (¬P ∧ ¬Q) [RHS of second demorgan's law] 515 | 516 | Exercise 14 517 | ----------- 518 | 519 | Note that the associative laws say only that parentheses are unnecessary 520 | when combining three statements with ∧ or ∨. In fact, these laws can be 521 | used to justify leaving parentheses out when more than three statements 522 | are combined. Use associative laws to show that [P ∧ (Q ∧ R)] ∧ S is 523 | equivalent to (P ∧ Q) ∧ (R ∧ S). 524 | 525 | So, the solution goes like this: 526 | 527 | [P ∧ (Q ∧ R)] ∧ S 528 | => ((P ∧ Q) ∧ R) ∧ S 529 | => (P ∧ Q) ∧ (R ∧ S) 530 | 531 | Exercise 15 532 | ----------- 533 | 534 | That's easy: 2^n 535 | 536 | Exercise 16 537 | ----------- 538 | 539 | Let's take conjugate of all the rows where it is true and then take 540 | the disjunctions of them. Or more simpler way would be to take 541 | conjugate of where it is false and inverse them in this case: 542 | 543 | ¬(¬P ∧ Q) 544 | => P ∨ ¬Q 545 | 546 | Exercise 17 547 | ----------- 548 | 549 | Solve using the same strategy as above: 550 | 551 | (¬P ∧ Q) ∨ (P ∧ ¬Q) 552 | => ((¬P ∧ Q) ∨ P) ∧ ((¬P ∧ Q) ∨ ¬Q) 553 | => (P ∨ Q) ∧ (¬P ∨ ¬Q) 554 | 555 | Exercise 18 556 | ----------- 557 | 558 | Suppose the conclusion of an argument is a tautology. What can you 559 | conclude about the validity of the argument? What if the conclusion is 560 | a contradiction? What if one of the premises is either a tautology or a 561 | contradiction? 562 | 563 | If the conclusion of an argument is a tautology, then no matter what 564 | the condition of the premise it is true for all the cases. Ain't that 565 | wonderful! Similarly if the conclusion is a contradiction, then no 566 | matter what the condition of the premises it is false for all the 567 | cases. How prejudiced it is! We cannot say anything about the validity 568 | of the argument in this case. 569 | 570 | What if one of the premises is either a tautology or a contradiction? 571 | Doesn't really matter as the conclusion of an argument is an 572 | tautology. 573 | -------------------------------------------------------------------------------- /chapter 6/section6-1.tex: -------------------------------------------------------------------------------- 1 | %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% 2 | % Author: Sibi 3 | %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% 4 | \documentclass{article} 5 | \usepackage{graphicx} 6 | \usepackage{verbatim} 7 | \usepackage{amsmath} 8 | \usepackage{amsfonts} 9 | \usepackage{amssymb} 10 | \usepackage{tabularx} 11 | \usepackage{mathtools} 12 | \newcommand{\BigO}[1]{\ensuremath{\operatorname{O}\bigl(#1\bigr)}} 13 | \setlength\parskip{\baselineskip} 14 | \begin{document} 15 | \title{Chapter 6 (Section 6.1)} 16 | \author{Sibi} 17 | \date{\today} 18 | \maketitle 19 | 20 | % See here: http://tex.stackexchange.com/a/43009/69223 21 | \DeclarePairedDelimiter\abs{\lvert}{\rvert}% 22 | \DeclarePairedDelimiter\norm{\lVert}{\rVert}% 23 | 24 | % Swap the definition of \abs* and \norm*, so that \abs 25 | % and \norm resizes the size of the brackets, and the 26 | % starred version does not. 27 | \makeatletter 28 | \let\oldabs\abs 29 | \def\abs{\@ifstar{\oldabs}{\oldabs*}} 30 | % 31 | \let\oldnorm\norm 32 | \def\norm{\@ifstar{\oldnorm}{\oldnorm*}} 33 | \makeatother 34 | \newpage 35 | 36 | \section{Solution 1} 37 | By mathematical induction, \\ 38 | Base case. When $n = 0$, $0 = \frac{n(n+1)}{2} = 0$. 39 | Induction step. Let $n$ be an arbitrary element and suppose $0 + 1 + 2 40 | + ... + n = \frac{n(n+1)}{2}$. Then, 41 | \begin{align*} 42 | 0 + 1 + 2 + ... + n + (n + 1) = \frac{n(n+1)}{2} + (n + 1) \\ 43 | = (n+1)(\frac{n}{2} + 1) \\ 44 | = \frac{(n+1)(n+2)}{2} 45 | \end{align*} 46 | 47 | \section{Solution 2} 48 | By mathematical induction, 49 | 50 | Base case. When $n = 0$, both sides of the equation becomes $0$. 51 | 52 | Induction step. Let $n$ be arbitrary and suppose $0^2 + 1^2 + 2^2 + 53 | ... + n^2 = \frac{n(n+1)(2n+1)}{6}$. Then, 54 | \begin{align*} 55 | 0^2 + 1^2 + 2^2 + ... + n^2 + (n+1)^2 = \frac{n(n+1)(n+2)}{6} + (n + 56 | 1)^2 \\ 57 | = (n+1)(\frac{n(2n + 1)}{6} + (n + 1)) \\ 58 | = (n+1)(\frac{2n^2 + n + 6n + 6}{6}) \\ 59 | = \frac{(n+1)(n+2)(n+3)}{6} \\ 60 | \end{align*} 61 | \section{Solution 3} 62 | By mathematical induction, 63 | 64 | Base case. When $n=0$, both sides of the equation becomes $0$. 65 | 66 | Induction step. Let $n$ be an arbitrary element in $N$. Suppose $0^3 + 67 | 1^3 + 2^3 + ... + n^3 = [\frac{n(n+1)}{2}]^2$. Then, 68 | 69 | \begin{align*} 70 | 0^3 + 1^3 + 2^3 + ... + n^3 + (n+1)^3= [\frac{n(n+1)}{2}]^2 + (n+1)^3 \\ 71 | = (n+1)^2(\frac{n^2 + 4n + 4}{4}) \\ 72 | = \frac{(n+1)^2(n+2)^2}{4} \\ 73 | = [\frac{(n+1)(n+2)}{2}]^2 74 | \end{align*} 75 | 76 | \section{Solution 4} 77 | By mathematical induction, 78 | 79 | Base case. When $n=1$, both sides of the equation becomes 1. 80 | 81 | Induction step. Let $n$ be an arbitrary element and suppose $1+3+5+... 82 | + (2n-1) = n^2$. Then, 83 | 84 | \begin{align*} 85 | 1 + 3 + 5 + ... + (2n - 1) = n^2 \\ 86 | 1 + 3 + 5 + ... + (2n - 1) + (n+1)^2 = n^2 + (n+1)^2 \\ 87 | 1 + 3 + 5 + ... + (2n - 1) + n^2 + 1 + 2n = n^2 + (n+1)^2 \\ 88 | 1 + 3 + 5 + ... + (2n + 1) = (n+1)^2. 89 | \end{align*} 90 | 91 | \section{Solution 5} 92 | By mathematical induction, 93 | 94 | Base case. When $n=0$, both sides of the equation becomes $0$. 95 | 96 | Inductive step. Let $n$ be an arbitrary element and suppose $0.1 + 1.2 97 | + 2.3 + ... + n(n+1) = \frac{n(n+1)(n+2)}{3}$. Then, 98 | 99 | \begin{align*} 100 | 0.1 + 1.2 + 2.3 + ... + n(n+1) + (n+1)(n+2) = \frac{n(n+1)(n+2)}{3} 101 | + (n+1)(n+2) \\ 102 | = (n+1)(n+2)(\frac{n}{3} + 1) \\ 103 | = \frac{(n+1)(n+2)(n+3)}{3} 104 | \end{align*} 105 | \section{Solution 6} 106 | Guess: $\frac{n(n+1)(n+2)(n+3)}{4}$ 107 | 108 | By mathematical induction, 109 | 110 | Base case. When $n=0$, both sides of the equation becomes $0$. 111 | 112 | Induction step. Let $n$ be an arbitrary element. Suppose $0.1.2 + 113 | 1.2.3 + 2.3.4 + ... + n(n+1)(n+2) = \frac{n(n+1)(n+2)(n+3)}{4}$. Then, 114 | 115 | \begin{align*} 116 | 0.1.2 + 1.2.3 + 2.3.4 + ... + n(n+1)(n+2) + (n+1)(n+2)(n+3) = 117 | \frac{n(n+1)(n+2)(n+3)}{4} + (n+1)(n+2)(n+3) \\ 118 | = (n+1)(n+2)(n+3)(\frac{n}{4} + 1) \\ 119 | = \frac{(n+1)(n+2)(n+3)(n+4)}{4} 120 | \end{align*} 121 | 122 | \section{Solution 7} 123 | Guess: $\frac{3^{n+1} - 1}{2}$ 124 | 125 | By mathematical induction, 126 | 127 | Base case. When $n=0$, both sides of the equation becomes 1. 128 | 129 | Induction case. Let $n$ be an arbitrary element. Suppose $3^0 + 3^1 + 130 | 3^2 + ... + 3^n = \frac{3^{n+1} - 1}{2}$. Then, 131 | 132 | \begin{align*} 133 | 3^0 + 3^1 + 3^2 + ... + 3^n + 3^{n+1} = \frac{3^{n+1} - 1}{2} + 134 | 3^{n+1} \\ 135 | = \frac{3^{n+1} - 1 + 2.3^{n+1}}{2} \\ 136 | = \frac{3^{n+2} - 1}{2} 137 | \end{align*} 138 | 139 | \section{Solution 8} 140 | By mathematical induction, 141 | 142 | Base case. When $n = 1$, both sides of the equation becomes $1/2$. 143 | 144 | Induction step. Let $n$ be an arbitrary element such that $n >= 1$ and 145 | suppose $1 - \frac{1}{2} + \frac{1}{3} + ... + \frac{1}{2n - 1} - 146 | \frac{1}{2n} = \frac{1}{n + 1} + \frac{1}{n+2} + \frac{1}{n+3} + ... 147 | + \frac{1}{2n} $. Then, 148 | \begin{align*} 149 | 1 - \frac{1}{2} + \frac{1}{3} + ... + \frac{1}{2n - 1} - 150 | \frac{1}{2n} + \frac{1}{2n+1} - \frac{1}{2n+2} \\ 151 | = \frac{1}{n+1} + \frac{1}{n+2} + \frac{1}{n+3} + ... + 152 | \frac{1}{2n} + \frac{1}{2n+1} - \frac{1}{2n+2} \\ 153 | = \frac{1}{n+2} + \frac{1}{n+3} + ... + \frac{1}{2n} + 154 | \frac{1}{2n+1} + \frac{1}{2n+2} 155 | \end{align*} 156 | 157 | as required. 158 | 159 | \section{Solution 9} 160 | \subsection{Solution (a)} 161 | By mathematical induction, 162 | 163 | Base case. When $n = 0$, $\exists k \in Z$ such that $2k = n^2 + 2$. 164 | Here if $k = 0$, then base case holds. 165 | 166 | Induction step. Let $n$ be an arbitrary element in $N$. Suppose $2k = 167 | n^2 + n$ for some $k \in Z$. Thus, 168 | \begin{align*} 169 | (n + 1)^2 (n + 1) = n^2 + 1 + 2n + n + 1 \\ 170 | = n^2 + 3n + 2 \\ 171 | = n^2 + n + 2n + 2 \\ 172 | = 2k + 2n + 2 \\ 173 | = 2(k + n + 2) 174 | \end{align*} 175 | 176 | Therefore $2 \mid (n+1)^2 + (n+1)$ as required. 177 | 178 | \subsection{Solution (b)} 179 | By mathematical induction, 180 | 181 | Base case. When $ n = 0$, it holds that $6 \mid (n^3 - n)$ for $0 \in 182 | Z$. 183 | 184 | Induction step. Let $n$ be an arbitrary element in $N$ such that $6 185 | \mid (n^3 - n)$. 186 | \begin{align*} 187 | (n+1)^3 - (n+1) \\ 188 | = (n+1)((n+1)^2 - 1) \\ 189 | = (n+1)(n^2 + 1 + 2n - 1) \\ 190 | = (n+1)(n^2 + 2n) \\ 191 | = n^3 + 3n^2 + 2n \\ 192 | = n^3 - n + 3n^2 + 2n \\ 193 | = 6k + 3n^2 + 3n \\ 194 | = 6(k + n^2/2 + n/2) \\ 195 | = 6(k + n(n+1)/2) 196 | \end{align*} 197 | Now $n(n+1)/2$ is the sum of number series upto $n$. So, $6 \mid 198 | (n+1)^3 - (n+1)$ holds. 199 | 200 | \section{Solution 10} 201 | By mathematical induction, 202 | 203 | Base case. When $n = 0$, $64 \mid (9^n - 8n - 1)$ holds for $0$. 204 | 205 | Induction step. Let $n$ be an arbitrary element in $N$ such that $64k 206 | = (9^n - 8n - 1)$. Then, 207 | \begin{align*} 208 | 9^{n+1} - 8(n+1) - 1 = 9^{n+1} - 8n - 9 \\ 209 | = 9^{n+1} - 9n - 9 + n \\ 210 | = 9(9^n - n - 1) + n \\ 211 | = 9(9^n - 8n - 1 + 7n) + n \\ 212 | = 9(64k + 7n) + n \\ 213 | = 64(9k + n) \\ 214 | \end{align*} 215 | So, $64 \mid (9^{n+1} - 8(n+1) - 1)$ as required. 216 | 217 | \section{Solution 11} 218 | By mathematical induction, 219 | 220 | Base case. When $n = 0$, $9 \mid (4^n + 6n - 1)$ holds for $0$. 221 | 222 | Induction step. Let $n$ be an arbitrary element in $N$ and suppose $9k 223 | = 4^n + 6n - 1$. Thus, 224 | \begin{align*} 225 | 4^{n+1} + 6(n+1) - 1 = 4^{n+1} + 6n + 5 \\ 226 | = 4^{n+1} + 4n + 2n + 5 \\ 227 | = 4(4^{n} + n) + 2n + 5 \\ 228 | = 4(4^{n} + 6n - 1 - 5n + 1) + 2n + 5 \\ 229 | = 4(9k - 5n + 1) + 2n + 5 \\ 230 | = 4.9k - 18n + 9 \\ 231 | = 9(4k - 2n + 1) 232 | \end{align*} 233 | So, $9 \mid 4^{n+1} + 6(n+1) - 1$ as required. 234 | 235 | \section{Solution 12} 236 | Let $a$ and $b$ be arbitrary integer. By mathematical induction, 237 | 238 | Base case. When $n = 0$, $(a-b) \mid (a^n - b^n)$ holds for $0$. 239 | Induction step. Let $n$ be arbitrary element in $N$ and suppose 240 | $(a-b)k = a^n - b^n$ . Thus, 241 | \begin{align*} 242 | a^{n+1} - b^{n+1} = a(a^n - b^n) + b^n(a-b) \\ 243 | = a(a-b)k + b^n(a-b) \\ 244 | = (a-b)(ak + b^n) \\ 245 | \end{align*} 246 | So, $(a-b) \mid (a^{n+1} - b^{n+1})$ as required. 247 | 248 | \section{Solution 13} 249 | Let $a$ and $b$ be arbitrary element. By mathematical induction, 250 | 251 | Base case. When $n=0$, $(a+b) \mid (a^{2n+1} + b^{2n + 1})$ holds for 252 | $1 \in Z$. 253 | 254 | Induction step. Let $n$ be an arbitrary element in $N$ such that 255 | $(a+b)k = a^{2n+1} + b^{2n+1}$. Thus, 256 | \begin{align*} 257 | a^{2(n+1) + 1} + b^{2(n+1) + 1} = a^{2n + 3} + b^{2n + 3} \\ 258 | = a^2(a^{2n + 1}+ b^{2n+1}) + b^{2n+1}(b^2 - a^2) \\ 259 | = a^2(a+b)k + b^{2n+1}(b-a)(b+a) \\ 260 | = (a+b)(a^2k + b^{2n+1}(b-a)) 261 | \end{align*} 262 | So, $(a+b) \mid a^{2(n+1) + 1} + b^{2(n+1) + 1}$ as required. 