├── .gitignore ├── LICENSE ├── ProofNetLean4.lean ├── README.md ├── formal ├── Artin.lean ├── Axler.lean ├── Cambridge-Tripos.lean ├── Dummit-Foote.lean ├── Herstein.lean ├── Ireland-Rosen.lean ├── Munkers.lean ├── Pough.lean ├── Putnam.lean ├── Rudin.lean └── Shakarachi.lean ├── informal ├── Artin.pdf ├── Artin.tex ├── Axler.pdf ├── Axler.tex ├── Cambridge-Tripos.pdf ├── Cambridge-Tripos.tex ├── Dummit-Foote.pdf ├── Dummit-Foote.tex ├── Herstein.pdf ├── Herstein.tex ├── Ireland-Rosen.pdf ├── Ireland-Rosen.tex ├── Munkres.pdf ├── Munkres.tex ├── Pugh.pdf ├── Pugh.tex ├── Putnam.pdf ├── Putnam.tex ├── Rudin.pdf ├── Rudin.tex ├── Shakarchi.pdf └── Shakarchi.tex ├── lake-manifest.json ├── lakefile.lean └── lean-toolchain /.gitignore: -------------------------------------------------------------------------------- 1 | /build 2 | /lake-packages/* 3 | /.lake -------------------------------------------------------------------------------- /LICENSE: -------------------------------------------------------------------------------- 1 | MIT License 2 | 3 | Copyright (c) 2024 Rahul Vishwakarma 4 | 5 | Permission is hereby granted, free of charge, to any person obtaining a copy 6 | of this software and associated documentation files (the "Software"), to deal 7 | in the Software without restriction, including without limitation the rights 8 | to use, copy, modify, merge, publish, distribute, sublicense, and/or sell 9 | copies of the Software, and to permit persons to whom the Software is 10 | furnished to do so, subject to the following conditions: 11 | 12 | The above copyright notice and this permission notice shall be included in all 13 | copies or substantial portions of the Software. 14 | 15 | THE SOFTWARE IS PROVIDED "AS IS", WITHOUT WARRANTY OF ANY KIND, EXPRESS OR 16 | IMPLIED, INCLUDING BUT NOT LIMITED TO THE WARRANTIES OF MERCHANTABILITY, 17 | FITNESS FOR A PARTICULAR PURPOSE AND NONINFRINGEMENT. IN NO EVENT SHALL THE 18 | AUTHORS OR COPYRIGHT HOLDERS BE LIABLE FOR ANY CLAIM, DAMAGES OR OTHER 19 | LIABILITY, WHETHER IN AN ACTION OF CONTRACT, TORT OR OTHERWISE, ARISING FROM, 20 | OUT OF OR IN CONNECTION WITH THE SOFTWARE OR THE USE OR OTHER DEALINGS IN THE 21 | SOFTWARE. 22 | -------------------------------------------------------------------------------- /ProofNetLean4.lean: -------------------------------------------------------------------------------- 1 | def hello := "world" -------------------------------------------------------------------------------- /README.md: -------------------------------------------------------------------------------- 1 | # ProofNet-lean4 2 | **This is the Lean 4 version of the ProofNet dataset, originally present in the repository https://github.com/zhangir-azerbayev/ProofNet.** -------------------------------------------------------------------------------- /formal/Artin.lean: -------------------------------------------------------------------------------- 1 | import Mathlib 2 | 3 | open Function Fintype Subgroup Ideal Polynomial Submodule Zsqrtd 4 | open scoped BigOperators 5 | noncomputable section 6 | 7 | theorem exercise_2_2_9 {G : Type*} [Group G] {a b : G} 8 | (h : a * b = b * a) : 9 | ∀ x y : closure {x | x = a ∨ x = b}, x*y = y*x := 10 | sorry 11 | 12 | theorem exercise_2_3_2 {G : Type*} [Group G] (a b : G) : 13 | ∃ g : G, b* a = g * a * b * g⁻¹ := 14 | sorry 15 | 16 | theorem exercise_2_4_19 {G : Type*} [Group G] {x : G} 17 | (hx : orderOf x = 2) (hx1 : ∀ y, orderOf y = 2 → y = x) : 18 | x ∈ center G := 19 | sorry 20 | 21 | instance exercise_2_8_6 {G H : Type*} [Group G] [Group H] : 22 | center (G × H) ≃* (center G) × (center H) := 23 | sorry 24 | 25 | theorem exercise_2_11_3 {G : Type*} [Group G] [Fintype G] 26 | (hG : Even (card G)) : ∃ x : G, orderOf x = 2 := 27 | sorry 28 | 29 | theorem exercise_3_2_7 {F : Type*} [Field F] {G : Type*} [Field G] 30 | (φ : F →+* G) : Injective φ := 31 | sorry 32 | 33 | theorem exercise_3_5_6 {K V : Type*} [Field K] [AddCommGroup V] 34 | [Module K V] {S : Set V} (hS : Set.Countable S) 35 | (hS1 : span K S = ⊤) {ι : Type*} (R : ι → V) 36 | (hR : LinearIndependent K R) : Countable ι := 37 | sorry 38 | 39 | theorem exercise_3_7_2 {K V : Type*} [Field K] [AddCommGroup V] 40 | [Module K V] {ι : Type*} [Fintype ι] (γ : ι → Submodule K V) 41 | (h : ∀ i : ι, γ i ≠ ⊤) : 42 | (⋂ (i : ι), (γ i : Set V)) ≠ ⊤ := 43 | sorry 44 | 45 | theorem exercise_6_1_14 (G : Type*) [Group G] 46 | (hG : IsCyclic $ G ⧸ (center G)) : 47 | center G = ⊤ := 48 | sorry 49 | 50 | theorem exercise_6_4_2 {G : Type*} [Group G] [Fintype G] {p q : ℕ} 51 | (hp : Prime p) (hq : Prime q) (hG : card G = p*q) : 52 | IsSimpleGroup G → false := 53 | sorry 54 | 55 | theorem exercise_6_4_3 {G : Type*} [Group G] [Fintype G] {p q : ℕ} 56 | (hp : Prime p) (hq : Prime q) (hG : card G = p^2 *q) : 57 | IsSimpleGroup G → false := 58 | sorry 59 | 60 | theorem exercise_6_4_12 {G : Type*} [Group G] [Fintype G] 61 | (hG : card G = 224) : 62 | IsSimpleGroup G → false := 63 | sorry 64 | 65 | theorem exercise_6_8_1 {G : Type*} [Group G] 66 | (a b : G) : closure ({a, b} : Set G) = Subgroup.closure {b*a*b^2, b*a*b^3} := 67 | sorry 68 | 69 | theorem exercise_10_1_13 {R : Type*} [Ring R] {x : R} 70 | (hx : IsNilpotent x) : IsUnit (1 + x) := 71 | sorry 72 | 73 | theorem exercise_10_2_4 : 74 | span ({2} : Set $ Polynomial ℤ) ⊓ (span {X}) = 75 | span ({2 * X} : Set $ Polynomial ℤ) := 76 | sorry 77 | 78 | theorem exercise_10_6_7 {I : Ideal GaussianInt} 79 | (hI : I ≠ ⊥) : ∃ (z : I), z ≠ 0 ∧ (z : GaussianInt).im = 0 := 80 | sorry 81 | 82 | theorem exercise_10_4_6 {R : Type*} [CommRing R] 83 | [NoZeroDivisors R] (I J : Ideal R) (x : ↑(I ⊓ J)) : 84 | IsNilpotent ((Ideal.Quotient.mk (I*J)) x) := 85 | sorry 86 | 87 | theorem exercise_10_4_7a {R : Type*} [CommRing R] [NoZeroDivisors R] 88 | (I J : Ideal R) (hIJ : I + J = ⊤) : I * J = I ⊓ J := 89 | sorry 90 | 91 | theorem exercise_10_7_10 {R : Type*} [Ring R] 92 | (M : Ideal R) (hM : ∀ (x : R), x ∉ M → IsUnit x) 93 | (hProper : ∃ x : R, x ∉ M) : 94 | IsMaximal M ∧ ∀ (N : Ideal R), IsMaximal N → N = M := 95 | sorry 96 | 97 | theorem exercise_11_2_13 (a b : ℤ) : 98 | (ofInt a : GaussianInt) ∣ ofInt b → a ∣ b := 99 | sorry 100 | 101 | theorem exercise_11_4_1b {F : Type*} [Field F] [Fintype F] (hF : card F = 2) : 102 | Irreducible (12 + 6 * X + X ^ 3 : Polynomial F) := 103 | sorry 104 | 105 | theorem exercise_11_4_6a {F : Type*} [Field F] [Fintype F] (hF : card F = 7) : 106 | Irreducible (X ^ 2 + 1 : Polynomial F) := 107 | sorry 108 | 109 | theorem exercise_11_4_6b {F : Type*} [Field F] [Fintype F] (hF : card F = 31) : 110 | Irreducible (X ^ 3 - 9 : Polynomial F) := 111 | sorry 112 | 113 | theorem exercise_11_4_6c : Irreducible (X^3 - 9 : Polynomial (ZMod 31)) := 114 | sorry 115 | 116 | theorem exercise_11_4_8 (p : ℕ) (hp : Prime p) (n : ℕ) : 117 | -- p ∈ ℕ can be written as p ∈ ℚ[X] 118 | Irreducible (X ^ n - (p : Polynomial ℚ) : Polynomial ℚ) := 119 | sorry 120 | 121 | theorem exercise_11_13_3 (N : ℕ): 122 | ∃ p ≥ N, Nat.Prime p ∧ p + 1 ≡ 0 [MOD 4] := 123 | sorry 124 | 125 | theorem exercise_13_4_10 126 | {p : ℕ} {hp : Nat.Prime p} (h : ∃ r : ℕ, p = 2 ^ r + 1) : 127 | ∃ (k : ℕ), p = 2 ^ (2 ^ k) + 1 := 128 | sorry 129 | 130 | theorem exercise_13_6_10 {K : Type*} [Field K] [Fintype Kˣ] : 131 | (∏ x : Kˣ, x) = -1 := 132 | sorry 133 | -------------------------------------------------------------------------------- /formal/Axler.lean: -------------------------------------------------------------------------------- 1 | import Mathlib 2 | 3 | open Fintype Complex Polynomial LinearMap FiniteDimensional Module Module.End 4 | open scoped BigOperators InnerProductSpace 5 | 6 | theorem exercise_1_2 : 7 | (⟨-1/2, Real.sqrt 3 / 2⟩ : ℂ) ^ 3 = -1 := 8 | sorry 9 | 10 | theorem exercise_1_3 {F V : Type*} [AddCommGroup V] [Field F] 11 | [Module F V] {v : V} : -(-v) = v := 12 | sorry 13 | 14 | theorem exercise_1_4 {F V : Type*} [AddCommGroup V] [Field F] 15 | [Module F V] (v : V) (a : F): a • v = 0 ↔ a = 0 ∨ v = 0 := 16 | sorry 17 | 18 | theorem exercise_1_6 : ∃ U : Set (ℝ × ℝ), 19 | (U ≠ ∅) ∧ 20 | (∀ (u v : ℝ × ℝ), u ∈ U ∧ v ∈ U → u + v ∈ U) ∧ 21 | (∀ (u : ℝ × ℝ), u ∈ U → -u ∈ U) ∧ 22 | (∀ U' : Submodule ℝ (ℝ × ℝ), U ≠ ↑U') := 23 | sorry 24 | 25 | theorem exercise_1_7 : ∃ U : Set (ℝ × ℝ), 26 | (U ≠ ∅) ∧ 27 | (∀ (c : ℝ) (u : ℝ × ℝ), u ∈ U → c • u ∈ U) ∧ 28 | (∀ U' : Submodule ℝ (ℝ × ℝ), U ≠ ↑U') := 29 | sorry 30 | 31 | theorem exercise_1_8 {F V : Type*} [AddCommGroup V] [Field F] 32 | [Module F V] {ι : Type*} (u : ι → Submodule F V) : 33 | ∃ U : Submodule F V, (⋂ (i : ι), (u i).carrier) = ↑U := 34 | by 35 | sorry 36 | 37 | theorem exercise_1_9 {F V : Type*} [AddCommGroup V] [Field F] 38 | [Module F V] (U W : Submodule F V): 39 | ∃ U' : Submodule F V, (U'.carrier = ↑U ∩ ↑W ↔ (U ≤ W ∨ W ≤ U)) := 40 | sorry 41 | 42 | theorem exercise_3_1 {F V : Type*} 43 | [AddCommGroup V] [Field F] [Module F V] [FiniteDimensional F V] 44 | (T : V →ₗ[F] V) (hT : finrank F V = 1) : 45 | ∃ c : F, ∀ v : V, T v = c • v:= 46 | sorry 47 | 48 | theorem exercise_3_8 {F V W : Type*} [AddCommGroup V] 49 | [AddCommGroup W] [Field F] [Module F V] [Module F W] 50 | (L : V →ₗ[F] W) : 51 | ∃ U : Submodule F V, U ⊓ (ker L) = ⊥ ∧ 52 | (range L = range (domRestrict L U)):= 53 | sorry 54 | 55 | theorem exercise_4_4 (p : Polynomial ℂ) : 56 | p.degree = @card (rootSet p ℂ) (rootSetFintype p ℂ) ↔ 57 | Disjoint 58 | (@card (rootSet (derivative p) ℂ) (rootSetFintype (derivative p) ℂ)) 59 | (@card (rootSet p ℂ) (rootSetFintype p ℂ)) := 60 | sorry 61 | 62 | theorem exercise_5_1 {F V : Type*} [AddCommGroup V] [Field F] 63 | [Module F V] {L : V →ₗ[F] V} {n : ℕ} (U : Fin n → Submodule F V) 64 | (hU : ∀ i : Fin n, Submodule.map L (U i) = U i) : 65 | Submodule.map L (∑ i : Fin n, U i : Submodule F V) = 66 | (∑ i : Fin n, U i : Submodule F V) := 67 | sorry 68 | 69 | theorem exercise_5_4 {F V : Type*} [AddCommGroup V] [Field F] 70 | [Module F V] (S T : V →ₗ[F] V) (hST : S ∘ T = T ∘ S) (c : F): 71 | Submodule.map S (ker (T - c • LinearMap.id)) = ker (T - c • LinearMap.id) := 72 | sorry 73 | 74 | theorem exercise_5_11 {F V : Type*} [AddCommGroup V] [Field F] 75 | [Module F V] (S T : End F V) : 76 | (S * T).Eigenvalues = (T * S).Eigenvalues := 77 | sorry 78 | 79 | theorem exercise_5_12 {F V : Type*} [AddCommGroup V] [Field F] 80 | [Module F V] {S : End F V} 81 | (hS : ∀ v : V, ∃ c : F, v ∈ eigenspace S c) : 82 | ∃ c : F, S = c • LinearMap.id := 83 | sorry 84 | 85 | theorem exercise_5_13 {F V : Type*} [AddCommGroup V] [Field F] 86 | [Module F V] [FiniteDimensional F V] {T : End F V} 87 | (hS : ∀ U : Submodule F V, finrank F U = finrank F V - 1 → 88 | Submodule.map T U = U) : ∃ c : F, T = c • LinearMap.id := 89 | sorry 90 | 91 | theorem exercise_5_20 {F V : Type*} [AddCommGroup V] [Field F] 92 | [Module F V] [FiniteDimensional F V] {S T : End F V} 93 | (h1 : card (T.Eigenvalues) = finrank F V) 94 | (h2 : ∀ v : V, ∃ c : F, v ∈ eigenspace S c ↔ ∃ c : F, v ∈ eigenspace T c) : 95 | S * T = T * S := 96 | sorry 97 | 98 | theorem exercise_5_24 {V : Type*} [AddCommGroup V] 99 | [Module ℝ V] [FiniteDimensional ℝ V] {T : End ℝ V} 100 | (hT : ∀ c : ℝ, eigenspace T c = ⊥) {U : Submodule ℝ V} 101 | (hU : Submodule.map T U = U) : Even (finrank U) := 102 | sorry 103 | 104 | theorem exercise_6_2 {V : Type*} [NormedAddCommGroup V] [Module ℂ V] 105 | [InnerProductSpace ℂ V] (u v : V) : 106 | ⟪u, v⟫_ℂ = 0 ↔ ∀ (a : ℂ), ‖u‖ ≤ ‖u + a • v‖ := 107 | sorry 108 | 109 | theorem exercise_6_3 {n : ℕ} (a b : Fin n → ℝ) : 110 | (∑ i, a i * b i) ^ 2 ≤ (∑ i : Fin n, i * a i ^ 2) * (∑ i, b i ^ 2 / i) := 111 | sorry 112 | 113 | theorem exercise_6_7 {V : Type*} [NormedAddCommGroup V] [InnerProductSpace ℂ V] (u v : V) : 114 | ⟪u, v⟫_ℂ = (‖u + v‖^2 - ‖u - v‖^2 + I*‖u + I•v‖^2 - I*‖u-I•v‖^2) / 4 := 115 | sorry 116 | 117 | theorem exercise_6_13 {V : Type*} [NormedAddCommGroup V] [InnerProductSpace ℂ V] {n : ℕ} 118 | {e : Fin n → V} (he : Orthonormal ℂ e) (v : V) : 119 | ‖v‖^2 = ∑ i : Fin n, ‖⟪v, e i⟫_ℂ‖^2 ↔ v ∈ Submodule.span ℂ (e '' Set.univ) := 120 | sorry 121 | 122 | theorem exercise_6_16 {K V : Type*} [RCLike K] [NormedAddCommGroup V] [InnerProductSpace K V] 123 | {U : Submodule K V} : 124 | U.orthogonal = ⊥ ↔ U = ⊤ := 125 | sorry 126 | 127 | theorem exercise_7_5 {V : Type*} [NormedAddCommGroup V] [InnerProductSpace ℂ V] 128 | [FiniteDimensional ℂ V] (hV : finrank V ≥ 2) : 129 | ∀ U : Submodule ℂ (End ℂ V), U.carrier ≠ 130 | {T | T * adjoint T = adjoint T * T} := 131 | sorry 132 | 133 | theorem exercise_7_6 {V : Type*} [NormedAddCommGroup V] [InnerProductSpace ℂ V] 134 | [FiniteDimensional ℂ V] (T : End ℂ V) 135 | (hT : T * adjoint T = adjoint T * T) : 136 | range T = range (adjoint T) := 137 | sorry 138 | 139 | theorem exercise_7_9 {V : Type*} [NormedAddCommGroup V] [InnerProductSpace ℂ V] 140 | [FiniteDimensional ℂ V] (T : End ℂ V) 141 | (hT : T * adjoint T = adjoint T * T) : 142 | IsSelfAdjoint T ↔ ∀ e : T.Eigenvalues, (e : ℂ).im = 0 := 143 | sorry 144 | 145 | theorem exercise_7_10 {V : Type*} [NormedAddCommGroup V] [InnerProductSpace ℂ V] 146 | [FiniteDimensional ℂ V] (T : End ℂ V) 147 | (hT : T * adjoint T = adjoint T * T) (hT1 : T^9 = T^8) : 148 | IsSelfAdjoint T ∧ T^2 = T := 149 | sorry 150 | 151 | theorem exercise_7_11 {V : Type*} [NormedAddCommGroup V] [InnerProductSpace ℂ V] 152 | [FiniteDimensional ℂ V] {T : End ℂ V} (hT : T*adjoint T = adjoint T*T) : 153 | ∃ (S : End ℂ V), S ^ 2 = T := 154 | sorry 155 | 156 | theorem exercise_7_14 {𝕜 V : Type*} [RCLike 𝕜] [NormedAddCommGroup V] 157 | [InnerProductSpace 𝕜 V] [FiniteDimensional 𝕜 V] 158 | {T : End 𝕜 V} (hT : IsSelfAdjoint T) 159 | {l : 𝕜} {ε : ℝ} (he : ε > 0) : ∃ v : V, ‖v‖= 1 ∧ (‖T v - l • v‖ < ε → 160 | (∃ l' : T.Eigenvalues, ‖l - l'‖ < ε)) := 161 | sorry 162 | -------------------------------------------------------------------------------- /formal/Cambridge-Tripos.lean: -------------------------------------------------------------------------------- 1 | import Mathlib 2 | 3 | open Filter Set 4 | 5 | theorem exercise_2022_IA_4_I_1E_a : ∀ N : ℕ, ∃ n ≥ N, (3*n+1).Prime ∧ (3*n+1) ≥ N := 6 | sorry 7 | 8 | theorem exercise_2022_IA_4_I_2D_a : Irrational (2^((1:ℝ)/3) + 3^((1:ℝ)/3)) := 9 | sorry 10 | 11 | theorem exercise_2022_IB_3_II_13G_a_i (U : Set ℂ) (hU : IsOpen U) 12 | (hU1 : Nonempty U) (hU2 : IsConnected U) (f : ℕ → ℂ → ℂ) (f' : ℂ → ℂ) 13 | (hf : ∀ n : ℕ, DifferentiableOn ℂ (f n) U) 14 | (hf1 : ∀ X ⊂ U, CompactSpace X → 15 | (TendstoUniformly (λ n => restrict X (f n)) (restrict X f') atTop)) : 16 | DifferentiableOn ℂ f' U := 17 | sorry 18 | -------------------------------------------------------------------------------- /formal/Dummit-Foote.lean: -------------------------------------------------------------------------------- 1 | import Mathlib 2 | 3 | open Fintype Subgroup Set Polynomial Ideal 4 | open scoped BigOperators 5 | noncomputable section 6 | 7 | theorem exercise_1_1_2a : ∃ a b : ℤ, a - b ≠ b - a := 8 | sorry 9 | 10 | theorem exercise_1_1_3 (n : ℤ) : 11 | ∀ (a b c : ℤ), (a+b)+c ≡ a+(b+c) [ZMOD n] := 12 | sorry 13 | 14 | theorem exercise_1_1_4 (n : ℕ) : 15 | ∀ (a b c : ℕ), (a * b) * c ≡ a * (b * c) [ZMOD n] := 16 | sorry 17 | 18 | theorem exercise_1_1_5 (n : ℕ) (hn : 1 < n) : 19 | IsEmpty (Group (ZMod n)) := 20 | sorry 21 | 22 | theorem exercise_1_1_15 {G : Type*} [Group G] (as : List G) : 23 | as.prod⁻¹ = (as.reverse.map (λ x => x⁻¹)).prod := 24 | sorry 25 | 26 | theorem exercise_1_1_16 {G : Type*} [Group G] 27 | (x : G) (hx : x ^ 2 = 1) : 28 | orderOf x = 1 ∨ orderOf x = 2 := 29 | sorry 30 | 31 | theorem exercise_1_1_17 {G : Type*} [Group G] {x : G} {n : ℕ} 32 | (hxn: orderOf x = n) : 33 | x⁻¹ = x ^ (n - 1 : ℤ) := 34 | sorry 35 | 36 | theorem exercise_1_1_18 {G : Type*} [Group G] 37 | (x y : G) : (x * y = y * x ↔ y⁻¹ * x * y = x) ↔ (x⁻¹ * y⁻¹ * x * y = 1) := 38 | sorry 39 | 40 | theorem exercise_1_1_20 {G : Type*} [Group G] {x : G} : 41 | orderOf x = orderOf x⁻¹ := 42 | sorry 43 | 44 | theorem exercise_1_1_22a {G : Type*} [Group G] (x g : G) : 45 | orderOf x = orderOf (g⁻¹ * x * g) := 46 | sorry 47 | 48 | theorem exercise_1_1_22b {G: Type*} [Group G] (a b : G) : 49 | orderOf (a * b) = orderOf (b * a) := 50 | sorry 51 | 52 | theorem exercise_1_1_25 {G : Type*} [Group G] 53 | (h : ∀ x : G, x ^ 2 = 1) : ∀ a b : G, a*b = b*a := 54 | sorry 55 | 56 | theorem exercise_1_1_29 {A B : Type*} [Group A] [Group B] : 57 | ∀ x y : A × B, x*y = y*x ↔ (∀ x y : A, x*y = y*x) ∧ 58 | (∀ x y : B, x*y = y*x) := 59 | sorry 60 | 61 | theorem exercise_1_1_34 {G : Type*} [Group G] {x : G} 62 | (hx_inf : orderOf x = 0) (n m : ℤ) : 63 | x ^ n ≠ x ^ m := 64 | sorry 65 | 66 | theorem exercise_1_3_8 : Infinite (Equiv.Perm ℕ) := 67 | sorry 68 | 69 | theorem exercise_1_6_4 : 70 | IsEmpty (Multiplicative ℝ ≃* Multiplicative ℂ) := 71 | sorry 72 | 73 | instance exercise_1_6_11 {A B : Type*} [Group A] [Group B] : 74 | A × B ≃* B × A := 75 | sorry 76 | 77 | theorem exercise_1_6_17 {G : Type*} [Group G] (f : G → G) 78 | (hf : f = λ g => g⁻¹) : 79 | ∀ x y : G, f x * f y = f (x*y) ↔ ∀ x y : G, x*y = y*x := 80 | sorry 81 | 82 | theorem exercise_1_6_23 {G : Type*} 83 | [Group G] (σ : MulAut G) (hs : ∀ g : G, σ g = 1 → g = 1) 84 | (hs2 : ∀ g : G, σ (σ g) = g) : 85 | ∀ x y : G, x*y = y*x := 86 | sorry 87 | 88 | theorem exercise_2_1_5 {G : Type*} [Group G] [Fintype G] 89 | (hG : card G > 2) (H : Subgroup G) [Fintype H] : 90 | card H ≠ card G - 1 := 91 | sorry 92 | 93 | theorem exercise_2_1_13 (H : AddSubgroup ℚ) {x : ℚ} 94 | (hH : x ∈ H → (1 / x) ∈ H): 95 | H = ⊥ ∨ H = ⊤ := 96 | sorry 97 | 98 | theorem exercise_2_4_4 {G : Type*} [Group G] (H : Subgroup G) : 99 | closure ((H : Set G) \ {1}) = ⊤ := 100 | sorry 101 | 102 | theorem exercise_2_4_16a {G : Type*} [Group G] {H : Subgroup G} 103 | (hH : H ≠ ⊤) : 104 | ∃ M : Subgroup G, M ≠ ⊤ ∧ 105 | ∀ K : Subgroup G, M ≤ K → K = M ∨ K = ⊤ ∧ 106 | H ≤ M := 107 | sorry 108 | 109 | theorem exercise_2_4_16b {n : ℕ} {hn : n ≠ 0} 110 | {R : Subgroup (DihedralGroup n)} 111 | (hR : R = Subgroup.closure {DihedralGroup.r 1}) : 112 | R ≠ ⊤ ∧ 113 | ∀ K : Subgroup (DihedralGroup n), R ≤ K → K = R ∨ K = ⊤ := 114 | sorry 115 | 116 | theorem exercise_2_4_16c {n : ℕ} (H : AddSubgroup (ZMod n)) : 117 | ∃ p : (ZMod n), Prime p ∧ H = AddSubgroup.closure {p} ↔ 118 | (H ≠ ⊤ ∧ ∀ K : AddSubgroup (ZMod n), H ≤ K → K = H ∨ K = ⊤) := 119 | sorry 120 | 121 | theorem exercise_3_1_3a {A : Type*} [CommGroup A] (B : Subgroup A) : 122 | ∀ a b : A ⧸ B, a*b = b*a := 123 | sorry 124 | 125 | theorem exercise_3_1_22a (G : Type*) [Group G] (H K : Subgroup G) 126 | [Normal H] [Normal K] : 127 | Normal (H ⊓ K) := 128 | sorry 129 | 130 | theorem exercise_3_1_22b {G : Type*} [Group G] (I : Type*) 131 | (H : I → Subgroup G) (hH : ∀ i : I, Normal (H i)) : 132 | Normal (⨅ (i : I), H i):= 133 | sorry 134 | 135 | theorem exercise_3_2_8 {G : Type*} [Group G] (H K : Subgroup G) 136 | [Fintype H] [Fintype K] 137 | (hHK : Nat.Coprime (card H) (card K)) : 138 | H ⊓ K = ⊥ := 139 | sorry 140 | 141 | theorem exercise_3_2_11 {G : Type*} [Group G] {H K : Subgroup G} 142 | (hHK : H ≤ K) : 143 | H.index = K.index * H.relindex K := 144 | sorry 145 | 146 | theorem exercise_3_2_16 (p : ℕ) (hp : Nat.Prime p) (a : ℕ) : 147 | Nat.Coprime a p → a ^ p ≡ a [ZMOD p] := 148 | sorry 149 | 150 | theorem exercise_3_2_21a (H : AddSubgroup ℚ) (hH : H ≠ ⊤) : H.index = 0 := 151 | sorry 152 | 153 | theorem exercise_3_3_3 {p : Nat.Primes} {G : Type*} [Group G] 154 | {H : Subgroup G} [hH : H.Normal] (hH1 : H.index = p) : 155 | ∀ K : Subgroup G, K ≤ H ∨ H ⊔ K = ⊤ ∨ (K ⊓ H).relindex K = p := 156 | sorry 157 | 158 | theorem exercise_3_4_1 (G : Type*) [CommGroup G] [IsSimpleGroup G] : 159 | IsCyclic G ∧ ∃ G_fin : Fintype G, Nat.Prime (@card G G_fin) := 160 | sorry 161 | 162 | theorem exercise_3_4_4 {G : Type*} [CommGroup G] [Fintype G] {n : ℕ} 163 | (hn : n ∣ (card G)) : 164 | ∃ (H : Subgroup G) (H_fin : Fintype H), @card H H_fin = n := 165 | sorry 166 | 167 | theorem exercise_3_4_5a {G : Type*} [Group G] 168 | (H : Subgroup G) [IsSolvable G] : IsSolvable H := 169 | sorry 170 | 171 | theorem exercise_3_4_5b {G : Type*} [Group G] [IsSolvable G] 172 | (H : Subgroup G) [Normal H] : 173 | IsSolvable (G ⧸ H) := 174 | sorry 175 | 176 | theorem exercise_3_4_11 {G : Type*} [Group G] [IsSolvable G] 177 | {H : Subgroup G} (hH : H ≠ ⊥) [H.Normal] : 178 | ∃ A ≤ H, A.Normal ∧ ∀ a b : A, a*b = b*a := 179 | sorry 180 | 181 | theorem exercise_4_2_8 {G : Type*} [Group G] {H : Subgroup G} 182 | {n : ℕ} (hn : n > 0) (hH : H.index = n) : 183 | ∃ K ≤ H, K.Normal ∧ K.index ≤ n.factorial := 184 | sorry 185 | 186 | theorem exercise_4_3_26 {α : Type*} [Fintype α] (ha : card α > 1) 187 | (h_tran : ∀ a b: α, ∃ σ : Equiv.Perm α, σ a = b) : 188 | ∃ σ : Equiv.Perm α, ∀ a : α, σ a ≠ a := 189 | sorry 190 | 191 | theorem exercise_4_2_9a {G : Type*} [Fintype G] [Group G] {p α : ℕ} 192 | (hp : p.Prime) (ha : α > 0) (hG : card G = p ^ α) : 193 | ∀ H : Subgroup G, H.index = p → H.Normal := 194 | sorry 195 | 196 | theorem exercise_4_2_14 {G : Type*} [Fintype G] [Group G] 197 | (hG : ¬ (card G).Prime) (hG1 : ∀ k : ℕ, k ∣ card G → 198 | ∃ (H : Subgroup G) (fH : Fintype H), @card H fH = k) : 199 | ¬ IsSimpleGroup G := 200 | sorry 201 | 202 | theorem exercise_4_4_2 {G : Type*} [Fintype G] [Group G] 203 | {p q : Nat.Primes} (hpq : p ≠ q) (hG : card G = p*q) : 204 | IsCyclic G := 205 | sorry 206 | 207 | theorem exercise_4_4_6a {G : Type*} [Group G] (H : Subgroup G) 208 | [Characteristic H] : Normal H := 209 | sorry 210 | 211 | theorem exercise_4_4_6b : 212 | ∃ (G : Type*) (hG : Group G) (H : @Subgroup G hG), @Characteristic G hG H ∧ ¬ @Normal G hG H := 213 | sorry 214 | 215 | theorem exercise_4_4_7 {G : Type*} [Group G] {H : Subgroup G} [Fintype H] 216 | (hH : ∀ (K : Subgroup G) (fK : Fintype K), card H = @card K fK → H = K) : 217 | H.Characteristic := 218 | sorry 219 | 220 | theorem exercise_4_4_8a {G : Type*} [Group G] (H K : Subgroup G) 221 | (hHK : H ≤ K) [hHK1 : (H.subgroupOf K).Normal] [hK : K.Normal] : 222 | H.Normal := 223 | sorry 224 | 225 | theorem exercise_4_5_1a {p : ℕ} {G : Type*} [Group G] 226 | {P : Subgroup G} (hP : IsPGroup p P) (H : Subgroup G) 227 | (hH : P ≤ H) : IsPGroup p H := 228 | sorry 229 | 230 | theorem exercise_4_5_13 {G : Type*} [Group G] [Fintype G] 231 | (hG : card G = 56) : 232 | ∃ (p : ℕ) (P : Sylow p G), P.Normal := 233 | sorry 234 | 235 | theorem exercise_4_5_14 {G : Type*} [Group G] [Fintype G] 236 | (hG : card G = 312) : 237 | ∃ (p : ℕ) (P : Sylow p G), P.Normal := 238 | sorry 239 | 240 | theorem exercise_4_5_15 {G : Type*} [Group G] [Fintype G] 241 | (hG : card G = 351) : 242 | ∃ (p : ℕ) (P : Sylow p G), P.Normal := 243 | sorry 244 | 245 | theorem exercise_4_5_16 {p q r : ℕ} {G : Type*} [Group G] 246 | [Fintype G] (hpqr : p < q ∧ q < r) 247 | (hpqr1 : p.Prime ∧ q.Prime ∧ r.Prime)(hG : card G = p*q*r) : 248 | Nonempty (Sylow p G) ∨ Nonempty (Sylow q G) ∨ Nonempty (Sylow r G) := 249 | sorry 250 | 251 | theorem exercise_4_5_17 {G : Type*} [Fintype G] [Group G] 252 | (hG : card G = 105) : 253 | Nonempty (Sylow 5 G) ∧ Nonempty (Sylow 7 G) := 254 | sorry 255 | 256 | theorem exercise_4_5_18 {G : Type*} [Fintype G] [Group G] 257 | (hG : card G = 200) : 258 | ∃ N : Sylow 5 G, N.Normal := 259 | sorry 260 | 261 | theorem exercise_4_5_19 {G : Type*} [Fintype G] [Group G] 262 | (hG : card G = 6545) : ¬ IsSimpleGroup G := 263 | sorry 264 | 265 | theorem exercise_4_5_20 {G : Type*} [Fintype G] [Group G] 266 | (hG : card G = 1365) : ¬ IsSimpleGroup G := 267 | sorry 268 | 269 | theorem exercise_4_5_21 {G : Type*} [Fintype G] [Group G] 270 | (hG : card G = 2907) : ¬ IsSimpleGroup G := 271 | sorry 272 | 273 | theorem exercise_4_5_22 {G : Type*} [Fintype G] [Group G] 274 | (hG : card G = 132) : ¬ IsSimpleGroup G := 275 | sorry 276 | 277 | theorem exercise_4_5_23 {G : Type*} [Fintype G] [Group G] 278 | (hG : card G = 462) : ¬ IsSimpleGroup G := 279 | sorry 280 | 281 | instance exercise_4_5_28 {G : Type*} [Group G] [Fintype G] 282 | (hG : card G = 105) {P : Sylow 3 G} (hP : P.Normal) : 283 | CommGroup G := 284 | sorry 285 | 286 | theorem exercise_4_5_33 {G : Type*} [Group G] [Fintype G] {p : ℕ} 287 | (P : Sylow p G) [hP : P.Normal] (H : Subgroup G) [Fintype H] : 288 | ∀ R : Sylow p H, R.toSubgroup = (H ⊓ P.toSubgroup).subgroupOf H ∧ 289 | Nonempty (Sylow p H) := 290 | sorry 291 | 292 | theorem exercise_5_4_2 {G : Type*} [Group G] (H : Subgroup G) : 293 | H.Normal ↔ ⁅(⊤ : Subgroup G), H⁆ ≤ H := 294 | sorry 295 | 296 | theorem exercise_7_1_2 {R : Type*} [Ring R] {u : R} 297 | (hu : IsUnit u) : IsUnit (-u) := 298 | sorry 299 | 300 | theorem exercise_7_1_11 {R : Type*} [Ring R] [IsDomain R] 301 | {x : R} (hx : x^2 = 1) : x = 1 ∨ x = -1 := 302 | sorry 303 | 304 | theorem exercise_7_1_12 {F : Type*} [Field F] {K : Subring F} 305 | (hK : (1 : F) ∈ K) : IsDomain K := 306 | sorry 307 | 308 | instance exercise_7_1_15 {R : Type*} [Ring R] (hR : ∀ a : R, a^2 = a) : 309 | CommRing R := 310 | sorry 311 | 312 | theorem exercise_7_2_2 {R : Type*} [Ring R] (p : Polynomial R) : 313 | p ∣ 0 ↔ ∃ b : R, b ≠ 0 ∧ b • p = 0 := 314 | sorry 315 | 316 | theorem exercise_7_2_12 {R G : Type*} [Ring R] [Group G] [Fintype G] : 317 | ∑ g : G, MonoidAlgebra.of R G g ∈ center (MonoidAlgebra R G) := 318 | sorry 319 | 320 | theorem exercise_7_3_16 {R S : Type*} [Ring R] [Ring S] 321 | {φ : R →+* S} (hf : Function.Surjective φ) : 322 | φ '' (center R) ⊂ center S := 323 | sorry 324 | 325 | theorem exercise_7_3_37 {p m : ℕ} (hp : p.Prime) : 326 | IsNilpotent (span ({↑p} : Set $ ZMod $ p^m) : Ideal $ ZMod $ p^m) := 327 | sorry 328 | 329 | theorem exercise_7_4_27 {R : Type*} [CommRing R] (hR : (0 : R) ≠ 1) 330 | {a : R} (ha : IsNilpotent a) (b : R) : 331 | IsUnit (1-a*b) := 332 | sorry 333 | 334 | theorem exercise_8_1_12 {N : ℕ} (hN : N > 0) {M M': ℤ} {d : ℕ} 335 | (hMN : M.gcd N = 1) (hMd : d.gcd N.totient = 1) 336 | (hM' : M' ≡ M^d [ZMOD N]) : 337 | ∃ d' : ℕ, d' * d ≡ 1 [ZMOD N.totient] ∧ 338 | M ≡ M'^d' [ZMOD N] := 339 | sorry 340 | 341 | theorem exercise_8_2_4 {R : Type*} [Ring R][NoZeroDivisors R] 342 | [CancelCommMonoidWithZero R] [GCDMonoid R] 343 | (h1 : ∀ a b : R, a ≠ 0 → b ≠ 0 → ∃ r s : R, gcd a b = r*a + s*b) 344 | (h2 : ∀ a : ℕ → R, (∀ i j : ℕ, i < j → a i ∣ a j) → 345 | ∃ N : ℕ, ∀ n ≥ N, ∃ u : R, IsUnit u ∧ a n = u * a N) : 346 | IsPrincipalIdealRing R := 347 | sorry 348 | 349 | theorem exercise_8_3_4 {R : Type*} {n : ℤ} {r s : ℚ} 350 | (h : r^2 + s^2 = n) : 351 | ∃ a b : ℤ, a^2 + b^2 = n := 352 | sorry 353 | 354 | theorem exercise_8_3_5a {n : ℤ} (hn0 : n > 3) (hn1 : Squarefree n) : 355 | Irreducible (2 : Zsqrtd $ -n) ∧ 356 | Irreducible (⟨0, 1⟩ : Zsqrtd $ -n) ∧ 357 | Irreducible (1 + ⟨0, 1⟩ : Zsqrtd $ -n) := 358 | sorry 359 | 360 | theorem exercise_8_3_6a {R : Type} [Ring R] 361 | (hR : R = (GaussianInt ⧸ span ({⟨0, 1⟩} : Set GaussianInt))) : 362 | IsField R ∧ ∃ finR : Fintype R, @card R finR = 2 := 363 | sorry 364 | 365 | theorem exercise_8_3_6b {q : ℕ} (hq0 : q.Prime) 366 | (hq1 : q ≡ 3 [ZMOD 4]) {R : Type} [Ring R] 367 | (hR : R = (GaussianInt ⧸ span ({↑q} : Set GaussianInt))) : 368 | IsField R ∧ ∃ finR : Fintype R, @card R finR = q^2 := 369 | sorry 370 | 371 | theorem exercise_9_1_6 : ¬ Submodule.IsPrincipal 372 | (span ({MvPolynomial.X 0, MvPolynomial.X 1} : Set (MvPolynomial (Fin 2) ℚ))) := 373 | sorry 374 | 375 | theorem exercise_9_1_10 {f : ℕ → MvPolynomial ℕ ℤ} 376 | (hf : f = λ i => MvPolynomial.X i * MvPolynomial.X (i+1)): 377 | Infinite (minimalPrimes (MvPolynomial ℕ ℤ ⧸ span (range f))) := 378 | sorry 379 | 380 | theorem exercise_9_3_2 {f g : Polynomial ℚ} (i j : ℕ) 381 | (hfg : ∀ n : ℕ, ∃ a : ℤ, (f*g).coeff = a) : 382 | ∃ a : ℤ, f.coeff i * g.coeff j = a := 383 | sorry 384 | 385 | theorem exercise_9_4_2a : Irreducible (X^4 - 4*X^3 + 6 : Polynomial ℤ) := 386 | sorry 387 | 388 | theorem exercise_9_4_2b : Irreducible 389 | (X^6 + 30*X^5 - 15*X^3 + 6*X - 120 : Polynomial ℤ) := 390 | sorry 391 | 392 | theorem exercise_9_4_2c : Irreducible 393 | (X^4 + 4*X^3 + 6*X^2 + 2*X + 1 : Polynomial ℤ) := 394 | sorry 395 | 396 | theorem exercise_9_4_2d {p : ℕ} (hp : p.Prime ∧ p > 2) 397 | {f : Polynomial ℤ} (hf : f = (X + 2)^p): 398 | Irreducible (∑ n ∈ (f.support \ {0}), (f.coeff n : Polynomial ℤ) * X ^ (n-1) : 399 | Polynomial ℤ) := 400 | sorry 401 | 402 | theorem exercise_9_4_9 : 403 | Irreducible (X^2 - C Zsqrtd.sqrtd : Polynomial (Zsqrtd 2)) := 404 | sorry 405 | 406 | theorem exercise_9_4_11 : 407 | Irreducible ((MvPolynomial.X 0)^2 + (MvPolynomial.X 1)^2 - 1 : MvPolynomial (Fin 2) ℚ) := 408 | sorry 409 | 410 | instance exercise_11_1_13 {ι : Type*} [Fintype ι] : 411 | (ι → ℝ) ≃ₗ[ℚ] ℝ := 412 | sorry 413 | -------------------------------------------------------------------------------- /formal/Herstein.lean: -------------------------------------------------------------------------------- 1 | import Mathlib 2 | 3 | open Fintype Set Real Ideal Polynomial 4 | open scoped BigOperators 5 | noncomputable section 6 | 7 | theorem exercise_2_1_18 {G : Type*} [Group G] 8 | [Fintype G] (hG2 : Even (card G)) : 9 | ∃ (a : G), a ≠ 1 ∧ a = a⁻¹ := 10 | sorry 11 | 12 | instance exercise_2_1_21 (G : Type*) [Group G] [Fintype G] 13 | (hG : card G = 5) : 14 | CommGroup G := 15 | sorry 16 | 17 | theorem exercise_2_1_26 {G : Type*} [Group G] 18 | [Fintype G] (a : G) : ∃ (n : ℕ), a ^ n = 1 := 19 | sorry 20 | 21 | theorem exercise_2_1_27 {G : Type*} [Group G] 22 | [Fintype G] : ∃ (m : ℕ), ∀ (a : G), a ^ m = 1 := 23 | sorry 24 | 25 | instance exercise_2_2_3 {G : Type*} [Group G] 26 | {P : ℕ → Prop} {hP : P = λ i => ∀ a b : G, (a*b)^i = a^i * b^i} 27 | (hP1 : ∃ n : ℕ, P n ∧ P (n+1) ∧ P (n+2)) : CommGroup G := 28 | sorry 29 | 30 | instance exercise_2_2_5 {G : Type*} [Group G] 31 | (h : ∀ (a b : G), (a * b) ^ 3 = a ^ 3 * b ^ 3 ∧ (a * b) ^ 5 = a ^ 5 * b ^ 5) : 32 | CommGroup G := 33 | sorry 34 | 35 | theorem exercise_2_2_6c {G : Type*} [Group G] {n : ℕ} (hn : n > 1) 36 | (h : ∀ (a b : G), (a * b) ^ n = a ^ n * b ^ n) : 37 | ∀ (a b : G), (a * b * a⁻¹ * b⁻¹) ^ (n * (n - 1)) = 1 := 38 | sorry 39 | 40 | theorem exercise_2_3_17 {G : Type*} [Mul G] [Group G] (a x : G) : 41 | centralizer {x⁻¹*a*x} = 42 | (λ g : G => x⁻¹*g*x) '' (centralizer {a}) := 43 | sorry 44 | 45 | theorem exercise_2_3_16 {G : Type*} [Group G] 46 | (hG : ∀ H : Subgroup G, H = ⊤ ∨ H = ⊥) : 47 | IsCyclic G ∧ ∃ (p : ℕ) (Fin : Fintype G), Nat.Prime p ∧ @card G Fin = p := 48 | sorry 49 | 50 | theorem exercise_2_4_36 {a n : ℕ} (h : a > 1) : 51 | n ∣ (a ^ n - 1).totient := 52 | sorry 53 | 54 | theorem exercise_2_5_23 {G : Type*} [Group G] 55 | (hG : ∀ (H : Subgroup G), H.Normal) (a b : G) : 56 | ∃ (j : ℤ) , b*a = a^j * b:= 57 | sorry 58 | 59 | theorem exercise_2_5_30 {G : Type*} [Group G] [Fintype G] 60 | {p m : ℕ} (hp : Nat.Prime p) (hp1 : ¬ p ∣ m) (hG : card G = p*m) 61 | {H : Subgroup G} [Fintype H] [H.Normal] (hH : card H = p): 62 | Subgroup.Characteristic H := 63 | sorry 64 | 65 | theorem exercise_2_5_31 {G : Type*} [CommGroup G] [Fintype G] 66 | {p m n : ℕ} (hp : Nat.Prime p) (hp1 : ¬ p ∣ m) (hG : card G = p^n*m) 67 | {H : Subgroup G} [Fintype H] (hH : card H = p^n) : 68 | Subgroup.Characteristic H := 69 | sorry 70 | 71 | instance exercise_2_5_37 (G : Type*) [Group G] [Fintype G] 72 | (hG : card G = 6) (hG' : IsEmpty (CommGroup G)) : 73 | G ≃* Equiv.Perm (Fin 3) := 74 | sorry 75 | 76 | instance exercise_2_5_43 (G : Type*) [Group G] [Fintype G] 77 | (hG : card G = 9) : 78 | CommGroup G := 79 | sorry 80 | 81 | theorem exercise_2_5_44 {G : Type*} [Group G] [Fintype G] {p : ℕ} 82 | (hp : Nat.Prime p) (hG : card G = p^2) : 83 | ∃ (N : Subgroup G) (Fin : Fintype N), @card N Fin = p ∧ N.Normal := 84 | sorry 85 | 86 | theorem exercise_2_5_52 {G : Type*} [Group G] [Fintype G] 87 | (φ : G ≃* G) {I : Finset G} (hI : ∀ x ∈ I, φ x = x⁻¹) 88 | (hI1 : (0.75 : ℚ) * card G ≤ card I) : 89 | ∀ x : G, φ x = x⁻¹ ∧ ∀ x y : G, x*y = y*x := 90 | sorry 91 | 92 | theorem exercise_2_6_15 {G : Type*} [CommGroup G] {m n : ℕ} 93 | (hm : ∃ (g : G), orderOf g = m) 94 | (hn : ∃ (g : G), orderOf g = n) 95 | (hmn : m.Coprime n) : 96 | ∃ (g : G), orderOf g = m * n := 97 | sorry 98 | 99 | theorem exercise_2_7_7 {G : Type*} [Group G] {G' : Type*} [Group G'] 100 | (φ : G →* G') (N : Subgroup G) [N.Normal] : 101 | (Subgroup.map φ N).Normal := 102 | sorry 103 | 104 | instance exercise_2_8_12 {G H : Type*} [Fintype G] [Fintype H] 105 | [Group G] [Group H] (hG : card G = 21) (hH : card H = 21) 106 | (hG1 : IsEmpty (CommGroup G)) (hH1 : IsEmpty (CommGroup H)) : 107 | G ≃* H := 108 | sorry 109 | 110 | instance exercise_2_8_15 {G H: Type*} [Fintype G] [Group G] [Fintype H] 111 | [Group H] {p q : ℕ} (hp : Nat.Prime p) (hq : Nat.Prime q) 112 | (h : p > q) (h1 : q ∣ p - 1) (hG : card G = p*q) (hH : card G = p*q) : 113 | G ≃* H := 114 | sorry 115 | 116 | theorem exercise_2_9_2 {G H : Type*} [Fintype G] [Fintype H] [Group G] 117 | [Group H] (hG : IsCyclic G) (hH : IsCyclic H) : 118 | IsCyclic (G × H) ↔ (card G).Coprime (card H) := 119 | sorry 120 | 121 | theorem exercise_2_10_1 {G : Type*} [Group G] (A : Subgroup G) 122 | [A.Normal] {b : G} (hp : Nat.Prime (orderOf b)) : 123 | A ⊓ (Subgroup.closure {b}) = ⊥ := 124 | sorry 125 | 126 | theorem exercise_2_11_6 {G : Type*} [Group G] {p : ℕ} (hp : Nat.Prime p) 127 | {P : Sylow p G} (hP : P.Normal) : 128 | ∀ (Q : Sylow p G), P = Q := 129 | sorry 130 | 131 | theorem exercise_2_11_7 {G : Type*} [Group G] {p : ℕ} (hp : Nat.Prime p) 132 | {P : Sylow p G} (hP : P.Normal) : 133 | Subgroup.Characteristic (P : Subgroup G) := 134 | sorry 135 | 136 | theorem exercise_2_11_22 {p : ℕ} {n : ℕ} {G : Type*} [Fintype G] 137 | [Group G] (hp : Nat.Prime p) (hG : card G = p ^ n) {K : Subgroup G} 138 | [Fintype K] (hK : card K = p ^ (n-1)) : 139 | K.Normal := 140 | sorry 141 | 142 | theorem exercise_3_2_21 {α : Type*} [Fintype α] {σ τ: Equiv.Perm α} 143 | (h1 : ∀ a : α, σ a = a ↔ τ a ≠ a) (h2 : τ ∘ σ = id) : 144 | σ = 1 ∧ τ = 1 := 145 | sorry 146 | 147 | theorem exercise_4_1_19 : Infinite {x : Quaternion ℝ | x^2 = -1} := 148 | sorry 149 | 150 | instance exercise_4_1_34 : Equiv.Perm (Fin 3) ≃* Matrix.GeneralLinearGroup (Fin 2) (ZMod 2) := 151 | sorry 152 | 153 | instance exercise_4_2_5 {R : Type*} [Ring R] 154 | (h : ∀ x : R, x ^ 3 = x) : CommRing R := 155 | sorry 156 | 157 | theorem exercise_4_2_6 {R : Type*} [Ring R] (a x : R) 158 | (h : a ^ 2 = 0) : a * (a * x + x * a) = (x + x * a) * a := 159 | sorry 160 | 161 | theorem exercise_4_2_9 {p : ℕ} (hp : Nat.Prime p) (hp1 : Odd p) : 162 | ∃ (a b : ℤ), (a / b : ℚ) = ∑ i ∈ Finset.range p, 1 / (i + 1) → ↑p ∣ a := 163 | sorry 164 | 165 | theorem exercise_4_3_1 {R : Type*} [CommRing R] (a : R) : 166 | ∃ I : Ideal R, {x : R | x*a=0} = I := 167 | sorry 168 | 169 | theorem exercise_4_3_25 (I : Ideal (Matrix (Fin 2) (Fin 2) ℝ)) : 170 | I = ⊥ ∨ I = ⊤ := 171 | sorry 172 | 173 | theorem exercise_4_4_9 (p : ℕ) (hp : Nat.Prime p) : 174 | (∃ S : Finset (ZMod p), S.card = (p-1)/2 ∧ ∃ x : ZMod p, x^2 = p) ∧ 175 | (∃ S : Finset (ZMod p), S.card = (p-1)/2 ∧ ¬ ∃ x : ZMod p, x^2 = p) := 176 | sorry 177 | 178 | theorem exercise_4_5_16 {p n: ℕ} (hp : Nat.Prime p) 179 | {q : Polynomial (ZMod p)} (hq : Irreducible q) (hn : q.degree = n) : 180 | ∃ is_fin : Fintype $ Polynomial (ZMod p) ⧸ span ({q} : Set (Polynomial $ ZMod p)), 181 | @card (Polynomial (ZMod p) ⧸ span {q}) is_fin = p ^ n ∧ 182 | IsField (Polynomial $ ZMod p):= 183 | sorry 184 | 185 | theorem exercise_4_5_23 {p q: Polynomial (ZMod 7)} 186 | (hp : p = X^3 - 2) (hq : q = X^3 + 2) : 187 | Irreducible p ∧ Irreducible q ∧ 188 | (Nonempty $ Polynomial (ZMod 7) ⧸ span ({p} : Set $ Polynomial $ ZMod 7) ≃+* 189 | Polynomial (ZMod 7) ⧸ span ({q} : Set $ Polynomial $ ZMod 7)) := 190 | sorry 191 | 192 | theorem exercise_4_5_25 {p : ℕ} (hp : Nat.Prime p) : 193 | Irreducible (∑ i : Finset.range p, X ^ p : Polynomial ℚ) := 194 | sorry 195 | 196 | theorem exercise_4_6_2 : Irreducible (X^3 + 3*X + 2 : Polynomial ℚ) := 197 | sorry 198 | 199 | theorem exercise_4_6_3 : 200 | Infinite {a : ℤ | Irreducible (X^7 + 15*X^2 - 30*X + (a : Polynomial ℚ) : Polynomial ℚ)} := 201 | sorry 202 | 203 | theorem exercise_5_1_8 {p m n: ℕ} {F : Type*} [Field F] 204 | (hp : Nat.Prime p) (hF : CharP F p) (a b : F) (hm : m = p ^ n) : 205 | (a + b) ^ m = a^m + b^m := 206 | sorry 207 | 208 | theorem exercise_5_2_20 {F V ι: Type*} [Infinite F] [Field F] 209 | [AddCommGroup V] [Module F V] {u : ι → Submodule F V} 210 | (hu : ∀ i : ι, u i ≠ ⊤) : 211 | (⋃ i : ι, (u i : Set V)) ≠ ⊤ := 212 | sorry 213 | 214 | theorem exercise_5_3_7 {K : Type*} [Field K] {F : Subfield K} 215 | {a : K} (ha : IsAlgebraic F (a ^ 2)) : IsAlgebraic F a := 216 | sorry 217 | 218 | theorem exercise_5_3_10 : IsAlgebraic ℚ (cos (Real.pi / 180)) := 219 | sorry 220 | 221 | theorem exercise_5_4_3 {a : ℂ} {p : ℂ → ℂ} 222 | (hp : p = λ (x : ℂ) => x^5 + sqrt 2 * x^3 + sqrt 5 * x^2 + sqrt 7 * x + 11) 223 | (ha : p a = 0) : 224 | ∃ p : Polynomial ℂ , p.degree < 80 ∧ a ∈ p.roots ∧ 225 | ∀ n : p.support, ∃ a b : ℤ, p.coeff n = a / b := 226 | sorry 227 | 228 | theorem exercise_5_5_2 : Irreducible (X^3 - 3*X - 1 : Polynomial ℚ) := 229 | sorry 230 | 231 | theorem exercise_5_6_14 {p m n: ℕ} (hp : Nat.Prime p) {F : Type*} 232 | [Field F] [CharP F p] (hm : m = p ^ n) : 233 | card (rootSet (X ^ m - X : Polynomial F) F) = m := 234 | sorry 235 | -------------------------------------------------------------------------------- /formal/Ireland-Rosen.lean: -------------------------------------------------------------------------------- 1 | import Mathlib 2 | 3 | open Real 4 | open scoped BigOperators 5 | noncomputable section 6 | 7 | theorem exercise_1_27 {n : ℕ} (hn : Odd n) : 8 ∣ (n^2 - 1) := 8 | sorry 9 | 10 | theorem exercise_1_30 {n : ℕ} : 11 | ¬ ∃ a : ℤ, ∑ i : Fin n, (1 : ℚ) / (n+2) = a := 12 | sorry 13 | 14 | theorem exercise_1_31 : (⟨1, 1⟩ : GaussianInt) ^ 2 ∣ 2 := 15 | sorry 16 | 17 | theorem exercise_2_4 {a : ℤ} (ha : a ≠ 0) 18 | (f_a := λ n m : ℕ => Int.gcd (a^(2^n) + 1) (a^(2^m)+1)) {n m : ℕ} 19 | (hnm : n > m) : 20 | (Odd a → f_a n m = 1) ∧ (Even a → f_a n m = 2) := 21 | sorry 22 | 23 | theorem exercise_2_21 {l : ℕ → ℝ} 24 | (hl : ∀ p n : ℕ, p.Prime → l (p^n) = log p ) 25 | (hl1 : ∀ m : ℕ, ¬ IsPrimePow m → l m = 0) : 26 | l = λ n => ∑ d : Nat.divisors n, ArithmeticFunction.moebius (n/d) * log d := 27 | sorry 28 | 29 | theorem exercise_2_27a : 30 | ¬ Summable (λ i : {p : ℤ // Squarefree p} => (1 : ℚ) / i) := 31 | sorry 32 | 33 | theorem exercise_3_1 : Infinite {p : Nat.Primes // p ≡ -1 [ZMOD 6]} := 34 | sorry 35 | 36 | theorem exercise_3_4 : ¬ ∃ x y : ℤ, 3*x^2 + 2 = y^2 := 37 | sorry 38 | 39 | theorem exercise_3_5 : ¬ ∃ x y : ℤ, 7*x^3 + 2 = y^3 := 40 | sorry 41 | 42 | theorem exercise_3_10 {n : ℕ} (hn0 : ¬ n.Prime) (hn1 : n ≠ 4) : 43 | Nat.factorial (n-1) ≡ 0 [MOD n] := 44 | sorry 45 | 46 | theorem exercise_3_14 {p q n : ℕ} (hp0 : p.Prime ∧ p > 2) 47 | (hq0 : q.Prime ∧ q > 2) (hpq0 : p ≠ q) (hpq1 : p - 1 ∣ q - 1) 48 | (hn : n.gcd (p*q) = 1) : 49 | n^(q-1) ≡ 1 [MOD p*q] := 50 | sorry 51 | 52 | theorem exercise_4_4 {p t: ℕ} (hp0 : p.Prime) (hp1 : p = 4*t + 1) 53 | (a : ZMod p) : 54 | IsPrimitiveRoot a p ↔ IsPrimitiveRoot (-a) p := 55 | sorry 56 | 57 | theorem exercise_4_5 {p t : ℕ} (hp0 : p.Prime) (hp1 : p = 4*t + 3) 58 | (a : ZMod p) : 59 | IsPrimitiveRoot a p ↔ ((-a) ^ ((p-1)/2) = 1 ∧ ∀ (k : ℕ), k < (p-1)/2 → (-a)^k ≠ 1) := 60 | sorry 61 | 62 | theorem exercise_4_6 {p n : ℕ} (hp : p.Prime) (hpn : p = 2^n + 1) : 63 | IsPrimitiveRoot 3 p := 64 | sorry 65 | 66 | theorem exercise_4_8 {p a : ℕ} (hp : Odd p) : 67 | IsPrimitiveRoot a p ↔ (∀ q : ℕ, q ∣ (p-1) → q.Prime → ¬ a^(p-1) ≡ 1 [MOD p]) := 68 | sorry 69 | 70 | theorem exercise_4_11 {p : ℕ} (hp : p.Prime) (k s: ℕ) 71 | (s := ∑ n : Fin p, (n : ℕ) ^ k) : 72 | ((¬ p - 1 ∣ k) → s ≡ 0 [MOD p]) ∧ (p - 1 ∣ k → s ≡ 0 [MOD p]) := 73 | sorry 74 | 75 | theorem exercise_5_13 {p x: ℤ} (hp : Prime p) 76 | (hpx : p ∣ (x^4 - x^2 + 1)) : p ≡ 1 [ZMOD 12] := 77 | sorry 78 | 79 | theorem exercise_5_28 {p : ℕ} (hp : p.Prime) (hp1 : p ≡ 1 [MOD 4]): 80 | ∃ x, x^4 ≡ 2 [MOD p] ↔ ∃ A B, p = A^2 + 64*B^2 := 81 | sorry 82 | 83 | theorem exercise_5_37 {p q : ℕ} [Fact (p.Prime)] [Fact (q.Prime)] {a : ℤ} 84 | (ha : a < 0) (h0 : p ≡ q [ZMOD 4*a]) (h1 : ¬ ((p : ℤ) ∣ a)) : 85 | legendreSym p a = legendreSym q a := 86 | sorry 87 | 88 | theorem exercise_12_12 : IsAlgebraic ℚ (sin (pi/12)) := 89 | sorry 90 | 91 | theorem exercise_18_4 {n : ℕ} (hn : ∃ x y z w : ℤ, 92 | x^3 + y^3 = n ∧ z^3 + w^3 = n ∧ x ≠ z ∧ x ≠ w ∧ y ≠ z ∧ y ≠ w) : 93 | n ≥ 1729 := 94 | sorry 95 | -------------------------------------------------------------------------------- /formal/Munkers.lean: -------------------------------------------------------------------------------- 1 | import Mathlib 2 | 3 | open Filter Set TopologicalSpace 4 | open Topology 5 | noncomputable section 6 | 7 | theorem exercise_13_1 (X : Type*) [TopologicalSpace X] (A : Set X) 8 | (h1 : ∀ x ∈ A, ∃ U : Set X, x ∈ U ∧ IsOpen U ∧ U ⊆ A) : 9 | IsOpen A := 10 | sorry 11 | 12 | theorem exercise_13_3b : ¬ ∀ X : Type, ∀s : Set (Set X), 13 | (∀ t : Set X, t ∈ s → (Set.Infinite tᶜ ∨ t = ∅ ∨ t = ⊤)) → 14 | (Set.Infinite (⋃₀ s)ᶜ ∨ (⋃₀ s) = ∅ ∨ (⋃₀ s) = ⊤) := 15 | sorry 16 | 17 | def is_topology (X : Type*) (T : Set (Set X)) := 18 | univ ∈ T ∧ 19 | (∀ s t, s ∈ T → t ∈ T → s ∩ t ∈ T) ∧ 20 | (∀s, (∀t ∈ s, t ∈ T) → sUnion s ∈ T) 21 | 22 | theorem exercise_13_4a1 (X I : Type*) (T : I → Set (Set X)) (h : ∀ i, is_topology X (T i)) : 23 | is_topology X (⋂ i : I, T i) := 24 | sorry 25 | 26 | theorem exercise_13_4a2 : 27 | ∃ (X I : Type*) (T : I → Set (Set X)), 28 | (∀ i, is_topology X (T i)) ∧ ¬ is_topology X (⋂ i : I, T i) := 29 | sorry 30 | 31 | theorem exercise_13_4b1 (X I : Type*) (T : I → Set (Set X)) (h : ∀ i, is_topology X (T i)) : 32 | ∃! T', is_topology X T' ∧ (∀ i, T i ⊆ T') ∧ 33 | ∀ T'', is_topology X T'' → (∀ i, T i ⊆ T'') → T'' ⊆ T' := 34 | sorry 35 | 36 | theorem exercise_13_4b2 (X I : Type*) (T : I → Set (Set X)) (h : ∀ i, is_topology X (T i)) : 37 | ∃! T', is_topology X T' ∧ (∀ i, T' ⊆ T i) ∧ 38 | ∀ T'', is_topology X T'' → (∀ i, T'' ⊆ T i) → T' ⊆ T'' := 39 | sorry 40 | 41 | theorem exercise_13_5a {X : Type*} 42 | [TopologicalSpace X] (A : Set (Set X)) (hA : IsTopologicalBasis A) : 43 | generateFrom A = generateFrom (sInter {T | is_topology X T ∧ A ⊆ T}) := 44 | sorry 45 | 46 | theorem exercise_13_5b {X : Type*} 47 | [t : TopologicalSpace X] (A : Set (Set X)) (hA : t = generateFrom A) : 48 | generateFrom A = generateFrom (sInter {T | is_topology X T ∧ A ⊆ T}) := 49 | sorry 50 | 51 | def lower_limit_topology (X : Type) [Preorder X] := 52 | generateFrom {S : Set X | ∃ a b, a < b ∧ S = Ico a b} 53 | 54 | def Rl := lower_limit_topology ℝ 55 | 56 | def K : Set