├── .gitignore ├── timetable.pdf ├── lecture-7-proofs ├── assignment-7.pdf └── solution-7.tex ├── lecture-3-implication ├── assignment-3.pdf └── solution-3.tex ├── lecture-4-equivalence ├── assignment-4.pdf └── solution-4.tex ├── lecture-5-quantifiers ├── assignment-5.pdf └── solution-5.tex ├── lecture-2-logical-combinators ├── assignment-2.pdf └── solution-2.tex ├── lecture-8-proofs-with-quantifiers ├── assignment-8.pdf └── solution-8.tex ├── lecture-0 └── background-reading-what-is-mathematics.pdf ├── lecture-1-introductionary-material ├── assignment-1.pdf └── solution-1.tex └── lecture-6-working-with-quantifiers ├── assignment-6.pdf └── solution-6.tex /.gitignore: -------------------------------------------------------------------------------- 1 | *.log 2 | solution*.pdf 3 | *.aux 4 | -------------------------------------------------------------------------------- /timetable.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/kyrylo/coursera-mathematical-thinking/HEAD/timetable.pdf -------------------------------------------------------------------------------- /lecture-7-proofs/assignment-7.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/kyrylo/coursera-mathematical-thinking/HEAD/lecture-7-proofs/assignment-7.pdf -------------------------------------------------------------------------------- /lecture-3-implication/assignment-3.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/kyrylo/coursera-mathematical-thinking/HEAD/lecture-3-implication/assignment-3.pdf -------------------------------------------------------------------------------- /lecture-4-equivalence/assignment-4.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/kyrylo/coursera-mathematical-thinking/HEAD/lecture-4-equivalence/assignment-4.pdf -------------------------------------------------------------------------------- /lecture-5-quantifiers/assignment-5.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/kyrylo/coursera-mathematical-thinking/HEAD/lecture-5-quantifiers/assignment-5.pdf -------------------------------------------------------------------------------- /lecture-2-logical-combinators/assignment-2.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/kyrylo/coursera-mathematical-thinking/HEAD/lecture-2-logical-combinators/assignment-2.pdf -------------------------------------------------------------------------------- /lecture-8-proofs-with-quantifiers/assignment-8.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/kyrylo/coursera-mathematical-thinking/HEAD/lecture-8-proofs-with-quantifiers/assignment-8.pdf -------------------------------------------------------------------------------- /lecture-0/background-reading-what-is-mathematics.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/kyrylo/coursera-mathematical-thinking/HEAD/lecture-0/background-reading-what-is-mathematics.pdf -------------------------------------------------------------------------------- /lecture-1-introductionary-material/assignment-1.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/kyrylo/coursera-mathematical-thinking/HEAD/lecture-1-introductionary-material/assignment-1.pdf -------------------------------------------------------------------------------- /lecture-6-working-with-quantifiers/assignment-6.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/kyrylo/coursera-mathematical-thinking/HEAD/lecture-6-working-with-quantifiers/assignment-6.pdf -------------------------------------------------------------------------------- /lecture-3-implication/solution-3.tex: -------------------------------------------------------------------------------- 1 | \documentclass{article} 2 | \title{Solution 3} 3 | \author{@kyrylo} 4 | 5 | \usepackage{enumerate} 6 | 7 | \begin{document} 8 | 9 | \section*{Solutions to assignment 3} 10 | 11 | \section{} 12 | 13 | \begin{enumerate}[(a)] 14 | \item $T \Rightarrow (D \wedge Y)$ 15 | \item $D \Rightarrow \neg Y$ 16 | \item $\neg D \Rightarrow \neg T$ 17 | \item $T \Rightarrow \neg (D \wedge Y)$ 18 | \item $\neg D \wedge Y \wedge T$ 19 | \item $T \Rightarrow (Y \Rightarrow \neg D)$ 20 | \item $T \Rightarrow (D \Leftrightarrow Y)$ 21 | \item $T \Rightarrow (D \wedge \neg Y) \vee (\neg D \wedge Y)$ 22 | \end{enumerate} 23 | 24 | \section{} 25 | 26 | \begin{tabular}{ | c | c | c | c | c | } 27 | \hline 28 | $\phi$ & $\neg \phi$ & $\psi$ & $\phi \Rightarrow \psi$ & $\neg \phi \vee \psi$ \\ 29 | \hline 30 | T & F & T & T & T \\ 31 | T & F & F & F & F \\ 32 | F & T & T & T & T \\ 33 | F & T & F & T & T \\ 34 | \hline 35 | \end{tabular} 36 | 37 | \section{} 38 | 39 | The conclusion is that $\phi \Rightarrow \psi$ means the same as $\neg \phi \vee \psi$. 