263 | 264 | \section{Solution 14} 265 | By mathematical induction, 266 | 267 | Base case. When $n = 10$, $2^n > n^3$ holds since $2^{10} > 2^3$. 268 | 269 | Induction step. Let $n$ be an arbitrary element and suppose $n >= 10$ 270 | and $2^n > n^3$. Then, 271 | \begin{align*} 272 | 2^{n+1} = 2.2^n \\ 273 | > 2.n^3 (\text{Inductive hypothesis}) \\ 274 | = n^3 + n^3 \\ 275 | >= n^3 + 10n^2 (\text{Since n >= 10}) \\ 276 | = n^3 + 3n^2 + 7n^2 \\ 277 | >= n^3 + 3n^2 + 70n (\text{Since n >= 10}) \\ 278 | = n^3 + 3n^2 + 3n + 67n \\ 279 | > n^3 + 3n^2 + 3n + 1 \\ 280 | > (n+1)^3 281 | \end{align*} 282 | So, $2^{n+1} > (n+1)^3$ as required. 283 | 284 | \section{Solution 15} 285 | By mathematical induction, 286 | 287 | Base case. When $n=0$, $n \equiv 0 (mod 3)$ holds for $0 \in Z$. 288 | 289 | Induction step. Let $n$ be an arbitrary element in $N$. Suppose 290 | $n \equiv 0 (\text{mod } 3)$ or $n \equiv 1 (\text{mod } 3)$ or 291 | $n \equiv 2 (\text{mod } 3)$. 292 | 293 | Case 1. $n \equiv 0 (\text{mod } 3)$. It follows that $n = 3a$ for 294 | some $a \in Z$. Adding $1$ on both sides, we get $n+1 = 3a + 1$. So, 295 | $(n+1) - 1 = 3a$. Therefore, $n+1 = 1(\text{mod } 3)$. So, either 296 | $n+1 = 0(\text{mod } 3)$ or $n+1 = 1(\text{mod } 3)$ or 297 | $n+1 = 2(\text{mod } 3)$. 298 | 299 | Case 2. $n \equiv 1 (\text{mod } 3)$. It follows that $n-1=3b$ for 300 | some $b \in Z$. We can rewrite it as $(n+1) - 2 = 3b$. So, 301 | $(n+1) = 2(\text{mod } 3)$. So, $(n+1) \equiv 2(\text{mod } 3)$. So, 302 | either $n+1 = 0(\text{mod } 3)$ or $n+1 = 1(\text{mod } 3)$ or 303 | $n+1 = 2(\text{mod } 3)$. 304 | 305 | Case 3. $n \equiv 2 (\text{mod } 3)$. It follows that $n-2 = 3c$ for 306 | some $c \in Z$. Rewriting it, we get $n+1 = 3(c+1)$. So, 307 | $n+1 \equiv 0 (\text{mod } 3)$. So, either $n+1 = 0(\text{mod } 3)$ or 308 | $n+1 = 1(\text{mod } 3)$ or $n+1 = 2(\text{mod } 3)$. 309 | 310 | \section{Solution 16} 311 | By mathematical induction, 312 | 313 | Base case. When $n = 1$, both sides of the equation becomes $8$. 314 | 315 | Induction step. Let $n >=1$ be an arbitrary element. Suppose $2.2^1 + 316 | 3.2^2 + 4.2^3 + ... + (n+1)2^n = n2^{n+1}$. Then, 317 | \begin{align*} 318 | 2.2^1 + 3.2^2 + 4.2^3 + ... + (n+2)2^{n+1} \\ 319 | = (2.2^1 + 3.2^2 + ... + (n+1)2^n) + (n+2)2^{n+1} \\ 320 | = n2^{n + 1} + (n+2)2^{n+1} \\ 321 | = 2^{n+1}(n + n + 2) \\ 322 | = 2^{n+2}(n+1) 323 | \end{align*} 324 | 325 | \section{Solution 17} 326 | \subsection{Solution (a)} 327 | Base case doesn't hold. (When $n = 0$). 328 | 329 | Let $f(n) = 1.3^0 + 3.3^1 + 5.3^2 + ... + (2n+1)3^n$ 330 | When $n = 0$, $f(n) = 1$, 331 | $n \neq 0, f(n) = n3^{n + 1}$. 332 | 333 | Proof. Copy the question of 17(a) :-) 334 | 335 | \section{Solution 18} 336 | Suppose $a$ is a real number and $a = 0$. Let $n$ be an arbitrary 337 | element in $N$. Let us consider the cases. 338 | 339 | Case 1. n is even. By mathematical induction, 340 | 341 | Base case. When $n = 0$, then $a^n > 0$ since $a^n = 1$. 342 | 343 | Induction step. Let $n$ be an even arbitrary number in $N$. Suppose 344 | $a^n > 0$. We have to prove $a^{n+2} > 0$. We know that $a^n > 0$. 345 | Since $a$ is a negative real number, $a^2$ will be a positive number. 346 | So, $a^n.a^2 > 0$. So, $a^{n+2} > 0$ as required. 347 | 348 | Case 2. n is odd. By mathematical induction, 349 | 350 | Base case. When $n=1$, $a^{n} < 0$ since $a$ is a negative number. 351 | 352 | Induction step. Let $n$ be an odd arbitrary number in $N$. Suppose 353 | $a^n < 0$. We have to prove that $a^{n+2} < 0$. Since $a < 0$, $a^2 > 354 | 0$ since $a$ is a negative number. Multiplying $a^2$ with $a^n < 0$, 355 | we get $a^{n+2} < 0$ as required. 356 | 357 | \section{Solution 19} 358 | Suppose $a$ and $b$ are real numbers and $0 < a < b$. 359 | \subsection{Solution (a)} 360 | By mathematical induction, 361 | 362 | Base case. When $n=1$, $0 < a^n < b^n$. 363 | 364 | Induction step. Let $n$ be an arbitrary element such that $n>= 1$. 365 | Suppose $0 < a^n < b^n$. Then, 366 | $0 < a^n < b^n$ \\ 367 | \begin{equation}\label{eq:1} 368 | 0 < a^{n+1} < ab^n 369 | \end{equation} 370 | We know that $a < b$. So, $ab^n < b^{n+1}$. From \ref{eq:1} , it follows 371 | that $0 < a^{n+1} < b^{n+1}$ as required. 372 | 373 | \subsection{Solution (b)} 374 | By mathematical induction, 375 | 376 | Base case. When $n = 2$, $0 < \sqrt[n]{a} < \sqrt[n]{b}$ since $0 < a 377 | < b$. 378 | 379 | Induction step. Let $n$ be an arbitrary element such that $n >= 2$. 380 | Suppose $0 < \sqrt[n]{a} < \sqrt[n]{b}$. Then, 381 | \begin{align*} 382 | 0 < \sqrt[n]{a} < \sqrt[n]{b} \\ 383 | 0 < \sqrt{a} x \sqrt[n]{a} < \sqrt{a} x \sqrt[n]{b} \\ 384 | \end{align*} 385 | \begin{equation} \label{eq:2} 386 | 0 < (n+1)\sqrt[n+1]{a} < \sqrt{a}.\sqrt[n]{b} 387 | \end{equation} 388 | 389 | We know that $a < b$. So, $\sqrt{a} < \sqrt{b}$. Multiplying it by 390 | $\sqrt[n]{b}$ on both sides, $\sqrt{n}. \sqrt[n]{b} < \sqrt[n+1]{b}$. 391 | From \ref{eq:2}, $0 < \sqrt[n+1]{a} < \sqrt[n+1]{b}$ as required. 392 | 393 | \subsection{Solution (c)} 394 | By mathematical induction, 395 | 396 | Base case. Since $a > 0$ and $b>0$, it follows that $(a-b)^2 > 0$. So, 397 | $a^2 + b^2 - 2ab > 0$. So, $a^2 + b^2 > 2ab$. 398 | 399 | Induction step. Let $n$ be an arbitrary element such that $n >= 1$. 400 | Suppose $ab^n + ba^n < a^{n+1} + b^{n+1}$. Then, 401 | \begin{align*} 402 | 0 < a^{n+1} + b^{n+1} - ba^n - ab^n \\ 403 | 0 < a(a^n - b^n) + b(b^n - a^n) \\ 404 | (a^n - b^n)(a - b) > 0 \\ 405 | \end{align*} 406 | From (a), it follows that $a^{n+1} < b^{n+1}$ . We know that $a - b > 407 | 0$. Multiplying both, 408 | \begin{align*} 409 | (a^{n+1} - b^{n+1})(a-b) > 0 \\ 410 | ab^{n+1} + ba^{n+1} < a^{n+2} + b^{n+2} \text{ as required} 411 | \end{align*} 412 | 413 | \subsection{Solution (d)} 414 | By mathematical induction, 415 | 416 | Base case. When $n = 2$, $(\frac{a+b}{2})^2 < \frac{a^2 + b^2}{2}$ since 417 | $2ab < a^2 + b^2$. (Apply n = 1, to question 19 (c)) 418 | 419 | Induction step. Let $n >= 2$ be an arbitrary element and suppose 420 | $(\frac{a+b}{2})^2 < \frac{a^2 + b^2}{2}$. Then, 421 | \begin{align*} 422 | (\frac{a+b}{2})^{n+1} < \frac{a^n + b^n}{2}.\frac{a+b}{2} \\ 423 | (\frac{a+b}{2})^{n+1} < \frac{a^{n+1} + b^{n+1} + ba^n + ab^n}{4} \\ 424 | \end{align*} 425 | From part (c), $ab^n + ba^n < a^{n+1} + b^{n+1}$ 426 | So, $\frac{a^{n+1} + b^{n+1} + ba^n + ab^n}{4} < \frac{2a^{n+1} + 427 | 2b^{n+1}}{4}$. Hence. $(\frac{a+b}{2})^{n+1} < \frac{a^{n+1} + b^{n+1}}{2}$. 428 | 429 | \end{document} 430 | -------------------------------------------------------------------------------- /chapter 6/section6-2.tex: -------------------------------------------------------------------------------- 1 | %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% 2 | % Author: Sibi 3 | %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% 4 | \documentclass{article} 5 | \usepackage{graphicx} 6 | \usepackage{verbatim} 7 | \usepackage{amsmath} 8 | \usepackage{amsfonts} 9 | \usepackage{amssymb} 10 | \usepackage{tabularx} 11 | \usepackage{mathtools} 12 | \newcommand{\BigO}[1]{\ensuremath{\operatorname{O}\bigl(#1\bigr)}} 13 | \setlength\parskip{\baselineskip} 14 | \begin{document} 15 | \title{Chapter 6 (Section 6.2)} 16 | \author{Sibi} 17 | \date{\today} 18 | \maketitle 19 | 20 | % See here: http://tex.stackexchange.com/a/43009/69223 21 | \DeclarePairedDelimiter\abs{\lvert}{\rvert}% 22 | \DeclarePairedDelimiter\norm{\lVert}{\rVert}% 23 | 24 | % Swap the definition of \abs* and \norm*, so that \abs 25 | % and \norm resizes the size of the brackets, and the 26 | % starred version does not. 27 | \makeatletter 28 | \let\oldabs\abs 29 | \def\abs{\@ifstar{\oldabs}{\oldabs*}} 30 | % 31 | \let\oldnorm\norm 32 | \def\norm{\@ifstar{\oldnorm}{\oldnorm*}} 33 | \makeatother 34 | \newpage 35 | 36 | \section{Solution 1} 37 | 38 | 39 | $R' = R \cap (A' \times A')$ 40 | $A' = A \setminus \{a\}$ 41 | 42 | \subsection{Solution (a)} 43 | Reflexive. 44 | 45 | Let $b$ be an arbitrary element on $A'$. Since $A' \subseteq A$, it 46 | follows that $(b,b) \in R$ since $R$ is reflexive. Also, 47 | $(b,b) \in A' \times A'$. So, $(b,b) \in R \cap (A' \times A')$. Since 48 | $b$ is arbitrary we can conclude that $R$ is reflexive. 49 | 50 | Transitive. 51 | 52 | Let $b,c,d$ be arbitrary element on $A'$ such that $(b,c) \in R'$ and 53 | $(c,d) \in R'$. Since $R$ is transitive, it follows that 54 | $(b,d) \in R$. Also, $(b,d) \in A' \times A'$. So, 55 | $(b,d) \in R \cap (A' \times A')$. 56 | 57 | Anti-symmetric. 58 | 59 | Let $x,y$ be arbitrary element on $A'$ such that $xR'y$ and $yR'x$. 60 | Since $R$ is anti-symmetric, it follows that $x=y$. 61 | 62 | \subsection{Solution (b)} 63 | Reflexive. 64 | 65 | Let $x$ be an arbitrary element in $A$. Then either $x \in A'$ or 66 | $x \in \{a\}$. Let us consider the cases: 67 | 68 | Case 1. $x \in A'$ Since $T'$ is a total order in $A'$, it follows 69 | that $(x,x) \in T$. 70 | Case 2. $x \in \{a\}$. Then $x = a$. It follows that 71 | $(a,a) \in \{a\} \times A$. So, $(x,x) \in T$. 72 | 73 | Transitive. 74 | 75 | Let $x,y$ and $z$ be arbitrary element in $A$ such that $(x,y) \in T$ 76 | and $(y,z) \in T$. Let us consider the cases. 77 | 78 | Case 1. $x \neq a$. Then $x \in A'$. Since 79 | $T = T' \cup (\{a\} \times A)$, it follows that $y \in A'$ and 80 | $z \in A'$. Since $T'$ is transitive, it follows that $(x,z) \in T$. 81 | Case 2. $x = a$ Since $z \in A$, it follows that 82 | $(a,z) \in \{a\} \times A$. So, $(x,z) \in T$. 83 | 84 | Anti-symmetric 85 | 86 | Let $x,y$ be arbitrary element on $A$ such that $xTy$ and $yTx$. Let 87 | us consider the cases: 88 | Case 1. $x \neq a$. Then $x \in A'$, $y \in A'$ since $(x,y) \in T' 89 | \cup ({a} \times A)$ and $x \in A'$. So, $(x,y) \in T'$. Similarly, 90 | $(y,x) \in T'$. Since $T'$ is a total order on $A'$, it follows that 91 | $x = y$. 92 | Case 2. $x = a$. So, $(y,a) \in T$. Therefore $(y,a) \in T' \cup \{a\} 93 | \times A$. Now $T'$ is a relation on $A'$ which doesn't have $a$. So, 94 | $(y,a) \notin T'$. So, $(y,a) \in \{a\} \times A$. Now, clearly $y = 95 | a$. So, $x = y$. 96 | 97 | $\forall x \in A \forall y \in A (xTy \lor yTx)$ 98 | 99 | Let $x,y$ be arbitrary element on $A$. Let us consider the cases: 100 | Case 1. $x = a$. Since $y \in A$, it follows that $(a,y) \in \{a\} 101 | \times A$. So, $xTy \lor yTx$. 102 | Case 2. $x \neq a$. Now if $y = a$, then $xRa$ will contradict the $R$ 103 | minimality of $A$. If $y \neq a$, then $y \in A'$. So, $(x,y) \in R'$. 104 | Since $R' \subseteq T' \subseteq T$, it follows that $(x,y) \in T$. 105 | 106 | \section{Solution 3} 107 | We will show by induction that for every natural number $n >= 1$, 108 | every subset of $A$ with $n$ elements has a smallest element. 109 | 110 | Base case. When $n =1$, then the single element present in it is the 111 | smallest element. 