ℝ := {r | ∃ n : ℕ, r = 1 / n} 57 | 58 | def K_topology := generateFrom 59 | ({S : Set ℝ | ∃ a b, a < b ∧ S = Ioo a b} ∪ {S : Set ℝ | ∃ a b, a < b ∧ S = Ioo a b \ K}) 60 | 61 | theorem exercise_13_6 : 62 | ¬ (∀ U, Rl.IsOpen U → K_topology.IsOpen U) ∧ ¬ (∀ U, K_topology.IsOpen U → Rl.IsOpen U) := 63 | sorry 64 | 65 | theorem exercise_13_8a : 66 | IsTopologicalBasis {S : Set ℝ | ∃ a b : ℚ, a < b ∧ S = Ioo ↑a ↑b} := 67 | sorry 68 | 69 | theorem exercise_13_8b : 70 | (generateFrom {S : Set ℝ | ∃ a b : ℚ, a < b ∧ S = Ico ↑a ↑b}).IsOpen ≠ 71 | (lower_limit_topology ℝ).IsOpen := 72 | sorry 73 | 74 | theorem exercise_16_1 {X : Type*} [TopologicalSpace X] 75 | (Y : Set X) 76 | (A : Set Y) : 77 | ∀ U : Set A, IsOpen U ↔ IsOpen (Subtype.val '' U) := 78 | sorry 79 | 80 | theorem exercise_16_4 {X Y : Type*} [TopologicalSpace X] [TopologicalSpace Y] 81 | (π₁ : X × Y → X) 82 | (π₂ : X × Y → Y) 83 | (h₁ : π₁ = Prod.fst) 84 | (h₂ : π₂ = Prod.snd) : 85 | IsOpenMap π₁ ∧ IsOpenMap π₂ := 86 | sorry 87 | 88 | def rational (x : ℝ) := x ∈ range ((↑) : ℚ → ℝ) 89 | 90 | theorem exercise_16_6 91 | (S : Set (Set (ℝ × ℝ))) 92 | (hS : ∀ s, s ∈ S → ∃ a b c d, (rational a ∧ rational b ∧ rational c ∧ rational d 93 | ∧ s = {x | ∃ x₁ x₂, x = (x₁, x₂) ∧ a < x₁ ∧ x₁ < b ∧ c < x₂ ∧ x₂ < d})) : 94 | IsTopologicalBasis S := 95 | sorry 96 | 97 | theorem exercise_17_4 {X : Type*} [TopologicalSpace X] 98 | (U A : Set X) (hU : IsOpen U) (hA : IsClosed A) : 99 | IsOpen (U \ A) ∧ IsClosed (A \ U) := 100 | sorry 101 | 102 | theorem exercise_18_8a {X Y : Type*} [TopologicalSpace X] [TopologicalSpace Y] 103 | [LinearOrder Y] [OrderTopology Y] {f g : X → Y} 104 | (hf : Continuous f) (hg : Continuous g) : 105 | IsClosed {x | f x ≤ g x} := 106 | sorry 107 | 108 | theorem exercise_18_8b {X Y : Type*} [TopologicalSpace X] [TopologicalSpace Y] 109 | [LinearOrder Y] [OrderTopology Y] {f g : X → Y} 110 | (hf : Continuous f) (hg : Continuous g) : 111 | Continuous (λ x => min (f x) (g x)) := 112 | sorry 113 | 114 | theorem exercise_18_13 115 | {X : Type*} [TopologicalSpace X] {Y : Type*} [TopologicalSpace Y] 116 | [T2Space Y] {A : Set X} {f : A → Y} (hf : Continuous f) 117 | (g : closure A → Y) 118 | (g_con : Continuous g) : 119 | ∀ (g' : closure A → Y), Continuous g' → (∀ (x : closure A), g x = g' x) := 120 | sorry 121 | 122 | theorem exercise_19_6a 123 | {n : ℕ} 124 | {f : Fin n → Type*} {x : ℕ → Πa, f a} 125 | (y : Πi, f i) 126 | [Πa, TopologicalSpace (f a)] : 127 | Tendsto x atTop (𝓝 y) ↔ ∀ i, Tendsto (λ j => (x j) i) atTop (𝓝 (y i)) := 128 | sorry 129 | 130 | theorem exercise_20_2 131 | [TopologicalSpace (ℝ ×ₗ ℝ)] [OrderTopology (ℝ ×ₗ ℝ)] 132 | : MetrizableSpace (ℝ ×ₗ ℝ) := 133 | sorry 134 | 135 | abbrev I : Set ℝ := Icc 0 1 136 | 137 | theorem exercise_21_6a 138 | (f : ℕ → I → ℝ ) 139 | (h : ∀ x n, f n x = x ^ n) : 140 | ∀ x, ∃ y, Tendsto (λ n => f n x) atTop (𝓝 y) := 141 | sorry 142 | 143 | theorem exercise_21_6b 144 | (f : ℕ → I → ℝ ) 145 | (h : ∀ x n, f n x = x ^ n) : 146 | ¬ ∃ f₀, TendstoUniformly f f₀ atTop := 147 | sorry 148 | 149 | theorem exercise_21_8 150 | {X : Type*} [TopologicalSpace X] {Y : Type*} [MetricSpace Y] 151 | {f : ℕ → X → Y} {x : ℕ → X} 152 | (hf : ∀ n, Continuous (f n)) 153 | (x₀ : X) 154 | (hx : Tendsto x atTop (𝓝 x₀)) 155 | (f₀ : X → Y) 156 | (hh : TendstoUniformly f f₀ atTop) : 157 | Tendsto (λ n => f n (x n)) atTop (𝓝 (f₀ x₀)) := 158 | sorry 159 | 160 | theorem exercise_22_2a {X Y : Type*} [TopologicalSpace X] 161 | [TopologicalSpace Y] (p : X → Y) (h : Continuous p) : 162 | IsQuotientMap p ↔ ∃ (f : Y → X), Continuous f ∧ p ∘ f = id := 163 | sorry 164 | 165 | theorem exercise_22_2b {X : Type*} [TopologicalSpace X] 166 | {A : Set X} (r : X → A) (hr : Continuous r) (h : ∀ x : A, r x = x) : 167 | IsQuotientMap r := 168 | sorry 169 | 170 | theorem exercise_22_5 {X Y : Type*} [TopologicalSpace X] 171 | [TopologicalSpace Y] (p : X → Y) (hp : IsOpenMap p) 172 | (A : Set X) (hA : IsOpen A) : IsOpenMap (p ∘ Subtype.val : A → Y) := 173 | sorry 174 | 175 | theorem exercise_23_2 {X : Type*} 176 | [TopologicalSpace X] {A : ℕ → Set X} (hA : ∀ n, IsConnected (A n)) 177 | (hAn : ∀ n, A n ∩ A (n + 1) ≠ ∅) : 178 | IsConnected (⋃ n, A n) := 179 | sorry 180 | 181 | theorem exercise_23_3 {X : Type*} [TopologicalSpace X] 182 | [TopologicalSpace X] {A : ℕ → Set X} 183 | (hAn : ∀ n, IsConnected (A n)) 184 | (A₀ : Set X) 185 | (hA : IsConnected A₀) 186 | (h : ∀ n, A₀ ∩ A n ≠ ∅) : 187 | IsConnected (A₀ ∪ (⋃ n, A n)) := 188 | sorry 189 | 190 | set_option checkBinderAnnotations false 191 | 192 | theorem exercise_23_4 {X : Type*} [TopologicalSpace X] [CofiniteTopology X] 193 | (s : Set X) : Infinite s → IsConnected s := 194 | sorry 195 | 196 | theorem exercise_23_6 {X : Type*} 197 | [TopologicalSpace X] {A C : Set X} (hc : IsConnected C) 198 | (hCA : C ∩ A ≠ ∅) (hCXA : C ∩ Aᶜ ≠ ∅) : 199 | C ∩ (frontier A) ≠ ∅ := 200 | sorry 201 | 202 | theorem exercise_23_9 {X Y : Type*} 203 | [TopologicalSpace X] [TopologicalSpace Y] 204 | (A₁ A₂ : Set X) 205 | (B₁ B₂ : Set Y) 206 | (hA : A₁ ⊂ A₂) 207 | (hB : B₁ ⊂ B₂) 208 | (hA : IsConnected A₂) 209 | (hB : IsConnected B₂) : 210 | IsConnected ({x | ∃ a b, x = (a, b) ∧ a ∈ A₂ ∧ b ∈ B₂} \ 211 | {x | ∃ a b, x = (a, b) ∧ a ∈ A₁ ∧ b ∈ B₁}) := 212 | sorry 213 | 214 | theorem exercise_23_11 {X Y : Type*} [TopologicalSpace X] [TopologicalSpace Y] 215 | (p : X → Y) (hq : QuotientMap p) 216 | (hY : ConnectedSpace Y) (hX : ∀ y : Y, IsConnected (p ⁻¹' {y})) : 217 | ConnectedSpace X := 218 | sorry 219 | 220 | theorem exercise_24_2 {f : (Metric.sphere 0 1 : Set ℝ) → ℝ} 221 | (hf : Continuous f) : ∃ x, f x = f (-x) := 222 | sorry 223 | 224 | theorem exercise_24_3a [TopologicalSpace I] [CompactSpace I] 225 | (f : I → I) (hf : Continuous f) : 226 | ∃ (x : I), f x = x := 227 | sorry 228 | 229 | theorem exercise_25_4 {X : Type*} [TopologicalSpace X] 230 | [LocPathConnectedSpace X] (U : Set X) (hU : IsOpen U) 231 | (hcU : IsConnected U) : IsPathConnected U := 232 | sorry 233 | 234 | theorem exercise_25_9 {G : Type*} [TopologicalSpace G] [Group G] 235 | [IsTopologicalGroup G] (C : Set G) (h : C = connectedComponent 1) : 236 | ∃ C' : Subgroup G, C' = C ∧ C'.Normal := 237 | sorry 238 | 239 | theorem exercise_26_11 240 | {X : Type*} [TopologicalSpace X] [CompactSpace X] [T2Space X] 241 | (A : Set (Set X)) (hA : ∀ (a b : Set X), a ∈ A → b ∈ A → a ⊆ b ∨ b ⊆ a) 242 | (hA' : ∀ a ∈ A, IsClosed a) (hA'' : ∀ a ∈ A, IsConnected a) : 243 | IsConnected (⋂₀ A) := 244 | sorry 245 | 246 | theorem exercise_26_12 {X Y : Type*} [TopologicalSpace X] [TopologicalSpace Y] 247 | (p : X → Y) (h : Function.Surjective p) (hc : Continuous p) (hp : ∀ y, IsCompact (p ⁻¹' {y})) 248 | (hY : CompactSpace Y) : CompactSpace X := 249 | sorry 250 | 251 | theorem exercise_27_4 252 | {X : Type*} [MetricSpace X] [ConnectedSpace X] (hX : ∃ x y : X, x ≠ y) : 253 | ¬ Countable (univ : Set X) := 254 | sorry 255 | 256 | def countably_compact (X : Type*) [TopologicalSpace X] := 257 | ∀ U : ℕ → Set X, 258 | (∀ i, IsOpen (U i)) ∧ ((univ : Set X) ⊆ ⋃ i, U i) → 259 | (∃ t : Finset ℕ, (univ : Set X) ⊆ ⋃ i ∈ t, U i) 260 | 261 | def limit_point_compact (X : Type*) [TopologicalSpace X] := 262 | ∀ U : Set X, Infinite U → ∃ x ∈ U, ClusterPt x (𝓟 U) 263 | 264 | theorem exercise_28_4 {X : Type*} 265 | [TopologicalSpace X] (hT1 : T1Space X) : 266 | countably_compact X ↔ limit_point_compact X := 267 | sorry 268 | 269 | theorem exercise_28_5 270 | (X : Type*) [TopologicalSpace X] : 271 | countably_compact X ↔ ∀ (C : ℕ → Set X), (∀ n, IsClosed (C n)) ∧ 272 | (∀ n, C n ≠ ∅) ∧ (∀ n, C n ⊆ C (n + 1)) → ∃ x, ∀ n, x ∈ C n := 273 | sorry 274 | 275 | theorem exercise_28_6 {X : Type*} [MetricSpace X] 276 | [CompactSpace X] {f : X → X} (hf : Isometry f) : 277 | Function.Bijective f := 278 | sorry 279 | 280 | theorem exercise_29_1 : ¬ LocallyCompactSpace ℚ := 281 | sorry 282 | 283 | theorem exercise_29_4 [TopologicalSpace (ℕ → I)] : 284 | ¬ LocallyCompactSpace (ℕ → I) := 285 | sorry 286 | 287 | theorem exercise_29_10 {X : Type*} 288 | [TopologicalSpace X] [T2Space X] (x : X) 289 | (hx : ∃ U : Set X, x ∈ U ∧ IsOpen U ∧ (∃ K : Set X, U ⊂ K ∧ IsCompact K)) 290 | (U : Set X) (hU : IsOpen U) (hxU : x ∈ U) : 291 | ∃ (V : Set X), IsOpen V ∧ x ∈ V ∧ IsCompact (closure V) ∧ closure V ⊆ U := 292 | sorry 293 | 294 | theorem exercise_30_10 295 | {X : ℕ → Type*} [∀ i, TopologicalSpace (X i)] 296 | (h : ∀ i, ∃ (s : Set (X i)), Countable s ∧ Dense s) : 297 | ∃ (s : Set (Π i, X i)), Countable s ∧ Dense s := 298 | sorry 299 | 300 | theorem exercise_30_13 {X : Type*} [TopologicalSpace X] 301 | (h : ∃ (s : Set X), Countable s ∧ Dense s) (U : Set (Set X)) 302 | (hU : ∀ (x y : Set X), x ∈ U → y ∈ U → x ≠ y → x ∩ y = ∅) : 303 | Countable U := 304 | sorry 305 | 306 | theorem exercise_31_1 {X : Type*} [TopologicalSpace X] 307 | (hX : RegularSpace X) (x y : X) : 308 | ∃ (U V : Set X), IsOpen U ∧ IsOpen V ∧ x ∈ U ∧ y ∈ V ∧ closure U ∩ closure V = ∅ := 309 | sorry 310 | 311 | theorem exercise_31_2 {X : Type*} 312 | [TopologicalSpace X] [NormalSpace X] {A B : Set X} 313 | (hA : IsClosed A) (hB : IsClosed B) (hAB : Disjoint A B) : 314 | ∃ (U V : Set X), IsOpen U ∧ IsOpen V ∧ A ⊆ U ∧ B ⊆ V ∧ closure U ∩ closure V = ∅ := 315 | sorry 316 | 317 | theorem exercise_31_3 {α : Type*} [PartialOrder α] 318 | [TopologicalSpace α] (h : OrderTopology α) : RegularSpace α := 319 | sorry 320 | 321 | theorem exercise_32_1 {X : Type*} [TopologicalSpace X] 322 | (hX : NormalSpace X) (A : Set X) (hA : IsClosed A) : 323 | NormalSpace {x // x ∈ A} := 324 | sorry 325 | 326 | theorem exercise_32_2a 327 | {ι : Type*} {X : ι → Type*} [∀ i, TopologicalSpace (X i)] 328 | (h : ∀ i, Nonempty (X i)) (h2 : T2Space (Π i, X i)) : 329 | ∀ i, T2Space (X i) := 330 | sorry 331 | 332 | theorem exercise_32_2b 333 | {ι : Type*} {X : ι → Type*} [∀ i, TopologicalSpace (X i)] 334 | (h : ∀ i, Nonempty (X i)) (h2 : RegularSpace (Π i, X i)) : 335 | ∀ i, RegularSpace (X i) := 336 | sorry 337 | 338 | theorem exercise_32_2c 339 | {ι : Type*} {X : ι → Type*} [∀ i, TopologicalSpace (X i)] 340 | (h : ∀ i, Nonempty (X i)) (h2 : NormalSpace (Π i, X i)) : 341 | ∀ i, NormalSpace (X i) := 342 | sorry 343 | 344 | theorem exercise_32_3 {X : Type*} [TopologicalSpace X] 345 | (hX : LocallyCompactSpace X) (hX' : T2Space X) : 346 | RegularSpace X := 347 | sorry 348 | 349 | theorem exercise_33_7 {X : Type*} [TopologicalSpace X] 350 | (hX : LocallyCompactSpace X) (hX' : T2Space X) : 351 | ∀ x A, IsClosed A ∧ ¬ x ∈ A → 352 | ∃ (f : X → I), Continuous f ∧ f x = 1 ∧ f '' A = {0} 353 | := 354 | sorry 355 | 356 | theorem exercise_33_8 357 | (X : Type*) [TopologicalSpace X] [RegularSpace X] 358 | (h : ∀ x A, IsClosed A ∧ ¬ x ∈ A → 359 | ∃ (f : X → I), Continuous f ∧ f x = (1 : I) ∧ f '' A = {0}) 360 | (A B : Set X) (hA : IsClosed A) (hB : IsClosed B) 361 | (hAB : Disjoint A B) 362 | (hAc : IsCompact A) : 363 | ∃ (f : X → I), Continuous f ∧ f '' A = {0} ∧ f '' B = {1} := 364 | sorry 365 | 366 | theorem exercise_34_9 367 | (X : Type*) [TopologicalSpace X] [CompactSpace X] 368 | (X1 X2 : Set X) (hX1 : IsClosed X1) (hX2 : IsClosed X2) 369 | (hX : X1 ∪ X2 = univ) (hX1m : MetrizableSpace X1) 370 | (hX2m : MetrizableSpace X2) : MetrizableSpace X := 371 | sorry 372 | 373 | theorem exercise_38_6 {X : Type*} 374 | (X : Type*) [TopologicalSpace X] [RegularSpace X] 375 | (h : ∀ x A, IsClosed A ∧ ¬ x ∈ A → 376 | ∃ (f : X → I), Continuous f ∧ f x = (1 : I) ∧ f '' A = {0}) : 377 | IsConnected (univ : Set X) ↔ IsConnected (univ : Set (StoneCech X)) := 378 | sorry 379 | 380 | theorem exercise_43_2 {X : Type*} [MetricSpace X] 381 | {Y : Type*} [MetricSpace Y] [CompleteSpace Y] (A : Set X) 382 | (f : X → Y) (hf : UniformContinuousOn f A) : 383 | ∃! (g : X → Y), ContinuousOn g (closure A) ∧ 384 | UniformContinuousOn g (closure A) ∧ ∀ (x : A), g x = f x := 385 | sorry 386 | -------------------------------------------------------------------------------- /formal/Pough.lean: -------------------------------------------------------------------------------- 1 | import Mathlib 2 | 3 | open Filter Real Function 4 | open scoped Topology 5 | noncomputable section 6 | 7 | theorem exercise_2_12a (f : ℕ → ℕ) (p : ℕ → ℝ) (a : ℝ) 8 | (hf : Injective f) (hp : Tendsto p atTop (𝓝 a)) : 9 | Tendsto (λ n => p (f n)) atTop (𝓝 a) := 10 | sorry 11 | 12 | theorem exercise_2_26 {M : Type*} [TopologicalSpace M] 13 | (U : Set M) : IsOpen U ↔ ∀ x ∈ U, ¬ ClusterPt x (𝓟 Uᶜ) := 14 | sorry 15 | 16 | theorem exercise_2_29 (M : Type*) [MetricSpace M] 17 | (O C : Set (Set M)) 18 | (hO : O = {s | IsOpen s}) 19 | (hC : C = {s | IsClosed s}) : 20 | ∃ f : O → C, Bijective f := 21 | sorry 22 | 23 | theorem exercise_2_32a (A : Set ℕ) : IsClopen A := 24 | sorry 25 | 26 | theorem exercise_2_41 (m : ℕ) {X : Type*} [NormedSpace ℝ ((Fin m) → ℝ)] : 27 | IsCompact (Metric.closedBall 0 1) := 28 | sorry 29 | 30 | theorem exercise_2_46 {M : Type*} [MetricSpace M] 31 | {A B : Set M} (hA : IsCompact A) (hB : IsCompact B) 32 | (hAB : Disjoint A B) (hA₀ : A ≠ ∅) (hB₀ : B ≠ ∅) : 33 | ∃ a₀ b₀, a₀ ∈ A ∧ b₀ ∈ B ∧ ∀ (a : M) (b : M), 34 | a ∈ A → b ∈ B → dist a₀ b₀ ≤ dist a b := 35 | sorry 36 | 37 | theorem exercise_2_57 {X : Type*} [TopologicalSpace X] 38 | : ∃ (S : Set X), IsConnected S ∧ ¬ IsConnected (interior S) := 39 | sorry 40 | 41 | theorem exercise_2_92 {α : Type*} [TopologicalSpace α] 42 | {s : ℕ → Set α} 43 | (hs : ∀ i, IsCompact (s i)) 44 | (hs : ∀ i, (s i).Nonempty) 45 | (hs : ∀ i, (s i) ⊃ (s (i + 1))) : 46 | (⋂ i, s i).Nonempty := 47 | sorry 48 | 49 | theorem exercise_2_126 {E : Set ℝ} 50 | (hE : ¬ Set.Countable E) : ∃ (p : ℝ), ClusterPt p (𝓟 E) := 51 | sorry 52 | 53 | theorem exercise_3_1 {f : ℝ → ℝ} 54 | (hf : ∀ x y, |f x - f y| ≤ |x - y| ^ 2) : 55 | ∃ c, f = λ x => c := 56 | sorry 57 | 58 | theorem exercise_3_4 (n : ℕ) : 59 | Tendsto (λ n => (sqrt (n + 1) - sqrt n)) atTop (𝓝 0) := 60 | sorry 61 | 62 | theorem exercise_3_63a (p : ℝ) (f : ℕ → ℝ) (hp : p > 1) 63 | (h : f = λ (k : ℕ) => (1 : ℝ) / (k * (log k) ^ p)) : 64 | ∃ l, Tendsto f atTop (𝓝 l) := 65 | sorry 66 | 67 | theorem exercise_3_63b (p : ℝ) (f : ℕ → ℝ) (hp : p ≤ 1) 68 | (h : f = λ (k : ℕ) => (1 : ℝ) / (k * (log k) ^ p)) : 69 | ¬ ∃ l, Tendsto f atTop (𝓝 l) := 70 | sorry 71 | 72 | theorem exercise_4_15a {α : Type*} 73 | (a b : ℝ) (F : Set (ℝ → ℝ)) : 74 | (∀ x : ℝ, ∀ ε > 0, ∃ U ∈ (𝓝 x), 75 | (∀ y z : U, ∀ f : ℝ → ℝ, f ∈ F → (dist (f y) (f z) < ε))) 76 | ↔ 77 | ∃ (μ : ℝ → ℝ), ∀ (x : ℝ), (0 : ℝ) ≤ μ x ∧ Tendsto μ (𝓝 0) (𝓝 0) ∧ 78 | (∀ (s t : ℝ) (f : ℝ → ℝ), f ∈ F → |(f s) - (f t)| ≤ μ (|s - t|)) := 79 | sorry 80 | -------------------------------------------------------------------------------- /formal/Putnam.lean: -------------------------------------------------------------------------------- 1 | import Mathlib 2 | 3 | open scoped BigOperators 4 | 5 | theorem exercise_2020_b5 (z : Fin 4 → ℂ) (hz0 : ∀ n, ‖z n‖ < 1) 6 | (hz1 : ∀ n : Fin 4, z n ≠ 1) : 7 | 3 - z 0 - z 1 - z 2 - z 3 + (z 0) * (z 1) * (z 2) * (z 3) ≠ 0 := 8 | sorry 9 | 10 | theorem exercise_2018_a5 (f : ℝ → ℝ) (hf : ContDiff ℝ ⊤ f) 11 | (hf0 : f 0 = 0) (hf1 : f 1 = 1) (hf2 : ∀ x, f x ≥ 0) : 12 | ∃ (n : ℕ) (x : ℝ), iteratedDeriv n f x = 0 := 13 | sorry 14 | 15 | theorem exercise_2018_b2 (n : ℕ) (hn : n > 0) (f : ℕ → ℂ → ℂ) 16 | (hf : ∀ n : ℕ, f n = λ (z : ℂ) => (∑ i : Fin n, (n-i)* z^(i : ℕ))) : 17 | ¬ (∃ z : ℂ, ‖z‖ ≤ 1 ∧ f n z = 0) := 18 | sorry 19 | 20 | theorem exercise_2018_b4 (a : ℝ) (x : ℕ → ℝ) (hx0 : x 0 = a) 21 | (hx1 : x 1 = a) 22 | (hxn : ∀ n : ℕ, n ≥ 2 → x (n+1) = 2*(x n)*(x (n-1)) - x (n-2)) 23 | (h : ∃ n, x n = 0) : 24 | ∃ c, Function.Periodic x c := 25 | sorry 26 | 27 | theorem exercise_2017_b3 (f : ℝ → ℝ) (c : ℕ → ℝ) 28 | (hf : f = λ x => (∑' (i : ℕ), (c i) * x^i)) 29 | (hc : ∀ n, c n = 0 ∨ c n = 1) 30 | (hf1 : f (2/3) = 3/2) : 31 | Irrational (f (1/2)) := 32 | sorry 33 | 34 | theorem exercise_2014_a5 (P : ℕ → Polynomial ℤ) 35 | (hP : ∀ n, P n = ∑ i : Fin n, (n+1) * Polynomial.X ^ n) : 36 | ∀ (j k : ℕ), j ≠ k → IsCoprime (P j) (P k) := 37 | sorry 38 | 39 | theorem exercise_2010_a4 (n : ℕ) : 40 | ¬ Nat.Prime (10^10^10^n + 10^10^n + 10^n - 1) := 41 | sorry 42 | 43 | theorem exercise_2001_a5 : 44 | ∃! a : ℕ × ℕ, a.1 > 0 ∧ a.2 > 0 ∧ a.1^(a.2+1) - (a.1+1)^a.2 = 2001 := 45 | sorry 46 | 47 | theorem exercise_2000_a2 : 48 | ∀ N : ℕ, ∃ n : ℕ, n > N ∧ ∃ i : Fin 6 → ℕ, n = (i 0)^2 + (i 1)^2 ∧ 49 | n + 1 = (i 2)^2 + (i 3)^2 ∧ n + 2 = (i 4)^2 + (i 5)^2 := 50 | sorry 51 | 52 | theorem exercise_1999_b4 (f : ℝ → ℝ) (hf: ContDiff ℝ 3 f) 53 | (hf1 : ∀ n ≤ 3, ∀ x : ℝ, iteratedDeriv n f x > 0) 54 | (hf2 : ∀ x : ℝ, iteratedDeriv 3 f x ≤ f x) : 55 | ∀ x : ℝ, deriv f x < 2 * f x := 56 | sorry 57 | 58 | theorem exercise_1998_a3 (f : ℝ → ℝ) (hf : ContDiff ℝ 3 f) : 59 | ∃ a : ℝ, (f a) * (deriv f a) * (iteratedDeriv 2 f a) * (iteratedDeriv 3 f a) ≥ 0 := 60 | sorry 61 | 62 | theorem exercise_1998_b6 (a b c : ℤ) : 63 | ∃ n : ℤ, n > 0 ∧ ¬ ∃ m : ℤ, Real.sqrt (n^3 + a*n^2 + b*n + c) = m := 64 | sorry 65 | -------------------------------------------------------------------------------- /formal/Rudin.lean: -------------------------------------------------------------------------------- 1 | import Mathlib 2 | 3 | open Topology Filter Real Complex TopologicalSpace Finset 4 | open scoped BigOperators 5 | noncomputable section 6 | 7 | theorem exercise_1_1a 8 | (x : ℝ) (y : ℚ) : 9 | ( Irrational x ) -> Irrational ( x + y ) := 10 | sorry 11 | 12 | theorem exercise_1_1b 13 | (x : ℝ) 14 | (y : ℚ) 15 | (h : y ≠ 0) 16 | : ( Irrational x ) -> Irrational ( x * y ) := 17 | sorry 18 | 19 | theorem exercise_1_2 : ¬ ∃ (x : ℚ), ( x ^ 2 = 12 ) := 20 | sorry 21 | 22 | theorem exercise_1_4 23 | (α : Type*) [PartialOrder α] 24 | (s : Set α) 25 | (x y : α) 26 | (h₀ : Set.Nonempty s) 27 | (h₁ : x ∈ lowerBounds s) 28 | (h₂ : y ∈ upperBounds s) 29 | : x ≤ y := 30 | sorry 31 | 32 | theorem exercise_1_5 (A minus_A : Set ℝ) (hA : A.Nonempty) 33 | (hA_bdd_below : BddBelow A) (hminus_A : minus_A = {x | -x ∈ A}) : 34 | Min A = Max minus_A := 35 | sorry 36 | 37 | theorem exercise_1_8 : ¬ ∃ (r : ℂ → ℂ → Prop), IsLinearOrder ℂ r := 38 | sorry 39 | 40 | theorem exercise_1_11a (z : ℂ) : 41 | ∃ (r : ℝ) (w : ℂ), norm w = 1 ∧ z = r * w := 42 | sorry 43 | 44 | theorem exercise_1_12 (n : ℕ) (f : ℕ → ℂ) : 45 | norm (∑ i ∈ range n, f i) ≤ ∑ i ∈ range n, norm (f i) := 46 | sorry 47 | 48 | theorem exercise_1_13 (x y : ℂ) : 49 | |(norm x) - (norm y)| ≤ norm (x - y) := 50 | sorry 51 | 52 | theorem exercise_1_14 53 | (z : ℂ) (h : norm z = 1) 54 | : (norm (1 + z)) ^ 2 + (norm (1 - z)) ^ 2 = 4 := 55 | sorry 56 | 57 | theorem exercise_1_16a 58 | (n : ℕ) 59 | (d r : ℝ) 60 | (x y z : EuclideanSpace ℝ (Fin n)) -- R^n 61 | (h₁ : n ≥ 3) 62 | (h₂ : ‖x - y‖ = d) 63 | (h₃ : d > 0) 64 | (h₄ : r > 0) 65 | (h₅ : 2 * r > d) 66 | : Set.Infinite {z : EuclideanSpace ℝ (Fin n) | ‖z - x‖ = r ∧ ‖z - y‖ = r} := 67 | sorry 68 | 69 | theorem exercise_1_17 70 | (n : ℕ) 71 | (x y : EuclideanSpace ℝ (Fin n)) -- R^n 72 | : ‖x + y‖^2 + ‖x - y‖^2 = 2*‖x‖^2 + 2*‖y‖^2 := 73 | sorry 74 | 75 | theorem exercise_1_18a 76 | (n : ℕ) 77 | (h : n > 1) 78 | (x : EuclideanSpace ℝ (Fin n)) -- R^n 79 | : ∃ (y : EuclideanSpace ℝ (Fin n)), y ≠ 0 ∧ (inner x y) = (0 : ℝ) := 80 | sorry 81 | 82 | theorem exercise_1_18b 83 | : ¬ ∀ (x : ℝ), ∃ (y : ℝ), y ≠ 0 ∧ x * y = 0 := 84 | sorry 85 | 86 | theorem exercise_1_19 87 | (n : ℕ) 88 | (a b c x : EuclideanSpace ℝ (Fin n)) 89 | (r : ℝ) 90 | (h₁ : r > 0) 91 | (h₂ : 3 • c = 4 • b - a) 92 | (h₃ : 3 * r = 2 * ‖x - b‖) 93 | : ‖x - a‖ = 2 * ‖x - b‖ ↔ ‖x - c‖ = r := 94 | sorry 95 | 96 | theorem exercise_2_19a {X : Type*} [MetricSpace X] 97 | (A B : Set X) (hA : IsClosed A) (hB : IsClosed B) (hAB : Disjoint A B) : 98 | SeparatedNhds A B := 99 | sorry 100 | 101 | theorem exercise_2_24 {X : Type*} [MetricSpace X] 102 | (hX : ∀ (A : Set X), Infinite A → ∃ (x : X), x ∈ closure A) : 103 | SeparableSpace X := 104 | sorry 105 | 106 | theorem exercise_2_25 {K : Type*} [MetricSpace K] [CompactSpace K] : 107 | ∃ (B : Set (Set K)), Set.Countable B ∧ IsTopologicalBasis B := 108 | sorry 109 | 110 | theorem exercise_2_27a (k : ℕ) (E P : Set (EuclideanSpace ℝ (Fin k))) 111 | (hE : E.Nonempty ∧ ¬ Set.Countable E) 112 | (hP : P = {x | ∀ U ∈ 𝓝 x, ¬ Set.Countable (P ∩ E)}) : 113 | IsClosed P ∧ P = {x | ClusterPt x (𝓟 P)} := 114 | sorry 115 | 116 | theorem exercise_2_27b (k : ℕ) (E P : Set (EuclideanSpace ℝ (Fin k))) 117 | (hE : E.Nonempty ∧ ¬ Set.Countable E) 118 | (hP : P = {x | ∀ U ∈ 𝓝 x, (P ∩ E).Nonempty ∧ ¬ Set.Countable (P ∩ E)}) : 119 | Set.Countable (E \ P) := 120 | sorry 121 | 122 | theorem exercise_2_28 (X : Type*) [MetricSpace X] [SeparableSpace X] 123 | (A : Set X) (hA : IsClosed A) : 124 | ∃ P₁ P₂ : Set X, A = P₁ ∪ P₂ ∧ 125 | IsClosed P₁ ∧ P₁ = {x | ClusterPt x (𝓟 P₁)} ∧ 126 | Set.Countable P₂ := 127 | sorry 128 | 129 | theorem exercise_2_29 (U : Set ℝ) (hU : IsOpen U) : 130 | ∃ (f : ℕ → Set ℝ), (∀ n, ∃ a b : ℝ, f n = {x | a < x ∧ x < b}) ∧ (∀ n, f n ⊆ U) ∧ 131 | (∀ n m, n ≠ m → f n ∩ f m = ∅) ∧ 132 | U = ⋃ n, f n := 133 | sorry 134 | 135 | theorem exercise_3_1a 136 | (f : ℕ → ℝ) 137 | (h : ∃ (a : ℝ), Tendsto (λ (n : ℕ) => f n) atTop (𝓝 a)) 138 | : ∃ (a : ℝ), Tendsto (λ (n : ℕ) => |f n|) atTop (𝓝 a) := 139 | sorry 140 | 141 | theorem exercise_3_2a 142 | : Tendsto (λ (n : ℝ) => (sqrt (n^2 + n) - n)) atTop (𝓝 (1/2)) := 143 | sorry 144 | 145 | noncomputable def f : ℕ → ℝ 146 | | 0 => sqrt 2 147 | | (n + 1) => sqrt (2 + sqrt (f n)) 148 | 149 | theorem exercise_3_3 150 | : ∃ (x : ℝ), Tendsto f atTop (𝓝 x) ∧ ∀ n, f n < 2 := 151 | sorry 152 | 153 | theorem exercise_3_5 154 | (a b : ℕ → ℝ) 155 | (h : limsup a + limsup b ≠ 0) : 156 | limsup (λ n => a n + b n) ≤ limsup a + limsup b := 157 | sorry 158 | 159 | def g (n : ℕ) : ℝ := sqrt (n + 1) - sqrt n 160 | 161 | theorem exercise_3_6a 162 | : Tendsto (λ (n : ℕ) => (∑ i ∈ range n, g i)) atTop atTop := 163 | sorry 164 | 165 | theorem exercise_3_7 166 | (a : ℕ → ℝ) 167 | (h : ∃ y, (Tendsto (λ n => (∑ i ∈ (range n), a i)) atTop (𝓝 y))) : 168 | ∃ y, Tendsto (λ n => (∑ i ∈ (range n), sqrt (a i) / n)) atTop (𝓝 y) := 169 | sorry 170 | 171 | theorem exercise_3_8 172 | (a b : ℕ → ℝ) 173 | (h1 : ∃ y, (Tendsto (λ n => (∑ i ∈ (range n), a i)) atTop (𝓝 y))) 174 | (h2 : Monotone b) 175 | (h3 : Bornology.IsBounded (Set.range b)) : 176 | ∃ y, Tendsto (λ n => (∑ i ∈ (range n), (a i) * (b i))) atTop (𝓝 y) := 177 | sorry 178 | 179 | theorem exercise_3_13 180 | (a b : ℕ → ℝ) 181 | (ha : ∃ y, (Tendsto (λ n => (∑ i ∈ (range n), |a i|)) atTop (𝓝 y))) 182 | (hb : ∃ y, (Tendsto (λ n => (∑ i ∈ (range n), |b i|)) atTop (𝓝 y))) : 183 | ∃ y, (Tendsto (λ n => (∑ i ∈ (range n), 184 | λ i => (∑ j ∈ range (i + 1), a j * b (i - j)))) atTop (𝓝 y)) := 185 | sorry 186 | 187 | theorem exercise_3_20 {X : Type*} [MetricSpace X] 188 | (p : ℕ → X) (l : ℕ) (r : X) 189 | (hp : CauchySeq p) 190 | (hpl : Tendsto (λ n => p (l * n)) atTop (𝓝 r)) : 191 | Tendsto p atTop (𝓝 r) := 192 | sorry 193 | 194 | theorem exercise_3_21 195 | {X : Type*} [MetricSpace X] [CompleteSpace X] 196 | (E : ℕ → Set X) 197 | (hE : ∀ n, E n ⊃ E (n + 1)) 198 | (hE' : Tendsto (λ n => Metric.diam (E n)) atTop (𝓝 0)) : 199 | ∃ a, Set.iInter E = {a} := 200 | sorry 201 | 202 | theorem exercise_3_22 (X : Type*) [MetricSpace X] [CompleteSpace X] 203 | (G : ℕ → Set X) (hG : ∀ n, IsOpen (G n) ∧ Dense (G n)) : 204 | ∃ x, ∀ n, x ∈ G n := 205 | sorry 206 | 207 | theorem exercise_4_1a 208 | : ∃ (f : ℝ → ℝ), (∀ (x : ℝ), Tendsto (λ y => f (x + y) - f (x - y)) (𝓝 0) (𝓝 0)) ∧ ¬ Continuous f := 209 | sorry 210 | 211 | theorem exercise_4_2a 212 | {α : Type} [MetricSpace α] 213 | {β : Type} [MetricSpace β] 214 | (f : α → β) 215 | (h₁ : Continuous f) 216 | : ∀ (x : Set α), f '' (closure x) ⊆ closure (f '' x) := 217 | sorry 218 | 219 | theorem exercise_4_3 220 | {α : Type} [MetricSpace α] 221 | (f : α → ℝ) (h : Continuous f) (z : Set α) (g : z = f⁻¹' {0}) 222 | : IsClosed z := 223 | sorry 224 | 225 | theorem exercise_4_4a 226 | {α : Type} [MetricSpace α] 227 | {β : Type} [MetricSpace β] 228 | (f : α → β) 229 | (s : Set α) 230 | (h₁ : Continuous f) 231 | (h₂ : Dense s) 232 | : f '' Set.univ ⊆ closure (f '' s) := 233 | sorry 234 | 235 | theorem exercise_4_4b 236 | {α : Type} [MetricSpace α] 237 | {β : Type} [MetricSpace β] 238 | (f g : α → β) 239 | (s : Set α) 240 | (h₁ : Continuous f) 241 | (h₂ : Continuous g) 242 | (h₃ : Dense s) 243 | (h₄ : ∀ x ∈ s, f x = g x) 244 | : f = g := 245 | sorry 246 | 247 | theorem exercise_4_5a 248 | (f : ℝ → ℝ) 249 | (E : Set ℝ) 250 | (h₁ : IsClosed E) 251 | (h₂ : ContinuousOn f E) 252 | : ∃ (g : ℝ → ℝ), Continuous g ∧ ∀ x ∈ E, f x = g x := 253 | sorry 254 | 255 | theorem exercise_4_5b 256 | : ∃ (E : Set ℝ) (f : ℝ → ℝ), (ContinuousOn f E) ∧ 257 | (¬ ∃ (g : ℝ → ℝ), Continuous g ∧ ∀ x ∈ E, f x = g x) := 258 | sorry 259 | 260 | theorem exercise_4_6 261 | (f : ℝ → ℝ) 262 | (E : Set ℝ) 263 | (G : Set (ℝ × ℝ)) 264 | (h₁ : IsCompact E) 265 | (h₂ : G = {(x, f x) | x ∈ E}) 266 | : ContinuousOn f E ↔ IsCompact G := 267 | sorry 268 | 269 | theorem exercise_4_8a 270 | (E : Set ℝ) (f : ℝ → ℝ) (hf : UniformContinuousOn f E) 271 | (hE : Bornology.IsBounded E) : Bornology.IsBounded (Set.image f E) := 272 | sorry 273 | 274 | theorem exercise_4_8b 275 | (E : Set ℝ) : 276 | ∃ f : ℝ → ℝ, UniformContinuousOn f E ∧ ¬ Bornology.IsBounded (Set.image f E) := 277 | sorry 278 | 279 | theorem exercise_4_11a 280 | {X : Type*} [MetricSpace X] 281 | {Y : Type*} [MetricSpace Y] 282 | (f : X → Y) (hf : UniformContinuous f) 283 | (x : ℕ → X) (hx : CauchySeq x) : 284 | CauchySeq (λ n => f (x n)) := 285 | sorry 286 | 287 | theorem exercise_4_12 288 | {α β γ : Type*} [UniformSpace α] [UniformSpace β] [UniformSpace γ] 289 | {f : α → β} {g : β → γ} 290 | (hf : UniformContinuous f) (hg : UniformContinuous g) : 291 | UniformContinuous (g ∘ f) := 292 | sorry 293 | 294 | theorem exercise_4_15 {f : ℝ → ℝ} 295 | (hf : Continuous f) (hof : IsOpenMap f) : 296 | Monotone f := 297 | sorry 298 | 299 | theorem exercise_4_19 300 | {f : ℝ → ℝ} (hf : ∀ a b c, a < b → f a < c → c < f b → ∃ x, a < x ∧ x < b ∧ f x = c) 301 | (hg : ∀ r : ℚ, IsClosed {x | f x = r}) : Continuous f := 302 | sorry 303 | 304 | theorem exercise_4_21a {X : Type*} [MetricSpace X] 305 | (K F : Set X) (hK : IsCompact K) (hF : IsClosed F) (hKF : Disjoint K F) : 306 | ∃ (δ : ℝ), δ > 0 ∧ ∀ (p q : X), p ∈ K → q ∈ F → dist p q ≥ δ := 307 | sorry 308 | 309 | theorem exercise_4_24 {f : ℝ → ℝ} 310 | (hf : Continuous f) (a b : ℝ) (hab : a < b) 311 | (h : ∀ x y : ℝ, a < x → x < b → a < y → y < b → f ((x + y) / 2) ≤ (f x + f y) / 2) : 312 | ConvexOn ℝ (Set.Ioo a b) f := 313 | sorry 314 | 315 | theorem exercise_5_1 316 | {f : ℝ → ℝ} (hf : ∀ x y : ℝ, |(f x - f y)| ≤ (x - y) ^ 2) : 317 | ∃ c, f = λ x => c := 318 | sorry 319 | 320 | theorem exercise_5_2 {a b : ℝ} 321 | {f g : ℝ → ℝ} (hf : ∀ x ∈ Set.Ioo a b, deriv f x > 0) 322 | (hg : g = f⁻¹) 323 | (hg_diff : DifferentiableOn ℝ g (Set.Ioo a b)) : 324 | DifferentiableOn ℝ g (Set.Ioo a b) ∧ 325 | ∀ x ∈ Set.Ioo a b, deriv g x = 1 / deriv f x := 326 | sorry 327 | 328 | theorem exercise_5_3 {g : ℝ → ℝ} (hg : Continuous g) 329 | (hg' : ∃ M : ℝ, ∀ x : ℝ, |deriv g x| ≤ M) : 330 | ∃ N, ∀ ε > 0, ε < N → Function.Injective (λ x : ℝ => x + ε * g x) := 331 | sorry 332 | 333 | theorem exercise_5_4 {n : ℕ} 334 | (C : ℕ → ℝ) 335 | (hC : ∑ i ∈ (range (n + 1)), (C i) / (i + 1) = 0) : 336 | ∃ x, x ∈ (Set.Icc (0 : ℝ) 1) ∧ ∑ i ∈ range (n + 1), (C i) * (x^i) = 0 := 337 | sorry 338 | 339 | theorem exercise_5_5 340 | {f : ℝ → ℝ} 341 | (hfd : Differentiable ℝ f) 342 | (hf : Tendsto (deriv f) atTop (𝓝 0)) : 343 | Tendsto (λ x => f (x + 1) - f x) atTop atTop := 344 | sorry 345 | 346 | theorem exercise_5_6 347 | {f : ℝ → ℝ} 348 | (hf1 : Continuous f) 349 | (hf2 : ∀ x, DifferentiableAt ℝ f x) 350 | (hf3 : f 0 = 0) 351 | (hf4 : Monotone (deriv f)) : 352 | MonotoneOn (λ x => f x / x) (Set.Ioi 0) := 353 | sorry 354 | 355 | theorem exercise_5_7 356 | {f g : ℝ → ℝ} {x : ℝ} 357 | (hf' : DifferentiableAt ℝ f 0) 358 | (hg' : DifferentiableAt ℝ g 0) 359 | (hg'_ne_0 : deriv g 0 ≠ 0) 360 | (f0 : f 0 = 0) (g0 : g 0 = 0) : 361 | Tendsto (λ x => f x / g x) (𝓝 x) (𝓝 (deriv f x / deriv g x)) := 362 | sorry 363 | 364 | theorem exercise_5_15 {f : ℝ → ℝ} (a M0 M1 M2 : ℝ) 365 | (hf' : DifferentiableOn ℝ f (Set.Ici a)) 366 | (hf'' : DifferentiableOn ℝ (deriv f) (Set.Ici a)) 367 | (hM0 : M0 = sSup {(|f x|) | x ∈ (Set.Ici a)}) 368 | (hM1 : M1 = sSup {(|deriv f x|) | x ∈ (Set.Ici a)}) 369 | (hM2 : M2 = sSup {(|deriv (deriv f) x|) | x ∈ (Set.Ici a)}) : 370 | (M1 ^ 2) ≤ 4 * M0 * M2 := 371 | sorry 372 | 373 | theorem exercise_5_17 374 | {f : ℝ → ℝ} 375 | (hf' : DifferentiableOn ℝ f (Set.Icc (-1) 1)) 376 | (hf'' : DifferentiableOn ℝ (deriv f) (Set.Icc 1 1)) 377 | (hf''' : DifferentiableOn ℝ (deriv (deriv f)) (Set.Icc 1 1)) 378 | (hf0 : f (-1) = 0) 379 | (hf1 : f 0 = 0) 380 | (hf2 : f 1 = 1) 381 | (hf3 : deriv f 0 = 0) : 382 | ∃ x, x ∈ Set.Ioo (-1 : ℝ) 1 ∧ deriv (deriv (deriv f)) x ≥ 3 := 383 | sorry 384 | -------------------------------------------------------------------------------- /formal/Shakarachi.lean: -------------------------------------------------------------------------------- 1 | import Mathlib 2 | 3 | open Complex Filter Function Metric Finset 4 | open scoped BigOperators Topology 5 | 6 | theorem exercise_1_13a {f : ℂ → ℂ} (Ω : Set ℂ) (a b : Ω) (h : IsOpen Ω) 7 | (hf : DifferentiableOn ℂ f Ω) (hc : ∃ (c : ℝ), ∀ z ∈ Ω, (f z).re = c) : 8 | f a = f b := 9 | sorry 10 | 11 | theorem exercise_1_13b {f : ℂ → ℂ} (Ω : Set ℂ) (a b : Ω) (h : IsOpen Ω) 12 | (hf : DifferentiableOn ℂ f Ω) (hc : ∃ (c : ℝ), ∀ z ∈ Ω, (f z).im = c) : 13 | f a = f b := 14 | sorry 15 | 16 | theorem exercise_1_13c {f : ℂ → ℂ} (Ω : Set ℂ) (a b : Ω) (h : IsOpen Ω) 17 | (hf : DifferentiableOn ℂ f Ω) (hc : ∃ (c : ℝ), ∀ z ∈ Ω, norm (f z) = c) : 18 | f a = f b := 19 | sorry 20 | 21 | theorem exercise_1_19a (z : ℂ) (hz : norm z = 1) (s : ℕ → ℂ) 22 | (h : s = (λ n => ∑ i ∈ (range n), i * z ^ i)) : 23 | ¬ ∃ y, Tendsto s atTop (𝓝 y) := 24 | sorry 25 | 26 | theorem exercise_1_19b (z : ℂ) (hz : norm z = 1) (s : ℕ → ℂ) 27 | (h : s = (λ n => ∑ i ∈ (range n), i * z / i ^ 2)) : 28 | ∃ y, Tendsto s atTop (𝓝 y) := 29 | sorry 30 | 31 | theorem exercise_1_19c (z : ℂ) (hz : norm z = 1) (hz2 : z ≠ 1) (s : ℕ → ℂ) 32 | (h : s = (λ n => ∑ i ∈ (range n), i * z / i)) : 33 | ∃ z, Tendsto s atTop (𝓝 z) := 34 | sorry 35 | 36 | theorem exercise_1_26 37 | (f F₁ F₂ : ℂ → ℂ) (Ω : Set ℂ) (h1 : IsOpen Ω) (h2 : IsConnected Ω) 38 | (hF₁ : DifferentiableOn ℂ F₁ Ω) (hF₂ : DifferentiableOn ℂ F₂ Ω) 39 | (hdF₁ : ∀ x ∈ Ω, deriv F₁ x = f x) (hdF₂ : ∀ x ∈ Ω, deriv F₂ x = f x) 40 | : ∃ c : ℂ, ∀ x, F₁ x = F₂ x + c := 41 | sorry 42 | 43 | theorem exercise_2_2 : 44 | Tendsto (λ y => ∫ x in (0 : ℝ)..y, Real.sin x / x) atTop (𝓝 (Real.pi / 2)) := 45 | sorry 46 | 47 | theorem exercise_2_9 48 | {f : ℂ → ℂ} (Ω : Set ℂ) (b : Bornology.IsBounded Ω) (h : IsOpen Ω) 49 | (hf : DifferentiableOn ℂ f Ω) (z : Ω) (hz : f z = z) (h'z : deriv f z = 1) : 50 | ∃ (f_lin : ℂ →L[ℂ] ℂ), ∀ x ∈ Ω, f x = f_lin x := 51 | sorry 52 | 53 | theorem exercise_2_13 {f : ℂ → ℂ} 54 | (hf : ∀ z₀ : ℂ, ∃ (s : Set ℂ) (c : ℕ → ℂ), IsOpen s ∧ z₀ ∈ s ∧ 55 | ∀ z ∈ s, Tendsto (λ n => ∑ i ∈ range n, (c i) * (z - z₀)^i) atTop (𝓝 (f z₀)) 56 | ∧ ∃ i, c i = 0) : 57 | ∃ (c : ℕ → ℂ) (n : ℕ), f = λ z => ∑ i ∈ range n, (c i) * z ^ n := 58 | sorry 59 | 60 | 61 | theorem exercise_3_3 (a : ℝ) (ha : 0 < a) : 62 | Tendsto (λ y => ∫ x in -y..y, Real.cos x / (x ^ 2 + a ^ 2)) 63 | atTop (𝓝 (Real.pi * (Real.exp (-a) / a))) := 64 | sorry 65 | 66 | theorem exercise_3_4 (a : ℝ) (ha : 0 < a) : 67 | Tendsto (λ y => ∫ x in -y..y, x * Real.sin x / (x ^ 2 + a ^ 2)) 68 | atTop (𝓝 (Real.pi * (Real.exp (-a)))) := 69 | sorry 70 | 71 | theorem exercise_3_9 : ∫ x in (0 : ℝ)..(1 : ℝ), Real.log (Real.sin (Real.pi * x)) = - Real.log 2 := 72 | sorry 73 | 74 | theorem exercise_3_14 {f : ℂ → ℂ} (hf : Differentiable ℂ f) 75 | (hf_inj : Function.Injective f) : 76 | ∃ (a b : ℂ), f = (λ z => a * z + b) ∧ a ≠ 0 := 77 | sorry 78 | 79 | theorem exercise_3_22 (D : Set ℂ) (hD : D = ball 0 1) (f : ℂ → ℂ) 80 | (hf : DifferentiableOn ℂ f D) (hfc : ContinuousOn f (closure D)) : 81 | ¬ ∀ z ∈ (sphere (0 : ℂ) 1), f z = 1 / z := 82 | sorry 83 | 84 | theorem exercise_5_1 (f : ℂ → ℂ) (hf : DifferentiableOn ℂ f (ball 0 1)) 85 | (hb : Bornology.IsBounded (Set.range f)) (h0 : f ≠ 0) (zeros : ℕ → ℂ) (hz : ∀ n, f (zeros n) = 0) 86 | (hzz : Set.range zeros = {z | f z = 0 ∧ z ∈ (ball (0 : ℂ) 1)}) : 87 | ∃ (z : ℂ), Tendsto (λ n => (∑ i ∈ range n, (1 - zeros i))) atTop (𝓝 z) := 88 | sorry 89 | -------------------------------------------------------------------------------- /informal/Artin.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/rahul3613/ProofNet-lean4/1893b2f054fddb1c4c90b91732173528c16f3a37/informal/Artin.pdf -------------------------------------------------------------------------------- /informal/Artin.tex: -------------------------------------------------------------------------------- 1 | \documentclass{article} 2 | 3 | \title{\textbf{ 4 | Exercises from \\ 5 | \textit{Algebra} \\ 6 | by Michael Artin 7 | }} 8 | 9 | \date{} 10 | 11 | \usepackage{amsmath} 12 | \usepackage{amssymb} 13 | \usepackage{amsthm} 14 | 15 | \begin{document} 16 | \maketitle 17 | 18 | 19 | \paragraph{Exercise 2.2.9} Let $H$ be the subgroup generated by two elements $a, b$ of a group $G$. Prove that if $a b=b a$, then $H$ is an abelian group. 20 | \begin{proof} 21 | Since $a$ and $b$ commute, for any $g, h\in H$ we can write $g=a^ib^j$ and $h = a^kb^l$. Then $gh = a^ib^ja^kb^l = a^kb^la^ib^j = hg$. Thus $H$ is abelian. 22 | \end{proof} 23 | 24 | 25 | 26 | \paragraph{Exercise 2.3.2} Prove that the products $a b$ and $b a$ are conjugate elements in a group. 27 | \begin{proof} 28 | We have that $(a^{-1})ab(a^{-1})^{-1} = ba$. 29 | \end{proof} 30 | 31 | 32 | 33 | \paragraph{Exercise 2.4.19} Prove that if a group contains exactly one element of order 2 , then that element is in the center of the group. 34 | \begin{proof} 35 | Let $x$ be the element of order two. Consider the element $z=y^{-1} x y$, we have: $z^2=\left(y^{-1} x y\right)^2=\left(y^{-1} x y\right)\left(y^{-1} x y\right)=e$. So: $z=x$, and $y^{-1} x y=x$. So: $x y=y x$. So: $x$ is in the center of $G$. 36 | \end{proof} 37 | 38 | 39 | 40 | \paragraph{Exercise 2.8.6} Prove that the center of the product of two groups is the product of their centers. 41 | \begin{proof} 42 | We have that $(g_1, g_2)\cdot (h_1, h_2) = (h_1, h_2)\cdot (g_1, g_2)$ if and only if $g_1h_1 = h_1g_1$ and $g_2h_2 = h_2g_2$. 43 | \end{proof} 44 | 45 | 46 | 47 | \paragraph{Exercise 2.11.3} Prove that a group of even order contains an element of order $2 .$ 48 | \begin{proof} 49 | Pair up if possible each element of $G$ with its inverse, and observe that 50 | $$ 51 | g^2 \neq e \Longleftrightarrow g \neq g^{-1} \Longleftrightarrow \text { there exists the pair }\left(g, g^{-1}\right) 52 | $$ 53 | Now, there is one element that has no pairing: the unit $e$ (since indeed $e=e^{-1} \Longleftrightarrow e^2=e$ ), so since the number of elements of $G$ is even there must be at least one element more, say $e \neq a \in G$, without a pairing, and thus $a=a^{-1} \Longleftrightarrow a^2=e$ 54 | \end{proof} 55 | 56 | 57 | 58 | \paragraph{Exercise 3.2.7} Prove that every homomorphism of fields is injective. 59 | \begin{proof} 60 | Suppose $f(a)=f(b)$, then $f(a-b)=0=f(0)$. If $u=(a-b) \neq 0$, then $f(u) f\left(u^{-1}\right)=f(1)=1$, but that means that $0 f\left(u^{-1}\right)=1$, which is impossible. Hence $a-b=0$ and $a=b$. 61 | \end{proof} 62 | 63 | 64 | 65 | \paragraph{Exercise 3.5.6} Let $V$ be a vector space which is spanned by a countably infinite set. Prove that every linearly independent subset of $V$ is finite or countably infinite. 66 | \begin{proof} 67 | Let $A$ be the countable generating set, and let $U$ be an uncountable linearly independent set. It can be extended to a basis $B$ of the whole space. Now consider the subset $C$ of elements of $B$ that appear in the $B$-decompositions of elements of $A$. 68 | Since only finitely many elements are involved in the decomposition of each element of $A$, the set $C$ is countable. But $C$ also clearly generates the vector space $V$. This contradicts the fact that it is a proper subset of the basis $B$ (since $B$ is uncountable). 69 | \end{proof} 70 | 71 | 72 | 73 | \paragraph{Exercise 3.7.2} Let $V$ be a vector space over an infinite field $F$. Prove that $V$ is not the union of finitely many proper subspaces. 74 | \begin{proof} 75 | If $V$ is the set-theoretic union of $n$ proper subspaces $W_i$ ( $1 \leq i \leq n$ ), then $|F| \leq n-1$. 76 | Proof. We may suppose no $W_i$ is contained in the union of the other subspaces. Let $u \in W_i, \quad u \notin \bigcup_{j \neq i} W_j$ and $v \notin W_i$. 77 | Then $(v+F u) \cap W_i=\varnothing$ and $(v+F u) \cap W_j(j \neq i)$ contains at most one vector since otherwise $W_j$ would contain $u$. Hence 78 | $$ 79 | |v+F u|=|F| \leq n-1 . 80 | $$ 81 | Corollary: Avoidance lemma for vector spaces. 82 | Let $E$ be a vector space over an infinite field. If a subspace is contained in a finite union of subspaces, it is contained in one of them. 83 | \end{proof} 84 | 85 | 86 | 87 | \paragraph{Exercise 6.1.14} Let $Z$ be the center of a group $G$. Prove that if $G / Z$ is a cyclic group, then $G$ is abelian and hence $G=Z$. 88 | \begin{proof} 89 | We have that $G / Z(G)$ is cyclic, and so there is an element $x \in G$ such that $G / Z(G)=\langle x Z(G)\rangle$, where $x Z(G)$ is the coset with representative $x$. Now let $g \in G$ 90 | We know that $g Z(G)=(x Z(G))^m$ for some $m$, and by definition $(x Z(G))^m=x^m Z(G)$. 91 | Now, in general, if $H \leq G$, we have by definition too that $a H=b H$ if and only if $b^{-1} a \in H$. 92 | In our case, we have that $g Z(G)=x^m Z(G)$, and this happens if and only if $\left(x^m\right)^{-1} g \in Z(G)$. 93 | Then, there's a $z \in Z(G)$ such that $\left(x^m\right)^{-1} g=z$, and so $g=x^m z$. 94 | 95 | $g, h \in G$ implies that $g=x^{a_1} z_1$ and $h=x^{a_2} z_2$, so 96 | $$ 97 | \begin{aligned} 98 | g h & =\left(x^{a_1} z_1\right)\left(x^{a_2} z_2\right) \\ 99 | & =x^{a_1} x^{a_2} z_1 z_2 \\ 100 | & =x^{a_1+a_2} z_2 z_1 \\ 101 | & =\ldots=\left(x^{a_2} z_2\right)\left(x^{a_1} z_1\right)=h g . 102 | \end{aligned} 103 | $$ 104 | Therefore, $G$ is abelian. 105 | \end{proof} 106 | 107 | 108 | 109 | \paragraph{Exercise 6.4.2} Prove that no group of order $p q$, where $p$ and $q$ are prime, is simple. 110 | \begin{proof} 111 | If $|G|=n=p q$ then the only two Sylow subgroups are of order $p$ and $q$. 112 | From Sylow's third theorem we know that $n_p \mid q$ which means that $n_p=1$ or $n_p=q$. 113 | If $n_p=1$ then we are done (by a corollary of Sylow's theorem) 114 | If $n_p=q$ then we have accounted for $q(p-1)=p q-q$ elements of $G$ and so there is only one group of order $q$ and again we are done. 115 | \end{proof} 116 | 117 | 118 | 119 | \paragraph{Exercise 6.4.3} Prove that no group of order $p^2 q$, where $p$ and $q$ are prime, is simple. 120 | \begin{proof} 121 | We may as well assume $p0$ and $b^2>0$, so $-a^2-b^2<0$. Furthermore, it is an integer. Thus, we have found a nonzero integer in $I$. 207 | \end{proof} 208 | 209 | 210 | 211 | \paragraph{Exercise 10.4.6} Let $I, J$ be ideals in a ring $R$. Prove that the residue of any element of $I \cap J$ in $R / I J$ is nilpotent. 212 | \begin{proof} 213 | If $x$ is in $I \cap J, x \in I$ and $x \in J . R / I J=\{r+a b: a \in I, b \in J, r \in R\}$. Then $x \in I \cap J \Rightarrow x \in I$ and $x \in J$, and so $x^2 \in I J$. Thus 214 | $$ 215 | [x]^2=\left[x^2\right]=[0] \text { in } R / I J 216 | $$ 217 | \end{proof} 218 | 219 | 220 | 221 | \paragraph{Exercise 10.4.7a} Let $I, J$ be ideals of a ring $R$ such that $I+J=R$. Prove that $I J=I \cap J$. 222 | \begin{proof} 223 | We have seen that $IJ \subset I \cap J$, so it remains to show that $I \cap J \subset IJ$. Since $I+J = (1)$, there are elements $i \in I$ and $j \in J$ such that $i+j = 1$. Let $k \in I \cap J$, and multiply $i+j=1$ through by $k$ to get $ki+kj = k$. Write this more suggestively as 224 | \[ k = ik+kj. \] 225 | The first term is in $IJ$ because $k \in J$, and the second term is in $IJ$ because $k \in I$, so $k \in IJ$ as desired. 226 | \end{proof} 227 | 228 | \paragraph{Exercise 10.7.10} Let $R$ be a ring, with $M$ an ideal of $R$. Suppose that every element of $R$ which is not in $M$ is a unit of $R$. Prove that $M$ is a maximal ideal and that moreover it is the only maximal ideal of $R$. 229 | \begin{proof} 230 | Suppose there is an ideal $M\subset I\subset R$. If $I\neq M$, then $I$ contains a unit, thus $I=R$. Therefore $M$ is a maximal ideal. 231 | 232 | Suppose we have an arbitrary maximal ideal $M^\prime$ of $R$. The ideal $M^\prime$ cannot contain a unit, otherwise $M^\prime =R$. Therefore $M^\prime \subset M$. But we cannot have $M^\prime \subsetneq M \subsetneq R$, therefore $M=M^\prime$. 233 | \end{proof} 234 | 235 | 236 | 237 | \paragraph{Exercise 11.2.13} If $a, b$ are integers and if $a$ divides $b$ in the ring of Gauss integers, then $a$ divides $b$ in $\mathbb{Z}$. 238 | \begin{proof} 239 | Suppose $a|b$ in $\mathbb{Z}[i]$ and $a,b\in\mathbb{Z}$. Then $a(x+yi)=b$ for $x,y\in\mathbb{Z}$. Expanding this we get $ax+ayi=b$, and equating imaginary parts gives us $ay=0$, implying $y=0$. 240 | \end{proof} 241 | 242 | 243 | \paragraph{Exercise 11.4.1b} Prove that $x^3 + 6x + 12$ is irreducible in $\mathbb{Q}$. 244 | \begin{proof} 245 | Apply Eisenstein's criterion with $p=3$. 246 | \end{proof} 247 | 248 | 249 | 250 | \paragraph{Exercise 11.4.6a} Prove that $x^2+x+1$ is irreducible in the field $\mathbb{F}_2$. 251 | \begin{proof} 252 | If $x^2+x+1$ were reducible in $\mathbb{F}_2$, its factors must be linear. But we neither have that $0^2+0+1=$ nor $1^2+1+1=0$, therefore $x^2+x+1$ is irreducible. 253 | \end{proof} 254 | 255 | 256 | \paragraph{Exercise 11.4.6b} Prove that $x^2+1$ is irreducible in $\mathbb{F}_7$ 257 | \begin{proof} 258 | If $p(x)=x^2+1$ were reducible, its factors must be linear. But no $p(a)$ for $a\in\mathbb{F}_7$ evaluates to 0, therefore $x^2+1$ is irreducible. 259 | \end{proof} 260 | 261 | 262 | 263 | \paragraph{Exercise 11.4.6c} Prove that $x^3 - 9$ is irreducible in $\mathbb{F}_{31}$. 264 | \begin{proof} 265 | If $p(x) = x^3-9$ were reducible, it would have a linear factor, since it either has a linear factor and a quadratic factor or three linear factors. We can then verify by brute force that $p(x)\neq 0$ for $x \in \mathbb{F}_31$. 266 | \end{proof} 267 | 268 | 269 | 270 | \paragraph{Exercise 11.4.8} Let $p$ be a prime integer. Prove that the polynomial $x^n-p$ is irreducible in $\mathbb{Q}[x]$. 