40 | 41 | \section{} 42 | 43 | \begin{tabular}{ | c | c | c | c | c | c | } 44 | \hline 45 | $\phi$ & $\psi$ & $\neg \psi$ & $\phi \Rightarrow \psi$ & $\phi \not \Rightarrow \psi$ & $\phi \wedge \neg \psi$ \\ 46 | \hline 47 | T & T & F & T & F & F \\ 48 | T & F & T & F & T & T \\ 49 | F & T & F & T & F & F \\ 50 | F & T & F & T & F & F \\ 51 | \hline 52 | \end{tabular} 53 | 54 | \section{} 55 | 56 | The conclusion is that $\phi \not \Rightarrow \psi$ means the same as $\phi \wedge \neg \psi$. 57 | 58 | \end{document} 59 | -------------------------------------------------------------------------------- /lecture-5-quantifiers/solution-5.tex: -------------------------------------------------------------------------------- 1 | \documentclass{article} 2 | \title{Solution 5} 3 | \author{@kyrylo} 4 | 5 | \usepackage{enumerate} 6 | \usepackage{amsmath} 7 | \usepackage{centernot} 8 | \usepackage{amsfonts} 9 | 10 | \begin{document} 11 | 12 | \section*{Solutions to assignment 5} 13 | 14 | \section{} 15 | 16 | \begin{enumerate}[(a)] 17 | \item $(\exists x \in \mathbb{N})[x^3 = 27]$ 18 | \item $(\exists x \in \mathbb{N})[x > 1000000]$ 19 | \item $(\exists p \in \mathbb{N})(\exists q \in \mathbb{N}[p > 1 \wedge q > 1 \wedge n = pq)$ 20 | \end{enumerate} 21 | 22 | \section{} 23 | 24 | \begin{enumerate}[(a)] 25 | \item $(\forall x \in \mathbb{N})\neg[x^3 = 28]$ 26 | \item $(\forall n \in \mathbb{N})[0 < n]$ 27 | \item $(\forall p \in \mathbb{N})(\forall q \in \mathbb{N})[(n = pq \implies (p = 1 \vee q = 1)]$ 28 | \end{enumerate} 29 | 30 | \section{} 31 | 32 | \begin{enumerate}[(a)] 33 | \item $\forall e\exists s[love(e, s)]$ 34 | \item $\forall e[tall(e) \vee short(s)]$ 35 | \item $(\forall e)[tall(e)] \vee (\forall e)[short(e)]$ 36 | \item $\neg(\forall e)[athome(e)]$ 37 | \item $Comes(John) \implies (\forall x)[Woman(x) \implies Leaves(x)]$ 38 | \item $(\exists x)[Man(x) \wedge Comes(x)] \implies (\forall x)[Woman(x) \implies Leaves(x)]$ 39 | \end{enumerate} 40 | 41 | \section{} 42 | 43 | \begin{enumerate}[(a)] 44 | \item $(\forall a \in \mathbb{R})(\exists x \in \mathbb{R})[x^2 + a = 0]$ 45 | \item $(\forall a \in \mathbb{R})[(a < 0) \implies (\exists x \in \mathbb(R)(x^2 + a = 0)]$ 46 | \item $(\forall x \in \mathbb{R})(\exists m \in \mathbb{N})(\exists n \in \mathbb{N})[m = nx \vee m = -nx \vee x = 0]$ 47 | \item $(\exists x \in \mathbb{R})(\forall m \in \mathbb{N})(\forall n \in \mathbb{N})[m \neq nx \wedge m \neq -nx]$ 48 | \item $(\forall y \in \mathbb{R})(\exists x \in \mathbb{R})[(x > y) \wedge (\forall m \in \mathbb{N})(\forall n \in \mathbb{N})(m \neq nx)]$ 49 | \end{enumerate} 50 | 51 | \section{} 52 | 53 | \begin{enumerate}[(a)] 54 | \item $(\forall x \in C)[D(x) \implies M(x)]$ 55 | \item $(\forall x \in C)[\neg D(x) \implies M(x)]$ 56 | \item $(\forall x \in C)[M(x) \implies D(x)]$ 57 | \item $(\exists x \in C)[D(x) \wedge \neg M(x)]$ 58 | \item $(\exists x \in C)[\neg D(x) \wedge M(x)]$ 59 | \end{enumerate} 60 | 61 | \section{} 62 | 63 | $\forall x \forall y[(x < y) \implies \exists z(Q(z) \wedge (x < z < y))]$ 64 | 65 | \section{} 66 | 67 | $(\exists t)(\forall p)[Fool(p, t)] \wedge (\exists p)(\forall t)[Fool(p, t)] \wedge (\forall p)(\forall t)\neg[Fool(p, t)]$ 68 | 69 | \section{} 70 | 71 | $\exists x \forall t A(x, t)$ 72 | 73 | 74 | \section{} 75 | 76 | $\forall t \exists t A(x, t)$ 77 | 78 | \end{document} 79 | -------------------------------------------------------------------------------- /lecture-7-proofs/solution-7.tex: -------------------------------------------------------------------------------- 1 | \documentclass{article} 2 | \title{Solution 7} 3 | \author{@kyrylo} 4 | 5 | \usepackage{enumerate} 6 | \usepackage{amsmath} 7 | \usepackage{centernot} 8 | \usepackage{amsfonts} 9 | 10 | \begin{document} 11 | 12 | \section*{Solutions to assignment 7} 13 | 14 | \section{} 15 | 16 | Statement: ``All birds can fly''. 