112 | 113 | Induction step. Suppose $n >= 1$ and suppose that every subset of $A$ 114 | has a smallest element. Now let $B$ be arbitrary subset of $A$ with 115 | $n+1$ elements. Let $b$ be an arbitrary element of $B$ and let 116 | $B' = B \setminus \{b\}$ with n elements. By inductive hypothesis, 117 | $B'$ has a smallest element $c \in B'$. Let us consider the cases. 118 | 119 | Case 1. $bRc$ We know that $\forall x \in B'(cRx)$ and $B' = B 120 | \setminus \{b\}$. 121 | Now from $bRc$, it is clear that $b$ is the smallest element. (Note 122 | that $R$ is transitive). 123 | 124 | Case 2. $cRb$ Then it follows that $\forall x \in B(cRx)$. So, $c$ is 125 | the smallest element. 126 | 127 | \section{Solution 4} 128 | \subsection{Solution (a)} 129 | We will show by mathematical induction that $\forall n >= 1 \forall B 130 | \subseteq A (\forall x \in A \forall y \in A(xRy \lor yRx) \implies 131 | \exists z \in B \forall y \in B((z,y) \in R \circ R))$. 132 | 133 | Base case. When $n=1$, then $B = \{b\}$ for some $b \in A$. Since 134 | $(b,b) \in R \circ R$, base case holds. 135 | 136 | Induction step. Suppose $n >= 1$ and suppose that every subset of $A$ 137 | has some elements $z \in B$ such that $\forall y \in B((z,y) \in R 138 | \circ R)$. Now let $C$ be arbitrary subset of $A$ with $n+1$ elements. 139 | Now let $c$ be some element in $C$ and let $C' = C \setminus \{c\}$. 140 | By inductive hypothesis, $\exists z \in C' \forall y \in C'((z,y) \in 141 | R \circ R)$. 142 | Case 1. $(z,c) \in R \circ R$. Then $\forall y \in C((z,y) \in R \circ 143 | R)$. 144 | Case 2. $(z,c) \notin R \circ R$. Then we will try to prove that 145 | $\forall y \in C((c,y) \in R \circ R)$. Let $y$ be arbitrary element 146 | in $C$. If $y = c$, then since $R$ is reflexive, $(c,y) \in R \circ 147 | R$. Now, if $y \neq c$, then $y \in C'$. So, $(z,y) \in R \circ R$. 148 | This means $\exists x \in A such that (z,x) \in R$ and $(x,y) \in R$. 149 | Now if $(z,x) \in R$and $(x,c) \in R$, then $(z,c) \in R \circ R$ 150 | contrary to assumption. So, $(x,c) \notin R$. From the hypothesis on 151 | $R$, $(c,x) \in R$. From $(c,x) \in R$ and $(x,y) \in R$, it follows 152 | that $(c,y) \in R \circ R$. 153 | 154 | \subsection{Solution (b)} 155 | \begin{align*} 156 | A = \text{Set of all contestants} \\ 157 | R = \{(x,y) \in A \times A \mid x beats y \} 158 | \end{align*} 159 | Now $R$ satisfies this: $\forall x \in A \forall y \in A(xRy \lor 160 | yRx)$. Now from $(a)$, it follows that $B \subseteq A$, $\exists x \in 161 | B(\forall y \in B((x,y) \in R \circ R))$. Let $B=A$, then $(x,y) \in R 162 | \circ R$. So there is at least one successful player. 163 | 164 | \section{Solution 5} 165 | By mathematical induction, 166 | 167 | Base case. When $n=1$, $F_1 = F_o.F_1.F_2$. Both the sides of the 168 | equation becomes $5$ and so the base case holds. 169 | 170 | Induction step. Let $n$ be arbitrary element and suppose $F_n = 171 | 2^{(2^n)} + 1$. We have to prove that $F_{n+1} = F_0.F_1.F_2 ... F_n + 172 | 2$. By inductive hypothesis, 173 | \begin{align*} 174 | F_n = F_0.F_1.F_2 ... F_{n-1} + 2 \\ 175 | F_0.F_1.F_2 ... F_{n-1}.F_n = (F_n - 2)F_n \\ 176 | F_0.F_1.F_2 ... F_{n-1}.F_n + 2 = (F_n - 2)F_n + 2 \\ 177 | = (F_n)^2 - 2F_n + 2 \\ 178 | = (2^{(2^n)} + 1)^2 - 2(2^{2^n} + 1) + 2 \\ 179 | = 2^{(2^n)} + 1 \\ 180 | = F_n + 1 181 | \end{align*} 182 | 183 | \section{Solution 6} 184 | By mathematical induction, 185 | 186 | Base case. When $n=1$, $|a_1| = |a_1|$ so base case holds. 187 | 188 | Induction step. Let $n$ be arbitrary element and suppose $a_1 + a_2 + 189 | ... + a_n <= |a_1| + |a_2| + ... + |a_n|$. Let $a_{n+1}$ be arbitrary 190 | real number. Let us consider the cases. 191 | 192 | Case 1. $a_{n+1}$ is positive: From our inductive hypothesis, it 193 | follows that $a_1 + a_2 + ... + a_n + a_{n+1} <= |a_1| + |a_2| + ... + 194 | |a_n| + |a_{n+1}|$. 195 | 196 | Case 2. $a_{n+1}$ is negative: Then it follows that 197 | $a_1 + a_2 + ... + a_n + a_{n+1} < |a_1| + |a_2| + ... + |a_n| + 198 | |a_{n+1}|$. So, 199 | $a_1 + a_2 + ... + a_n + a_{n+1} <= |a_1| + |a_2| + ... + |a_n| + 200 | |a_{n+1}|$. 201 | 202 | \section{Solution 7} 203 | \subsection{Solution (a)} 204 | Let $a$ and $b$ be arbitrary positive real numbers. We know that 205 | $(a-b)^2 >= 0$. Then, 206 | \begin{align*} 207 | (a-b)^2 >= 0 \\ 208 | a^2 + b^2 >= 2ab \\ 209 | \frac{a}{b} + \frac{b}{a} >= 2 210 | \end{align*} 211 | 212 | QED. 213 | 214 | \subsection{Solution (b)} 215 | Suppose $a,b$ and $c$ be arbitrary real number and suppose $0 < a <= b 216 | <= c$. Then, $(c-b)(c-a) >= 0$ since $c >= b$ and $c >= a$. Then, 217 | \begin{align*} 218 | c(c-b)-a(c-b) >= 0 \\ 219 | c(c-b) + ab >= ac \\ 220 | \frac{ab + c(c-b)}{ac} >= 1 \\ 221 | \frac{b}{c} + \frac{c-b}{a} >= 1 \\ 222 | \frac{b}{c} + \frac{c}{a} - \frac{b}{a} >= 1 223 | \end{align*} 224 | QED. 225 | 226 | \subsection{Solution (c)} 227 | By mathematical induction, 228 | 229 | Base case. When $n = 2$, from 7(a), it follows that $\frac{a_1}{a_2} + 230 | \frac{a_2}{a_1} >= 2$. So base case holds. 231 | 232 | Induction step. Let $n >= 2$ be arbitrary element and suppose 233 | $\frac{a_1}{a_2} + \frac{a_2}{a_3} + \frac{a_3}{a_4} + ... + 234 | \frac{a_{n-1}}{a_n} + \frac{a_n}{a_1} >= n$. We have to prove 235 | $\frac{a_1}{a_2} + \frac{a_2}{a_3} + \frac{a_3}{a_4} + ... + 236 | \frac{a_{n-1}}{a_n} + \frac{a_n}{a_{n+1}} + \frac{a_{n+1}}{a_1} >= n + 237 | 1$. From $7(b)$, it follows that $\frac{a_n}{a_{n+1}} + 238 | \frac{a_{n+1}}{a_1} - \frac{a_n}{a_1} >= 1$. 239 | From inductive hypothesis, it follows that: 240 | $\frac{a_1}{a_2} + \frac{a_2}{a_3} + \frac{a_3}{a_4} + ... + 241 | \frac{a_{n-1}}{a_n} + \frac{a_n}{a_{n+1}} + 242 | \frac{a_{n+1}}{a_1} >= n + 1$. 243 | 244 | QED. 245 | 246 | \section{Solution 8} 247 | \subsection{Solution (a)} 248 | Let $a$ and $b$ be arbitrary positive real number. We know that 249 | $(a-b)^2 >= 0$. So, 250 | \begin{align*} 251 | (a-b)^2 >= 0 \\ 252 | a^2 + b^2 >= 2ab \\ 253 | a^2 + b^2 + 2ab >= 4ab \\ 254 | (a+b)^2 >= 4ab \\ 255 | \frac{(a+b)}{2} >= \sqrt{ab} 256 | \end{align*} 257 | QED. 258 | 259 | \subsection{Solution (b)} 260 | By mathematical induction on $n$, 261 | 262 | Base case. When $n=1$, From 8(a), base case holds. 263 | 264 | Induction step. Suppose $n>=1$ and the arithmetic geometric mean 265 | inequality holds for lists of length $2^n$. Now let $a_1,a_2,..., 266 | a_{a^{n+1}}$ be a list of $2^{n+1}$ positive real numbers. Let 267 | $m_1 = \frac{a_1 + a_2 + ... + a_{2^n}}{2^n} and m_2 = \frac{a_{2^n + 268 | 1} + a_{2^n + 2} + ... + a_{2^{n+1}}}{2^n}$. 269 | 270 | So, $a_1 + a_2 + ... + a_{2^n} = m_12^n$ and similarly $a_{2^n + 1} + 271 | a_{2^n + 2} + ... + a_{2^{n+1}} = m_22^n$. Also, by inductive 272 | hypothesis, we know that $m_1 >= \sqrt[2^n]{a_1a_2...a_{2^n}}$ and 273 | $m_2 >= \sqrt[2^n]{a_{2^n + 1}a_{2^n + 2}...a_{2^{n+1}} }$. Therefore, 274 | \begin{align*} 275 | \frac{a_1 + a_2 + ... + a_{2^n + 1}}{2^{n+1}} \\ 276 | = \frac{m_12^n + m_22^n}{2^{n+1}} \\ 277 | = \frac{m_1 + m_2}{2} \\ 278 | >= \sqrt{m_1m_2} (\text{from Solution (a)}) \\ 279 | >= \sqrt{\sqrt[2^n]{a_1a_2...a_{2^n}}\sqrt[2^n]{a_{2^n + 1}a_{2^n+2}...a_{2^{n+1}}}} \\ 280 | >= \sqrt[2^{n+1}]{a_1a_2...a_{a^{n+1}}} 281 | \end{align*} 282 | 283 | \section{Solution (c)} 284 | By mathematical induction on $n$, 285 | 286 | Base case. When $n=n_0$, then base case holds because of inductive 287 | hypothesis. 288 | 289 | Induction step. Suppose $n >= n_0$ and there are positive real numbers 290 | $a_1, a_2, ..., a_n$ such that 291 | \begin{align*} 292 | \frac{a_1 + a_2 + ... + a_n}{n} < \sqrt[n]{a_1a_2...a_n} 293 | \end{align*} 294 | Let $m = \frac{a_1 + a_2 + ... + a_n}{n}$ and $a_{n+1} = m$. Then we 295 | have $m < \sqrt[n]{a_1a_2...a_n}$, so $m^n < a_1a_2...a_n$. So, 296 | $m^{n+1} < a_1a_2...a_na_{n+1}$. So, $m < 297 | \sqrt[n+1]{a_1a_2...a_{n+1}}$. Now, 298 | \begin{align*} 299 | \frac{a_1 + a_2 + ... + a_{n+1}}{n+1} \\ 300 | = \frac{mn + m}{n+1} \\ 301 | = m 302 | \end{align*} 303 | So, $\frac{a_1 + a_2 + ... + a_{n+1}}{n+1} < 304 | \sqrt[n+1]{a_1a_2...a_{n+1}}$ 305 | 306 | \subsection{Solution (d)} 307 | $(b)$ and $(c)$ give contradiction for a list of length $n_0 < 2^n$ 308 | where $n >= 1$. So the inequality must hold. 309 | 310 | \section{Solution 9} 311 | Let $\frac{1}{a_n} = k_n$. From arithmetic geometric mean inequality, 312 | 313 | \begin{align*} 314 | \frac{k_1 + k_2 + ... + k_n }{n} <= \sqrt[n]{k_1K_2...K_n} \\ 315 | \frac{n}{k_1 + k_2 + ... + k_n } <= (k_1k_2...k_n)^{-1/n} \\ 316 | <= (a_1a_2...a_n)^{1/n} \\ 317 | \end{align*} 318 | QED. 319 | 320 | \section{Solution 10} 321 | By mathematical induction on $n$, 322 | 323 | Base case. When $n=0$, $P(A)$ has 1 element which is the empty set. 324 | 325 | Induction step. Let $n$ be an arbitrary element and suppose if $A$ has 326 | $n$ elements then $P(A)$ has $2^n$ elements. Let $A$ has $n+1$ 327 | elements. Then $P(A)$ can be divided into two sets $A_1$ and $A_2$. 328 | 329 | $A_1$ = Set which doesn't have $(n+1)$th element in it. 330 | $A_2$ = Set which has $(n+1)$th element in it. 331 | 332 | By inductive hypothesis, $A_1$ has $2^n$ elements. Now $A_2 = A_1 333 | \times \{x\}$ where $x$ is the $(n+1)$th element. 334 | 335 | So, $2^n + 2^n = 2^{n+1}$ elements. 336 | 337 | \section{Solution 11} 338 | By mathematical induction, 339 | Base case. When $n=0$, $P_2(A)$ has zero elements, so base case holds. 340 | 341 | Induction step. Let $n$ be an arbitrary element and suppose if $A$ has 342 | $n$ elements then $P_2(A)$ has $n(n-1)/2$ elements. Suppose $A$ has 343 | $n+1$ elements. Let us pick an arbitrary element $a$ from $A$ such 344 | that $A' = A \setminus \{a\}$. Now $A'$ has $n$ elements and $P_2(A')$ 345 | has $n(n-1)/2$ elements. There are two kinds of subsets of $A$ those 346 | that contain $a$ as an element and those that don't. The subset hat 347 | don't contain $a$ are just the subsets of $A'$ and by inductive 348 | hypothesis there are $n(n-1)/2$ of them. Now the subset of $A$ that 349 | contain $A$ and have a length of two are $n$. Summing them, we have 350 | $n(n+1)/2$ elements as required. 351 | 352 | \section{Solution 12} 353 | By mathematical induction, 354 | 355 | Base case. When $n=1$, the remaining area is a trapezoidal. 356 | 357 | Induction step. Let $n$ be an arbitrary element and suppose if an 358 | equilateral triangle is cut into $4^n$ congruent equilateral triangles 359 | and one corner is removed, then the remaining area is covered by 360 | trapezoidal tiles. Suppose equilateral triangle is cut into $4^{n+1}$ 361 | congruent triangles with $4^n$ triangles in each of them. Now you can 362 | view this as $4$ congruent equilateral triangles. Now since one corner 363 | has been removed, one triangle out of the $4$ has remaining area of it 364 | is covered by trapezoidal tiles. Now for the remaining three triangle, 365 | cut a trapezoidal area such that a corner is removed from each of 366 | them. Then again applying inductive hypothesis on each of them, we get 367 | trapezoidal area in each of them. 368 | 369 | \section{Solution 13} 370 | By mathematical induction on $n$, 371 | 372 | Base case. When $n = 1$, the chord divides the circle into $2$. 