271 | \begin{proof} 272 | Straightforward application of Eisenstein's criterion with $p$. 273 | \end{proof} 274 | 275 | \paragraph{Exercise 11.13.3} Prove that there are infinitely many primes congruent to $-1$ (modulo $4$). 276 | \begin{proof} 277 | First we show a lemma: if $a \equiv 3(\bmod 4)$ then there exists a prime $p$ such that $p \mid a$ and $p \equiv 3(\bmod 4)$. 278 | 279 | Clearly, all primes dividing $a$ are odd. Suppose all of them would be $\equiv 1(\bmod 4)$. Then their product would also be $a \equiv 1(\bmod 4)$, which is a contradiction. 280 | 281 | To prove the main claim, suppose that $p_1, \ldots, p_n$ would be all such primes. (In particular, we have $p_1=3$.) Consider $a=4 p_2 \cdots p_n+3$. (Or you can take $a=4 p_2 \cdots p_n-1$.) Show that $p_i \nmid a$ for $i=1, \ldots, n$. (The case $3 \nmid a$ is solved differently than the other primes - this is the reason for omitting $p_1$ in the definition of $a$.) Then use the above lemma to get a contradiction. 282 | \end{proof} 283 | 284 | 285 | 286 | \paragraph{Exercise 13.4.10} Prove that if a prime integer $p$ has the form $2^r+1$, then it actually has the form $2^{2^k}+1$. 287 | \begin{proof} 288 | In particular, we have 289 | $$ 290 | \frac{x^a+1}{x+1}=\frac{(-x)^a-1}{(-x)-1}=1-x+x^2-\cdots+(-x)^{a-1} 291 | $$ 292 | by the geometric sum formula. In this case, specialize to $x=2^{2^m}$ and we have a nontrivial divisor. 293 | \end{proof} 294 | 295 | 296 | 297 | \paragraph{Exercise 13.6.10} Let $K$ be a finite field. Prove that the product of the nonzero elements of $K$ is $-1$. 298 | \begin{proof} 299 | Since we are working with a finite field with $q$ elements, anyone of them is a root of the following polynomial 300 | $$ 301 | x^q-x=0 . 302 | $$ 303 | In particular if we rule out the 0 element, any $a_i \neq 0$ is a root of 304 | $$ 305 | x^{q-1}-1=0 . 306 | $$ 307 | This polynomial splits completely in $\mathbb{F}_q$ so we find 308 | $$ 309 | \left(x-a_1\right) \cdots\left(x-a_{q-1}\right)=0 310 | $$ 311 | in particular 312 | $$ 313 | x^{q-1}-1=\left(x-a_1\right) \cdots\left(x-a_{q-1}\right) 314 | $$ 315 | Thus $a_1 \cdots a_{q-1}=-1$. 316 | \end{proof} 317 | \end{document} 318 | -------------------------------------------------------------------------------- /informal/Axler.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/rahul3613/ProofNet-lean4/1893b2f054fddb1c4c90b91732173528c16f3a37/informal/Axler.pdf -------------------------------------------------------------------------------- /informal/Axler.tex: -------------------------------------------------------------------------------- 1 | \documentclass{article} 2 | 3 | \title{\textbf{ 4 | Exercises from \\ 5 | \textit{Linear Algebra Done Right} \\ 6 | by Sheldon Axler 7 | }} 8 | 9 | \date{} 10 | 11 | \usepackage{amsmath, amsthm} 12 | \usepackage{amssymb} 13 | 14 | 15 | \theoremstyle{definition} 16 | \newtheorem{lem}{Lemma}[section] 17 | \newtheorem{cor}[lem]{Corollary} 18 | \newtheorem{prop}[lem]{Proposition} 19 | \newtheorem{thm}[lem]{Theorem} 20 | \newtheorem{remark}[lem]{Remark} 21 | \newtheorem{defn}[lem]{Definition} 22 | 23 | \newtheorem*{prop*}{Proposition} 24 | \newtheorem*{thm*}{Theorem} 25 | \newtheorem*{defn*}{Definition} 26 | \newtheorem*{lem*}{Lemma} 27 | 28 | 29 | \begin{document} 30 | \maketitle 31 | 32 | 33 | \paragraph{Exercise 1.2} Show that $\frac{-1 + \sqrt{3}i}{2}$ is a cube root of 1 (meaning that its cube equals 1). 34 | \begin{proof} 35 | $$ 36 | \left(\frac{-1+\sqrt{3} i}{2}\right)^2=\frac{-1-\sqrt{3} i}{2}, 37 | $$ 38 | hence 39 | $$ 40 | \left(\frac{-1+\sqrt{3} i}{2}\right)^3=\frac{-1-\sqrt{3} i}{2} \cdot \frac{-1+\sqrt{3} i}{2}=1 41 | $$ 42 | This means $\frac{-1+\sqrt{3} i}{2}$ is a cube root of 1. 43 | \end{proof} 44 | 45 | 46 | 47 | \paragraph{Exercise 1.3} Prove that $-(-v) = v$ for every $v \in V$. 48 | \begin{proof} 49 | By definition, we have 50 | $$ 51 | (-v)+(-(-v))=0 \quad \text { and } \quad v+(-v)=0 . 52 | $$ 53 | This implies both $v$ and $-(-v)$ are additive inverses of $-v$, by the uniqueness of additive inverse, it follows that $-(-v)=v$. 54 | \end{proof} 55 | 56 | 57 | 58 | \paragraph{Exercise 1.4} Prove that if $a \in \mathbf{F}$, $v \in V$, and $av = 0$, then $a = 0$ or $v = 0$. 59 | \begin{proof} 60 | If $a=0$, then we immediately have our result. So suppose $a \neq 0$. Then, because $a$ is some nonzero real or complex number, it has a multiplicative inverse $\frac{1}{a}$. Now suppose that $v$ is some vector such that 61 | $$ 62 | a v=0 63 | $$ 64 | Multiply by $\frac{1}{a}$ on both sides of this equation to get 65 | $$ 66 | \begin{aligned} 67 | \frac{1}{a}(a v) & =\frac{1}{a} 0 & & \\ 68 | \frac{1}{a}(a v) & =0 & & \\ 69 | \left(\frac{1}{a} \cdot a\right) v & =0 & & \text { (associativity) } \\ 70 | 1 v & =0 & & \text { (definition of } 1/a) \\ 71 | v & =0 & & \text { (multiplicative identity) } 72 | \end{aligned} 73 | $$ 74 | Hence either $a=0$ or, if $a \neq 0$, then $v=0$. 75 | \end{proof} 76 | 77 | 78 | 79 | \paragraph{Exercise 1.6} Give an example of a nonempty subset $U$ of $\mathbf{R}^2$ such that $U$ is closed under addition and under taking additive inverses (meaning $-u \in U$ whenever $u \in U$), but $U$ is not a subspace of $\mathbf{R}^2$. 80 | \begin{proof} 81 | \[U=\mathbb{Z}^2=\left\{(x, y) \in \mathbf{R}^2: x, y \text { are integers }\right\}\] 82 | $U=\mathbb{Z}^2$ satisfies the desired properties. To come up with this, note by assumption, $U$ must be closed under addition and subtraction, so in particular, it must contain 0 . We need to find a set which fails scalar multiplication. A discrete set like $\mathbb{Z}^2$ does this. 83 | \end{proof} 84 | 85 | 86 | 87 | \paragraph{Exercise 1.7} Give an example of a nonempty subset $U$ of $\mathbf{R}^2$ such that $U$ is closed under scalar multiplication, but $U$ is not a subspace of $\mathbf{R}^2$. 88 | \begin{proof} 89 | $$ 90 | U=\left\{(x, y) \in \mathbf{R}^2:|x|=|y|\right\} 91 | $$ 92 | For $(x, y) \in U$ and $\lambda \in \mathbb{R}$, it follows $\lambda(x, y)=$ $(\lambda x, \lambda y)$, so $|\lambda x|=|\lambda||x|=|\lambda||y|=|\lambda y|$. Therefore, $\lambda(x, y) \in U$. 93 | 94 | On the other hand, consider $a=(1,-1), b=$ $(1,1) \in U$. Then, $a+b=(1,-1)+(1,1)=$ $(2,0) \notin U$. So, $U$ is not a subspace of $\mathbb{R}^2$. 95 | \end{proof} 96 | 97 | 98 | 99 | \paragraph{Exercise 1.8} Prove that the intersection of any collection of subspaces of $V$ is a subspace of $V$. 100 | \begin{proof} 101 | Let $V_1, V_2, \ldots, V_n$ be subspaces of the vector space $V$ over the field $F$. We must show that their intersection $V_1 \cap V_2 \cap \ldots \cap V_n$ is also a subspace of $V$. 102 | 103 | To begin, we observe that the additive identity $0$ of $V$ is in $V_1 \cap V_2 \cap \ldots \cap V_n$. This is because $0$ is in each subspace $V_i$, as they are subspaces and hence contain the additive identity. 104 | 105 | Next, we show that the intersection of subspaces is closed under addition. Let $u$ and $v$ be vectors in $V_1 \cap V_2 \cap \ldots \cap V_n$. By definition, $u$ and $v$ belong to each of the subspaces $V_i$. Since each $V_i$ is a subspace and therefore closed under addition, it follows that $u+v$ belongs to each $V_i$. Thus, $u+v$ belongs to the intersection $V_1 \cap V_2 \cap \ldots \cap V_n$. 106 | 107 | Finally, we show that the intersection of subspaces is closed under scalar multiplication. Let $a$ be a scalar in $F$ and let $v$ be a vector in $V_1 \cap V_2 \cap \ldots \cap V_n$. Since $v$ belongs to each $V_i$, we have $av$ belongs to each $V_i$ as well, as $V_i$ are subspaces and hence closed under scalar multiplication. Therefore, $av$ belongs to the intersection $V_1 \cap V_2 \cap \ldots \cap V_n$. 108 | 109 | Thus, we have shown that $V_1 \cap V_2 \cap \ldots \cap V_n$ is a subspace of $V$. 110 | \end{proof} 111 | 112 | 113 | 114 | \paragraph{Exercise 1.9} Prove that the union of two subspaces of $V$ is a subspace of $V$ if and only if one of the subspaces is contained in the other. 115 | \begin{proof} 116 | To prove this one way, suppose for purposes of contradiction that for $U_1$ and $U_2$, which are subspaces of $V$, that $U_1 \cup U_2$ is a subspace and neither is completely contained within the other. In other words, $U_1 \nsubseteq U_2$ and $U_2 \nsubseteq U_1$. We will show that you can pick a vector $v \in U_1$ and a vector $u \in U_2$ such that $v+u \notin U_1 \cup U_2$, proving that if $U_1 \cup U_2$ is a subspace, one must be completely contained inside the other. 117 | 118 | If $U_1 \nsubseteq U_2$, we can pick a $v \in U_1$ such that $v \notin U_2$. Since $v$ is in the subspace $U_1$, then $(-v)$ must also be, by definition. Similarly, if $U_2 \nsubseteq U_1$, then we can pick a $u \in U_2$ such that $u \notin U_1$. Since $u$ is in the subspace $U_2$, then $(-u)$ must also be, by definition. 119 | 120 | If $v+u \in U_1 \cup U_2$, then $v+u$ must be in $U_1$ or $U_2$. But, $v+u \in U_1 \Rightarrow v+u+(-v) \in U_1 \Rightarrow u \in U_1$ 121 | Similarly, 122 | $$ 123 | v+u \in U_2 \Rightarrow v+u+(-u) \in U_2 \Rightarrow v \in U_2 124 | $$ 125 | This is clearly a contradiction, as each element was defined to not be in these subspaces. Thus our initial assumption must have been wrong, and $U_1 \subseteq U_2$ or $U_2 \subseteq U_1$ 126 | To prove the other way, Let $U_1 \subseteq U_2$ (WLOG). $U_1 \subseteq U_2 \Rightarrow U_1 \cup U_2=U_2$. Since $U_2$ is a subspace, $U_1 \cup U_2$ is as well. QED. 127 | \end{proof} 128 | 129 | 130 | 131 | \paragraph{Exercise 3.1} Show that every linear map from a one-dimensional vector space to itself is multiplication by some scalar. More precisely, prove that if $\operatorname{dim} V=1$ and $T \in \mathcal{L}(V, V)$, then there exists $a \in \mathbf{F}$ such that $T v=a v$ for all $v \in V$. 132 | \begin{proof} 133 | If $\operatorname{dim} V=1$, then in fact, $V=\mathbf{F}$ and it is spanned by $1 \in \mathbf{F}$. 134 | Let $T$ be a linear map from $V$ to itself. Let $T(1)=\lambda \in V(=\mathbf{F})$. 135 | Step 2 136 | 2 of 3 137 | Every $v \in V$ is a scalar. Therefore, 138 | $$ 139 | \begin{aligned} 140 | T(v) & =T(v \cdot 1) \\ 141 | & =v T(1) \ldots .(\text { By the linearity of } T) \\ 142 | & =v \lambda 143 | \end{aligned} 144 | $$ 145 | Hence, $T v=\lambda v$ for every $v \in V$. 146 | \end{proof} 147 | 148 | 149 | 150 | \paragraph{Exercise 3.8} Suppose that $V$ is finite dimensional and that $T \in \mathcal{L}(V, W)$. Prove that there exists a subspace $U$ of $V$ such that $U \cap \operatorname{null} T=\{0\}$ and range $T=\{T u: u \in U\}$. 151 | \begin{proof} 152 | The point here is to note that every subspace of a vector space has a complementary subspace. 153 | In this example, $U$ will precisely turn out to be the complementary subspace of null $T$. That is, $V=U \oplus \operatorname{null} T$ 154 | How should we characterize $U$ ? This can be achieved by extending a basis $B_1=\left\{v_1, v_2, \ldots, v_m\right\}$ of null $T$ to a basis of $V$. Let $B_2=\left\{u_1, u_2, \ldots, u_n\right\}$ be such that $B=B_1 \cup B_2$ is a basis of $V$. 155 | 156 | Define $U=\operatorname{span}\left(B_2\right)$. Now, since $B_1$ and $B_2$ are complementary subsets of the basis $B$ of $V$, their spans will turn out to be complementary subspaces of $V$. Let's prove that $V=U \oplus$ null $T$. 157 | 158 | Let $v \in V$. Then, $v$ can be expressed as a linear combination of the vectors in $B$. 159 | Let $v=a_1 u_1+\cdots+a_n u_n+c_1 v_1+\cdots+c_m v_m$. However, since $\left\{u_1, u_2, \ldots, u_n\right\}$ is a basis of $U, a_1 u_1+$ $\cdots+a_n u_n=u \in U$ and since $\left\{v_1, v_2, \ldots, v_m\right\}$ is a basis of null $T, c_1 v_1+\cdots+c_m v_m=w \in$ null $T$. 160 | Hence, $v=u+w \in U+\operatorname{null} T$. This shows that 161 | $$ 162 | V=U+\operatorname{null} T 163 | $$ 164 | Now, let $v \in U \cap \operatorname{null} T$. 165 | Since $v \in U, u$ can be expressed as a linear combination of basis vectors of $U$. Let 166 | $$ 167 | v=a_1 u_1+\cdots+a_n u_n 168 | $$ 169 | Similarly, since $v \in \operatorname{null} T$, it can also be expressed as a linear combination of the basis vectors of null $T$. Let 170 | $$ 171 | v=c_1 v_1+\cdots+c_m v_m 172 | $$ 173 | The left hand sides of the above two equations are equal. Therefore, we can equate the right hand sides. 174 | $$ 175 | \begin{aligned} 176 | & a_1 u_1+\cdots+a_n u_n=v=c_1 v_1+\cdots+c_m v_m \\ 177 | & a_1 u_1+\cdots+a_n u_n-c_1 v_1-\cdots-c_m v_m=0 178 | \end{aligned} 179 | $$ 180 | We have found a linear combination of $u_i^{\prime}$ 's and $v_i$ 's which is equal to zero. However, they are basis vectors of $V$. Hence, all the multipliers $c_i$ 's and $a_i$ 's must be zero implying that $v=0$. 181 | Therefore, if $v \in U \cap$ null $T$, then $v=0$. this means that 182 | $$ 183 | U \cap \operatorname{null} T=\{0\} 184 | $$ 185 | The above shows that $U$ satisfies the first of the required conditions. 186 | Now let $w \in$ range $T$. Then, there exists $v \in V$ such that $T v=w$. This allows us to write $v=u+w$ where $u \in U$ and $w \in$ null $T$. This implies 187 | $$ 188 | \begin{aligned} 189 | w & =T v \\ 190 | & =T(u+w) \\ 191 | & =T u+T w \\ 192 | & =T u+0 \quad \quad(\text { since } w \in \operatorname{null} T) \\ 193 | & =T u 194 | \end{aligned} 195 | $$ 196 | This shows that if $w \in$ range $T$ then $w=T u$ for some $u \in U$. Therefore, range $T \subseteq\{T u \mid u \in U\}$. 197 | Since $U$ is a subspace of $V$, it follows that $T u \in$ range $T$ for all $u \in U$. Thus, $\{T u \mid u \in U\} \subseteq$ range $T$. 198 | Therefore, range $T=\{T u \mid u \in U\}$. 199 | This shows that $U$ satisfies the second required condition as well. 200 | \end{proof} 201 | 202 | 203 | \paragraph{Exercise 4.4} Suppose $p \in \mathcal{P}(\mathbf{C})$ has degree $m$. Prove that $p$ has $m$ distinct roots if and only if $p$ and its derivative $p^{\prime}$ have no roots in common. 204 | \begin{proof} 205 | First, let $p$ have $m$ distinct roots. Since $p$ has the degree of $m$, then this could imply that $p$ can be actually written in the form of $p(z)=c\left(z-\lambda_1\right) \ldots\left(z-\lambda_m\right)$, which you have $\lambda_1, \ldots, \lambda_m$ being distinct. 206 | To prove that both $p$ and $p^{\prime}$ have no roots in commons, we must now show that $p^{\prime}\left(\lambda_j\right) \neq 0$ for every $j$. So, to do so, just fix $j$. The previous expression for $p$ shows that we can now write $p$ in the form of $p(z)=\left(z-\lambda_j\right) q(z)$, which $q$ is a polynomial such that $q\left(\lambda_j\right) \neq 0$. 207 | 208 | When you differentiate both sides of the previous equation, then you would then have $p^{\prime}(z)=(z-$ $\left.\lambda_j\right) q^{\prime}(z)+q(z)$ 209 | 210 | Therefore: $\left.=p^{\prime}\left(\lambda_j\right)=q \lambda_j\right)$ 211 | Equals: $p^{\prime}\left(\lambda_j\right) \neq 0$ 212 | 213 | Now, to prove the other direction, we would now prove the contrapositive, which means that we will be proving that if $p$ has actually less than $m$ distinct roots, then both $p$ and $p^{\prime}$ have at least one root in common. 214 | 215 | Now, for some root of $\lambda$ of $p$, we can write $p$ is in the form of $\left.p(z)=(z-\lambda)^n q(z)\right)$, which is where both $n \geq 2$ and $q$ is a polynomial. When differentiating both sides of the previous equations, we would then have $p^{\prime}(z)=(z-\lambda)^n q^{\prime}(z)+n(z-\lambda)^{n-1} q(z)$. 216 | Therefore, $p^{\prime}(\lambda)=0$, which would make $\lambda$ is a common root of both $p$ and $p^{\prime}$. 217 | \end{proof} 218 | 219 | 220 | 221 | \paragraph{Exercise 5.1} Suppose $T \in \mathcal{L}(V)$. Prove that if $U_{1}, \ldots, U_{m}$ are subspaces of $V$ invariant under $T$, then $U_{1}+\cdots+U_{m}$ is invariant under $T$. 222 | \begin{proof} 223 | First off, assume that $U_1, \ldots, U_m$ are subspaces of $V$ invariant under $T$. Now, consider a vector $u \in$ $U_1+\ldots+U_m$. There does exist $u_1 \in U_1, \ldots, u_m \in U_m$ such that $u=u_1+\ldots+u_m$. 224 | 225 | Once you apply $T$ towards both sides of the previous equation, we would then get $T u=T u_1+\ldots+$ $T u_m$. 226 | 227 | Since each $U_j$ is invariant under $T$, then we would have $T u_1 \in U_1+\ldots+T u_m$. This would then make the equation shows that $T u \in U_1+\ldots+T u_m$, which does imply that $U_1+. .+U_m$ is invariant under $T$ 228 | \end{proof} 229 | 230 | 231 | 232 | \paragraph{Exercise 5.4} Suppose that $S, T \in \mathcal{L}(V)$ are such that $S T=T S$. Prove that $\operatorname{null} (T-\lambda I)$ is invariant under $S$ for every $\lambda \in \mathbf{F}$. 233 | \begin{proof} 234 | First off, fix $\lambda \in F$. Secondly, let $v \in \operatorname{null}(T-\lambda I)$. If so, then $(T-\lambda I)(S v)=T S v-\lambda S v=$ $S T v-\lambda S v=S(T v-\lambda v)=0$. Therefore, $S v \in \operatorname{null}(T-\lambda I)$ since $n u l l(T-\lambda I)$ is actually invariant under $S$. 235 | \end{proof} 236 | 237 | 238 | 239 | \paragraph{Exercise 5.11} Suppose $S, T \in \mathcal{L}(V)$. Prove that $S T$ and $T S$ have the same eigenvalues. 240 | \begin{proof} 241 | To start, let $\lambda \in F$ be an eigenvalue of $S T$. Now, we would want $\lambda$ to be an eigenvalue of $T S$. Since $\lambda$, by itself, is an eigenvalue of $S T$, then there has to be a nonzero vector $v \in V$ such that $(S T) v=\lambda v$. 242 | Now, With a given reference that $(S T) v=\lambda v$, you will then have the following: $(T S)(T v)=$ $T(S T v)=T(\lambda v)=\lambda T v$ 243 | If $T v \neq 0$, then the listed equation above shows that $\lambda$ is an eigenvalue of $T S$. 244 | If $T v=0$, then $\lambda=0$, since $S(T v)=\lambda T v$. This also means that $T$ isn't invertible, which would imply that $T S$ isn't invertible, which can also be implied that $\lambda$, which equals 0 , is an eigenvalue of $T S$. 245 | Step 3 246 | 3 of 3 247 | Now, regardless of whether $T v=0$ or not, we would have shown that $\lambda$ is an eigenvalue of $T S$. Since $\lambda$ (was) an arbitrary eigenvalue of $S T$, we have shown that every single eigenvalue of $S T$ is an eigenvalue of $T S$. When you do reverse the roles of both $S$ and $T$, then we can conclude that that every single eigenvalue of $T S$ is also an eigenvalue of $S T$. Therefore, both $S T$ and $T S$ have the exact same eigenvalues. 248 | \end{proof} 249 | 250 | 251 | 252 | \paragraph{Exercise 5.12} Suppose $T \in \mathcal{L}(V)$ is such that every vector in $V$ is an eigenvector of $T$. Prove that $T$ is a scalar multiple of the identity operator. 253 | \begin{proof} 254 | For every single $v \in V$, there does exist $a_v \in F$ such that $T v=a_v v$. Since $T 0=0$, then we have to make $a_0$ be the any number in F. However, for every single $v \in V\{0\}$, then the value of $a_V$ is uniquely determined by the previous equation of $T v=a_v v$. 255 | 256 | Now, to show that $T$ is a scalar multiple of the identity, then me must show that $a_v$ is independent of $v$ for $v \in V\{0\}$. We would now want to show that $a_v=a_w$. 257 | 258 | First, just make the case of where $(v, w)$ is linearly dependent. Then, there does exist $b \in F$ such that $w=b v$. Now, you would have the following: $a_W w=T w=T(b v)=b T v=b\left(a_v v\right)=a_v w$. This is showing that $a_v=a_w$. 259 | Finally, make the consideration to make $(v, w)$ be linearly independent. Now, we would have the following: $\left.a_{(} v+w\right)(v+w)=T(v+w)=T v+T w=a_v v+a_w w$. 260 | 261 | That previous equation implies the following: $\left.\left.\left(a_{(} v+w\right)-a_v\right) v+\left(a_{(} v+w\right)-a_w\right) w=0$. Since $(v, w)$ is linearly independent, this would imply that both $\left.a_{(} v+w\right)=a_v$ and $\left.a_{(} v+w\right)=a_w$. Therefore, $a_v=a_w$. 262 | \end{proof} 263 | 264 | 265 | 266 | \paragraph{Exercise 5.13} Suppose $T \in \mathcal{L}(V)$ is such that every subspace of $V$ with dimension $\operatorname{dim} V-1$ is invariant under $T$. Prove that $T$ is a scalar multiple of the identity operator. 267 | \begin{proof} 268 | First off, let $T$ isn't a scalar multiple of the identity operator. So, there does exists that $v \in V$ such that $u$ isn't an eigenvector of $T$. Therefore, $(u, T u)$ is linearly independent. 269 | 270 | Next, you should extend $(u, T u)$ to a basis of $\left(u, T u, v_1, \ldots, v_n\right)$ of $V$. So, let $U=\operatorname{span}\left(u, v_1, \ldots, v_n\right)$. Then, $U$ is a subspace of $V$ and $\operatorname{dim} U=\operatorname{dim} V-1$. However, $U$ isn't invariant under $T$ since both $u \in U$ and $T u \in U$. This given contradiction to our hypothesis about $T$ actually shows us that our guess that $T$ is not a scalar multiple of the identity must have been false. 271 | \end{proof} 272 | 273 | 274 | 275 | \paragraph{Exercise 5.20} Suppose that $T \in \mathcal{L}(V)$ has $\operatorname{dim} V$ distinct eigenvalues and that $S \in \mathcal{L}(V)$ has the same eigenvectors as $T$ (not necessarily with the same eigenvalues). Prove that $S T=T S$. 