17 | \\ 18 | Converting it to the language of mathematics, it means the following: 19 | \\\\ 20 | $(\forall x \in Birds)[CanFly(x)]$ 21 | \\\\ 22 | Disproof: ostrich is a bird and it can't fly. 23 | \\ 24 | QED. 25 | 26 | \section{} 27 | 28 | Statement: $(\forall x,y \in \mathbb{R})[(x - y)^2 > 0]$ 29 | \\\\ 30 | Disproof: if $x = p$ and $y = p$, then $x - y = p - p = 0$. Hence, $0^2 = 0$. 31 | The statement is false. 32 | \\ 33 | QED. 34 | 35 | \section{} 36 | 37 | Statement: ``Between any two unequal rationals there is a third rational.'' 38 | \\\\ 39 | Proof: let $x, y$ be unequal rational numbers. Then $z = (x + y) / 2$ is a 40 | rational number, such that $z > x \wedge z < y$. 41 | \\ 42 | QED. 43 | 44 | \section{} 45 | 46 | If $\phi \implies \psi$ and vice versa, it means that $\phi$ and $\psi$ are 47 | equal. $\phi \Leftrightarrow \psi$ is just a shorthand for $(\phi \implies \psi) \wedge 48 | (\psi \implies \phi)$. 49 | 50 | \section{} 51 | 52 | Because $(\neg\phi) \implies (\neg\psi)$ is equal to $\psi \implies \phi$. 53 | 54 | \section{} 55 | 56 | Statement: ``If five investors split a payout of \$2M, at least one investor 57 | receives at least \$400,000.'' 58 | \\\\ 59 | Proof: in order to split the payout evenly it is needed to divide it by the 60 | number of investors. Thus, $x = 2,000,000 / 5$. That is, $x$ is equal to 61 | 400,000, which proves the statement. 62 | \\ 63 | QED. 64 | 65 | \section{} 66 | 67 | Theorem: ``$\sqrt{3}$ is irrational.'' 68 | \\\\ 69 | Proof: Assume, on the contrary, that $\sqrt{3}$ were rational. Then there are 70 | natural numbers $p, q$ with no common factors, such that $\sqrt{3} = p / q$. 71 | \\ 72 | \\ 73 | Squaring the equation gives $3 = p^2 / q^2$. Rearranging: $3q^2 = p^2$. If $q$ 74 | is odd, then $p^2$ is odd, otherwise $p^2$ is even. If it is even, then it is 75 | rational. If it is odd, then... 76 | 77 | \section{} 78 | 79 | \begin{enumerate}[(a)] 80 | \item $(\neg D) \implies Y$\\ 81 | $(\neg D) \wedge (\neg Y)$ 82 | \item $(\exists x \in \mathbb{R})(\exists y \in \mathbb{R})[x < y \implies -y < x]$\\ 83 | $(\forall x \in \mathbb{R})(\forall y \in \mathbb{R})[x < y \wedge -y \geq x]$ 84 | \item $(\exists x)(\exists y)[Congruent(x) \wedge Congruent(y) \implies SameArea(x, y)]$\\ 85 | $(\forall x)(\forall y)[Congruent(x) \wedge Congruent(y) \wedge \neg SameArea(x, y)]$ 86 | \item $(\exists a \in \mathbb{R})(\exists b \in \mathbb{R})(\exists c \in \mathbb{R})(\exists x \in \mathbb{R})[a \neq 0 \wedge (b^2 \geq 4ac) \implies ax^2 + bx + c = 0]$\\ 87 | $(\forall a \in \mathbb{R})(\forall b \in \mathbb{R})(\forall c \in \mathbb{R})(\forall x \in \mathbb{R})[a \neq 0 \wedge (b^2 \geq 4ac) \wedge ax^2 + bx + c \neq 0]$ 88 | \item $?$ 89 | \item $?$ 90 | \item $(\exists n \in \mathbb{N})[\neg D3(n) \implies ?]$ 91 | \end{enumerate} 92 | 93 | \end{document} 94 | -------------------------------------------------------------------------------- /lecture-1-introductionary-material/solution-1.tex: -------------------------------------------------------------------------------- 1 | \documentclass{article} 2 | \title{Solution 1} 3 | \author{@kyrylo} 4 | 5 | \begin{document} 6 | 7 | \section*{Solutions to assignment 1} 8 | 9 | \section{} 10 | 11 | Elimination of the ambiguity of the {\it The man saw the woman with a telescope} 12 | sentence can be avoided with these two natural sounding equivalents: 13 | 14 | \begin{enumerate} 15 | \item The man with a telescope saw the woman. 16 | \item The man saw the woman carrying a telescope. 17 | \end{enumerate} 18 | 19 | \section{} 20 | 21 | This is how you can remove the ambiguity, retainging the brevity of a typical 22 | headline: 23 | 24 | \begin{enumerate} 25 | \item Sisters reunited in checkout line at Safeway after ten years 26 | \item Large hole appears in High Street. City authorities are investinating it 27 | \item Mayor says bus passengers should wear belts 28 | \end{enumerate} 29 | 30 | \section{} 31 | 32 | {\it No head injury is too trivial to ignore} can be paraphrased as {\it None 33 | of head injuries are too trivial to ignore} or {\it Every head injury is vital 34 | and can't be ignored}. 35 | 36 | \section{} 37 | 38 | Since ``Fire'' has multiple different meanings, a user of this word must supply 39 | a context. In case of the following notice the context is clear: {\it In case of 40 | fire, do not use elevator}. However, without context this phrase can have 41 | another meaning. It can mean that in case of any fire (such as a cigarette 42 | lighter or even shooting) you are not advised to use the elevator. So, it can be 43 | paraphrased as {\it Do not use elevator, if the building is on fire}. 