373 | Calculating $\frac{n^2 + n + 2}{2}$, we know that base case holds. 374 | 375 | Induction step. Let $n$ be an arbitrary element such that $n$ chords 376 | cut the circle into $\frac{n^2 + n + 2}{2}$ regions. Now suppose 377 | $(n+1)$ chords are drawn in the circle. Now for $n$ chords, we have 378 | $\frac{n^2 + n + 2}{2}$ regions. And for the next chord, we have 379 | $(n+1)$ regions. 380 | So, 381 | \begin{align*} 382 | \frac{n^2 + n + 2}{2} + (n+1) \\ 383 | = \frac{n^2 + 3n + 4}{2} \\ 384 | = \frac{(n+1)^2 + (n+1) + 2}{2} 385 | \end{align*} 386 | 387 | \section{Solution 15} 388 | You cannot make any assumptions about $n$. $n \in A$ doesn't imply 389 | $(n+1) \in A$. 390 | 391 | \section{Solution 16} 392 | Induction case assumes that there is at least three elements of $A$ 393 | which is not the case always. $n >= 0$, so $n$ can be $2$. 394 | \end{document} 395 | -------------------------------------------------------------------------------- /chapter 4/section4-3.tex: -------------------------------------------------------------------------------- 1 | %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% 2 | % Author: Sibi 3 | %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% 4 | \documentclass{article} 5 | \usepackage{graphicx} 6 | \usepackage{verbatim} 7 | \usepackage{amsmath} 8 | \usepackage{amsfonts} 9 | \usepackage{amssymb} 10 | \usepackage{tabularx} 11 | \setlength\parskip{\baselineskip} 12 | \begin{document} 13 | \title{Chapter 4 (Section 4.3)} 14 | \author{Sibi} 15 | \date{\today} 16 | \maketitle 17 | \newpage 18 | 19 | \section{Problem 1} 20 | 21 | $R = \{(a,bad), (a,cab), (b,bad), (b,bed), (b,cab), (c,cab), (d,bad), 22 | (d, bed), (e,bed), (e,bed)\}$ 23 | 24 | \section{Problem 2} 25 | 26 | $A = \{cat, dog, bird, rat\}$ 27 | $R = \{(x,y) \in A \times A \mid there is at least one letter that 28 | occurs in both of the words x and y\}$ 29 | 30 | $R = \{(cat, cat), (cat, rat), (dog, dog), (dog, birds), (bird, dog), 31 | (bird, bird), (bird, rat), (rat, cat), (rat, bird), (rat, rat)\}$ 32 | 33 | \begin{itemize} 34 | \item Reflexive 35 | \item Symmetric 36 | \item Not transitive: there is an edge from bird to rat and rat to 37 | cat. But there is no edge from bird to cat. 38 | \end{itemize} 39 | 40 | \section{Problem 3} 41 | 42 | $i_{A} = \{(x,y) \in A \times A \mid x = y\}$ 43 | $i_{A} = \{(1,1), (2,2), (3,3), (4,4)\}$ 44 | 45 | \section{Problem 4} 46 | 47 | \subsection(Solution (a)) 48 | 49 | $R = \{(a,c), (c,c), (d,a), (d,b), (b,d)\}$ 50 | 51 | \begin{itemize} 52 | \item Not reflexive since $(a,a) \notin R$ 53 | \item Not symmetric since $(a,c) \implies (c,a)$ doesn't hold 54 | \item Not transitive since $(d,a) \land (d,c) \implies (d,c)$ doesn't hold 55 | \end{itemize} 56 | 57 | \subsection{Solution (b)} 58 | 59 | $R = \{(a,b), (b,a), (a,d), (b,d)\}$ 60 | 61 | \begin{itemize} 62 | \item Not reflexive since $(a,a) \notin R$ 63 | \item Not symmetric since $(a,d) \implies (d,a)$ doesn't hold 64 | \item Not transitive since $(a,b) \land (b,a) \implies (a,a)$ doesn't hold 65 | \end{itemize} 66 | 67 | \subsection{Solution (c)} 68 | 69 | $R = \{(a,a), (c,c), (b,b), (b,d), (d,b), (d,d)\}$ 70 | 71 | \begin{itemize} 72 | \item Reflexive 73 | \item Symmetric 74 | \item Transitive 75 | \end{itemize} 76 | 77 | \subsection{Solution (d)} 78 | 79 | $R = \{(a,c), (a,b), (a,d), (b,d), (c,d)\}$ 80 | 81 | \begin{itemize} 82 | \item Not reflexive 83 | \item Not symmetric 84 | \item Transitive 85 | \end{itemize} 86 | 87 | \section{Solution 5} 88 | 89 | $S \circ R = \{(a,c) \in A \times C \mid \exists b \in B((a,b) \in R 90 | \land (b,c) \in S\}$ 91 | 92 | $S \circ R = \{(b,x), (a,y), (c,y), (a,z), (c,z)\}$ 93 | 94 | \section{Solution 6} 95 | 96 | $ D_r = \{(x,y) \in R \times R \mid |x - y| < r \}$ 97 | $ D_s = \{(x,y) \in R \times R \mid |x - y| < s \}$ 98 | 99 | $ D_r \circ D_s = \{(a,b) \in R \times R \mid \exists c \in R((a,c) 100 | \in D_s \land (c,b) \in D_r)\}$ 101 | $= \{(a,b) \in R \times R \mid \exists c \in R (|a - c| < s \land |c - 102 | b| < r)\}$ 103 | 104 | From triangle inequality, $|a + b| < |a| + |b|$ 105 | $|(a-c) + (c-b)| < |a-c| + |c-b|$ 106 | $|a + (-b)| < s + r$ 107 | $ D_r \circ D_s = \{(a,b) \in R \times R \mid |a-b| < s + r\}$ 108 | 109 | \section{Problem 7} 110 | 111 | ($\Rightarrow$) Suppose $R$ is reflexive. Then from the definition of 112 | reflexive it follows that $\forall. x \in A((x,x) \in R)$. Then 113 | clearly, $i_A \subseteq R$. 114 | 115 | ($\Leftarrow$) Suppose $i_A \subseteq R$. Let $x$ be an arbitrary 116 | element in $A$. Then $(x,x) \in i_A$. From $i_A \subseteq R$, it 117 | follows that $(x,x) \in R$. Since $x$ was arbitrary we can conclude 118 | that $R$ is reflexive. 119 | 120 | \section{Problem 8} 121 | 122 | ($\Rightarrow$) Suppose $R$ is transitive. Let $(x,z)$ be an arbitrary 123 | element in $R \circ R$. Then by existential instantiation, $(x,c) \in 124 | R$ and $(c,z) \in R$. Since $R$ is transitive, it follows that $(x,z) 125 | \in R$. So $R \circ R \subseteq R$. 126 | 127 | ($\Leftarrow$) Suppose $R \circ R \subseteq R$. Let $x$, $y$, $z$ be 128 | an arbitrary element in $A$ such that $(x,y) \in R$ and $(y,z) \in R$. 129 | So, $(x,z) \in R \circ R$. It follows that $(x,z) \in R$. Since $x,z$ 130 | are arbitrary, $R$ is transitive. 131 | 132 | \section{Problem 9} 133 | 134 | \subsection{Solution (a)} 135 | 136 | ($\Rightarrow$) Suppose $R \circ i_A$. Let $(a,b)$ be an arbitrary 137 | element in $R \circ i_A$. By existential instantiation, it follows 138 | that $(a,c) \in i_A$ and $(c,b) \in R$. From $(a,c) \in i_A$, it 139 | follows that $a = c$. So, $(a,b) \in R$. Since $(a,b)$ is arbitrary it 140 | follows that $R \circ i_A \subseteq R$. 141 | 142 | ($\Leftarrow$) Suppose $R$. Let $(a,b)$ be an arbitrary element in 143 | $R$. Now $a \in A$. So, $(a,a) \in i_A$. Therefore $(a,b) \in R \circ 144 | i_A$. Since $(a,b)$ is arbitrary, we can conclude that $R \subseteq R 145 | \circ i_A$. 146 | 147 | \subsection{Solution (b)} 148 | 149 | ($\Rightarrow$) Suppose $i_B \circ R$. Let $(a,b)$ be an arbitrary element in $i_B 150 | \circ R$. By existential instantiation, $(a,c) \in R$ and $(c,b) \in 151 | i_B$. From the definition of identity relation it follows that $c = 152 | b$. So, $(a,b) \in R$. Therefore, $i_B \circ R \subseteq R$. 153 | 154 | ($\Leftarrow$) Let $(a,b)$ be an arbitrary element in $R$. Now $b \in 155 | B$. So, $(b,b) \in i_B$. So, $(a,b) \in i_B \circ R$. Since $(a,b)$ is 156 | arbitrary it follows that $R \subseteq i_B \circ R$. 157 | 158 | \section{Problem 10} 159 | 160 | Let $(a,a)$ be an arbitrary element in $i_D$. It follows that $a \in 161 | D$. By existential instantiation, $(a,b) \in S$. Also, $(b,a) \in 162 | S^{-1}$. So, $(a,a) \in S^{-1} \circ S$. Since $(a,a)$ was an 163 | arbitrary element it follows that $i_D \subseteq S^{-1} \circ S$. 164 | 165 | Let $(a,a)$ be an arbitrary element in $i_R$. It follows that $a \in 166 | R$. By existential instantiation, it follows that $(b,a) \in S$. Also, 167 | $(a,b) \in S^{-1}$. So $(a,a) \in S \circ S^{-1}$. Since $(a,a)$ was 168 | an arbitrary element it $i_R \subseteq S \circ S^{-1}$. 169 | 170 | \section{Problem 11} 171 | 172 | Suppose $R$ is reflexive. Let $(a,b)$ be an arbitrary element in $R$. 173 | It follows that both $a \in A$ and $b \in A$. Now since $R$ is 174 | reflexive it follows that $(b,b) \in R$. Therefore $(a,b) \in R \circ 175 | R$. Since $(a,b)$ is arbitrary it follows that $R \subseteq R \circ R$. 176 | 177 | \section{Problem 12} 178 | 179 | \subsection{Solution (a)} 180 | Suppose $R$ is reflexive. Let $a$ be an arbitrary element in $A$. Then 181 | it follows that $(a,a) \in R$ . Then clearly $(a,a) \in R^{-1}$. Since 182 | $a$ was arbitrary it follows that $\forall a \in A((a,a) \in R^{-1})$. 183 | Therefore, $R^{-1}$ is reflexive. 184 | 185 | \subsection{Solution (b)} 186 | Suppose $R$ is symmetric. Let $a$ and $b$ be an arbitrary element in 187 | $A$ such that $(a,b) \in R^{-1}$. From $(a,b) \in R^{-1}$ it follows 188 | that $(b,a) \in R$. Since $R$ is symmetric, it follows that $(a,b) \in 189 | R$. So, $(b,a) \in R^{-1}$. Therefore, if $(a,b) \in R^{-1}$ then 190 | $(b,a) \in R^{-1}$. Since $a$ and $b$ are arbitrary, it follows that 191 | $R^{-1}$ is symmetric. 192 | 193 | \subsection{Solution (c)} 194 | Suppose $R$ is transitive. Let $a$, $b$ and $c$ be an arbitrary 195 | element in $A$ such that $(a,b) \in R^{-1}$ and $(b,c) \in R^{-a}$. It 196 | follows that $(b,a) \in R$ and $(c,b) \in R$. Since $R$ is transitive 197 | we get $(c,a) \in R$. So, clearly $(a,c) \in R^{-1}$. Therefore 198 | $R^{-1}$ is transitive. 199 | 200 | \section{Problem 13} 201 | 202 | \subsection{Solution (a)} 203 | 204 | Let $a$ be an arbitrary element in $A$. Suppose $R_1$ and $R_2$ are 205 | reflexive. Then it follows that $(a,a) \in R_1$ and $(a,a) \in R_2$. 206 | Therefore $R_1 \cup R_2$ is reflexive. 207 | 208 | \subsection{Solution (b)} 209 | 210 | Let $a, b$ be an arbitrary element in $A$. Suppose $a (R_1 \cup R_2)b$ 211 | or $(a,b) \in R_1 \cup R_2$. Let us consider the cases separately: 212 | 213 | Case 1. $(a,b) \in R_1$. Since $R_1$ is symmetric it follows that 214 | $(b,a) \in R_1$. Therefore $(b,a) \in R_1 \cup R_2$. 215 | Case 2. $(a,b) \in R_2$. Since $R_2$ is symmetric it follows that 216 | $(b,a) \in R_2$. Therefore $(b,a) \in R_1 \cup R_2$. 217 | 218 | Since a and b are arbitrary, it follows that $R_1 \cup R_2$ is symmetric. 219 | 220 | \subsection{Solution (c)} 221 | 222 | $R_1 = \{(1,2), (2,1), (1,1)$ 223 | $R_2 = \{(4,2), (2,9), (4,9)\}$ 224 | 225 | It's not transitive since $(4,2) \in R_1 \cup R_2 \land (2,1) \in R_1 226 | \cup R_2 \implies (4,1) \in R_1 \cup R_2$ doesn't hold. 227 | 228 | \section{Problem 14} 229 | 230 | \subsection{Solution (a)} 231 | 232 | Suppose $R_1$ and $R_2$ are reflexive. Let $a$ be an arbitrary element 233 | in $A$. Since $R_1$ and $R_2$ are reflexive, it follows that $(a,a) 234 | \in R_1 \cap R_2$. Since $a$ is arbitrary, we can conclude that $R_1 235 | \cap R_2$ is reflexive. 236 | 237 | \subsection{Solution (b)} 238 | 239 | Suppose $R_1$ and $R_2$ are symmetric. Let $x$ and $y$ be arbitrary 240 | elements in $A$ such that $(x,y) \in R_1 \cap R_2$. Since both $R_1$ 241 | and $R_2$ are symmetric, it follows that $(y,x) \in R_1 \cap R_2$. 242 | Since $x$ and $y$ are arbitrary, we can conclude that $R_1 \cap R_2$ 243 | is symmetric. 244 | 245 | \subsection{Solution (c)} 246 | 247 | Suppose $R_1$ and $R_2$ are transitive. Let $x, y$ and $z$ e an 248 | arbitrary element in $A$ such that $(x,y) \in R_1 \cap R_2$ and $(y,z) 249 | \in R_1 \cap R_2$. Since $R_1$ and $R_2$ are transitive, it follows 250 | that $(x,z) \in R_1 \cap R_2$. Therefore $R_1 \cap R_2$ is transitive. 