276 | \begin{proof} 277 | First off, let $n=\operatorname{dim} V$. so, there is a basis of $\left(v_1, \ldots, v_j\right)$ of $V$ that consist of eigenvectors of $T$. Now, let $\lambda_1, \ldots, \lambda_n$ be the corresponding eigenvalues, then we would have $T v_j=\lambda_1 v_j$ for every single $j$. 278 | 279 | Now, for every $v_j$ is also an eigenvector of S, so $S v_j=a_j v_j$ for some $a_j \in F$. For each $j$, we would then have $(S T) v_j=S\left(T v_j\right)=\lambda_j S v_j=a_j \lambda_j v_j$ and $(T S) v_j=T\left(S v_j\right)=a_j T v_j=a_j \lambda_j v_j$. Since both operators, which are $S T$ and $T S$, agree on a basis, then both are equal. 280 | \end{proof} 281 | 282 | 283 | 284 | \paragraph{Exercise 5.24} Suppose $V$ is a real vector space and $T \in \mathcal{L}(V)$ has no eigenvalues. Prove that every subspace of $V$ invariant under $T$ has even dimension. 285 | \begin{proof} 286 | First off, let us assume that $U$ is a subspace of $V$ that is invariant under $T$. Therefore, $\left.T\right|_U \in \mathcal{L}(U)$. If $\operatorname{dim}$ $U$ were odd, then $\left.T\right|_U$ would have an eigenvalue $\lambda \in \mathbb{R}$, so there would exist a nonzero vector $u \in U$ such that 287 | $$ 288 | \left.T\right|_U u=\lambda u . 289 | $$ 290 | So, this would imply that $T_u=\lambda u$, which would imply that $\lambda$ is an eigenvalue of $T$. But $T$ has no eigenvalues, so $\operatorname{dim} U$ must be even. 291 | \end{proof} 292 | 293 | 294 | 295 | \paragraph{Exercise 6.2} Suppose $u, v \in V$. Prove that $\langle u, v\rangle=0$ if and only if $\|u\| \leq\|u+a v\|$ for all $a \in \mathbf{F}$. 296 | \begin{proof} 297 | First off, let us suppose that $(u, v)=0$. 298 | Now, let $a \in \mathbb{F}$. Next, $u, a v$ are orthogonal. 299 | The Pythagorean theorem thus implies that 300 | $$ 301 | \begin{aligned} 302 | \|u+a v\|^2 & =\|u\|^2+\|a v\|^2 \\ 303 | & \geq\|u\|^2 304 | \end{aligned} 305 | $$ 306 | So, by taking the square roots, this will now give us $\|u\| \leq\|u+a v\|$. 307 | Now, to prove the implication in the other direction, we must now let $\|u\| \leq$ $\|u+a v\|$ for all $a \in \mathbb{F}$. Squaring this inequality, we get both: 308 | $$ 309 | \begin{gathered} 310 | \|u\|^2 a n d \leq\|u+a v\|^2 \\ 311 | =(u+a v, u+a v) \\ 312 | =(u, u)+(u, a v)+(a v, u)+(a v, a v) \\ 313 | =\|u\|^2+\bar{a}(u, v)+a \overline{(u, v)}+|a|^2\|v\|^2 \\ 314 | \|u\|^2+2 \Re \bar{a}(u, v)+|a|^2\|v\|^2 315 | \end{gathered} 316 | $$ 317 | for all $a \in \mathbb{F}$. 318 | Therefore, 319 | $$ 320 | -2 \Re \bar{a}(u, v) \leq|a|^2\|v\|^2 321 | $$ 322 | for all $a \in \mathbb{F}$. In particular, we can let $a$ equal $-t(u, v)$ for $t>0$. Substituting this value for $a$ into the inequality above gives 323 | $$ 324 | 2 t|(u, v)|^2 \leq t^2|(u, v)|^2\|v\|^2 325 | $$ 326 | for all $t>0$. 327 | Step 4 328 | 4 of 4 329 | Divide both sides of the inequality above by $t$, getting 330 | $$ 331 | 2|(u, v)|^2 \leq t \mid(u, v)^2\|v\|^2 332 | $$ 333 | for all $t>0$. If $v=0$, then $(u, v)=0$, as desired. If $v \neq 0$, set $t$ equal to $1 /\|v\|^2$ in the inequality above, getting 334 | $$ 335 | 2|(u, v)|^2 \leq|(u, v)|^2, 336 | $$ 337 | which implies that $(u, v)=0$. 338 | \end{proof} 339 | 340 | 341 | \paragraph{Exercise 6.3} Prove that $\left(\sum_{j=1}^{n} a_{j} b_{j}\right)^{2} \leq\left(\sum_{j=1}^{n} j a_{j}{ }^{2}\right)\left(\sum_{j=1}^{n} \frac{b_{j}{ }^{2}}{j}\right)$ for all real numbers $a_{1}, \ldots, a_{n}$ and $b_{1}, \ldots, b_{n}$. 342 | \begin{proof} 343 | Let $a_1, a_2, \ldots, a_n, b_1, b_2, \ldots, b_n \in R$. 344 | We have that 345 | $$ 346 | \left(\sum_{j=1}^n a_j b_j\right)^2 347 | $$ 348 | is equal to the 349 | $$ 350 | \left(\sum_{j=1}^n a_j b_j \frac{\sqrt{j}}{\sqrt{j}}\right)^2=\left(\sum_{j=1}^n\left(\sqrt{j} a_j\right)\left(b_j \frac{1}{\sqrt{j}}\right)\right)^2 351 | $$ 352 | This can be observed as an inner product, and using the Cauchy-Schwarz Inequality, we get 353 | $$ 354 | \begin{aligned} 355 | &\left(\sum_{j=1}^n a_j b_j\right)^2=\left(\sum_{j=1}^n\left(\sqrt{j} a_j\right)\left(b_j \frac{1}{\sqrt{j}}\right)\right)^2 \\ 356 | &=\left\langle\left(a, \sqrt{2} a_2, \ldots, \sqrt{n} a_n\right),\left(b_1, \frac{b_2}{\sqrt{2}}, \ldots, \frac{b_n}{\sqrt{n}}\right)\right\rangle \\ 357 | & \leq\left\|\left(a, \sqrt{2} a_2, \ldots, \sqrt{n} a_n\right)\right\|^2\left\|\left(b_1, \frac{b_2}{\sqrt{2}}, \ldots, \frac{b_n}{\sqrt{n}}\right)\right\|^2 \\ 358 | &=\left(\sum_{j=1}^n j a_j^2\right)\left(\sum_{j=1}^n \frac{b_j^2}{j}\right) \\ 359 | & \text { Hence, }\left(\sum_{j=1}^n a_j b_j\right)^2=\left(\sum_{j=1}^n j a_j^2\right)\left(\sum_{j=1}^n \frac{b_j^2}{j}\right) . 360 | \end{aligned} 361 | $$ 362 | \end{proof} 363 | 364 | 365 | 366 | \paragraph{Exercise 6.7} Prove that if $V$ is a complex inner-product space, then $\langle u, v\rangle=\frac{\|u+v\|^{2}-\|u-v\|^{2}+\|u+i v\|^{2} i-\|u-i v\|^{2} i}{4}$ for all $u, v \in V$. 367 | \begin{proof} 368 | Let $V$ be an inner-product space and $u, v\in V$. Then 369 | $$ 370 | \begin{aligned} 371 | \|u+v\|^2 & =\langle u+v, v+v\rangle \\ 372 | & =\|u\|^2+\langle u, v\rangle+\langle v, u\rangle+\|v\|^2 \\ 373 | -\|u-v\|^2 & =-\langle u-v, u-v\rangle \\ 374 | & =-\|u\|^2+\langle u, v\rangle+\langle v, u\rangle-\|v\|^2 \\ 375 | i\|u+i v\|^2 & =i\langle u+i v, u+i v\rangle \\ 376 | & =i\|u\|^2+\langle u, v\rangle-\langle v, u\rangle+i\|v\|^2 \\ 377 | -i\|u-i v\|^2 & =-i\langle u-i v, u-i v\rangle \\ 378 | & =-i\|u\|^2+\langle u, v\rangle-\langle v, u\rangle-i\|v\|^2 . 379 | \end{aligned} 380 | $$ 381 | Thus $\left(\|u+v\|^2\right)-\|u-v\|^2+\left(i\|u+i v\|^2\right)-i\|u-i v\|^2=4\langle u, v\rangle.$ 382 | \end{proof} 383 | 384 | 385 | 386 | \paragraph{Exercise 6.13} Suppose $\left(e_{1}, \ldots, e_{m}\right)$ is an orthonormal list of vectors in $V$. Let $v \in V$. Prove that $\|v\|^{2}=\left|\left\langle v, e_{1}\right\rangle\right|^{2}+\cdots+\left|\left\langle v, e_{m}\right\rangle\right|^{2}$ if and only if $v \in \operatorname{span}\left(e_{1}, \ldots, e_{m}\right)$. 387 | \begin{proof} 388 | If $v \in \operatorname{span}\left(e_1, \ldots, e_m\right)$, it means that 389 | $$ 390 | v=\alpha_1 e_1+\ldots+\alpha_m e_m . 391 | $$ 392 | for some scalars $\alpha_i$. We know that $\alpha_k=\left\langle v, e_k\right\rangle, \forall k \in\{1, \ldots, m\}$. Therefore, 393 | $$ 394 | \begin{aligned} 395 | \|v\|^2 & =\langle v, v\rangle \\ 396 | & =\left\langle\alpha_1 e_1+\ldots+\alpha_m e_m, \alpha_1 e_1+\ldots+\alpha_m e_m\right\rangle \\ 397 | & =\left|\alpha_1\right|^2\left\langle e_1, e_1\right\rangle+\ldots+\left|\alpha_m\right|^2\left\langle e_m, e_m\right\rangle \\ 398 | & =\left|\alpha_1\right|^2+\ldots+\left|\alpha_m\right|^2 \\ 399 | & =\left|\left\langle v, e_1\right\rangle\right|^2+\ldots+\left|\left\langle v, e_m\right\rangle\right|^2 . 400 | \end{aligned} 401 | $$ 402 | $\Rightarrow$ Assume that $v \notin \operatorname{span}\left(e_1, \ldots, e_m\right)$. Then, we must have 403 | $$ 404 | v=v_{m+1}+\frac{\left\langle v, v_0\right\rangle}{\left\|v_0\right\|^2} v_0, 405 | $$ 406 | where $v_0=\alpha_1 e_1+\ldots+\alpha_m e_m, \alpha_k=\left\langle v, e_k\right\rangle, \forall k \in\{1, \ldots, m\}$, and $v_{m+1}=v-$ $\frac{\left\langle v, v_0\right\rangle}{\left\|v_0\right\|^2} v_0 \neq 0$. 407 | 408 | We have $\left\langle v_0, v_{m+1}\right\rangle=0$ (from which we get $\left\langle v, v_0\right\rangle=\left\langle v_0, v_0\right\rangle$ and $\left\langle v, v_{m+1}\right\rangle=$ $\left.\left\langle v_{m+1}, v_{m+1}\right\rangle\right)$. Now, 409 | $$ 410 | \begin{aligned} 411 | \|v\|^2 & =\langle v, v\rangle \\ 412 | & =\left\langle v, v_{m+1}+\frac{\left\langle v, v_0\right\rangle}{\left\|v_0\right\|^2} v_0\right\rangle \\ 413 | & =\left\langle v, v_{m+1}\right\rangle+\left\langle v, \frac{\left\langle v, v_0\right\rangle}{\left\|v_0\right\|^2} v_0\right\rangle \\ 414 | & =\left\langle v_{m+1}, v_{m+1}\right\rangle+\frac{\left\langle v_0, v_0\right\rangle}{\left\|v_0\right\|^2}\left\langle v_0, v_0\right\rangle \\ 415 | & =\left\|v_{m+1}\right\|^2+\left\|v_0\right\|^2 \\ 416 | & >\left\|v_0\right\|^2 \\ 417 | & =\left|\alpha_1\right|^2+\ldots+\left|\alpha_m\right|^2 \\ 418 | & =\left|\left\langle v, e_1\right\rangle\right|^2+\ldots+\left|\left\langle v, e_m\right\rangle\right|^2 . 419 | \end{aligned} 420 | $$ 421 | By contrapositive, if $\left\|v_1\right\|^2=\left|\left\langle v, e_1\right\rangle\right|^2+\ldots+\left|\left\langle v, e_m\right\rangle\right|^2$, then $v \in \operatorname{span}\left(e_1, \ldots, e_m\right)$. 422 | \end{proof} 423 | 424 | 425 | 426 | \paragraph{Exercise 6.16} Suppose $U$ is a subspace of $V$. Prove that $U^{\perp}=\{0\}$ if and only if $U=V$ 427 | \begin{proof} 428 | $V=U \bigoplus U^{\perp}$, therefore $U^\perp = \{0\}$ iff $U=V$. 429 | \end{proof} 430 | 431 | \paragraph{Exercise 7.5} Show that if $\operatorname{dim} V \geq 2$, then the set of normal operators on $V$ is not a subspace of $\mathcal{L}(V)$. 432 | \begin{proof} 433 | First off, suppose that $\operatorname{dim} V \geq 2$. Next let $\left(e_1, \ldots, e_n\right)$ be an orthonormal basis of $V$. Now, define $S, T \in L(V)$ by both $S\left(a_1 e_1+\ldots+a_n e_n\right)=a_2 e_1-a_1 e_2$ and $T\left(a_1 e_1+\ldots+\right.$ $\left.a_n e_n\right)=a_2 e_1+a_1 e_2$. So, just by now doing a simple calculation verifies that $S^*\left(a_1 e_1+\right.$ $\left.\ldots+a_n e_n\right)=-a_2 e_1+a_1 e_2$ 434 | 435 | Now, based on this formula, another calculation would show that $S S^*=S^* S$. Another simple calculation would that that $T$ is self-adjoint. Therefore, both $S$ and $T$ are normal. However, $S+T$ is given by the formula of $(S+T)\left(a_1 e_1+\ldots+a_n e_n\right)=2 a_2 e_1$. In this case, a simple calculator verifies that $(S+T)^*\left(a_1 e_1+\ldots+a_n e_n\right)=2 a_1 e_2$. 436 | 437 | Therefore, there is a final simple calculation that shows that $(S+T)(S+T)^* \neq(S+$ $T)^*(S+T)$. So, in other words, $S+T$ isn't normal. Therefore, the set of normal operators on $V$ isn't closed under addition and hence isn't a subspace of $L(V)$. 438 | \end{proof} 439 | 440 | 441 | 442 | \paragraph{Exercise 7.6} Prove that if $T \in \mathcal{L}(V)$ is normal, then $\operatorname{range} T=\operatorname{range} T^{*}.$ 443 | \begin{proof} 444 | Let $T \in \mathcal{L}(V)$ to be a normal operator. 445 | Suppose $u \in \operatorname{null} T$. Then, by $7.20$, 446 | $$ 447 | 0=\|T u\|=\left\|T^* u\right\|, 448 | $$ 449 | which implies that $u \in \operatorname{null} T^*$. 450 | Hence 451 | $$ 452 | \operatorname{null} T=\operatorname{null} T^* 453 | $$ 454 | because $\left(T^*\right)^*=T$ and the same argument can be repeated. 455 | Now we have 456 | $$ 457 | \begin{aligned} 458 | \text { range } T & =\left(\text { null } T^*\right)^{\perp} \\ 459 | & =(\text { null } T)^{\perp} \\ 460 | & =\operatorname{range} T^*, 461 | \end{aligned} 462 | $$ 463 | where the first and last equality follow from items (d) and (b) of 7.7. 464 | Hence, range $T=$ range $T^*$. 465 | \end{proof} 466 | 467 | 468 | 469 | \paragraph{Exercise 7.9} Prove that a normal operator on a complex inner-product space is self-adjoint if and only if all its eigenvalues are real. 470 | \begin{proof} 471 | First off, suppose $V$ is a complex inner product space and $T \in L(V)$ is normal. If $T$ is self-adjoint, then all its eigenvalues are real. So, conversely, let all of the eigenvalues of $T$ be real. By the complex spectral theorem, there's an orthonormal basis $\left(e_1, \ldots, e_n\right)$ of $V$ consisting of eigenvectors of $T$. Thus, there exists real numbers $\lambda_1, \ldots, \lambda_n$ such that $T e_j=\lambda_j e_j$ for $j=$ $1, \ldots, n$. 472 | The matrix of $T$ with respect to the basis of $\left(e_1, \ldots, e_n\right)$ is the diagonal matrix with $\lambda_1, \ldots, \lambda_n$ on the diagonal. So, the matrix equals its conjugate transpose. Therefore, $T=T^*$. In other words, $T$ s self-adjoint. 473 | \end{proof} 474 | 475 | 476 | 477 | \paragraph{Exercise 7.10} Suppose $V$ is a complex inner-product space and $T \in \mathcal{L}(V)$ is a normal operator such that $T^{9}=T^{8}$. Prove that $T$ is self-adjoint and $T^{2}=T$. 478 | \begin{proof} 479 | Based on the complex spectral theorem, there is an orthonormal basis of $\left(e_1, \ldots, e_n\right)$ of $V$ consisting of eigenvectors of $T$. Now, let $\lambda_1, \ldots, \lambda_n$ be the corresponding eigenvalues. Therefore, 480 | $$ 481 | T e_1=\lambda_j e_j 482 | $$ 483 | for $j=1 \ldots n$. 484 | 485 | Next, by applying $T$ repeatedly to both sides of the equation above, we get $T^9 e_j=\left(\lambda_j\right)^9 e_j$ and rei =8ej. Thus $T^8 e_j=\left(\lambda_j\right)^8 e_j$, which implies that $\lambda_j$ equals 0 or 1 . In particular, all the eigenvalues of $T$ are real. This would then imply that $T$ is self-adjoint. 486 | 487 | Now, by applying $T$ to both sides of the equation above, we get 488 | $$ 489 | \begin{aligned} 490 | T^2 e_j & =\left(\lambda_j\right)^2 e_j \\ 491 | & =\lambda_j e_j \\ 492 | & =T e_j 493 | \end{aligned} 494 | $$ 495 | which is where the second equality holds because $\lambda_j$ equals 0 or 1 . Because $T^2$ and $T$ agree on a basis, they must be equal. 496 | \end{proof} 497 | 498 | 499 | 500 | \paragraph{Exercise 7.11} Suppose $V$ is a complex inner-product space. Prove that every normal operator on $V$ has a square root. (An operator $S \in \mathcal{L}(V)$ is called a square root of $T \in \mathcal{L}(V)$ if $S^{2}=T$.) 501 | \begin{proof} 502 | Let $V$ be a complex inner product space. 503 | It is known that an operator $S \in \mathcal{L}(V)$ is called a square root of $T \in \mathcal{L}(V)$ if 504 | $$ 505 | S^2=T 506 | $$ 507 | Now, suppose that $T$ is a normal operator on $V$. 508 | By the Complex Spectral Theorem, there is $e_1, \ldots, e_n$ an orthonormal basis of $V$ consisting of eigenvalues of $T$ and let $\lambda_1, \ldots, \lambda_n$ denote their corresponding eigenvalues. 509 | Define $S$ by 510 | $$ 511 | S e_j=\sqrt{\lambda_j} e_j, 512 | $$ 513 | for each $j=1, \ldots, n$. 514 | Obviously, $S^2 e_j=\lambda_j e_j=T e_j$. 515 | Hence, $S^2=T$ so there exist a square root of $T$. 516 | \end{proof} 517 | 518 | 519 | 520 | \paragraph{Exercise 7.14} Suppose $T \in \mathcal{L}(V)$ is self-adjoint, $\lambda \in \mathbf{F}$, and $\epsilon>0$. Prove that if there exists $v \in V$ such that $\|v\|=1$ and $\|T v-\lambda v\|<\epsilon,$ then $T$ has an eigenvalue $\lambda^{\prime}$ such that $\left|\lambda-\lambda^{\prime}\right|<\epsilon$. 521 | \begin{proof} 522 | Let $T \in \mathcal{L}(V)$ be a self-adjoint, and let $\lambda \in \mathbf{F}$ and $\epsilon>0$. 523 | By the Spectral Theorem, there is $e_1, \ldots, e_n$ an orthonormal basis of $V$ consisting of eigenvectors of $T$ and let $\lambda_1, \ldots, \lambda_n$ denote their corresponding eigenvalues. 524 | Choose an eigenvalue $\lambda^{\prime}$ of $T$ such that $\left|\lambda^{\prime}-\lambda\right|^2$ is minimized. 525 | There are $a_1, \ldots, a_n \in \mathbb{F}$ such that 526 | $$ 527 | v=a_1 e_1+\cdots+a_n e_n . 528 | $$ 529 | Thus, we have 530 | $$ 531 | \begin{aligned} 532 | \epsilon^2 & >|| T v-\left.\lambda v\right|^2 \\ 533 | & =\left|\left\langle T v-\lambda v, e_1\right\rangle\right|^2+\cdots+\left|\left\langle T v-\lambda v, e_n\right\rangle\right|^2 \\ 534 | & =\left|\lambda_1 a_1-\lambda a_1\right|^2+\cdots+\left|\lambda_n a_n-\lambda a_n\right|^2 \\ 535 | & =\left|a_1\right|^2\left|\lambda_1-\lambda\right|^2+\cdots+\left|a_n\right|^2\left|\lambda_n-\lambda\right|^2 \\ 536 | & \geq\left|a_1\right|^2\left|\lambda^{\prime}-\lambda\right|^2+\cdots+\left|a_n\right|^2\left|\lambda^{\prime}-\lambda\right|^2 \\ 537 | & =\left|\lambda^{\prime}-\lambda\right|^2 538 | \end{aligned} 539 | $$ 540 | where the second and fifth lines follow from $6.30$ (the fifth because $\|v\|=1$ ). Now, we taking the square root. 541 | Hence, $T$ has an eigenvalue $\lambda^{\prime}$ such that $\left|\lambda^{\prime}-\lambda\right|<\epsilon$ 542 | \end{proof} 543 | 544 | 545 | 546 | \end{document} 547 | -------------------------------------------------------------------------------- /informal/Cambridge-Tripos.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/rahul3613/ProofNet-lean4/1893b2f054fddb1c4c90b91732173528c16f3a37/informal/Cambridge-Tripos.pdf -------------------------------------------------------------------------------- /informal/Cambridge-Tripos.tex: -------------------------------------------------------------------------------- 1 | \documentclass{article} 2 | 3 | \title{\textbf{ 4 | Exercises from \\ 5 | \textit{Cambridge Tripos} 6 | }} 7 | 8 | \date{} 9 | 10 | \usepackage{amsmath} 11 | \usepackage{amssymb} 12 | \usepackage{amsthm} 13 | 14 | \begin{document} 15 | \maketitle 16 | 17 | 18 | \paragraph{Exercise 2022.IA.4-I-1E-a} Show that there are infinitely many primes of the form $3 n+2$ with $n \in \mathbb{N}$. 19 | \begin{proof} 20 | The general strategy is to find a (large) number $n$ that is relatively prime to each of the existing list of such primes, and is also congruent to 2 modulo 3 . The prime factorization of $n$ cannot consist only of primes congruent to 1 modulo 3 , since the product of any number of such is still 1 modulo 3 . Hence there must be some prime factor of $n$ that is congruent to 2 modulo 3 , which must be not on our list by the construction of $n$. 21 | Now, how to construct such an $n$ ? Suppose the finite list is $\left\{p_1, p_2, \ldots, p_k\right\}$. If $k$ is even, then take $n=p_1 p_2 \cdots p_k+1$. If $k$ is odd, then take $n=\left(p_1 p_2 \cdots p_k\right) p_k+1$. 22 | \end{proof} 23 | 24 | 25 | \paragraph{Exercise 2022.IA.4-I-2D-a} Prove that $\sqrt[3]{2}+\sqrt[3]{3}$ is irrational. 26 | \begin{proof} 27 | Suppose $\frac{a}{b} = \sqrt[3]{2}+\sqrt[3]{3}$ for $a,b\in\mathbb{Z}$. Cubing both sides, we get $a^3/b^3 = 2 + 3 \sqrt[3]{12} + 3\sqrt[3]{18}+3$. Therefore we have $\frac{c}{d} = \sqrt[3]{12} + \sqrt[3]{18}$ for some rational $c/d\in\mathbb{Q}$. Cubing both sides we get $c^3/d^3 = 81000\sqrt{3}$, which is a contradiction. 28 | \end{proof} 29 | 30 | 31 | \paragraph{Exercise 2022.IB.3-II-13G-a-i} Let $U \subset \mathbb{C}$ be a (non-empty) connected open set and let $f_n$ be a sequence of holomorphic functions defined on $U$. Suppose that $f_n$ converges uniformly to a function $f$ on every compact subset of $U$. Show that $f$ is holomorphic in $U$. 32 | \begin{proof} 33 | Let $\Delta \subset D$ be a closed triangle. Since each $f_n$ is holomorphic, by Cauchy's theorem, you have $\int_{\partial \Delta} f_n(z) d z=0$ for all $n$. 34 | $\partial \Delta$ is a compact subset of $D$, so you know that $f_n \rightarrow f$ uniformly on $\partial \Delta$. 35 | So you get, for all $n$, 36 | $$ 37 | \left|\int_{\partial \Delta} f(z) d z\right|=\left|\int_{\partial \Delta}\left(f(z)-f_n(z)\right) d z\right| \leq \operatorname{length}(\partial \Delta) 38 | $$ 39 | By letting $n \rightarrow \infty$, you find that $\int_{\partial \Delta} f(z) d z=0$. 40 | By Morera's theorem, $f$ is holomorphic. 41 | \end{proof} 42 | 43 | \end{document} 44 | -------------------------------------------------------------------------------- /informal/Dummit-Foote.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/rahul3613/ProofNet-lean4/1893b2f054fddb1c4c90b91732173528c16f3a37/informal/Dummit-Foote.pdf -------------------------------------------------------------------------------- /informal/Herstein.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/rahul3613/ProofNet-lean4/1893b2f054fddb1c4c90b91732173528c16f3a37/informal/Herstein.pdf -------------------------------------------------------------------------------- /informal/Ireland-Rosen.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/rahul3613/ProofNet-lean4/1893b2f054fddb1c4c90b91732173528c16f3a37/informal/Ireland-Rosen.pdf -------------------------------------------------------------------------------- /informal/Ireland-Rosen.tex: -------------------------------------------------------------------------------- 1 | 2 | \documentclass{article} 3 | 4 | \title{\textbf{ 5 | Exercises from \\ 6 | \textit{A Classical Introduction to Modern Number Theory} \\ 7 | by Kenneth Ireland and Michael Rosen 8 | }} 9 | 10 | \date{} 11 | 12 | \usepackage{amsmath} 13 | \usepackage{amssymb} 14 | \usepackage{amsthm} 15 | 16 | \begin{document} 17 | \maketitle 18 | 19 | 20 | \paragraph{Exercise 1.27} For all odd $n$ show that $8 \mid n^{2}-1$. 21 | \begin{proof} 22 | We have $n^2-1=(n+1)(n-1)$. Since $n$ is odd, both $n+1, n-1$ are even, and moreso, one of these must be divisible by 4 , as one of the two consecutive odd numbers is divisible by 4 . Thus, their product is divisible by 8 . Similarly, if 3 does not divide $n$, it must divide one of $n-1, n+1$, otherwise it wouldn't divide three consecutive integers, which is impossible. As $n$ is odd, $n+1$ is even, so $(n+1)(n-1)$ is divisible by both 2 and 3 , so it is divisible by 6 . 23 | \end{proof} 24 | 25 | 26 | 27 | \paragraph{Exercise 1.30} Prove that $\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}$ is not an integer. 28 | \begin{proof} 29 | Let $2^s$ be the largest power of 2 occurring as a denominator in $H_n$, say $2^s=k \leqslant n$. Write $H_n=$ $\frac{1}{2^s}+\left(1+1 / 2+\ldots+1 /(k-1)+1 /(k+1)+\ldots+1 / n\right.