44 | 45 | \section{} 46 | 47 | The sentence does not make a true statement, because the page is already not 48 | blank. If it was really blank, then it wouldn't have anything printed on it. 49 | The purpose of making such a statement is to show to the reader that the page 50 | is not the result of an error at the printing house, but an intentional deed. 51 | 52 | It can be reformulated as this: {\it No material on this page}. 53 | 54 | \section{} 55 | 56 | \begin{enumerate} 57 | \item Pakistan talks with Taliban 58 | \item \dots 59 | \item \dots 60 | \end{enumerate} 61 | 62 | \section{} 63 | 64 | {\it The temperature is hot today} is a very inaccurate mark of the temperature, 65 | since every human being perceives it differently.``hot'' for a resident 66 | of Siberia, could be the same as``mild'' or ``normal'' for a resident of Africa. 67 | 68 | \section{} 69 | 70 | We show that not every number of the form $N = (p_1 \cdot p_2 \cdot p_3 \cdot 71 | \dots \cdot p_n) + 1$ is prime, where $p_1, p_2, p_3, \cdots, p_n, \cdots$ is 72 | the list of all prime numbers. If $N$ divided by any of $p_i$ factors leaves the 73 | remainder of 1, it is a composite number. Otherwise, it is a prime number. 74 | 75 | \section*{JUST FOR FUN} 76 | \setcounter{section}{0} 77 | 78 | \section{} 79 | 80 | Context: robotics. 81 | 82 | The robot said ``I have a message for you and, and, and, and'', and then he 83 | broke. 84 | 85 | \section{} 86 | 87 | Context: recursion. 88 | 89 | Provide a context and a sentence within that context, where the words and, or, 90 | and, or, and occur in that order, with no other word between them. 91 | 92 | \end{document} 93 | -------------------------------------------------------------------------------- /lecture-8-proofs-with-quantifiers/solution-8.tex: -------------------------------------------------------------------------------- 1 | \documentclass{article} 2 | \title{Solution 8} 3 | \author{@kyrylo} 4 | 5 | \usepackage{enumerate} 6 | \usepackage{amsmath} 7 | \usepackage{centernot} 8 | \usepackage{amsfonts} 9 | 10 | \begin{document} 11 | 12 | \section*{Solutions to assignment 8} 13 | 14 | \section{} 15 | 16 | Theorem: ``There are integers $m, n$ such that $m^2 + mn + n^2$ is a perfect 17 | square''. 18 | \\ 19 | Proof: Let $A(m, n)$ denote ``$m^2 + mn + n^2$ is a perfect square''. 20 | Hence, we're proving $(\exists m \in \mathbb{Z})(\exists n \in \mathbb{Z})[A(m, n)]$ 21 | Let's pick arbitrarily $m = 1, n = 0$. We are proving $A(1, 0)$, which is 22 | $1^2 + 1 \cdot 0 + 0^2 = 1 + 0 + 0 = 1$. The acquired number $1$ is a perfect square and this 23 | proves $A(m, n)$. 24 | 25 | \section{} 26 | 27 | Theorem: ``For any positive integer $m$ there is a positive integer $n$ such 28 | that $mn + 1$ is a perfect square'' 29 | \\ 30 | Proof: Let $A(m, n)$ denote ``$mn + 1$ is a perfect square". Hence, we're 31 | proving $(\forall m \in \mathbb{Z})(\exists n \in \mathbb{Z})[A(m, n)]$. Let $m$ 32 | be an arbitrary positive integer. Set $n = m + 2$. Then $mn + 1 = m \cdot (m + 33 | 2) + 1$ can be expressed as 34 | \\ 35 | $m \cdot (m + 2) + 1 = m^2 + 2m + 1 = m^2 + m + m + 1 = (m + 1)(m + 1) = (m + 1)^2$ 36 | \\ 37 | This proves that $(m + 1)^2$ is a perfect square. 38 | 39 | \section{} 40 | 41 | Theorem: ``There is a quadratic $f(n) = n^2 + bn + c$ with positive integer 42 | coefficients $b, c$, such that $f(n)$ is composite for all positive integers 43 | $n$.'' 44 | \\ 45 | Proof: we are proving 46 | \\ 47 | $(\exists b \in \mathbb{Z})(\exists c \in \mathbb{Z})(\forall n \in 48 | \mathbb{Z})[(c > 0) \wedge (b > 0) \wedge (n > 0) \wedge \neg Prime(n^2 + bn + 49 | c)]$ Let $b = 1, c = 2$, then $f(n) = n^2 + 2n + 1 = (n + 1)^2$. Since $(n + 50 | 1)^2$ always gives a perfect square, it means that the result is composite, 51 | because it can be divided by $n + 1$ (and prime numbers can be divided only by 1 and 52 | by itself). For example: $n = 36, f(n) = (n + 1)^2 = (36 + 1)^2 = 37^2 = 53 | 1369$. So, $1369 / (n + 1)= 1369 / (36 + 1) = 37$. 