251 | 252 | \section{Problem 15} 253 | 254 | \subsection{Solution (a)} 255 | 256 | Counterexample: 257 | 258 | $A = \{1\}$ \\ 259 | $R_1 = \{(1,1)\}$ \\ 260 | $R_2 = \{(1,1)\}$ \\ 261 | $R_1 \setminus R_2 = \emptyset$ 262 | 263 | \subsection{Solution (b)} 264 | 265 | Suppose $R_1$ and $R_2$ are symmetric. Let $x$ and $y$ be arbitrary 266 | element in $A$ such that $(x,y) \in R_1 \setminus R_2$. So, 267 | $(x,y) \in R_1$. Since $R_1$ is symmetric, it follows that 268 | $(y,x) \in R_1$. Also $(x,y) \notin R_2$. Using contrapositive law and 269 | symmetric property, then $(y,x) \notin R_2$. Therefore $(y,x) \in 270 | R_1 \setminus R_2$. 271 | 272 | \subsection{Solution (c)} 273 | 274 | Counterexample: \\ 275 | $R_1 = \{(1,2),(2,4),(1,4)\}$ \\ 276 | $R_2 = \{(1,3),(3,4),(1,4)\}$ \\ 277 | $R_1 \setminus R_2 = \{(1,2), (3,4)\}$ 278 | 279 | \section{Problem 16} 280 | 281 | Suppose $R$ and $S$ are reflexive. Let $a$ be an arbitrary element in 282 | $A$. Since $R$ and $S$ are reflexive, it follows that $(a,a) \in R$ 283 | and $(a,a) \in S$. It follows that, $(a,a) \in R \circ S$. Since $a$ 284 | is arbitrary, we can conclude that $R \circ S$ is reflexive. 285 | 286 | \section{Problem 17} 287 | 288 | Suppose R and S are symmetric. 289 | 290 | ($\Rightarrow$) Suppose $R \circ S$ is symmetric. Let $(x,y)$ be an 291 | arbitrary element in $R \circ S$. Then by existential instantiation, 292 | it follows that $(x,a) \in S$ and $(a,y) \in R$. Now since $R$ and $S$ 293 | are symmetric it follows that $(a,x) \in S$ and $(y,a) \in R$. So, 294 | $(y,x) \in S \circ R$. Since $R \circ S$ is symmetric, 295 | $(y,x) \in R \circ S$. Since $x$ and $y$ are symmetric, it follows 296 | that $R \circ S \subseteq S \circ R$. Similarly we can prove that $S 297 | \circ R \subseteq R \circ S$. So, $S \circ R = R \circ S$. 298 | 299 | ($\Leftarrow$) Suppose $S \circ R = R \circ S$. Let $a,b$ be an 300 | arbitrary element in $A$ such that $(a,b) \in R \circ S$. From our 301 | assumption, it follows that $(a,b) \in S \circ R$. By existential 302 | instantiation, $(a,c) \in R$ and $(c,b) \in S$. Since $R$ and $S$ are 303 | symmetric, it follows that $(c,a) \in R$ and $(b,a) \in S$. So 304 | $(b,a) \in R \circ S$. Now since $a$ and $b$ are arbitrary it follows 305 | that $R \circ S$ is symmetric. 306 | 307 | \section{Problem 18} 308 | 309 | Suppose $S \circ R \subseteq R \circ S$. Let $a,b$ and $c$ be 310 | arbitrary elements of $A$ such that $(a,b) \in R \circ S$ and 311 | $(b,c) \in R \circ S$. By existential instantiation, it follows that 312 | $(a,d) \in S$ and $(d,b) \in R$. Similarly, $(b,e) \in S$ and 313 | $(e,c) \in R$. It follows that $(d,e) \in S \circ R$. From our 314 | assumption, it follows that $(d,e) \in R \circ S$. By existential 315 | instantiation, $(d,f) \in S$ and $(f,e) \in R$. From $(e,c) \in R$ and 316 | $(f,e) \in R$, it follows that $(f,c) \in R \circ R$. Similarly from 317 | $(d,f) \in S$ and $(a,d) \in S$, it follows that 318 | $(a,f) \in S \circ S$. Using Theorem 4.3.4(3), $(f,c) \in R$ and 319 | $(a,f) \in S$. So $(a,c) \in R \circ S$. Since $a$ and $c$ are 320 | arbitrary, $R \circ S$ is transitive. 321 | 322 | \section{Problem 19} 323 | 324 | \subsection{Solution (a)} 325 | 326 | Existential instantiation uses same variable. 327 | 328 | \subsection{Solution (b)} 329 | 330 | Counterexample: \\ 331 | $R = \{(1,3), (2,9)\}$ \\ 332 | $S = \{(\{1\},\{2,3\}), (\{(2,3\}, \{9\})\}$ 333 | 334 | \section{Problem 20} 335 | 336 | Suppose $R$ is transitive. Let $X,Y$ and $Z$ be arbitrary elements in 337 | $B$ such that $(X,Y) \in S$ and $(Y,Z) \in S$. From $(X,Y) \in S$, it 338 | follows that $\forall x \in X \forall y \in Y xRy$. Similarly, 339 | $\forall y \in Y \forall z \in Z yRz$. Now since $R$ is transitive, 340 | $\forall x \in X \forall z \in Z xRz$. Therefore $(X,Z) \in S$. Since 341 | $X,Y$ and $Z$ are arbitrary, $S$ is transitive. 342 | 343 | \section{Problem 21} 344 | 345 | \subsection{Solution (a)} 346 | 347 | Suppose $R$ is symmetric. Let $X$ be an arbitrary element in $P(A)$. 348 | We have to prove that $(X,X) \in S$. Let $x$ be an arbitrary element 349 | in $X$. From $x \in X$ and $X \in P(A)$, it follows that $x \in A$. 350 | Since $R$ is symmetric $(x,x) \in R$. Since $x$ is arbitrary, it 351 | follows that $\forall x \in X (x,x) \in R$. Applying universal 352 | instantiation and existential generalization gives us $\forall x \in X 353 | \exists y \in X(x,y) \in R$ 354 | 355 | Q.E.D 356 | 357 | \subsection{Solution(b)} 358 | 359 | Counter example: 360 | 361 | $R = \{(1,2), (2,1), (3,1), (1,3)\}$ \\ 362 | $X = \{1\}, Y = \{2,3\}$ \\ 363 | $(Y,X) \notin S$ 364 | 365 | \subsection{Solution (c)} 366 | 367 | Suppose $R$ is transitive. Let $X,Y$ and $Z$ be arbitrary element in 368 | $P(A)$ such that $(X,Y) \in R$ and $(Y,Z) \in R$. Let $x$ be an 369 | arbitrary element in $X$ and $z$ be some element in $Z$. We have to 370 | prove that $xRz$. From $(X,Y) \in R$ and $x \in X$, it follows that 371 | $\exists y \in Y (xRy)$. From $\exists y \in Y (xRy)$ and 372 | $(Y,Z) \in R$ and applying the transitive property, we get $xRz$. 373 | 374 | \section{Problem 22} 375 | 376 | Flaw is in the assumption of $xRy$. No where it is given to be 377 | assumed. \\ 378 | 379 | Theorem is false. 380 | 381 | %Phew! this took some time. Finally worked out this solution. 382 | %with help from lots of nice people.% 383 | \section{Problem 23} 384 | Suppose $F \subseteq P(A)$. Let $a,b,c$ be arbitrary elements of A 385 | such that $(a,b) \in R$ and $(b,c) \in R$. We have to prove that 386 | $(a,c) \in R$. Expanding $(a,c) \in R$, we get $\forall X(X \subseteq 387 | A \setminus \{a,c\}) \implies X \cup \{a\} \in F \implies X \cup \{C\} 388 | \in F)$. Let $X$ be arbitrary and suppose $X \subseteq A \setminus 389 | \{a,c\}$ and $X \cup \{a\} \in F$. We have to prove that $X \cup \{c\} 390 | \in F$. We can consider two cases: 391 | 392 | Case 1. $b \notin X$. From $X \subseteq A \setminus \{a,c\}$, it 393 | follows that $a \notin X$ and $c \notin X$. From $(a,b) \in R$ and 394 | $(b,c) \in R$, it follows that $X \cup \{c\} \in F$. 395 | 396 | Case 2. $b \in X$. Let $X_o = X \setminus \{b\}, X' = X_o \cup \{a\}, 397 | X'' = X_o \cup \{c\}$. We know that $c \notin X$, so $c \notin X'$. So 398 | $b,c \notin X'$. Also, $X' \cup \{b\} = X_o \cup \{a\} \cup \{b\} = X 399 | \setminus \{b\} \cup \{a\} \cup \{b\}$. Now $X \setminus \{b\} \cup 400 | \{a\} \cup \{b\} = X \cup \{a\} $ since $b \in X$. From $X \cup \{a\} 401 | \in F$, it follows that $X' \cup \{b\} \in F$. It follows that $X' 402 | \cup \{c\} \in F$ from $(b,c) \in R$. $X' \cup \{c\} = X_o \cup \{a\} 403 | \cup \{c\} = X'' \cup \{a\}$, so $X'' \cup \{a\} \in F$. We know that 404 | $a,b \notin X$, so $a,b \notin X_o$. So $a,b \notin X''$. We also know 405 | that $X'' \cup \{a\} \in F$. So from $(a,b) \in R$, it follows that 406 | $X'' \cup \{b\} \in F$. $X'' \cup \{b\} = X_o \cup \{b\} \cup \{c\} = 407 | X \setminus \{b\} \cup \{b\} \cup \{c\}$. $X \setminus \{b\} \cup 408 | \{b\} \cup \{c\} = X \cup \{c\}$ since $b \in X$. Therefore $X \cup 409 | \{c\} \in X$. 410 | 411 | \end{document} 412 | 413 | 414 | 415 | -------------------------------------------------------------------------------- /chapter 4/section4-5.tex: -------------------------------------------------------------------------------- 1 | %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% 2 | % Author: Sibi 3 | %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% 4 | \documentclass{article} 5 | \usepackage{graphicx} 6 | \usepackage{verbatim} 7 | \usepackage{amsmath} 8 | \usepackage{amsfonts} 9 | \usepackage{amssymb} 10 | \usepackage{tabularx} 11 | \setlength\parskip{\baselineskip} 12 | \begin{document} 13 | \title{Chapter 4 (Section 4.3)} 14 | \author{Sibi} 15 | \date{\today} 16 | \maketitle 17 | \newpage 18 | 19 | \section{Problem 1} 20 | \subsection{Solution (a)} 21 | 22 | Reflexive closure of $R = \{(b,b), (c,c), (a,a), (a,b), (b,c), 23 | (c,b)\}$ \\ 24 | Symmetric closure of $R = (a,a), (a,b), (b,a), (b,c), (c,b)$ \\ 25 | Transitive closure of $R = (a,a), (a,b), (a,c), (a,b), (b,c), (c,b), 26 | (b,b)$ 27 | 28 | \subsection{Solution (b)} 29 | 30 | Reflexive closure of $R = \{(x,y \in R \times R \mid x < y \lor x = 31 | y)\}$ \\ 32 | $= \{(x,y) \in R \times R \mid x \leq y\}$ \\ \\ 33 | Symmetric closure of $R = \{(x,y) \in R \times R \mid x < y \lor y < 34 | x\}$ \\ \\ 35 | Transitive closure of $R = R$ 36 | 37 | \subsection{Solution (c)} 38 | 39 | Reflexive and symmetric closure of $D_r = D_r$ 40 | Transitive closure of $D_r = \mathbb{R} \times \mathbb{R}$ 41 | 42 | \section{Problem 3} 43 | \subsection{Solution (a)} 44 | 45 | Suppose $R$ is asymmetric. Let $x$ and $y$ be an arbitrary element in 46 | $A$ such that $xRy$ and $yRx$. Since $R$ is asymmetric, from $xRy$ it 47 | follows that $(y,x) \notin R$. But this contradicts with out 48 | assumption that $yRx$. So, it is vacuously true. 49 | 50 | \subsection{Solution (b)} 51 | Suppose $R$ is a strict partial order on $A$. Let $x$ and $y$ be 52 | arbitrary element in $A$ such that $(x,y) \in R$. Let us try to prove 53 | by contradiction. Suppose $(y,x) \in R$. Then from transitive property 54 | of $R$ it follows that $(x,x) \in R$. But this contradicts the fact 55 | that $R$ is irreflexive. So $(y,x) \notin R$. 56 | 57 | \section{Problem 4} 58 | \subsection{Solution (a)} 59 | Suppose $R$ is a strict partial order on $A$. Let $S$ be the reflexive 60 | closure on $R$. That means $S = R \cup i_A$. 61 | \subsubsection{Proof of reflexive} 62 | By the definition of reflexive closure, $S$ is reflexive. 63 | 64 | \subsubsection{Proof of transitive} 65 | Let $x,y$ and $z$ be arbitrary elements in $A$ such that $(x,y) \in S$ 66 | and $(y,z) \in S$. We have to prove that $(x,z) \in S$. Now we know 67 | that $S = R \cup i_A$. Let us consider the cases separately: 68 | Case 1. $(x,y) \in R$ Now this itself has two sub cases within it. 69 | \begin{itemize} 70 | \item $y,z \in R$ We know that $R$ is transitive, so $x,z \in R$. 71 | So $(x,z) \in R \cup i_A$. 72 | \item $y,z \in i_A$ From the property of identity set it follows that 73 | $y = z$. Substituting it in $(x,y) \in R$, we get $(x,z) \in R$. So 74 | $(x,z) \in R \cup i_A$. 75 | \end{itemize} 76 | Case 2. $(x,y) \in i_A$ From the property of identity set it follows 77 | that $x = y$. Substituting that in $(y,z) \in S$ gives $(x,z) \in S$. 78 | 79 | \subsubsection{Proof of Anti-symmetric} 80 | Let $x$ and $y$ be arbitrary elements in $A$ such that $(x,y) \in S$ 81 | and $(y,x) \in S$. We have to prove that $x = y$. Now we know 82 | that $S = R \cup i_A$. Let us consider the cases separately: 83 | Case 1. $(x,y) \in R$ Now this itself has two sub cases within it. 84 | \begin{itemize} 85 | \item $y,x in R$. Since $R$ is transitive, it follows that $(x,x) \in 86 | R$. But we know that $R$ is irreflexive, so this contradicts the 87 | fact that $(x,x) \notin R$. So $(y,x) \notin R$. 88 | \item $y,x \in i_A$ From the property of identity set it follows that 89 | $y = x$. 90 | \end{itemize} 91 | Case 2. $(x,y) \in i_A$. From the property of identity set, it follows 92 | that $x = y$ 93 | 94 | \subsection{Solution (b)} 95 | Suppose $R$ is total strict order. Let $x$ and $y$ be arbitrary 96 | elements in $A$. We have to prove that $xSy \lor ySx$. We know that R 97 | satisfies trichotomy, so $xRy \lor yRx \lor x = y$. Let us consider 98 | the cases separately: 99 | \begin{itemize} 100 | \item Case 1. $xRy$ Since $S = R \cup i_A$, it follows that $(x,y) \in 101 | S$. So $xSy \lor ySx$. 