$. The sum in parentheses can be written as $1 / 2^{s-1}$ times sum of fractions with odd denominators, so the denominator of the sum in parentheses will not be divisible by $2^s$, but it must equal $2^s$ by Ex $1.29$. 30 | \end{proof} 31 | 32 | 33 | 34 | \paragraph{Exercise 1.31} Show that 2 is divisible by $(1+i)^{2}$ in $\mathbb{Z}[i]$. 35 | \begin{proof} 36 | We have $(1+i)^2=1+2 i-1=2 i$, so $2=-i(1+i)^2$. 37 | \end{proof} 38 | 39 | 40 | 41 | \paragraph{Exercise 2.4} If $a$ is a nonzero integer, then for $n>m$ show that $\left(a^{2^{n}}+1, a^{2^{m}}+1\right)=1$ or 2 depending on whether $a$ is odd or even. 42 | \begin{proof} 43 | \begin{align*} 44 | \operatorname{ord}_p\, n! &= \sum_{k\geq 1} \left \lfloor \frac{n}{p^{k}}\right \rfloor \leq \sum_{k\geq 1} \frac{n}{p^{k}} = \frac{n}{p} \frac{1}{1 - \frac{1}{p}} = \frac{n}{p-1} 45 | \end{align*} 46 | 47 | The decomposition of $n!$ in prime factors is 48 | 49 | $n! = p_1^{\alpha_1}p_2^{\alpha_2}\cdots p_k^{\alpha_k}$ 50 | where $\alpha_i = \operatorname{ord}_{p_i}\, n! \leq \frac{n}{p_i-1}$, and $p_i \leq n, \ i=1,2,\cdots,k$. 51 | 52 | Then 53 | \begin{align*} 54 | n! &\leq p_1^{\frac{n}{p_1-1}}p_2^{\frac{n}{p_2-1}}\cdots p_k^{\frac{n}{p_n-1}}\\ 55 | \sqrt[n]{n!} &\leq p_1^{\frac{1}{p_1-1}}p_2^{\frac{1}{p_2-1}}\cdots p_k^{\frac{1}{p_n-1}}\\ 56 | &\leq \prod_{p\leq n} p^{\frac{1}{p-1}} 57 | \end{align*} 58 | (the values of $p$ in this product describe all prime numbers $p\leq n$.) 59 | \end{proof} 60 | 61 | 62 | 63 | \paragraph{Exercise 2.21} Define $\wedge(n)=\log p$ if $n$ is a power of $p$ and zero otherwise. Prove that $\sum_{A \mid n} \mu(n / d) \log d$ $=\wedge(n)$. 64 | \begin{proof} 65 | $$ 66 | \left\{ 67 | \begin{array}{cccl} 68 | \land(n)& = & \log p & \mathrm{if}\ n =p^\alpha,\ \alpha \in \mathbb{N}^* \\ 69 | & = & 0 & \mathrm{otherwise }. 70 | \end{array} 71 | \right. 72 | $$ 73 | Let $n = p_1^{\alpha_1}\cdots p_t^{\alpha_t}$ the decomposition of $n$ in prime factors. As $\land(d) = 0$ for all divisors of $n$, except for $d = p_j^i, i>0, j=1,\ldots t$, 74 | \begin{align*} 75 | \sum_{d \mid n} \land(d)&= \sum_{i=1}^{\alpha_1} \land(p_1^{i}) + \cdots+ \sum_{i=1}^{\alpha_t} \land(p_t^{i})\\ 76 | &= \alpha_1 \log p_1+\cdots + \alpha_t \log p_t\\ 77 | &= \log n 78 | \end{align*} 79 | By Mobius Inversion Theorem, 80 | $$\land(n) = \sum_{d \mid n} \mu\left (\frac{n}{d}\right ) \log d.$$ 81 | \end{proof} 82 | 83 | 84 | 85 | \paragraph{Exercise 2.27a} Show that $\sum^{\prime} 1 / n$, the sum being over square free integers, diverges. 86 | \begin{proof} 87 | 88 | Let $S \subset \mathbb{N}^*$ the set of square free integers. 89 | 90 | Let $N \in \mathbb{N}^*$. Every integer $n, \, 1\leq n \leq N$ can be written as $n = a b^2$, where $a,b$ are integers and $a$ is square free. Then $1\leq a \leq N$, and $1\leq b \leq \sqrt{N}$, so 91 | $$\sum_{n\leq N} \frac{1}{n} \leq \sum_{a \in S, a\leq N}\ \sum_{1\leq b \leq \sqrt{N}} \frac{1}{ab^2} \leq \sum_{a \in S, a\leq N}\ \frac{1}{a} \, \sum_{b=1}^\infty \frac{1}{b^2} = \frac{\pi^2}{6} \sum_{a \in S, a\leq N}\ \frac{1}{a}.$$ 92 | So $$\sum_{a \in S, a\leq N} \frac{1}{a} \geq \frac{6}{\pi^2} \sum_{n\leq N} \frac{1}{n}.$$ 93 | As $\sum_{n=1}^\infty \frac{1}{n}$ diverges, $\lim\limits_{N \to \infty} \sum\limits_{a \in S, a\leq N} \frac{1}{a} = +\infty$, so the family $\left(\frac{1}{a}\right)_{a\in S}$ of the inverse of square free integers is not summable. 94 | 95 | Let $S_N = \prod_{p0$, so 115 | $$\log S_N = \sum_{k=1}^{l(N)} \log\left(1+\frac{1}{p_k}\right) \leq \sum_{k=1}^{l(N)} \frac{1}{p_k}.$$ 116 | $\lim\limits_{N\to \infty} \log S_N = +\infty$ and $\lim\limits_{N\to \infty} l(N) = +\infty$, so 117 | $$\lim_{N\to \infty} \sum_{p1$. 125 | 126 | Then $N \equiv -1 \pmod 6$. As $6k +2, 6k +3, 6k +4$ are composite for all integers $k$, every prime factor of $N$ is congruent to $1$ or $-1$ modulo $6$. If every prime factor of $N$ was congruent to 1, then $N \equiv 1 \pmod 6$ : this is a contradiction because $-1 \not \equiv 1 \pmod 6$. So there exists a prime factor $p$ of $N$ such that $p\equiv -1 \pmod 6$. 127 | 128 | If $p\leq n$, then $p \mid n!$, and $p \mid N = n!-1$, so $p \mid 1$. As $p$ is prime, this is a contradiction, so $p>n$. 129 | 130 | Conclusion : 131 | 132 | for any integer $n$, there exists a prime $p >n$ such that $p \equiv -1 \pmod 6$ : there are infinitely many primes congruent to $-1$ modulo $6$. 133 | \end{proof} 134 | 135 | 136 | 137 | \paragraph{Exercise 3.4} Show that the equation $3 x^{2}+2=y^{2}$ has no solution in integers. 138 | \begin{proof} 139 | If $3x^2+2 = y^2$, then $\overline{y}^2 = \overline{2}$ in $\mathbb{Z}/3\mathbb{Z}$. 140 | 141 | 142 | As $\{-1,0,1\}$ is a complete set of residues modulo $3$, the squares in $\mathbb{Z}/3\mathbb{Z}$ are $\overline{0} = \overline{0}^2$ and $\overline{1} = \overline{1}^2 = (\overline{-1})^2$, so $\overline{2}$ is not a square in $\mathbb{Z}/3\mathbb{Z}$ : $\overline{y}^2 = \overline{2}$ is impossible in $\mathbb{Z}/3\mathbb{Z}$. 143 | 144 | Thus $3x^2+2 = y^2$ has no solution in integers. 145 | \end{proof} 146 | 147 | 148 | 149 | \paragraph{Exercise 3.5} Show that the equation $7 x^{3}+2=y^{3}$ has no solution in integers. 150 | \begin{proof} 151 | If $7x^2 + 2 = y^3,\ x,y \in \mathbb{Z}$, then $y^3 \equiv 2 \pmod 7$ (so $y \not \equiv 0 \pmod 7$) 152 | 153 | From Fermat's Little Theorem, $y^6 \equiv 1 \pmod 7$, so $2^2 \equiv y^6 \equiv 1 \pmod 7$, which implies $7 \mid 2^2-1 = 3$ : this is a contradiction. Thus the equation $7x^2 + 2 = y^3$ has no solution in integers. 154 | \end{proof} 155 | 156 | 157 | 158 | \paragraph{Exercise 3.10} If $n$ is not a prime, show that $(n-1) ! \equiv 0(n)$, except when $n=4$. 159 | \begin{proof} 160 | Suppose that $n >1$ is not a prime. Then $n = uv$, where $2 \leq u \leq v \leq n-1$. 161 | 162 | $\bullet$ If $u \neq v$, then $n = uv \mid (n-1)! = 1\times 2 \times\cdots \times u \times\cdots \times v \times \cdots \times (n-1)$ (even if $u\wedge v \neq 1$ !). 163 | 164 | $\bullet$ If $u=v$, $n = u^2$ is a square. 165 | 166 | If $u$ is not prime, $u =st,\ 2\leq s \leq t \leq u-1 \leq n-1$, and $n = u' v'$, where $u' =s,v' =st^2$ verify $2 \leq u' < v' \leq n-1$. As in the first case, $n = u'v' \mid (n-1)!$. 167 | 168 | If $u = p$ is a prime, then $n =p^2$. 169 | 170 | In the case $p = 2$, $n = 4$ and $n=4 \nmid (n-1)! = 6$. In the other case $p >2$, and $(n-1)! = (p^2 - 1)!$ contains the factors $p < 2p < p^2$, so $p^2 \mid (p^2-1)!, n \mid (n-1)!$. 171 | 172 | Conclusion : if $n$ is not a prime, $(n - 1)! \equiv 0 \pmod n$, except when $n=4$. 173 | \end{proof} 174 | 175 | 176 | 177 | \paragraph{Exercise 3.14} Let $p$ and $q$ be distinct odd primes such that $p-1$ divides $q-1$. If $(n, p q)=1$, show that $n^{q-1} \equiv 1(p q)$. 178 | \begin{proof} 179 | As $n \wedge pq = 1, n\wedge p=1, n \wedge q = 1$, so from Fermat's Little Theorem 180 | $$n^{q-1} \equiv 1 \pmod q,\qquad n^{p-1} \equiv 1 \pmod p.$$ 181 | $p-1 \mid q-1$, so there exists $k \in \mathbb{Z}$ such that $q-1 = k(p-1)$. 182 | Thus 183 | $$n^{q-1} = (n^{p-1})^k \equiv 1 \pmod p.$$ 184 | $p \mid n^{q-1} - 1, q \mid n^{q-1} - 1$, and $p\wedge q = 1$, so $pq \mid n^{q-1} - 1$ : 185 | $$n^{q-1} \equiv 1 \pmod{pq}.$$ 186 | \end{proof} 187 | 188 | \paragraph{Exercise 4.4} Consider a prime $p$ of the form $4 t+1$. Show that $a$ is a primitive root modulo $p$ iff $-a$ is a primitive root modulo $p$. 189 | \begin{proof} 190 | Suppose that $a$ is a primitive root modulo $p$. As $p-1$ is even, $(-a)^{p-1}=a^{p-1} \equiv 1$ $(\bmod p)$ 191 | If $(-a)^n \equiv 1(\bmod p)$, with $n \in \mathbb{N}$, then $a^n \equiv(-1)^n(\bmod p)$. 192 | Therefore $a^{2 n} \equiv 1(\bmod p)$. As $a$ is a primitive root modulo $p, p-1|2 n, 2 t| n$, so $n$ is even. 193 | 194 | Hence $a^n \equiv 1(\bmod p)$, and $p-1 \mid n$. So the least $n \in \mathbb{N}^*$ such that $(-a)^n \equiv 1$ $(\bmod p)$ is $p-1:$ the order of $-a$ modulo $p$ is $p-1,-a$ is a primitive root modulo $p$. Conversely, if $-a$ is a primitive root modulo $p$, we apply the previous result at $-a$ to to obtain that $-(-a)=a$ is a primitive root. 195 | \end{proof} 196 | 197 | 198 | 199 | \paragraph{Exercise 4.5} Consider a prime $p$ of the form $4 t+3$. Show that $a$ is a primitive root modulo $p$ iff $-a$ has order $(p-1) / 2$. 200 | \begin{proof} 201 | Let $a$ a primitive root modulo $p$. 202 | As $a^{p-1} \equiv 1(\bmod p), p \mid\left(a^{(p-1) / 2}-1\right)\left(a^{(p-1) / 2}+1\right)$, so $p \mid a^{(p-1) / 2}-1$ or $p \mid$ $a^{(p-1) / 2}+1$. As $a$ is a primitive root modulo $p, a^{(p-1) / 2} \not \equiv 1(\bmod p)$, so 203 | $$ 204 | a^{(p-1) / 2} \equiv-1 \quad(\bmod p) . 205 | $$ 206 | Hence $(-a)^{(p-1) / 2}=(-1)^{2 t+1} a^{(p-1) / 2} \equiv(-1) \times(-1)=1(\bmod p)$. 207 | Suppose that $(-a)^n \equiv 1(\bmod p)$, with $n \in \mathbb{N}$. 208 | Then $a^{2 n}=(-a)^{2 n} \equiv 1(\bmod p)$, so $p-1\left|2 n, \frac{p-1}{2}\right| n$. 209 | So $-a$ has order $(p-1) / 2$ modulo $p$. 210 | Conversely, suppose that $-a$ has order $(p-1) / 2=2 t+1$ modulo $p$. Let $2, p_1, \ldots p_k$ the prime factors of $p-1$, where $p_i$ are odd. 211 | $a^{(p-1) / 2}=a^{2 t+1}=-(-a)^{2 t+1}=-(-a)^{(p-1) / 2} \equiv-1$, so $a^{(p-1) / 2} \not \equiv 1(\bmod 2)$. 212 | As $p-1$ is even, $(p-1) / p_i$ is even, so $a^{(p-1) / p_i}=(-a)^{(p-1) / p_i} \not \equiv 1(\bmod p)($ since $-a$ has order $p-1)$. 213 | So the order of $a$ is $p-1$ (see Ex. 4.8) : $a$ is a primitive root modulo $p$. 214 | \end{proof} 215 | 216 | 217 | 218 | \paragraph{Exercise 4.6} If $p=2^{n}+1$ is a Fermat prime, show that 3 is a primitive root modulo $p$. 219 | \begin{proof} 220 | \newcommand{\legendre}[2]{\genfrac{(}{)}{}{}{#1}{#2}} 221 | Write $p = 2^k + 1$, with $k = 2^n$. 222 | 223 | We suppose that $n>0$, so $k\geq 2, p \geq 5$. As $p$ is prime, $3^{p-1} \equiv 1 \pmod p$. 224 | 225 | In other words, $3^{2^k} \equiv 1 \pmod p$ : the order of $3$ is a divisor of $2^k$, a power of $2$. 226 | 227 | $3$ has order $2^k$ modulo $p$ iff $3^{2^{k-1}} \not \equiv 1 \pmod p$. As $\left (3^{2^{k-1}} \right)^2 \equiv 1 \pmod p$, where $p$ is prime, this is equivalent to $3^{2^{k-1}} \equiv -1 \pmod p$, which remains to prove. 228 | 229 | $3^{2^{k-1}} = 3^{(p-1)/2} \equiv \legendre{3}{p} \pmod p$. 230 | 231 | As the result is true for $p=5$, we can suppose $n\geq 2$. 232 | From the law of quadratic reciprocity : 233 | $$\legendre{3}{p} \legendre{p}{3} = (-1)^{(p-1)/2} = (-1)^{2^{k-1}} = 1.$$ 234 | So $\legendre{3}{p} = \legendre{p}{3}$ 235 | 236 | \begin{align*} 237 | p = 2^{2^n}+1 &\equiv (-1)^{2^n} + 1 \pmod 3\\ 238 | &\equiv 2 \equiv -1 \pmod 3, 239 | \end{align*} 240 | so $\legendre{3}{p} = \legendre {p}{3} = -1$, that is to say 241 | $$3^{2^{k-1}} \equiv -1 \pmod p.$$ 242 | The order of $3$ modulo $p = 2^{2^n} + 1$ is $p-1 = 2^{2^n}$ : $3$ is a primitive root modulo $p$. 243 | 244 | (On the other hand, if $3$ is of order $p-1$ modulo $p$, then $p$ is prime, so 245 | $$ F_n = 2^{2^n} + 1 \ \mathrm{is}\ \mathrm{prime}\ \iff 3^{(F_n-1)/2} = 3^{2^{2^n - 1}} \equiv -1 \pmod {F_n}.)$$ 246 | \end{proof} 247 | 248 | 249 | 250 | \paragraph{Exercise 4.8} Let $p$ be an odd prime. Show that $a$ is a primitive root modulo $p$ iff $a^{(p-1) / q} \not \equiv 1(p)$ for all prime divisors $q$ of $p-1$. 251 | \begin{proof} 252 | $\bullet$ If $a$ is a primitive root, then $a^k \not \equiv 1$ for all $k, 1\leq k < p-1$, so $a^{(p-1)/q} \not \equiv 1 \pmod p$ for all prime divisors $q$ of $p - 1$. 253 | 254 | $\bullet$ In the other direction, suppose $a^{(p-1)/q} \not \equiv 1 \pmod p$ for all prime divisors $q$ of $p - 1$. 255 | 256 | Let $\delta$ the order of $a$, and $p-1 = q_1^{a_1}q_2^{a_2}\cdots q_k^{a_k}$ the decomposition of $p-1$ in prime factors. As $\delta \mid p-1, \delta = q_1^{b_1}p_2^{b_2}\cdots q_k^{b_k}$, with $b_i \leq a_i, i=1,2,\ldots,k$. If $b_i < a_i$ for some index $i$, then $\delta \mid (p-1)/q_i$, so $a^{(p-1)/q_i} \equiv 1 \pmod p$, which is in contradiction with the hypothesis. Thus $b_i = a_i$ for all $i$, and $\delta = q-1$ : $a$ is a primitive root modulo $p$. 257 | \end{proof} 258 | 259 | 260 | 261 | \paragraph{Exercise 4.11} Prove that $1^{k}+2^{k}+\cdots+(p-1)^{k} \equiv 0(p)$ if $p-1 \nmid k$ and $-1(p)$ if $p-1 \mid k$. 262 | \begin{proof} 263 | Let $S_k = 1^k+2^k+\cdots+(p-1)^k$. 264 | 265 | Let $g$ a primitive root modulo $p$ : $\overline{g}$ a generator of $\mathbb{F}_p^*$. 266 | 267 | As $(\overline{1},\overline{g},\overline{g}^{2}, \ldots, \overline{g}^{p-2}) $ is a permutation of $ (\overline{1},\overline{2}, \ldots,\overline{p-1})$, 268 | \begin{align*} 269 | \overline{S_k} &= \overline{1}^k + \overline{2}^k+\cdots+ \overline{p-1}^k\\ 270 | &= \sum_{i=0}^{p-2} \overline{g}^{ki} = 271 | \left\{ 272 | \begin{array}{ccc} 273 | \overline{ p-1} = -\overline{1} & \mathrm{if} & p-1 \mid k \\ 274 | \frac{ \overline{g}^{(p-1)k} -1}{ \overline{g}^k -1} = \overline{0}& \mathrm{if} & p-1 \nmid k 275 | \end{array} 276 | \right. 277 | \end{align*} 278 | since $p-1 \mid k \iff \overline{g}^k = \overline{1}$. 279 | 280 | Conclusion : 281 | \begin{align*} 282 | 1^k+2^k+\cdots+(p-1)^k&\equiv 0 \pmod p\ \mathrm{if} \ p-1 \nmid k\\ 283 | 1^k+2^k+\cdots+(p-1)^k&\equiv -1 \pmod p\ \mathrm{if} \ p-1 \mid k\\ 284 | \end{align*} 285 | \end{proof} 286 | 287 | 288 | 289 | \paragraph{Exercise 5.13} Show that any prime divisor of $x^{4}-x^{2}+1$ is congruent to 1 modulo 12 . 290 | \begin{proof} 291 | \newcommand{\legendre}[2]{\genfrac{(}{)}{}{}{#1}{#2}} 292 | $\bullet$ As $a^6 +1 = (a^2+1)(a^4-a^2+1)$, $p\mid a^4 - a^2+1$ implies $p \mid a^6 + 1$, so $\legendre{-1}{p} = 1$ and $p\equiv 1 \pmod 4$. 293 | 294 | $\bullet$ $p \mid 4a^4 - 4 a^2 +4 = (2a-1)^2 + 3$, so $\legendre{-3}{p} = 1$. 295 | 296 | As $-3 \equiv 1 \pmod 4$, $\legendre{-3}{p} = \legendre{p}{3}$, so $\legendre{p}{3} = 1$, thus $p \equiv 1 \pmod 3$. 297 | 298 | $4 \mid p-1$ and $3 \mid p-1$, thus $12 \mid p-1$ : $$p \equiv 1 \pmod {12}.$$ 299 | \end{proof} 300 | 301 | 302 | 303 | \paragraph{Exercise 5.28} Show that $x^{4} \equiv 2(p)$ has a solution for $p \equiv 1(4)$ iff $p$ is of the form $A^{2}+64 B^{2}$. 304 | \begin{proof} 305 | If $p\equiv 1\ [4]$ and if there exists $x \in \mathbb{Z}$ such that $x^4 \equiv 2\ [p]$, then 306 | $$2^{\frac{p-1}{4} }\equiv x^{p-1} \equiv 1 \ [p].$$ 307 | 308 | From Ex. 5.27, where $p = a^2 +b^2, a$ odd, we know that $$f^{\frac{ab}{2}} \equiv 2^{\frac{p-1}{4} } \equiv 1 \ [p].$$ 309 | 310 | Since $f^2 \equiv -1\ [p]$, the order of $f$ modulo $p$ is 4, thus $4 \mid \frac{ab}{2}$, so $8\mid ab$. 311 | 312 | As $a$ is odd, $8 | b$, then $p = A^2 + 64 B^2$ (with $A = a, B = b/8$). 313 | 314 | \bigskip 315 | 316 | Conversely, if $p=A^2+64 B^2$, then $p\equiv A^2 \equiv 1 \ [4]$. 317 | 318 | Let $a=A,b=8B$. Then $$2^{\frac{p-1}{4} } \equiv f^{\frac{ab}{2}} \equiv f^{4AB} \equiv (-1)^{2AB} \equiv 1 \ [p].$$ 319 | 320 | As $2^{\frac{p-1}{4} } \equiv 1 \ [p]$, $x^4 \equiv 2 \ [p]$ has a solution in $\mathbb{Z}$ (Prop. 4.2.1) : $2$ is a biquadratic residue modulo $p$. 321 | 322 | Conclusion : 323 | 324 | $$\exists A \in \mathbb{Z}, \exists B \in \mathbb{Z}\,, p = A^2+64 B^2 \iff( p\equiv 1 \ [4] \ \mathrm{and} \ \exists x \in \mathbb{Z}, \, x^4 \equiv 2 \ [p]).$$ 325 | \end{proof} 326 | 327 | 328 | 329 | \paragraph{Exercise 5.37} Show that if $a$ is negative then $p \equiv q(4 a) together with p\not | a$ imply $(a / p)=(a / q)$. 330 | \begin{proof} 331 | \newcommand{\legendre}[2]{\genfrac{(}{)}{}{}{#1}{#2}} 332 | Write $a = -A, A>0$. As $p \equiv q \pmod {4a}$, we know from Prop. 5.3.3. (b) that $(A/p) = (A/q)$. 333 | 334 | Moreover, 335 | \begin{align*} 336 | \legendre{a}{p}&= \legendre{-A}{p} = (-1)^{(p-1)/2} \legendre{A}{p}\\ 337 | \legendre{a}{q}&= \legendre{-A}{q} = (-1^{(q-1)/2} \legendre{A}{q} 338 | \end{align*} 339 | As $p \equiv q \pmod {4a}$, $ p = q + 4ak, k\in \mathbb{Z}$, so 340 | $$(-1)^{(p-1)/2} = (-1)^{(q+4ak-1)/2} = (-1)^{(q-1)/2},$$ 341 | so $(a/p) = (a/q)$. 342 | \end{proof} 343 | 344 | 345 | 346 | \paragraph{Exercise 12.12} Show that $\sin (\pi / 12)$ is an algebraic number. 347 | \begin{proof} 348 | $$ 349 | \begin{aligned} 350 | \sin \pi/12=\sin \left(\pi/4-\pi/6\right) & =\sin \pi/4 \cos \pi/6-\cos \pi/4 \sin \pi/6 \\ 351 | & =\frac{\sqrt{3}}{2 \sqrt{2}}-\frac{1}{2 \sqrt{2}} \\ 352 | & =\frac{\sqrt{3}-1}{2 \sqrt{2}} 353 | \end{aligned} 354 | $$ 355 | \end{proof} 356 | 357 | \paragraph{Exercise 18.4} Show that 1729 is the smallest positive integer expressible as the sum of two different integral cubes in two ways. 358 | \begin{proof} 359 | Let $n=a^3+b^3$, and suppose that $\operatorname{gcd}(a, b)=1$. If a prime $p \mid a^3+b^3$, then 360 | $$ 361 | \left(a b^{-1}\right)^3 \equiv_p-1 362 | $$ 363 | Thus $3 \mid \frac{p-1}{2}$, that is, $p \equiv_6 1$. 364 | If we have $n=a^3+b^3=c^3+d^3$, then we can factor $n$ as 365 | $$ 366 | \begin{aligned} 367 | & n=(a+b)\left(a^2-a b+b^2\right) \\ 368 | & n=(c+d)\left(c^2-c d+d^2\right) 369 | \end{aligned} 370 | $$ 371 | Thus we need $n$ to have at least 3 distinct prime factors, and so the smallest taxicab number is on the form 372 | $$ 373 | n=(6 k+1)(12 k+1)(18 k+1) 374 | $$ 375 | \end{proof} 376 | \end{document} 377 | -------------------------------------------------------------------------------- /informal/Munkres.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/rahul3613/ProofNet-lean4/1893b2f054fddb1c4c90b91732173528c16f3a37/informal/Munkres.pdf -------------------------------------------------------------------------------- /informal/Pugh.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/rahul3613/ProofNet-lean4/1893b2f054fddb1c4c90b91732173528c16f3a37/informal/Pugh.pdf -------------------------------------------------------------------------------- /informal/Pugh.tex: -------------------------------------------------------------------------------- 1 | \documentclass{article} 2 | 3 | \title{\textbf{ 4 | Exercises from \\ 5 | \textit{Real Mathematical Analysis} \\ 6 | by Charles Pugh 7 | }} 8 | 9 | \date{} 10 | 11 | \usepackage{amsmath} 12 | \usepackage{amssymb} 13 | \usepackage{amsthm} 14 | 15 | \begin{document} 16 | \maketitle 17 | 18 | 19 | 20 | \paragraph{Exercise 2.12a} Let $(p_n)$ be a sequence and $f:\mathbb{N}\to\mathbb{N}$. The sequence $(q_k)_{k\in\mathbb{N}}$ with $q_k=p_{f(k)}$ is called a rearrangement of $(p_n)$. Show that if $f$ is an injection, the limit of a sequence is unaffected by rearrangement. 21 | \begin{proof} 22 | Let $\varepsilon>0$. Since $p_n \rightarrow L$, we have that, for all $n$ except $n \leq N$, $d\left(p_n, L\right)<\epsilon$. Let $S=\{n \mid f(n) \leq N\}$, let $n_0$ be the largest $n \in S$, we know there is such a largest $n$ because $f(n)$ is injective. Now we have that $\forall n>n_0 f(n)>N$ which implies that $p_{f(n)} \rightarrow L$, as required. 23 | \end{proof} 24 | 25 | 26 | \paragraph{Exercise 2.26} Prove that a set $U \subset M$ is open if and only if none of its points are limits of its complement. 27 | \begin{proof} 28 | Assume that none of the points of $U$ are limits of its complement, and let us prove that $U$ is open. Assume by contradiction that $U$ is not open, so there exists $p \in M$ so that $\forall r>0$ there exists $q \in M$ with $d(p, q)0$ such that $d(p, x)0$ such that 52 | $$ 53 | C_1\|x\|_E \leq\|x\| \leq C_2\|x\|_E, \forall x \in \mathbb{R}^m . 54 | $$ 55 | Assuming (1) to be true, let us finish the problem. First let us show that $B$ is bounded w.r.t. $d_E$, which is how we call the Euclidean distance in $\mathbb{R}^m$. Indeed, given $x \in B,\|x\|_E \leq \frac{1}{C_1}\|x\| \leq \frac{1}{C_1}$. Hence $B \subset\left\{x \in \mathbb{R}^m: d_E(x, 0)<\frac{1}{C_1}+1\right\}$, which means $B$ is bounded w.r.t $d_E$. 56 | Now let us show that $B$ is closed w.r.t. $d_E$. Let $x_n \rightarrow x$ w.r.t. $d_E$, where $x_n \in B$. Notice that this implies that $x_n \rightarrow x$ w.r.t. $d(x, y)=\|x-y\|$, the distance coming from $\|\cdot\|$, since by (1) we have 57 | $$ 58 | d\left(x_n, x\right)=\left\|x_n-x\right\| \leq C_2\left\|x_n-x\right\|_E \rightarrow 0 . 59 | $$ 60 | Also, notice that 61 | $$ 62 | \|x\| \leq\left\|x_n-x\right\|+\left\|x_n\right\| \leq\left\|x_n-x\right\|+1, 63 | $$ 64 | hence passing to the limit we obtain that $\|x\| \leq 1$, therefore $x \in B$ and so $B$ is closed w.r.t. $d_E$. Since $B$ is closed and bounded w.r.t. $d_E$, it must be compact. Now we claim that the identity function, $i d:\left(\mathbb{R}^m, d_E\right) \rightarrow\left(\mathbb{R}^m, d\right)$ where $\left(\mathbb{R}^m, d_E\right)$ means we are using the distance $d_E$ in $\mathbb{R}^m$ and $\left(\mathbb{R}^m, d\right)$ means we are using the distance $d$ in $\mathbb{R}^m$, is a homeomorphism. This follows by (1), since $i d$ is always a bijection, and it is continuous and its inverse is continuous by (1) (if $x_n \rightarrow x$ w.r.t. $d_E$, then $x_n \rightarrow x$ w.r.t. $d$ and vice-versa, by (1)). By a result we saw in class, since $B$ is compact in $\left(\mathbb{R}^m, d_E\right)$ and $i d$ is a homeomorphism, then $i d(B)=B$ is compact w.r.t. $d$. 65 | 66 | We are left with proving (1). Notice that it suffices to prove that $C_1 \leq\|x\| \leq$ $C_2, \forall x \in \mathbb{R}^m$ with $\|x\|_E=1$. Indeed, if this is true, given $x \in \mathbb{R}^m$, either $\|x\|_E=0$ (which implies $x=0$ and (1) holds in this case), or $x /\|x\|_E=y$ is such that $\|y\|_E=1$, so $C_1 \leq\|y\| \leq C_2$, which implies $C_1\|x\|_E \leq\|x\| \leq C_2\|x\|_E$. 