54 | 55 | \section{} 56 | 57 | Theorem: ``If every even natural number greater than 2 is a sum of two primes 58 | (the Goldbach Conjecture), then every odd natural number greater than 5 is a sum 59 | of three primes.'' 60 | \\ 61 | Proof: let $A(x)$ denote that $x$ can be expressed as a sum of two primes and 62 | let $B(x)$ denote that $x$ can be expressed as a sum of three primes. We're 63 | proving $(\forall x \in \mathbb{N})[x > 2 \wedge Even(x) \implies A(x)] \implies 64 | (\forall y \in \mathbb{N})[y > 5 \wedge Odd(y) \implies B(y)]$. Assume that 65 | $(\forall x \in \mathbb{N})[x > 2 \wedge Even(x) \implies A(x)]$ is true. Since 66 | the next odd number after 5 is 7, we can state that $y \geq 7$ and since the 67 | next even number after 2 is 4 we can state that $x \geq 4$. The difference between 68 | 7 and 4 is 3, which is a prime number. So by the Goldbach Conjecture we get 69 | $y - 3 = p + q$, where $p, q$ are prime numbers. Thus, $y = p + q + 3$ is a proof. 70 | 71 | \section{} 72 | 73 | Theorem: ``The sum of the first $n$ odd numbers is equal to $n^2$.'' 74 | \\ 75 | Proof: by mathematical induction. 76 | \\ 77 | For $n = 1$ the first and the last number is 1, so $1^2 = 1$, which is true. 78 | \\ 79 | For $n = 2$ the first odd numbers are $1, 3$ and their sum is 4, so $2^2 = 4$, 80 | which is also true. 81 | \\ 82 | For $n = 3$ the first odd numbers are $1, 3, 5$ and their sum is 9, so $3^2 = 83 | 9$, which is also true. 84 | \\ 85 | Let $A(n)$ be the statement $\sum\limits_{i=1}^n i = n^2$. Assume the identity 86 | holds for $n$, i.e. $A(n)$. Let's deduce $A(n + 1)$. So, $(n + 1)^2 = n^2 + 2n + 87 | 1$. 88 | \\ 89 | ? 90 | \end{document} 91 | -------------------------------------------------------------------------------- /lecture-2-logical-combinators/solution-2.tex: -------------------------------------------------------------------------------- 1 | \documentclass{article} 2 | \title{Solution 2} 3 | \author{@kyrylo} 4 | 5 | \usepackage{enumerate} 6 | 7 | \begin{document} 8 | 9 | \section*{Solutions to assignment 2} 10 | 11 | \section{} 12 | 13 | \begin{enumerate}[(a)] 14 | \item $(\pi > 0) \wedge (\pi < 10)$ [Answer: $0 < \pi < 10$.] 15 | \item $(p \geq 7) \wedge (p < 12)$ [Answer: $7 \leq p < 12$.] 16 | \item $(x > 5) \wedge (x < 7)$ [Answer: $5 < x < 7$.] 17 | \item $(x < 4) \wedge (x < 6)$ [Answer: $x < 4 < 6 = x < 4$.] 18 | \item $(y < 4) \wedge (y^2 < 9)$ [Answer: $(y < 4) \wedge (-3 < y < 3) = -3 < y < 3$.] 19 | \item $(x \geq 0) \wedge (x \leq 0)$ [Answer: $x = 0$.] 20 | \end{enumerate} 21 | 22 | \section{} 23 | 24 | \begin{enumerate}[(a)] 25 | \item $\pi$ is between 0 and 10. 26 | \item $p$ is greater than or equal to 7 and less than 12. 27 | \item $x$ is between 5 and 7. 28 | \item $x$ is less than 4. 29 | \item $y$ is between -3 and 4. 30 | \item $x$ is equal to 0. 31 | \end{enumerate} 32 | 33 | \section{} 34 | 35 | \begin{tabular}{ | c | c | c | c | } 36 | \hline 37 | $\phi_1$ & $\phi_2$ & $\phi_n$ & $\phi_1 \wedge \phi_2 \wedge \dots \wedge \phi_n$ \\ 38 | \hline 39 | T & T & T & T \\ 40 | \hline 41 | \end{tabular} 42 | 43 | \section{} 44 | 45 | If any $\phi$ is false, then the entire conjunction is false, too. 46 | 47 | \begin{tabular}{ | c | c | c | c | } 48 | \hline 49 | $\phi_1$ & $\phi_2$ & $\phi_n$ & $\phi_1 \wedge \phi_2 \wedge \dots \wedge \phi_n$ \\ 50 | \hline 51 | T & F & T & F \\ 52 | F & T & T & F \\ 53 | T & T & F & F \\ 54 | F & F & F & F \\ 55 | \hline 56 | \end{tabular} 57 | 58 | \section{} 59 | 60 | \begin{enumerate}[(a)] 61 | \item $(\pi > 3) \vee (\pi > 10)$ [Answer: $\pi > 3$.] 62 | \item $(x < 0) \vee (x > 0)$ [Answer: $x \neq 0$.] 63 | \item $(x = 0) \vee (x > 0)$ [Answer: $x \geq 0$.] 64 | \item $(x > 0) \vee (x \geq 0)$ [Answer: $x \geq 0$.] 65 | \item $(x > 3) \vee (x^2 > 9)$ [Answer: $(x > 3) \vee (x > -3) = x^2 > 9$.] 66 | \end{enumerate} 67 | 68 | \section{} 69 | 70 | \begin{enumerate}[(a)] 71 | \item $\pi$ is greater than 3. 72 | \item $x$ is not equal to 0. 73 | \item $x$ is greater than or equal to 0. 74 | \item $x$ is greater than or equal to 0. 75 | \item $x$ is greater than 3. 