102 | \item Case 2. $yRx$ Similar to the case 1 proof. 103 | \item Case 3. $x = y$ So, $(x,y) \in i_A$. It follows that $(x,y) \in 104 | R \cup i_A$. So $xSy \lor ySx$. 105 | \end{itemize} 106 | 107 | \section{Problem 5} 108 | Suppose $R$ is a relation on $A$. Let $S = R \setminus i_A$. 109 | \subsection{Solution (a)} 110 | Let 111 | $F = \{T \subset A \times A \mid T \subseteq R and T is 112 | irreflexive\}$. We have to prove things here. 113 | \subsubsection{Proof $S \in F$} 114 | Let $x,y$ be arbitrary elements in $A$ such that $(x,y) \in S$. From 115 | $S = R \setminus i_A$, it follows that $(x,y) \in R$. So, $S \subseteq 116 | R$. It also follows that $(x,y) \notin i_A$. So $\forall x \forall y 117 | (x,y) \notin R$. That can be re-stated as $\forall x (x,x) \notin R$. 118 | So $R$ is irreflexive. So, we can conclude that $S \in F$. 119 | \subsubsection{S is the largest element in F} 120 | Let $T$ be an arbitrary element in $F$. We have to prove that $T 121 | \subseteq S$. Let $(x,y)$ be arbitrary element in $T$. Since $T 122 | \subseteq R$, it follows that $(x,y) \in R$. Also since $T$ is 123 | irreflexive, it follows that $x \neq y$. So $(x,y) \notin i_A$. Hence 124 | $(x,y) \in R \setminus i_A$. So, $T \subseteq S$. 125 | 126 | \subsection{Solution (b)} 127 | Suppose $R$ is a partial order on $A$. 128 | \subsubsection{Proof that S is irreflexive} 129 | Let $x$ be an arbitrary element in $A$. Since $R$ is reflexive it 130 | follows that $(x,x) \in R$. From the identity property, it follows 131 | that $(x,x) \in i_A$. So, it can be concluded that $(x,x) \notin R 132 | \setminus i_A$. Therefore $S$ is irreflexive. 133 | \subsubsection{Proof that S is transitive} 134 | Let $x,y$ and $z$ be arbitrary elements in $A$ such that $(x,y) \in S$ 135 | and $(y,z) \in S$. We have to prove that $xSz$. We know that 136 | $S = R \setminus i_A$. Then we have $(x,y) \in R \land (x,y) \in i_A$ 137 | and $(y,z) \in R \land (y,z) \in i_A$. From 138 | $(x,y) \in R \land (y,z) \in R$, it follows that $(x,z) \in R$ since 139 | $R$ is transitive. Now there can be two cases here. If 140 | $(x,z) \notin i_A$ then $xSz$. But if $(x,z) \in i_A$, then $x = z$. 141 | From $xRy$ and $yRx$, it follows that $x = y$ since 142 | $R is antisymmetric$. But we know that $(x,y) \in S$ and so this 143 | contradicts with the fact that $(x,y) \notin i_A$. Therefore $x \neq 144 | z$. Hence $xSz$. 145 | 146 | \section{Problem 6} 147 | \subsection{Solution (a)} 148 | $S = \{(p,q) \in P \times P \mid q is the descendant of P \}$ 149 | \subsection{Solution (b)} 150 | $S^{-1} = \{(a,b) \in P \times P \mid a is the descendant of b\}$ 151 | $S \circ S^{-1} = \{(p,q) \in P \times P \mid \exists z \in P((p,z) 152 | \in S^{-1} \land (z,q) \in S\}$ \\ 153 | ${(p,q) \in P \times P \mid p is the descendant of z and q is the 154 | descendant of z}$ \\ 155 | ${(p,q) \in P \times P \mid p and q are descendant of some same man}$ 156 | 157 | \section{Problem 7} 158 | \subsection{Solution (a)} 159 | Let $S$ be the reflexive closure of $R$. 160 | $\Rightarrow$ Suppose $R$ is reflexive. From clause $1$ of reflexive 161 | closure it follows that $R \subseteq S$. From clause 3, it follows 162 | that $S \subseteq R$. So, $S = R$. 163 | $\Leftarrow$ Suppose $R = S$. By clause 2, $R$ is reflexive. 164 | 165 | \subsection{Solution (b)} 166 | Yes, it holds. 167 | 168 | \section{Problem 8} 169 | \begin{align*} 170 | a \in Dom(S) iff \exists b \in B((a,b) \in S) \\ 171 | iff \exists b \in B((a,b) \in R \lor (a,b) \in R^{-1}) \\ 172 | iff \exists b \in B((b,a) \in R^{-1} \lor (b,a) \in R) \\ 173 | iff \exists b \in B((b,a) \in R^{-1} \lor R) \\ 174 | iff \exists b \in B((b,a) \in S) \\ 175 | iff a \in Ran(S) 176 | \end{align*} 177 | 178 | \begin{align*} 179 | a \in Dom(S) iff \exists b \in B((a,b) \in S) \\ 180 | iff \exists b \in B((a,b) \in R \lor R^{-1}) \\ 181 | iff \exists b \in B((a,b) \in R \lor (a,b) \in R^{-1}) \\ 182 | iff \exists b \in B((a,b) \in R \lor (b,a) \in R) \\ 183 | iff \exists b \in B((a,b) \in R) \lor \exists b \in B((b,a) \in R) \\ 184 | iff a \in Ran(R) \cup Dom(R) 185 | \end{align*} 186 | 187 | \section{Problem 9} 188 | Let $T = \{(x,y) \in S \mid x \in Dom(R) \land y \in Ran(R)\}$ \\ 189 | 190 | \subsection{Proof of $R \subseteq T$} 191 | Let $a,b$ be arbitrary element in $A$ such that $(a,b) \in R$. Since 192 | $a \in Dom(R)$ and $b \in Ran(R)$ it follows that $(a,b) \in T$. Since 193 | $a$ and $b$ are arbitrary, it follows that $R \subseteq T$. 194 | 195 | \subsection{Proof of transitivity of T} 196 | Let $a,b$ and $c$ be arbitrary element in $A$ such that $(a,b) \in T$ 197 | and $(b,c) \in T$. So, it follows that $(a,b) \in S$ and $(b,c) \in 198 | S$. Since $S$ is transitive closure, it follows that $(a,c) \in S$. 199 | So, $(a,c) \in T$. Hence T is transitive. 200 | 201 | Proof. Since $T$ is transitive and $R \subseteq T$ it follows that $S 202 | \subseteq T$. Since $S$ is transitive closure of $R$, we know that $R 203 | \subseteq S$. So $Dom(R) \subseteq Dom(S)$. 204 | Let $a,b$ be arbitrary elements in $A$ such that $(a,b) \in S$ . Since 205 | $S \subseteq T$, it follows that $a \in Dom(R)$ and $b \in Dom(R)$. 206 | Since $a$ and $b$ were arbitrary it follows that $S \subseteq R$. So, 207 | $Dom(S) \subseteq Dom(R)$. Therefore $Dom(S) = Dom(R)$. 208 | 209 | \section{Problem 10} 210 | \subsection{Solution (a)} 211 | We know that $A \times A$ is symmetric. Also $R \subseteq A \times A$. 212 | So from the definition of $F$, it follows that $A \times A \in F$. 213 | Therefore $F \neq \emptyset$. 214 | 215 | \subsection{Solution (b)} 216 | \subsubsection{Proof of $R \subseteq \cap F$} 217 | Let $a,b$ be arbitrary elements in $A$ such that $(a,b) \in R$. Let 218 | $T$ be arbitrary elements of $F$. Then by the definition of $F$, $R 219 | \subseteq T$, so $(a,b) \in T$. Since $T$ was arbitrary this shows 220 | that $\forall T \in F((a,b) \in T)$. So $(a,b) \in \cap F$. Thus $R 221 | \subseteq \cap F$. 222 | 223 | \subsubsection{Proof of $\cap F$ is symmetric} 224 | Let $a,b$ be arbitrary element in $A$ such that $(a,b) \in S$. Let $T$ 225 | be arbitrary element in $F$. Then by the definition of $F$, $T$ is 226 | symmetric. So $(a,b) \in S = (a,b) \in \cap F$. So $(a,b) \in T$. 227 | Since $T$ is symmetric, $(b,a) \in T$. Now since $F$ is arbitrary, 228 | $(b,a) \in \cap F$. Therefore $\cap F$ is symmetric. 229 | 230 | \subsection{$\cap F$ is smallest element in F} 231 | Let $T$ be arbitrary element in $F$. We have to prove that $\cap F 232 | \subseteq T$. Let $a,b$ be arbitrary element in $A$ such that $(a,b) 233 | \in \cap F$. Since $T$ is arbitrary it follows that $\forall T \in 234 | F((a,b) \in T)$. So, $(a,b) \in T$. Q.E.D. 235 | 236 | \section{Problem 11} 237 | Suppose $R_1$ and $R_2$ are relations on $A$ and $R_1 \subseteq R_2$. 238 | \subsection{Solution (a)} 239 | Let $S_1$ and $S_2$ be reflexive closure of $R_1$ and $R_2$. Then it 240 | follows that $R_1 \subseteq S_1$ and $R_2 \subseteq S_2$. Let $a,b$ be 241 | arbitrary element in $A$ such that $(a,b) \in S_1$. There can be two 242 | cases now: 243 | 244 | Case 1. $(a,b) \in R_1$. Since $R_1 \subseteq R_2$, it follows that 245 | $(a,b) \in R_2$. From $R_2 \subseteq S_2$, it follows that $S_1 246 | \subseteq S_2$. 247 | Case 2. $(a,b) \notin R_1$ Then it follows that $(a,b)$ was built as 248 | part of reflexive closure, so $a = b$. Now $R_2$ is a relation on $A$. 249 | So, reflexive closure of $R_2$ will have $(a,b)$. So $(a,b) \in S_2$. 250 | 251 | \subsection{Transitive closure} 252 | Let $S_1$ and $S_2$ be transitive closure of $R_1$ and $R_2$. Since 253 | the transitive closure of a relation is the smallest transitive 254 | relation containing the original relation, we know $R_1 \subseteq R_2$ 255 | and $R_2 \subseteq S_2$. So, $R_1 \subseteq S_2$. $S_2$ is transitive 256 | and contains $R_1$, so it must contain the transitive closure of $R_1$ 257 | ie $S_1$. Since the transitive closure is the smallest transitive 258 | relation containing $R_1$. So $S_1 \subseteq S_2$. 259 | 260 | \subsubsection{Symmetric closure} 261 | Let $S_1$ and $S_2$ be symmetric closure of $R_1$ and $R_2$. Let $a,b$ 262 | be arbitrary element in $A$ such that $(a,b) \in S_1$. We know that 263 | $R_1 \subseteq S_1$. Let us consider the cases: 264 | Case 1. $(a,b) \in R_1$ Since $R_1 \subseteq R_2$, it follows that 265 | $(a,b) \in R_2$. From $R_2 \subseteq S_2$, it follows that $(a,b) \in 266 | S_2$. 267 | Case 2. $(a,b) \notin R_1$ This means $(b,a) \in R_1$. Since $R_1 268 | \subseteq R_2$, it follows that $(b,a) \in R_2$. Now symmetric closure 269 | of $R_2$ will have $(a,b)$. So $(a,b) \in S_2$. 270 | 271 | \section{Problem 12} 272 | \subsection{Solution (a)} 273 | $S_1 = R_1 \cup i_A$ 274 | $S_2 = R_2 \cup i_A$ 275 | $S_1 \cup S_2 = R_1 \cup i_A \cup R_2 \cup i_A$ 276 | $R_1 \cup R_2 \cup i_A$ 277 | $R \cup i_A$ 278 | $S$ 279 | 280 | \subsection{Solution (b)} 281 | $S_1 = R_1 \cup (R_1)^{-1}$ 282 | $S_2 = R_2 \cup (R_2)^{-1}$ 283 | \begin{align*} 284 | S_1 \cup S_2 = R_1 \cup R_2 \cup (R_1)^{-1} \cup (R_2)^{-1} \\ 285 | = R \cup R^{-1} \\ 286 | = S 287 | \end{align*} 288 | 289 | \section{Problem 13} 290 | Suppose $R_1$ and $R_2$ are relations on $A$. Let $R = R_1 \cap R_2$. 291 | 292 | \subsection{Solution (a)} 293 | Let $S_1, S_2$ and $S$ be the reflexive clsoure of $R_1, R_2$ and $R$. 294 | 295 | We know that $R \subseteq R_1$and $R \subseteq R_2$. From exercise 11, 296 | it follows that $S \subseteq S_1 $ and $S \subseteq S_2$. Let $(a,b)$ 297 | be arbitrary element in $S$. Then it follows that $(a,b) \in S_1 \land 298 | S_2$. So, $S \subseteq S_1 \cap S_2$. 299 | 300 | \subsection{Solution (b)} 301 | We know that $R \subseteq R_1$ and $R \subseteq R_2$. From exercise 302 | 11, it follows that $S \subseteq S_1$ and $S \subseteq S_2$. So, $S 303 | \subseteq S_1 \cap S_2$. 304 | 305 | \subsection{Solution (c)} 306 | Similar proof. 307 | 308 | \section{Solution 14} 309 | \begin{align*} 310 | R_1 = \{(1,2),(2,5),(2,9)\} \\ 311 | R_2 = \{(1,2)\} \\ 312 | S_1 = \{(1,2), (2,5), (1,5), (1,9)\} \\ 313 | S_2 = \{(1,2)\} \\ 314 | R = R_1 \setminus R_2 = \{(2,5),(2,9)\} \\ 315 | S_1 \setminus S_2 = \{(2,5), (1,5), (1,9)\} \\ 316 | S = \{(2,5), (2,9)\} \\ 317 | S_1 \setminus S_2 \nsubseteq S \land S \nsubseteq S_1 \setminus S_2 \\ 318 | \end{align*} 319 | 320 | \section{Solution 15} 321 | Suppose $R$ is a relation on $A$. Let $S = R \cup i_A \cup R^{-1}$. We 322 | know that $S$ is symmetric and reflexive. Let $F = \{ T \subseteq A 323 | \times A \mid T \text{is reflexive and symmetric and} R \subseteq 324 | T\}$. Let $T$ be an arbitrary element in $F$. We have to prove that $S 325 | \subseteq T$. Let $(a,b)$ be arbitrary element in $S$. Now let us 326 | consider the cases: 327 | 328 | Case 1. $(a,b) \in R$. Since $R \subseteq T$, it follows that 329 | $(a,b)\in T$. So $S \subseteq T$. 