76 | \end{enumerate} 77 | 78 | \section{} 79 | 80 | If any $\pi$ is true, then the disjunction is true. 81 | 82 | \begin{tabular}{ | c | c | c | c | } 83 | \hline 84 | $\phi_1$ & $\phi_2$ & $\phi_n$ & $\phi_1 \vee \phi_2 \vee \dots \vee \phi_n$ \\ 85 | \hline 86 | T & F & T & T \\ 87 | F & T & T & T \\ 88 | T & T & F & T \\ 89 | T & T & F & T \\ 90 | T & T & T & T \\ 91 | \hline 92 | \end{tabular} 93 | 94 | \section{} 95 | 96 | If each $\pi$ is false, then the disjunction is false. Otherwise, it's true. 97 | 98 | \begin{tabular}{ | c | c | c | c | } 99 | \hline 100 | $\phi_1$ & $\phi_2$ & $\phi_n$ & $\phi_1 \vee \phi_2 \vee \dots \vee \phi_n$ \\ 101 | \hline 102 | F & F & F & F \\ 103 | \hline 104 | \end{tabular} 105 | 106 | 107 | \section{} 108 | 109 | \begin{enumerate}[(a)] 110 | \item $\neg(\pi > 3.2)$ [Answer: $\pi \leq 3.2$.] 111 | \item $\neg(x < 0)$ [Answer: $x \geq 0$.] 112 | \item $\neg(x^2 > 0)$ [Answer: $\neg((x < 0) \vee (x > 0)) = \neg(x \neq 0) = 0$.] 113 | \item $\neg(x = 1)$ [Answer: $(x < 1) \vee (x > 1) = x \neq 1$.] 114 | \item $\neg\neg\phi$ [Answer: $\phi$.] 115 | \end{enumerate} 116 | 117 | \section{} 118 | 119 | \begin{enumerate}[(a)] 120 | \item $\pi$ is less than or equal to 3.2. 121 | \item $x$ is greater than or equal to 0. 122 | \item $x$ is equal to 0. 123 | \item $x$ is not equal to 0. 124 | \item $phi$. 125 | \end{enumerate} 126 | 127 | \section{} 128 | 129 | \begin{enumerate}[(a)] 130 | \item $D \wedge Y$ 131 | \item $D \wedge (\neg Y) \wedge T$ 132 | \item $\neg(D \wedge Y)$ 133 | \item $T \wedge (\neg D) \wedge (\neg Y)$ 134 | \item $\neg T \wedge D \wedge Y$ 135 | \end{enumerate} 136 | 137 | \section*{TWO TO THINK ABOUT AND DISCUSS WITH OTHER STUDENTS} 138 | \setcounter{section}{0} 139 | 140 | \section{} 141 | 142 | The ``not'' in mathematical sense does not correspond to the ``not'' from natural 143 | language. The reason is that natural language is not precise enough, while 144 | mathematics is a synonym for ``preciseness''. 145 | 146 | \section{} 147 | 148 | First of all, it's recommended to avoid double negations in natural language. 149 | We can capture ``I was not displeased with the movie'' like this: $\neg DISPLEASED$ 150 | 151 | \end{document} 152 | -------------------------------------------------------------------------------- /lecture-4-equivalence/solution-4.tex: -------------------------------------------------------------------------------- 1 | \documentclass{article} 2 | \title{Solution 4} 3 | \author{@kyrylo} 4 | 5 | \usepackage{enumerate} 6 | \usepackage{amsmath} 7 | \usepackage{centernot} 8 | 9 | \begin{document} 10 | 11 | \section*{Solutions to assignment 4} 12 | 13 | \section{} 14 | 15 | \begin{tabular}{ | c | c | c | } 16 | \hline 17 | $\phi$ & $\psi$ & $\phi \Leftrightarrow \psi$ \\ 18 | \hline 19 | T & T & T \\ 20 | T & F & F \\ 21 | F & T & F \\ 22 | F & F & T \\ 23 | \hline 24 | \end{tabular} 25 | 26 | \section{} 27 | 28 | \begin{tabular}{ | c | c | c | c | c | c | } 29 | \hline 30 | $\phi$ & $\psi$ & $\neg \phi$ & $\phi \implies \psi$ & $\neg \phi \vee \psi$ & $(\phi \implies \psi) \Leftrightarrow (\neg \phi \vee \psi)$ \\ 31 | \hline 32 | T & T & F & T & T & T \\ 33 | T & F & F & F & F & T \\ 34 | F & T & T & T & T & T \\ 35 | F & F & T & T & T & T \\ 36 | \hline 37 | \end{tabular} 38 | 39 | \section{} 40 | 41 | \begin{tabular}{ | c | c | c | c | c | c | } 42 | \hline 43 | $\phi$ & $\psi$ & $\neg \psi$ & $\phi \centernot\implies \psi$ & $\phi \wedge \neg\psi$ & $(\phi \centernot\implies \psi) \Leftrightarrow (\phi \wedge \neg\psi)$ \\ 44 | \hline 45 | T & T & F & F & F & T \\ 46 | T & F & T & T & T & T \\ 47 | F & T & F & F & F & T \\ 48 | F & F & T & F & F & T \\ 49 | \hline 50 | \end{tabular} 51 | 52 | 53 | \section{} 54 | 55 | \begin{enumerate}[(a)] 56 | \item 57 | \begin{tabular}{ | c | c | c | c | c | } 58 | \hline 59 | $\phi$ & $\psi$ & $\phi \implies \psi$ & $\phi \wedge (\phi \implies \psi)$ & $[\phi \wedge (\phi \implies \psi)] \implies \psi$ \\ 60 | \hline 61 | T & T & T & T & T \\ 62 | T & F & F & F & T \\ 63 | F & T & T & F & T \\ 64 | F & F & T & F & T \\ 65 | \hline 66 | \end{tabular} 67 | \item 68 | The statement gives truth in every case. Any combination of $\phi$ and $\psi$ is true. 69 | \end{enumerate} 70 | 71 | \section{} 72 | 73 | \begin{enumerate} 74 | \item $\phi \vee \psi$ means at least one from $\phi$ or $\psi$ must be true. 75 | \item Thus $\neg(\phi \vee \psi)$ means none of $\phi$ or $\psi$ is true. 76 | \item The conjoin $(\neg\phi) \wedge (\neg\psi)$ indicates that both $\phi$ and 77 | $\psi$ must be false. 78 | \item Thus $(\neg\phi) \wedge (\neg\psi)$ is equivalent to $\neg(\phi \vee \psi)$. 79 | \end{enumerate} 80 | 81 | \section{} 82 | 83 | \begin{enumerate}[(a)] 84 | \item 34,159 is not a prime number. 85 | \item Roses are not red or violets are not blue. 86 | \item There are no hamburgers, but I'll not have a hot dog. 87 | \item Fred will not go or he will play. 88 | \item The number $x$ is either positive or not greater than 10. 89 | \item We will not win the first game and the second. 90 | \end{enumerate} 91 | 92 | \section{} 93 | 94 | \begin{tabular}{ | c | c | c | c | c | c | } 95 | \hline 96 | $\phi$ & $\psi$ & $\phi \Leftrightarrow \psi $ & $\neg\phi$ & $\neg\psi$ & $(\neg\phi) \Leftrightarrow (\neg\psi)$ \\ 97 | \hline 98 | T & T & T & F & F & T \\ 99 | T & F & F & F & T & F \\ 100 | F & T & F & T & F & F \\ 101 | F & F & T & T & T & T \\ 102 | \hline 103 | \end{tabular} 104 | 105 | \section{} 106 | 107 | \begin{enumerate}[(a)] 108 | \item 109 | \begin{tabular}{ | c | c | c | } 110 | \hline 111 | $\phi$ & $\psi$ & $\phi \Leftrightarrow \psi $ \\ 112 | \hline 113 | T & T & T \\ 114 | T & F & F \\ 115 | F & T & F \\ 116 | F & F & T \\ 117 | \hline 118 | \end{tabular} 119 | 120 | \item 121 | \begin{tabular}{ | c | c | c | c | c | } 122 | \hline 123 | $\phi$ & $\psi$ & $\theta$ & $\psi \vee \theta$ & $\phi \implies (\psi \vee \theta)$ \\ 124 | \hline 125 | T & T & T & T & T \\ 126 | T & T & F & T & T \\ 127 | T & F & T & T & T \\ 128 | T & F & F & F & F \\ 129 | F & T & T & T & T \\ 130 | F & T & F & T & T \\ 131 | F & F & T & T & T \\ 132 | F & F & F & F & T \\ 133 | \hline 134 | \end{tabular} 135 | \end{enumerate} 136 | 137 | \section{} 138 | 139 | \begin{tabular}{ | c | c | c | c | c | } 140 | \hline 141 | $\phi$ & $\psi$ & $\theta$ & $\psi \wedge \theta$ & $\phi \implies (\psi \wedge \theta)$ \\ 142 | \hline 143 | T & T & T & T & T \\ 144 | T & T & F & F & F \\ 145 | T & F & T & F & F \\ 146 | T & F & F & F & F \\ 147 | F & T & T & F & T \\ 148 | F & T & F & F & T \\ 149 | F & F & T & F & T \\ 150 | F & F & F & F & T \\ 151 | \hline 152 | \end{tabular} 153 | \quad 154 | \begin{tabular}{ | c | c | c | c | c | c | } 155 | \hline 156 | $\phi$ & $\psi$ & $\theta$ & $\phi \implies \psi$ & $\phi \implies \theta$ & $(\phi \implies \psi) \wedge (\phi \implies \theta)$ \\ 157 | \hline 158 | T & T & T & T & T & T \\ 159 | T & T & F & T & F & F \\ 160 | T & F & T & F & T & F \\ 161 | T & F & F & F & F & F \\ 162 | F & T & T & T & T & T \\ 163 | F & T & F & T & T & T \\ 164 | F & F & T & T & T & T \\ 165 | F & F & F & T & T & T \\ 166 | \hline 167 | \end{tabular} 168 | 169 | \section{} 170 | 171 | Suppose $\psi, \theta$ are true. Conjuction of them is also true, hence it is 172 | obvious that $\phi \implies (\psi \wedge \theta)$ is true. Implication of $\psi, 173 | \theta$ from $\phi$ is obvious, too. I suck at explanations. 174 | 175 | \end{document} 176 | -------------------------------------------------------------------------------- /lecture-6-working-with-quantifiers/solution-6.tex: -------------------------------------------------------------------------------- 1 | \documentclass{article} 2 | \title{Solution 6} 3 | \author{@kyrylo} 4 | 5 | \usepackage{enumerate} 6 | \usepackage{amsmath} 7 | \usepackage{centernot} 8 | \usepackage{amsfonts} 9 | 10 | \begin{document} 11 | 12 | \section*{Solutions to assignment 6} 13 | 14 | \section{} 15 | 16 | In plain English $\neg[\exists x A(x)]$ means ``It is not the case that there 17 | exists $x$ such that $A(x)$ is true. 18 | \\\\ 19 | In plain English $\forall x[\neg A(x)]$ means ``The function $A(x)$ is false for 20 | any given $x$''. 21 | \\\\ 22 | The latter implies the former, it illustrates their equality. 23 | 24 | \section{} 25 | 26 | {\it There is an even prime bigger than 2} 27 | \\\\ 28 | Assertion: the statement is false. 29 | \\\\ 30 | Let $P(x)$: x is prime, $E(x)$: x is even. The sentence can be written 31 | symbolically as: 32 | 33 | $(\exists x > 2)[P(x) \implies E(x)]$. 34 | \\\\ 35 | When we negate this we get: 36 | 37 | $(\forall x > 2)[P(x) \wedge \neg E(x)]$. 