330 | Case 2. $(a,b) \notin R$. Then it means either $(a,b) \in i_A$ ir 331 | $(a,b) \in R^{-1}$. If $(a,b) \in i_A$ then $a = b$. Since $T$ is 332 | reflexive closure, $(a,b) \in T$. If $(a,b) \in R^{-1}$, then $(b,a) 333 | \in R$. Since $R \subseteq T$, it follows that $(b,a) \in T$. Now $T$ 334 | is symmetric closure, so $(a,b) \in T$. So, we can conclude that $S 335 | \subseteq T$. 336 | 337 | \section{Solution 16} 338 | Suppose $R$ is a relation on $A$ and $S$ be the reflexive closure of 339 | $R$. 340 | \subsection{Solution (a)} 341 | Suppose $R$ is symmetric. Let $a,b$ be arbitrary elements in $A$ such 342 | that $(a,b) \in S$. We know that $R \subseteq S$. Let us consider the 343 | cases: 344 | Case 1. $(a,b) \in S$. Since $R$ is symmetric, it follows that $(b,a) 345 | \in R \land (b,a) \in S$. 346 | Case 2. $(a,b) \notin R$ Then it means $a = b$. And $(b,a) \in S$.j 347 | 348 | \subsection{Solution (b)} 349 | Suppose $R$ is transitive. Let $a,b,c$ be arbitrary element in $A$ 350 | such that $(a,b) \in S$ and $(b,c) \in S$. We know that $R \subseteq 351 | S$. Let us consider the cases: 352 | 353 | Case 1. $(a,b) \in R \land (b,c) \in R$. Since $R$ is transitive it 354 | follows that $(a,c) \in R$. From $R \subseteq S$, it follows that 355 | $(a,c) \in S$. 356 | Case 2. $(a,b) \in R \land (b,c) \notin R$. Then it follows that $b = 357 | c$. So, $(a,c) \in R$. From $R \subseteq S$, it follows that $(a,c) 358 | \in S$. 359 | Case 3. $(a,b) \notin R \land (b,c) \in R$. Then it follows that $a = 360 | b$. So, $(a,c) \in R$. From $R \subseteq S$, it follows that $(a,c) 361 | \in S$. 362 | Case 4. $(a,b) \notin R \land (b,c) \notin R$. Then it follows that $a 363 | = b$ and $b = c$. Since $S$ is reflexive closure of $R$, it follows 364 | that $(a,c) \in S$. 365 | 366 | \section{Solution 17} 367 | 368 | Lemma. Suppose $a S b$. Then there must be an $R$ chain $a R ... R b$. 369 | 370 | Suppose $aSb$. Then there's an R-chain a R x1 R ... R xn b (n >=0 ). 371 | Since $R$ is symmetric, we can reverse every link in the chain: b xn R 372 | ... R x1 R a. Since $S$ includes $R$, this also an $S$ chain b xn S 373 | ... S x1 S a. Since $S$ is transitive $b S a$. QED. 374 | \end{document} 375 | 376 | 377 | -------------------------------------------------------------------------------- /chapter 3/section3.4.tex: -------------------------------------------------------------------------------- 1 | %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% 2 | % Author: Sibi 3 | %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% 4 | \documentclass{article} 5 | \usepackage{graphicx} 6 | \usepackage{verbatim} 7 | \usepackage{amsmath} 8 | \usepackage{amsfonts} 9 | \usepackage{amssymb} 10 | \usepackage{tabularx} 11 | \setlength\parskip{\baselineskip} 12 | \begin{document} 13 | \title{Chapter 3 (Section 3.4)} 14 | \author{Sibi} 15 | \date{\today} 16 | \maketitle 17 | \newpage 18 | 19 | \section{Problem 1} 20 | 21 | ( $\Rightarrow$ ) Suppose $\forall x ( P(x) \land Q(x))$. Let y be 22 | arbitrary. Then since $\forall x (P(x) \land Q(x))$, $P(y) \land Q(y)$ 23 | and so in particular $P(y)$. Since y is arbitrary, this shows that 24 | $\forall x P(x)$. Similarly, $\forall x Q(x)$ for arbitrary y. Thus, 25 | $\forall x P(x) \land \forall x Q(x)$. 26 | 27 | ( $\Leftarrow$ ) Suppose $\forall x P(x) \land \forall x Q(x)$. Let y 28 | be arbitrary. Then since $\forall x P(x)$, $P(y)$ and similarly since 29 | $\forall x Q(x)$, $Q(y)$. Thus $P(y) \land Q(y)$ and since y is 30 | arbitrary, it follows that $\forall x(P(x) \land Q(x))$. 31 | 32 | \section{Problem 2} 33 | 34 | Suppose $A \subseteq B \land A \subseteq C$. Let x be an arbitrary 35 | element in $A$. From $x \in A$ and $A \subseteq B$, it follows that $x 36 | \in B$. Similarly, from $x \in A$ and $A \subseteq C$, it follows that 37 | $x \in C$. Therefore, $x \in (B \cap C)$. Since x is arbitrary, we can 38 | conclude that $A \subseteq B \cap C$. 39 | 40 | \section{Problem 3} 41 | 42 | Suppose $A \subseteq B$. Let $C$ be an arbitrary set and x be an 43 | arbitrary element in $C \setminus B$. It follows that $x \in C$ and $x 44 | \notin B$. From $x \notin B$ and $A \subseteq B$, it follows that $x 45 | \notin A$. Therefore $x \in C$ and $x \notin A$. Since x is arbitrary, 46 | $C/B \subseteq C/A$. 47 | 48 | \section{Problem 4} 49 | 50 | Suppose $A \subseteq B \cap A \nsubseteq C$. Let x be an arbitrary 51 | element in A. From $x \in A$ and $A \subseteq B$, it follows that $x 52 | \in B$. Let there be some $y \in A$ and from $A \nsubseteq C$, it 53 | follows that $y \notin C$. Therefore, $x \in B$ and $y \notin C$. 54 | Since x is arbitrary, $y \in B$. Therefore, $y \nsubseteq C$. 55 | 56 | \section{Problem 5} 57 | 58 | Suppose $A \subseteq B \setminus C$ and $A \neq \emptyset$. Let x be 59 | some element in A. From $x \in A$ and $A \subseteq B \setminus C$, it 60 | follows that $x \in B$ and $x \notin C$. Therefore $x \in B \land x 61 | \notin C$, so $B \nsubseteq C$. 62 | 63 | \section{Problem 6} 64 | 65 | Let x be arbitrary, then 66 | \begin{align*} 67 | x \in A \setminus (B \cap C) \iff x \in A \land x \notin (B \cap C) \\ 68 | \iff x \in A \land \neg (x \in B \land 69 | C) \\ 70 | \iff x \in A \land \neg(x \in B \land x 71 | \in C) \\ 72 | \iff x \in A \land (x \notin B \lor x 73 | \notin C) \\ 74 | \iff (x \in A \land x \notin B) \lor (x 75 | \in A \land x \notin C) \\ 76 | \iff x \in (A \setminus B) \lor x \in 77 | (A \setminus C) \\ 78 | \iff x \in (A \setminus B) \cup (A 79 | \setminus C) 80 | \end{align*} 81 | 82 | Thus, $\forall x (x \in A \setminus (B \cap C)) \iff x \in (A 83 | \setminus B) \cup (A \setminus C)$, so $A \setminus (B \cap C) = (A 84 | \setminus B) \cup (A \setminus C)$. 85 | 86 | \section{Problem 7} 87 | 88 | ( $\Rightarrow$ ) Let A and B be arbitrary set. Suppose $x \in P(A 89 | \cap B)$. Then by definition, $x \subseteq A \cap B$. Let y be an 90 | arbitrary element such that $y \in x$. From $y \in x$ and $x \subseteq 91 | A \cap B$, it follows that $y \in A \cap B$. Similarly, $x \in P(B)$. 92 | Therefore $x \in (P(A) \cap P(B))$. 93 | 94 | ( $\Leftarrow$ ) Suppose $x \in P(A) \cap P(B)$. Then $x \in P(A)$ and 95 | $x \in P(B)$, so $x \subseteq A$ and $x \subseteq B$. Let y be an 96 | arbitrary element in x, then $y \in A$ and $y \in B$. Therefore, $y 97 | \in A \cap B$. Since $y \in x$ and $y \in A \cap B$, it follows that 98 | $x \subseteq A \cap B$. Thus, $x \in P(A \cap B)$. 99 | 100 | \section{Problem 8} 101 | 102 | ( $\Rightarrow$ ) Suppose $A \subseteq B$. Let x be an arbitrary 103 | element in $A$. It follows that $x \in A$ and $x \in B$, therefore $x 104 | \subseteq P(A)$ and $x \subseteq P(B)$. Let y be an arbitrary element 105 | in x. From, $y \in x$ and $x \subseteq P(A)$, it follows that $y \in 106 | P(A)$. Similarly $y \in P(B)$. Since y is arbitrary, $P(A) \subseteq 107 | P(B)$. 108 | 109 | ( $\Leftarrow$ ) Let x be an arbitrary element in $P(A)$. Suppose 110 | $P(A) \subseteq P(B)$. It follows that $x \in P(A)$ and $x \in P(B)$, 111 | therefore $x \subseteq A$ and $x \subseteq B$. From here, we can 112 | conclude that $A \subseteq B$. 113 | 114 | \section{Problem 9} 115 | 116 | Suppose x and y are odd integers. Let k be some number such that $x = 117 | 2k + 1$. Similarly, let j be some number such that $y = 2j + 1$. 118 | Multiplying x and y, we get $2(2kj + k + j) + 1$. Since $2kj + k + j$ 119 | is an integer, we can conclude that $xy$ is an odd number. 120 | 121 | \section{Problem 10} 122 | 123 | ( $\Rightarrow$ ) Let n be an arbitrary number in $ \mathbb{Z}$. We will prove 124 | the contrapositive. Suppose n is a odd number. Then there exists some 125 | k such that $n = 2k + 1$. Multiplying n with $n^2$ we get $2(2k^3 + 126 | 6k^2 + 3k) + 1$. Since $2k^3 + 6k^2 + 3k + 1$ is an integer, it 127 | follows that $n^3$ is an odd number. 128 | 129 | ( $\Leftarrow$ ) Let n be an arbitrary number in $ \mathbb{Z} $. Suppose n is 130 | even. Then there is an some number k such that $n = 2k$. Multiplying n 131 | by $n^2$, we get $n(n^2) = 2k(2k)^2 = 2k(4k^2) = 2(4k^3)$. Since 132 | $4k^3$ is an integer, it follows that $n^3$ is an even number. 133 | 134 | \section{Problem 11} 135 | 136 | Solution(a) 137 | 138 | You cannot introduce the same integer $k$ in both cases. 139 | 140 | Solution(b) 141 | 142 | \begin{align*} 143 | n = 0 \land m = 1 \\ 144 | n^2 - m^2 = 0 - 1 = -1 \\ 145 | n + m = 0 + 1 \\ 146 | \end{align*} 147 | 148 | \section{Problem 12} 149 | 150 | ( $\Rightarrow$ ) Let x be an arbitrary element in R. Let there be 151 | some $y$ in $\mathbb{R}$. We will prove the contrapositive. Suppose $x 152 | = 1$, then $x + y = 1 + y$ and $xy = y$. So, $x + y \neq xy$. 153 | Therefore, if $x + y = xy$, then $x \neq 1$. 154 | 155 | ( $\Leftarrow$ ) Let x be an arbitrary element in $R$. Suppose $x \neq 156 | 1$. Let $y = x / x - 1$. Summing it with $x$, $x+y = xy$. Therefore if 157 | $ x \neq 1$, then $x + y = xy$. 158 | 159 | \section{Problem 13} 160 | 161 | ( $\Rightarrow$ ) Let $z = 1$. Let x be an arbitrary element in 162 | $\mathbb{R^+}$. Let $y$ be an element in $\mathbb{R}$. Suppose $y - x 163 | = y / x$. We will prove the contradiction. Let $x = z$, it follows 164 | that $x = 1$. Putting it in $y - x$, we get $y -x = y - 1$. Similarly, 165 | putting it in $y - x$, we get $y - x = y - 1$. Similarly, putting it 166 | in $y / x$, we get $y / x = y$. Therefore $y - x \neq y/x$. But it 167 | contradicts the fact that $y - x = y / x$, so $x \neq z$. Therefore, 168 | if $y -x = y/x$, then $x \neq z$. 169 | 170 | ( $\Leftarrow$ ) Let $z = 1$. Let $x$ be an arbitrary element in 171 | $\mathbb{R^+}$. Let $y = x^2 / x - 1$. Suppose $x \neq z$. Solving $ y 172 | - x = \frac{x^2}{x - 1} = \frac{x}{x - 1} * \frac{x}{x} = 173 | \frac{y}{x}$. So, $y - x = \frac{y}{a}$. Therefore if $x \neq z$, then 174 | $y - x = \frac{y}{x}$. 175 | 176 | \section{Problem 14} 177 | 178 | Let x be an arbitrary element in $\cup \{ A \setminus B | A \in F \}$. 179 | It follows that there exists some $A$ in $F$ such that $x \in A 180 | \setminus B$. So, $x \in A$ and $x \notin B$. From $x \in A$ and $x 181 | \notin B$, it follows that $A \nsubseteq B$. Since $A \nsubseteq B$, 182 | it follows that $A \notin P(B)$. So $A \in F$ and $A \notin P(B)$, $A 183 | \in F \setminus P(B)$. Since $x \in A$, it follows that $x \in \cup(F 184 | \setminus P(B))$. Since x is arbitrary, we can conclude that $\cup \{ 185 | A \setminus B | A \in F \} \subseteq \cup (F \setminus P(B))$. 186 | 187 | \section{Problem 15} 188 | 189 | Let $A$ be an arbitrary element in $F$ and $B$ be some element in $G$. 190 | Suppose $\forall A \in F \exists B \in G (A \cap B = \emptyset)$. Let 191 | $x$ be an arbitrary element in $A$. From $x \in A$ and $A \in F$, it 192 | follows that $x \in \cup F$. Let us assume that $x \in \cap G$. Then 193 | for all element in $G$, x is present in it. But this contradicts the 194 | fact that there is some element $B in G$ such that $x \notin B$. 195 | Therefore $x \notin \cap G$, So $\cup F$ and $\cap G$ are disjoint 196 | sets. 197 | 198 | \section{Problem 16} 199 | 200 | ( $\Leftarrow$ ) Let $x$ be an arbitrary element in $A$. Since $A \in 201 | P(A)$, then by definition of $\cup P(A)$, $x \in \cup P(A)$. 202 | 203 | ( $\Rightarrow$ ) Let $x$ be an arbitrary element in $\cup P(A)$. Then 204 | by definition, there exists some element $B$ in $P(A)$ such that $x 205 | \in B$. Since $B \in P(A)$, it follows that $B \subseteq A$. From $x 206 | \in B$ and $B \subseteq A$, it follows that $x \in A$. Since x is 207 | arbitrary, $\cup P(A) \subseteq A$. 208 | 209 | \section{Problem 17} 210 | 211 | Solution (a) 212 | 213 | ( $\Leftarrow$ ) Let $x$ be an arbitrary element in $\cup(F \cap G)$. 214 | Then there exists some element $A$ in $F \cap G$ such that $x \in A$. 215 | Since $A \in F$ and $A \in G$, from $x \in A$, we can conclude that $x 216 | \in \cup F$ and $x \in \cup G$. Therefore, $x \in (\cup F) \cap (\cup 217 | G)$. Since $x$ is arbitrary, we can conclude that $(F \cap G) 218 | \subseteq (\cup F) \cap (\cup G)$. 219 | 220 | Solution (b) 221 | 222 | Both $A$ cannot be same. 223 | 224 | Solution (c) 225 | 226 | \begin{align*} 227 | F = \{\{1\}\} \\ 228 | G = \{\{1\}, \{2\}\} \\ 229 | \cup (F \cap G) = \emptyset \\ 230 | \cup (F) \cap \cup (G) = \{1\} \\ 231 | \end{align*} 232 | 233 | \section{Problem 18} 234 | 235 | ( $\Rightarrow$ ) Suppose $(\cup F) \cap (\cup G) \subseteq \cup (F 236 | \cap G)$. Let $x$ be an arbitrary element in $\forall A \in F \forall 237 | B \in G (A \cap B)$, so $x \in A$ and $x \in B$. Suppose $A \in F$ and 238 | $x \in A$, it follows that $x \in \cup F$. Similarly, $x \in \cup G$, 239 | therefore $x \in (\cup F) \cap (\cup G)$. From, $x \in (\cup F) \cap 240 | (\cup G)$ and $(\cup F) \cap (\cup G) \subseteq \cup (F \cap G)$ it 241 | follows that $x \in \cup (F \cap G).$ Since $x$ is arbitrary $\forall 242 | A \in F \forall B \in G (A \cap B) \subseteq \cup (F \cap G)$. 243 | 244 | ( $\Leftarrow$ ) Suppose $\forall A \in F \forall B \in G (A \cap B) 245 | \subseteq \cup (F \cap G)$. Let $x$ be an arbitrary element in $(\cup 246 | F) \cap (\cup G)$. Therefore $x \in \cup F$, it follows that there 247 | exists a set $f \in F$ such that $x \in f$. Similarly there exists a 248 | set $g \in G$ such that $x \in g$. Since $\forall A \in F \forall B \in G (A \cap B) 249 | \subseteq \cup (F \cap G)$, so $(f \cap g) \subseteq \cup (F \cap G)$, 250 | it follows $x \in (F \cap G)$. Since $x$ is arbitrary, we can conclude 251 | that $(\cup F) \cap (\cup G) \subseteq \cup(F \cap G)$. 252 | 253 | \section{Problem 19} 254 | 255 | ( $\Rightarrow$ ) Suppose $(\cup F) \cap (\cup G) = \emptyset$. Let 256 | $A$ be an arbitrary element in $F$ and $B$ be an arbitrary element in 257 | $G$. Let $x$ be an arbitrary element in $A$. From $x \in A$ and $A \in 258 | F$, it follows that $x \in \cup F$. Also, from $x \in \cup F$ and 259 | $(\cup F) \cap (\cup G)$, it follows that $x \notin \cup G$, so $x 260 | \notin B$. Since $x$ is arbitrary and $x \in A$ and $x \notin B$ it 261 | follows that $A \cap B = \emptyset$. Therefore if $(\cup F) \cap (\cup 262 | G) = \emptyset$, then $\forall A \in F \forall B \in G (A \cap B = 263 | \emptyset)$. 264 | 265 | ( $\Leftarrow$ ) Suppose $\forall A \in F \forall B \in G (A \cap B = 266 | \emptyset)$. Let $x$ be an arbitrary element in $\cup F$. Then there 267 | is some element $f$ in $F$ such that $x \in F$. From $\forall A \in F 268 | \forall B \in G (A \cap B = \emptyset)$, it follows $f \cap B = 269 | \emptyset$. Since $x \in f$, it follows that $x \notin B$. Since $x 270 | \notin B$, it follows that $x \notin \cup G$. Since $x$ is arbitrary 271 | and $x \in \cup F$ and $x \notin \cup G$, it follows that $(\cup F) 272 | \cap (\cup G) = \emptyset$. Therefore, if $\forall A \in F \forall B 273 | \in G (A \cap B = \emptyset)$ then $(\cup F) \cap (\cup G)$. 274 | 275 | \section{Problem 20} 276 | 277 | Solution (a) 278 | 279 | Let $x$ be an arbitrary element in $(\cup F) \setminus (\cup G)$, so 280 | $x \in \cup F$ and $x \notin \cup G$. From $x \in \cup F$, we can 281 | conclude that there is some $A$ in $F$ such that $x \in A$. From $x 282 | \notin \cup G$, we can conclude that for all elements $B$ in $G$, $x 283 | \notin B$. $A \notin G$ because if it was in $G$, it will lead to 284 | contradiction. Therefore, $A \in F \setminus G$. From $x \in F$ and $A 285 | \in F \setminus G$ it follows that $x \in \cup (F \setminus G)$. Since 286 | $x$ is arbitrary it follows that $(\cup F) \setminus (\cup G) 287 | \subseteq \cup (F \setminus G)$. 288 | 289 | Solution (b) 290 | 291 | Problem is with this line: Since $x \in A$ and $A \notin G$, $x \notin 292 | \cup G$. 293 | 294 | Example: $x = 1 \land A = \{1\} \land G = \{\{1,2\}\} \land x 295 | \in \cup G$. 296 | 297 | Solution (c) 298 | 299 | ( $\Rightarrow$ ) Suppose $\cup(F \setminus G) \subseteq (\cup F) 300 | \setminus (\cup G)$. Let $A$ be an arbitrary element in $F \setminus 301 | G$ and $B$ be an arbitrary element in $G$. Let $x$ be an arbitrary 302 | element in $A$. From $A \in F \setminus G$ and $x \in A$, it follows 303 | that $x \in \cup (F \setminus G)$. Since $x \in \cup(F \setminus G)$ 304 | and from $\cup(F \setminus G) \subseteq (\cup F) \setminus (\cup G)$, 305 | it follows that $x \in (\cup F) \setminus (\cup G)$. So $x \in \cup F$ 306 | and $x \notin \cup G$. Since $x \notin \cup G$ and $B \in G$, it 307 | follows that $x \notin B$. So, $x \in A$ and $x \notin B$. Since $x$ 308 | is arbitrary $A \cap B = \emptyset$. 309 | 310 | ( $\Leftarrow$ ) Suppose $\forall A \in (F \setminus G) \forall B \in 311 | G(A \cap B = \emptyset)$. Let $x$ be an arbitrary element in $\cup(F 312 | \setminus G)$. Then there is some element $Y$ in $F \setminus G$ such 313 | that $x \in Y$. From $Y \in F \setminus G$, it follows that $Y \in F$ 314 | and $Y \notin G$. Since $x \in Y$ and $Y \in F$ it follows that $x \in 315 | \cup F$. We will prove the contradiction. Let us assume $x \in \cup 316 | G$. Since $x \in \cup G$, it follows that there is some element $g$ in 317 | $G$ such that $x \in g$. But this contradicts the fact that for all 318 | element $B$ in $G$ if $x$ is in $A$ and $x \notin B$, so $x \notin \cup 319 | G$. Since $x$ is arbitrary, we can conclude $\cup (F \setminus G) 320 | \subseteq (\cup F) \setminus (\cup G)$. 321 | 322 | Solution (d) 323 | \begin{align*} 324 | F = \{\{1,2\}\} \\ 325 | G = \{\{2\}\} 326 | \end{align*} 327 | 328 | \section{Problem 21} 329 | 330 | Suppose $\cup F \nsubseteq \cup G$. Let there be some element $A \in 331 | F$ and let $B$ be an arbitrary element in $G$. Let $x$ be an arbitrary 332 | element in $A$. From $x \in A$ and $A \in F$, it follows that $x \in 333 | \cup F$. From $x \in \cup F$ and $\cup F \nsubseteq \cup G$, it 334 | follows that $x \notin \cup G$. This means that for all element in 335 | $G$, $x$ is not present in element. Therefore $x \notin B$. Since x is 336 | an arbitrary element, $A \nsubseteq B$. So, if $\cup F \nsubseteq \cup 337 | G$, then $\exists A \in F \forall B \in G (A \nsubseteq B)$. 338 | 339 | \section{Problem 22} 340 | 341 | Solution(a) 342 | 343 | Proof strategy using Binconditional and Conjunction. 344 | 345 | Solution(b) 346 | 347 | ($\Rightarrow$) Let $x$ be an arbitrary element in $B \setminus 348 | (\cup_{i \in I} A_i)$, so $x \in B$ and $x \notin \cup_{i \in I}A_i$. 349 | From $x \notin \cup_{i \in I}A_i$, it follows that for all element $i$ 350 | in $I$, $x \notin A_i$. So. $x \in B$ and $x \notin A_i$, therefore $x 351 | \in B \setminus A_i$. So, $ x \in \cap_{i \in I}(B \setminus A_i)$. 352 | 353 | ($\Leftarrow$) Let $x$ be an arbitrary element in $\cap_{i \in I}(B 354 | \setminus A_i)$. This means for all element $i \in I$, $x \in B 355 | \setminus A_i$. So $x \in B$ and $x \notin A_i$. Since i is arbitrary 356 | and $x \notin A_i$, it follows that $\forall i \in I(x \notin A_i)$. 357 | So, $x \notin \cup_{i \in I}A_i$. Therefore, $x \in B \setminus 358 | (\cup_{i \in I}A_i)$. 359 | 360 | Solution(c) 361 | 362 | $\cup_{i \in I}(B \setminus A_i)$ 363 | 364 | \section{Problem 23} 365 | 366 | Solution(a) 367 | 368 | Let $x$ be an arbitrary element in $\cup_{i \in I}(A_i \setminus 369 | B_i)$. Then there exists an element $i$ in $I$ such that $x \in A_i 370 | \setminus B_i$. So $x \in A_i$ and $x \notin B_i$. From $x \in A_i$, 371 | it follows $x \in \cup_{i \in I}A_i$. From $x \notin B_i$, it follows 372 | that $x \notin \cap_{i \in I}B_i$. Therefore, $x \in (\cup_{i \in 373 | I}A_i) \setminus (\cap_{i \in I}B_i)$. Since $x$ is arbitrary, 374 | $\cup_{i \in I}(A_i \setminus B_i) \subseteq (\cup_{i \in I}A_i) 375 | \setminus (\cap_{i \in I}B_i)$. 376 | 377 | Solution (b) 378 | 379 | \begin{align*} 380 | I = \{1,2\} \\ 381 | A_i = \{\{0,1\},\{2\}\} \\ 382 | B_i = \{\{2\}\} 383 | \end{align*} 384 | 385 | \section{Problem 24} 386 | 387 | Solution(a) 388 | 389 | Let $x$ be an arbitrary element in $\cup_{i \in I}(A_i \cap B_i)$. Then 390 | there exists some element $i \in I$ such that $x \in A_i \cap B_i$, so 391 | $x \in A_i$ and $x \in B_i$. From $x \in A_i$, it follows that $x \in 392 | \cup_{i \in I}A_i$. Similarly from $x \in B_i$, it follows that $x \in 393 | \cup_{i \in I}B_i$. So, $x \in (\cup_{i \in I}A_i \cap \cup_{i \in 394 | I}B_i)$. Since x is arbitrary, we can conclude that $\cup_{i \in I} 395 | (A_i \cap B_i) \subseteq (\cup_{i \in I} A_i) \cap (\cup_{i \in 396 | I}B_i)$. 397 | 398 | Solution(b) 399 | 400 | \begin{align*} 401 | i = \{1,2\} \\ 402 | A_i = \{\{1\},\{2\}\} \\ 403 | B_i = \{\{1,2\}, \{3\}\} 404 | \end{align*} 405 | 406 | \section{Problem 25} 407 | 408 | Let $a$ and $b$ be any arbitrary element in $\mathbb{Z}$. Let $c$ be 409 | some element in $\mathbb{Z}$. Since $a$ and $c$ are integer, then 410 | there exists some integer $k$ such that $ak = c$, so $a | c$. 411 | Similarly we can prove that $ b | c$. 412 | 413 | \section{Problem 26} 414 | 415 | Solution(a) 416 | 417 | ($\Rightarrow$) Let $n$ be an arbitrary element in $\mathbb{Z}$. 418 | Suppose $15 | n$. Then there exists an integer $k$ such that $15k = 419 | n$. From $15k = n$, it follows that $3(5k) = n$, so $3 | n$. 420 | Similarly, from $15k = n$, it follows that $5(3k) = n$, so $5|n$. 421 | Therefore, if $15 | n$ then $3|n \land 5|n$. 422 | 423 | ($\Leftarrow$)(incorrect. why?) Suppose $3|n$ and $5|n$. We will prove the 424 | contradiction. Suppose $15 is not divisible by n$, then there exists 425 | some $k$ such that $15 k \neq n$. From $15k \neq n$, it follows $3(5k) 426 | \neq n$. But this contradicts the fact that $3|n$. Therefore, $15 | 427 | n$. 428 | 429 | ($\Leftarrow$)Suppose $3|n$ and $5|n$. Then we can choose 430 | integers $p$ and $q$ s.t. $3p = n$ and $5q = n$. 431 | now $n = 6n-5n = 2(3(5q)) - 5(3p) = 15(2q-p)$. 432 | It follows since $2q-p$ is an integer $15|n$. 433 | 434 | Solution(b) 435 | 436 | $n = 30$ 437 | 438 | \end{document} 439 | 440 | 441 | 442 | --------------------------------------------------------------------------------