38 | \\\\ 39 | In words, each prime number $x$ bigger than 2 is odd. An odd number is a number 40 | that cannot be divided by 2 without getting a fraction. Let's observe the next 41 | prime number, 3. In this case the original statement is false and its negation 42 | is true. Every other next prime number behaves the same. 43 | 44 | \section{} 45 | 46 | \begin{enumerate}[(a)] 47 | \item $\forall x [Student(x) \implies LikesPizza(x)]$ 48 | \item $\exists x [Friend(x) \wedge \neg HasCar(x)]$ 49 | \item $\exists x [Elephant(x) \wedge \neg LikesMuffin(x)]$ 50 | \item $\forall x [Triangle(x) \implies Isosceles(x)]$ 51 | \item $\exists x [Student(x) \wedge \neg Here(x)]$ 52 | \item $\exists x \forall y Loves(x, y)$ 53 | \item $\forall x \forall y \neg Loves(x, y)$ 54 | \item $\forall x \forall y[Man(x) \wedge Comes(x) \implies WomenLeave(y)]$ 55 | \item $\forall x [Tall(x) \vee Short(x)]$ 56 | \item $\forall x Tall(x) \vee \forall x Short(x)$ 57 | \item $\exists x [Precious(x) \wedge \neg Beautiful(x)]$ 58 | \item $\exists x \forall y \neg Loves(x, y)$ 59 | \item $\exists x [AmericanSnake(x) \wedge Poisonous(x)]$ 60 | \item $\exists x [Snake(x) \wedge AmericanSnake(x) \wedge Poisonous(x)]$ 61 | \end{enumerate} 62 | 63 | \section{} 64 | 65 | \begin{enumerate}[(a)] 66 | \item $\exists x [Student(x) \wedge \neg LikesPizza(x)]$ --- There is a student who doesn't like pizza. 67 | \item $\forall x [Friend(x) \implies HasCar(x)]$ --- All my friends have cars. 68 | \item $\forall x [Elephant(x) \implies LikesMuffin(x)]$ --- All elephants like muffins. 69 | \item $\exists x [Triangle(x) \wedge \neg Isosceles(x)]$ --- There's a triangle which is not isosceles. 70 | \item $\forall x [Student(x) \implies Here(x)]$ --- All the students are here today. 71 | \item $\exists x \forall y \neg Loves(x, y)$ --- Everyone dislikes somebody. 72 | \item $\forall x \forall y Loves(x, y)$ --- Everybody loves everybody. 73 | \item $\forall x \forall y[Man(x) \wedge Comes(x) \wedge \neg WomenLeave(y)]$ --- If a man comes, no women leave. 74 | \item $\exists x \neg [Tall(x) \vee Short(x)]$ --- There is a person that is neither tall nor short. 75 | \item $\exists x \neg [Tall(x) \wedge Short(x)]$ --- All people are not tall and short. 76 | \item $\forall x [Precious(x) \implies Beautiful(x)]$ --- All precious stones are beatuiful. 77 | \item $\exists x \forall y Loves(x, y)$ --- Everybody loves me. 78 | \item $\forall x [AmericanSnake(x) \implies Poisonous(x)]$ --- All American snakes are poisonous. 79 | \item $\forall x [Snake(x) \implies AmericanSnake(x) \implies Poisonous(x)]$ --- All American snakes are poisonous. 80 | \end{enumerate} 81 | 82 | \section{} 83 | 84 | \begin{enumerate}[(a)] 85 | \item false 86 | \item false 87 | \item true 88 | \item false 89 | \item false 90 | \item false 91 | \item false 92 | \item false 93 | \end{enumerate} 94 | 95 | \section{} 96 | 97 | \begin{enumerate}[(a)] 98 | \item $\forall x [2x + 3 \neq 5x + 1]$ 99 | \item $\forall x [x^2 \neq 2]$ 100 | \item $\exists x \forall y [y \neq x^2]$ 101 | \item $\exists x \forall y [y \neq x^2]$ 102 | \item $\exists x \forall y \exists z [xy \neq xz]$ 103 | \item $\exists x \forall y \exists z [xy \neq xz]$ 104 | \item $\exists x [x < 0 \wedge \neg \exists y (y^2 = x)]$ 105 | \item $\exists x [x < 0 \wedge \neg \exists y (y^2 = x)]$ 106 | \end{enumerate} 107 | 108 | \section{} 109 | 110 | \begin{enumerate}[(a)] 111 | \item $(\exists x \in \mathbb{N})(\forall y \in \mathbb{N})[x + y \neq 1]$ or $(\exists x \in \mathbb{N})(\forall y \in \mathbb{N})[(x + y > 1) \vee (x + y < 1)]$ 112 | \item $(\exists x > 0)(\forall y < 0)[x + y \neq 0]$ or $(\exists x > 0)(\forall y < 0)[(x + y > 0) \vee (x + y < 0)]$ 113 | \item $\forall x(\exists \epsilon> 0)\neg[-\epsilon < x < \epsilon]$ or $\forall x(\exists \epsilon> 0)[(x \leq -\epsilon) \vee (x \geq \epsilon)]$ 114 | \item $(\exists x \in \mathbb{N})(\exists y \in \mathbb{N})(\forall z \in \mathbb{N})[x + y \neq z^2]$ or $?$ 115 | \end{enumerate} 116 | 117 | \section{} 118 | 119 | Statement: $\exists t \forall p Fool(p, t) \wedge \exists p \forall t Fool(p, t) \wedge \neg \forall p \forall t Fool(p, t)$ 120 | \\\\ 121 | Negation: $\forall t \exists p \neg Fool(p, t) \vee \forall p \exists t \neg Fool(p, t) \vee \forall p \forall t Fool(p, t)$ 122 | 123 | \section{} 124 | 125 | $(\exists \epsilon > 0)(\forall \delta > 0)(\exists x)[|x - a| < \delta \wedge |f(x) - f(a)| \geq \epsilon)$ 126 | 127 | \end{document} 128 | --------------------------------------------------------------------------------