67 | We want to show now that $\|\cdot\|$ is continuous w.r.t. $d_E$, that is, given $\varepsilon>0$ and $x \in \mathbb{R}^m$, there exists $\delta>0$ such that if $d_E(x, y)<\delta$, then $\|\mid x\|-\|y\| \|<\varepsilon$. 68 | 69 | By the triangle inequality, $\|x\|-\|y\| \leq\|x-y\|$, and $\|y\|-\|x\| \leq\|x-y\|$, therefore 70 | $$ 71 | |\|x||-\| y|\|\leq\| x-y \| . 72 | $$ 73 | Writing now $x=\sum_{i=1}^m a_i e_i, y=\sum_{i=1}^m b_i e_i$, where $e_i=(0, \ldots, 1,0, \ldots, 0)$ (with 1 in the i-th component), we obtain by the triangle inequality, 74 | $$ 75 | \begin{aligned} 76 | \|x-y\| & =\left\|\sum_{i=1}^m\left(a_i-b_i\right) e_i\right\| \leq \sum_{i=1}^m\left|a_i-b_i\left\|\left|\left\|e_i\right\| \leq \max _{i=1, \ldots, m}\left\|e_i\right\| \sum_{i=1}^m\right| a_i-b_i \mid\right.\right. \\ 77 | & =\max _{i=1, \ldots, m}\left\|e_i\right\| d_{s u m}(x, y) \leq \max _{i=1, \ldots, m}\left\|e_i\right\| m d_{\max }(x, y) \\ 78 | & \leq \max _{i=1, \ldots, m}\left\|e_i\right\| m d_E(x, y) . 79 | \end{aligned} 80 | $$ 81 | Let $\delta=\frac{\varepsilon}{m \max _{i=1, \ldots, m}\left\|e_i\right\|}$. Then if $d_E(x, y)<\delta,\|x\|-\|y\|||<\varepsilon$. 82 | Since $\|\cdot\|$ is continuous w.r.t. $d_E$ and $K=\left\{x \in \mathbb{R}^m:\|x\|_E=1\right\}$ is compact w.r.t. $d_E$, then the function $\|\cdot\|$ achieves a maximum and a minimum value on $K$. Call $C_1=\min _{x \in K}\|x\|, C_2=\max _{x \in K}\|x\|$. Then 83 | $$ 84 | C_1 \leq\|x\| \leq C_2, \forall x \in \mathbb{R}^m \text { such that }\|x\|_E=1, 85 | $$ 86 | which is what we needed. 87 | \end{proof} 88 | 89 | 90 | 91 | \paragraph{Exercise 2.46} Assume that $A, B$ are compact, disjoint, nonempty subsets of $M$. Prove that there are $a_0 \in A$ and $b_0 \in B$ such that for all $a \in A$ and $b \in B$ we have $d(a_0, b_0) \leq d(a, b)$. 92 | \begin{proof} 93 | Let $A$ and $B$ be compact, disjoint and non-empty subsets of $M$. We want to show that there exist $a_0 \in A, b_0 \in B$ such that for all $a \in A, b \in B$, 94 | $$ 95 | d\left(a_0, b_0\right) \leq d(a, b) . 96 | $$ 97 | We saw in class that the distance function $d: M \times M \rightarrow \mathbb{R}$ is continuous. We also saw in class that any continuous, real-valued function assumes maximum and minimum values on a compact set. Since $A$ and $B$ are compact, $A \times B$ is (non-empty) compact in $M \times M$. Therefore there exists $\left(a_0, b_0\right) \in A \times B$ such that $d\left(a_0, b_0\right) \leq d(a, b), \forall(a, b) \in A \times B$. 98 | \end{proof} 99 | 100 | 101 | 102 | \paragraph{Exercise 2.57} Show that if $S$ is connected, it is not true in general that its interior is connected. 103 | \begin{proof} 104 | Consider $X=\mathbb{R}^2$ and 105 | $$ 106 | A=([-2,0] \times[-2,0]) \cup([0,2] \times[0,2]) 107 | $$ 108 | which is connected, while $\operatorname{int}(A)$ is not connected. 109 | To see this consider the continuous function $f: \mathbb{R}^2 \rightarrow \mathbb{R}$ is defined by $f(x, y)=x+y$. Let $U=f^{-1}(0,+\infty)$ which is open in $\mathbb{R}^2$ and so $U \cap \operatorname{int}(A)$ is open in $\operatorname{int}(A)$. Also, since $(0,0) \notin \operatorname{int}(A)$, so for all $(x, y) \in \operatorname{int}(A), f(x, y) \neq 0$ and $U \cap \operatorname{int}(A)=f^{-1}[0,+\infty) \cap \operatorname{int}(A)$ is closed in $\operatorname{int}(A)$. Furthermore, $(1,1)=f^{-1}(2) \in U \cap \operatorname{int}(A)$ shows that $U \cap \operatorname{int}(A) \neq \emptyset$ while $(-1,-1) \in \operatorname{int}(A)$ and $(-1,-1) \notin U$ shows that $U \cap \operatorname{int}(A) \neq \operatorname{int}(A)$. 110 | \end{proof} 111 | 112 | 113 | \paragraph{Exercise 2.92} Give a direct proof that the nested decreasing intersection of nonempty covering compact sets is nonempty. 114 | \begin{proof} 115 | Let 116 | $$ 117 | A_1 \supset A_2 \supset \cdots \supset A_n \supset \cdots 118 | $$ 119 | be a nested decreasing sequence of compacts. Suppose that $\bigcap A_n=\emptyset$. Take $U_n=A_n^c$, then 120 | $$ 121 | \bigcup U_n=\bigcup A_n^c=\left(\bigcap A_n\right)^c=A_1 . 122 | $$ 123 | Here, I'm thinking of $A_1$ as the main metric space. Since $\left\{U_n\right\}$ is an open covering of $A_1$, we can extract a finite subcovering, that is, 124 | $$ 125 | A_{\alpha_1}^c \cup A_{\alpha_2}^c \cup \cdots \cup A_{\alpha_m}^c \supset A_1 126 | $$ 127 | or 128 | $$ 129 | \left(A_1 \backslash A_{\alpha_1}\right) \cup\left(A_1 \backslash A_{\alpha_2}\right) \cup \cdots \cup\left(A_1 \backslash A_{\alpha_m}\right) \supset A_1 . 130 | $$ 131 | But, this is true only if $A_{\alpha_i}=\emptyset$ for some $i$, a contradiction. 132 | \end{proof} 133 | 134 | \paragraph{Exercise 2.126} Suppose that $E$ is an uncountable subset of $\mathbb{R}$. Prove that there exists a point $p \in \mathbb{R}$ at which $E$ condenses. 135 | \begin{proof} 136 | I think this is the proof by contrapositive that you were getting at. 137 | Suppose that $E$ has no limit points at all. Pick an arbitrary point $x \in E$. Then $x$ cannot be a limit point, so there must be some $\delta>0$ such that the ball of radius $\delta$ around $x$ contains no other points of $E$ : 138 | $$ 139 | B_\delta(x) \cap E=\{x\} 140 | $$ 141 | Call this "point 1 ". For the next point, take the closest element to $x$ and on its left; that is, choose the point 142 | $$ 143 | \max [E \cap(-\infty, x)] 144 | $$ 145 | if it exists (that is important - if not, skip to the next step). Note that by the argument above, this supremum, should it exist, cannot equal $x$ and is therefore a new point in $E$. 146 | 147 | Call this "point 2 ". Now take the first point to the right of $x$ for "point 3 ". Take the first point to the left of point 2 for "point 4 ". And so on, ad infinitum. 148 | 149 | This gives a countable list of unique points; we must show that it exhausts the entire set $E$. Suppose not. Suppose there is some element $aa_1 \in E$ such that $a_2$ is not in the list. Repeating, we generate an infinite monotone-increasing sequence $\left\{a_i\right\}$ of elements in $E$ and not in the list, which is clearly bounded above by $x$. By the Monotone 150 | Convergence Theorem this sequence has a limit. But that means the sequence $\left\{a_i\right\} \subset E$ converges to a limit, and hence $E$ has a limit point, contradicting the assumption. Therefore our list exhausts $E$, and we have enumerated all its elements. 151 | \end{proof} 152 | 153 | 154 | \paragraph{Exercise 3.1} Assume that $f \colon \mathbb{R} \rightarrow \mathbb{R}$ satisfies $|f(t)-f(x)| \leq|t-x|^{2}$ for all $t, x$. Prove that $f$ is constant. 155 | \begin{proof} 156 | We have $|f(t)-f(x)| \leq|t-x|^2, \forall t, x \in \mathbb{R}$. Fix $x \in \mathbb{R}$ and let $t \neq x$. Then 157 | $$ 158 | \left|\frac{f(t)-f(x)}{t-x}\right| \leq|t-x| \text {, hence } \lim _{t \rightarrow x}\left|\frac{f(t)-f(x)}{t-x}\right|=0 \text {, } 159 | $$ 160 | so $f$ is differentiable in $\mathbb{R}$ and $f^{\prime}=0$. This implies that $f$ is constant, as seen in class. 161 | \end{proof} 162 | 163 | 164 | 165 | \paragraph{Exercise 3.4} Prove that $\sqrt{n+1}-\sqrt{n} \rightarrow 0$ as $n \rightarrow \infty$. 166 | \begin{proof} 167 | $$ 168 | \sqrt{n+1}-\sqrt{n}=\frac{(\sqrt{n+1}-\sqrt{n})(\sqrt{n+1}+\sqrt{n})}{\sqrt{n+1}+\sqrt{n}}=\frac{1}{\sqrt{n+1}+\sqrt{n}}<\frac{1}{2 \sqrt{n}} 169 | $$ 170 | \end{proof} 171 | 172 | 173 | \paragraph{Exercise 3.63a} Prove that $\sum 1/k(\log(k))^p$ converges when $p > 1$. 174 | \begin{proof} 175 | Using the integral test, for a set $a$, we see 176 | $$ 177 | \lim _{b \rightarrow \infty} \int_a^b \frac{1}{x \log (x)^c} d x=\lim _{b \rightarrow \infty}\left(\frac{\log (b)^{1-c}}{1-c}-\frac{\log (a)^{1-c}}{1-c}\right) 178 | $$ 179 | which goes to infinity if $c \leq 1$ and converges if $c>1$. Thus, 180 | $$ 181 | \sum_{n=2}^{\infty} \frac{1}{n \log (n)^c} 182 | $$ 183 | converges if and only if $c>1$. 184 | \end{proof} 185 | 186 | 187 | 188 | \paragraph{Exercise 3.63b} Prove that $\sum 1/k(\log(k))^p$ diverges when $p \leq 1$. 189 | \begin{proof} 190 | Using the integral test, for a set $a$, we see 191 | $$ 192 | \lim _{b \rightarrow \infty} \int_a^b \frac{1}{x \log (x)^c} d x=\lim _{b \rightarrow \infty}\left(\frac{\log (b)^{1-c}}{1-c}-\frac{\log (a)^{1-c}}{1-c}\right) 193 | $$ 194 | which goes to infinity if $c \leq 1$ and converges if $c>1$. Thus, 195 | $$ 196 | \sum_{n=2}^{\infty} \frac{1}{n \log (n)^c} 197 | $$ 198 | converges if and only if $c>1$. 199 | \end{proof} 200 | 201 | 202 | 203 | \paragraph{Exercise 4.15a} A continuous, strictly increasing function $\mu \colon (0, \infty) \rightarrow (0, \infty)$ is a modulus of continuity if $\mu(s) \rightarrow 0$ as $s \rightarrow 0$. A function $f \colon [a, b] \rightarrow \mathbb{R}$ has modulus of continuity $\mu$ if $|f(s) - f(t)| \leq \mu(|s - t|)$ for all $s, t \in [a, b]$. Prove that a function is uniformly continuous if and only if it has a modulus of continuity. 204 | \begin{proof} 205 | Suppose there exists a modulus of continuity $w$ for $f$, then fix $\varepsilon>0$, since $\lim _{s \rightarrow 0} w(s)=0$, there exists $\delta>0$ such that for any $|s|<\delta$, we have $w(s)<\varepsilon$, then we have for any $x, z \in X$ such that $d_X(x, z)<\delta$, we have $d_Y(f(x), f(z)) \leq w\left(d_X(x, z)\right)<\varepsilon$, which means $f$ is uniformly continuous. 206 | 207 | Suppose $f:\left(X, d_X\right) \rightarrow\left(Y, d_Y\right)$ is uniformly continuous. 208 | Let $\delta_1>0$ be such that $d_X(a, b)<\delta_1$ implies $d_Y(f(a), f(b))<1$. 209 | Define $w:[0, \infty) \rightarrow[0, \infty]$ by 210 | $$ 211 | w(s)= \begin{cases}\left.\sup \left\{d_Y(f(a), f(b))\right\} \mid d_X(a, b) \leq s\right\} & \text { if } s \leq \delta_1 \\ \infty & \text { if } s>\delta_1\end{cases} 212 | $$ 213 | We'll show that $w$ is a modulus of continuity for $f \ldots$ 214 | By definition of $w$, it's immediate that $w(0)=0$ and it's clear that 215 | $$ 216 | d_Y(f(a), f(b)) \leq w\left(d_X(a, b)\right) 217 | $$ 218 | for all $a, b \in X$. 219 | It remains to show $\lim _{s \rightarrow 0^{+}} w(s)=0$. 220 | It's easily seen that $w$ is nonnegative and non-decreasing, hence $\lim _{s \rightarrow 0^{+}}=L$ for some $L \geq 0$, where $L=\inf w((0, \infty))$ 221 | Let $\epsilon>0$. 222 | By uniform continuity of $f$, there exists $\delta>0$ such that $d_X(a, b)<\delta$ implies $d_Y(f(a), f(b))<\epsilon$, hence by definition of $w$, we get $w(\delta) \leq \epsilon$. 223 | Thus $L \leq \epsilon$ for all $\epsilon>0$, hence $L=0$. 224 | This completes the proof. 225 | \end{proof} 226 | \end{document} 227 | -------------------------------------------------------------------------------- /informal/Putnam.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/rahul3613/ProofNet-lean4/1893b2f054fddb1c4c90b91732173528c16f3a37/informal/Putnam.pdf -------------------------------------------------------------------------------- /informal/Putnam.tex: -------------------------------------------------------------------------------- 1 | 2 | \documentclass{article} 3 | 4 | \title{\textbf{ 5 | Exercises from \\ 6 | \textit{Putnam Competition} \\ 7 | }} 8 | 9 | \date{} 10 | 11 | \usepackage{amsmath} 12 | \usepackage{amssymb} 13 | \usepackage{amsthm} 14 | 15 | \begin{document} 16 | \maketitle 17 | 18 | 19 | \paragraph{Exercise 2020.b5} For $j \in\{1,2,3,4\}$, let $z_{j}$ be a complex number with $\left|z_{j}\right|=1$ and $z_{j} \neq 1$. Prove that $3-z_{1}-z_{2}-z_{3}-z_{4}+z_{1} z_{2} z_{3} z_{4} \neq 0 .$ 20 | \begin{proof} 21 | It will suffice to show that for any $z_1, z_2, z_3, z_4 \in \mathbb{C}$ of modulus 1 such that $|3-z_1-z_2-z_3-z_4| = |z_1z_2z_3z_4|$, at least one of $z_1, z_2, z_3$ is equal to 1. 22 | 23 | To this end, let $z_1=e^{\alpha i}, z_2=e^{\beta i}, z_3=e^{\gamma i}$ and 24 | \[ 25 | f(\alpha, \beta, \gamma)=|3-z_1-z_2-z_3|^2-|1-z_1z_2z_3|^2. 26 | \] 27 | A routine calculation shows that 28 | \begin{align*} 29 | f(\alpha, \beta, \gamma)&= 30 | 10 - 6\cos(\alpha) - 6\cos(\beta) - 6\cos(\gamma) \\ 31 | &\quad + 2\cos(\alpha + \beta + \gamma) + 2\cos(\alpha - \beta) \\ 32 | &\quad + 2\cos(\beta - \gamma) + 2\cos(\gamma - \alpha). 33 | \end{align*} 34 | Since the function $f$ is continuously differentiable, and periodic in each variable, $f$ has a maximum and a minimum and it attains these values only at points where $\nabla f=(0,0,0)$. A routine calculation now shows that 35 | \begin{align*} 36 | \frac{\partial f}{\partial \alpha} + \frac{\partial f}{\partial \beta} + \frac{\partial f}{\partial \gamma} &= 37 | 6(\sin(\alpha) +\sin(\beta)+\sin(\gamma)- \sin(\alpha + \beta + \gamma)) \\ 38 | &= 39 | 24\sin\left(\frac{\alpha+\beta}{2}\right) \sin\left(\frac{\beta+\gamma}{2}\right) 40 | \sin\left(\frac{\gamma+\alpha}{2}\right). 41 | \end{align*} 42 | Hence every critical point of $f$ must satisfy one of $z_1z_2=1$, $z_2z_3=1$, or $z_3z_1=1$. By symmetry, let us assume that $z_1z_2=1$. Then 43 | \[ 44 | f = |3-2\mathrm{Re}(z_1)-z_3|^2-|1-z_3|^2; 45 | \] 46 | since $3-2\mathrm{Re}(z_1)\ge 1$, $f$ is nonnegative and can be zero only if the real part of $z_1$, and hence also $z_1$ itself, is equal to $1$. 47 | \end{proof} 48 | 49 | 50 | 51 | \paragraph{Exercise 2018.a5} Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be an infinitely differentiable function satisfying $f(0)=0, f(1)=1$, and $f(x) \geq 0$ for all $x \in$ $\mathbb{R}$. Show that there exist a positive integer $n$ and a real number $x$ such that $f^{(n)}(x)<0$. 52 | \begin{proof} 53 | Call a function $f\colon \mathbb{R} \to \mathbb{R}$ \textit{ultraconvex} if $f$ is infinitely differentiable and $f^{(n)}(x) \geq 0$ for all $n \geq 0$ and all $x \in \mathbb{R}$, where $f^{(0)}(x) = f(x)$; 54 | note that if $f$ is ultraconvex, then so is $f'$. 55 | Define the set 56 | \[ 57 | S = \{ f :\thinspace \mathbb{R} \to \mathbb{R} \,|\,f \text{ ultraconvex and } f(0)=0\}. 58 | \] 59 | For $f \in S$, we must have $f(x) = 0$ for all $x < 0$: if $f(x_0) > 0$ for some $x_0 < 0$, then 60 | by the mean value theorem there exists $x \in (0,x_0)$ for which $f'(x) = \frac{f(x_0)}{x_0} < 0$. 61 | In particular, $f'(0) = 0$, so $f' \in S$ also. 62 | 63 | We show by induction that for all $n \geq 0$, 64 | \[ 65 | f(x) \leq \frac{f^{(n)}(1)}{n!} x^n \qquad (f \in S, x \in [0,1]). 66 | \] 67 | We induct with base case $n=0$, which holds because any $f \in S$ is nondecreasing. Given the claim for $n=m$, 68 | we apply the induction hypothesis to $f' \in S$ to see that 69 | \[ 70 | f'(t) \leq \frac{f^{(n+1)}(1)}{n!} t^n \qquad (t \in [0,1]), 71 | \] 72 | then integrate both sides from $0$ to $x$ to conclude. 73 | 74 | Now for $f \in S$, we have $0 \leq f(1) \leq \frac{f^{(n)}(1)}{n!}$ for all $n \geq 0$. 75 | On the other hand, by Taylor's theorem with remainder, 76 | \[ 77 | f(x) \geq \sum_{k=0}^n \frac{f^{(k)}(1)}{k!}(x-1)^k \qquad (x \geq 1). 78 | \] 79 | Applying this with $x=2$, we obtain $f(2) \geq \sum_{k=0}^n \frac{f^{(k)}(1)}{k!}$ for all $n$; 80 | this implies that $\lim_{n\to\infty} \frac{f^{(n)}(1)}{n!} = 0$. 81 | Since $f(1) \leq \frac{f^{(n)}(1)}{n!}$, we must have $f(1) = 0$. 82 | 83 | For $f \in S$, we proved earlier that $f(x) = 0$ for all $x\leq 0$, as well as for $x=1$. Since 84 | the function $g(x) = f(cx)$ is also ultraconvex for $c>0$, we also have $f(x) = 0$ for all $x>0$; 85 | hence $f$ is identically zero. 86 | 87 | To sum up, if $f\colon \mathbb{R} \to \mathbb{R}$ is infinitely differentiable, $f(0)=0$, and $f(1) = 1$, 88 | then $f$ cannot be ultraconvex. This implies the desired result. 89 | \end{proof} 90 | 91 | 92 | \paragraph{Exercise 2018.b2} Let $n$ be a positive integer, and let $f_{n}(z)=n+(n-1) z+$ $(n-2) z^{2}+\cdots+z^{n-1}$. Prove that $f_{n}$ has no roots in the closed unit disk $\{z \in \mathbb{C}:|z| \leq 1\}$. 93 | \begin{proof} 94 | Note first that $f_n(1) > 0$, so $1$ is not a root of $f_n$. 95 | Next, note that 96 | \[ 97 | (z-1)f_n(z) = z^n + \cdots + z - n; 98 | \] 99 | however, for $\left| z \right| \leq 1$, we have 100 | $\left| z^n + \cdots + z \right| \leq n$ by the triangle inequality; 101 | equality can only occur if $z,\dots,z^n$ have norm 1 and the same argument, which only happens for $z=1$. 102 | Thus there can be no root of $f_n$ with $|z| \leq 1$. 103 | \end{proof} 104 | 105 | 106 | 107 | \paragraph{Exercise 2018.b4} Given a real number $a$, we define a sequence by $x_{0}=1$, $x_{1}=x_{2}=a$, and $x_{n+1}=2 x_{n} x_{n-1}-x_{n-2}$ for $n \geq 2$. Prove that if $x_{n}=0$ for some $n$, then the sequence is periodic. 108 | \begin{proof} 109 | We first rule out the case $|a|>1$. In this case, we prove that $|x_{n+1}| \geq |x_n|$ for all $n$, meaning that we cannot have $x_n = 0$. We proceed by induction; the claim is true for $n=0,1$ by hypothesis. To prove the claim for $n \geq 2$, write 110 | \begin{align*} 111 | |x_{n+1}| &= |2x_nx_{n-1}-x_{n-2}| \\ 112 | &\geq 2|x_n||x_{n-1}|-|x_{n-2}| \\ 113 | &\geq |x_n|(2|x_{n-1}|-1) \geq |x_n|, 114 | \end{align*} 115 | where the last step follows from $|x_{n-1}| \geq |x_{n-2}| \geq \cdots \geq |x_0| = 1$. 116 | 117 | We may thus assume hereafter that $|a|\leq 1$. We can then write $a = \cos b$ for some $b \in [0,\pi]$. 118 | Let $\{F_n\}$ be the Fibonacci sequence, defined as usual by $F_1=F_2=1$ and $F_{n+1}=F_n+F_{n-1}$. We show by induction that 119 | \[ 120 | x_n = \cos(F_n b) \qquad (n \geq 0). 121 | \] 122 | Indeed, this is true for $n=0,1,2$; given that it is true for $n \leq m$, then 123 | \begin{align*} 124 | 2x_mx_{m-1}&=2\cos(F_mb)\cos(F_{m-1}b) \\ 125 | &= \cos((F_m-F_{m-1})b)+\cos((F_m+F_{m-1})b) \\ 126 | &= \cos(F_{m-2}b)+\cos(F_{m+1}b) 127 | \end{align*} 128 | and so 129 | $x_{m+1} = 2x_mx_{m-1}-x_{m-2} = \cos(F_{m+1}b)$. This completes the induction. 130 | 131 | 132 | Since $x_n = \cos(F_n b)$, if $x_n=0$ for some $n$ then $F_n b = \frac{k}{2} \pi$ for some odd integer $k$. In particular, we can write $b = \frac{c}{d}(2\pi)$ where $c = k$ and $d = 4F_n$ are integers. 133 | 134 | 135 | Let $x_n$ denote the pair $(F_n,F_{n+1})$, where each entry in this pair is viewed as an element of $\mathbb{Z}/d\mathbb{Z}$. Since there are only finitely many possibilities for $x_n$, there must be some $n_2>n_1$ such that $x_{n_1}=x_{n_2}$. Now $x_n$ uniquely determines both $x_{n+1}$ and $x_{n-1}$, and it follows that the sequence $\{x_n\}$ is periodic: for $\ell = n_2-n_1$, $x_{n+\ell} = x_n$ for all $n \geq 0$. In particular, $F_{n+\ell} \equiv F_n \pmod{d}$ for all $n$. But then $\frac{F_{n+\ell}c}{d}-\frac{F_n c}{d}$ is an integer, and so 136 | \begin{align*} 137 | x_{n+\ell} &= \cos\left(\frac{F_{n+\ell}c}{d}(2\pi)\right)\\ 138 | & = \cos\left(\frac{F_n c}{d}(2\pi)\right) = x_n 139 | \end{align*} 140 | for all $n$. Thus the sequence $\{x_n\}$ is periodic, as desired. 141 | \end{proof} 142 | 143 | 144 | 145 | \paragraph{Exercise 2017.b3} Suppose that $f(x)=\sum_{i=0}^{\infty} c_{i} x^{i}$ is a power series for which each coefficient $c_{i}$ is 0 or 1 . Show that if $f(2 / 3)=3 / 2$, then $f(1 / 2)$ must be irrational. 146 | \begin{proof} 147 | Suppose by way of contradiction that $f(1/2)$ is rational. Then $\sum_{i=0}^{\infty} c_i 2^{-i}$ is the binary expansion of a rational number, and hence must be eventually periodic; that is, there exist some integers $m,n$ such that 148 | $c_i = c_{m+i}$ for all $i \geq n$. We may then write 149 | \[ 150 | f(x) = \sum_{i=0}^{n-1} c_i x^i + \frac{x^n}{1-x^m} \sum_{i=0}^{m-1} c_{n+i} x^i. 151 | \] 152 | Evaluating at $x = 2/3$, we may equate $f(2/3) = 3/2$ with 153 | \[ 154 | \frac{1}{3^{n-1}} \sum_{i=0}^{n-1} c_i 2^i 3^{n-i-1} + \frac{2^n 3^m}{3^{n+m-1}(3^m-2^m)} \sum_{i=0}^{m-1} c_{n+i} 2^i 3^{m-1-i}; 155 | \] 156 | since all terms on the right-hand side have odd denominator, the same must be true of the sum, a contradiction. 157 | \end{proof} 158 | 159 | 160 | 161 | \paragraph{Exercise 2014.a5} Let 162 | $P_n(x)=1+2 x+3 x^2+\cdots+n x^{n-1} .$ Prove that the polynomials $P_j(x)$ and $P_k(x)$ are relatively prime for all positive integers $j$ and $k$ with $j \neq k$. 163 | \begin{proof} 164 | Suppose to the contrary that there exist positive integers $i \neq j$ and a complex number $z$ such that $P_i(z) = P_j(z) = 0$. Note that $z$ cannot be a nonnegative real number or else $P_i(z), P_j(z) > 0$; we may put $w = z^{-1} \neq 0,1$. For $n \in \{i+1,j+1\}$ we compute that 165 | \[ 166 | w^n = n w - n + 1, 167 | \qquad \overline{w}^n = n \overline{w} - n + 1; 168 | \] 169 | note crucially that these equations also hold for $n \in \{0,1\}$. 170 | Therefore, the function $f: [0, +\infty) \to \mathbb{R}$ given by 171 | \[ 172 | f(t) = \left| w \right|^{2t} - t^2 \left| w \right|^2 + 2t(t-1)\mathrm{Re}(w) - (t-1)^2 173 | \] 174 | satisfies $f(t) = 0$ for $t \in \{0,1,i+1,j+1\}$. On the other hand, for all $t \geq 0$ we have 175 | \[ 176 | f'''(t) = (2 \log \left| w \right|)^3 \left| w \right|^{2t} > 0, 177 | \] 178 | so by Rolle's theorem, the equation $f^{(3-k)}(t) = 0$ has at most $k$ distinct solutions for $k=0,1,2,3$. This yields the desired contradiction. 179 | \end{proof} 180 | 181 | 182 | 183 | \paragraph{Exercise 2010.a4} Prove that for each positive integer $n$, the number $10^{10^{10^n}}+10^{10^n}+10^n-1$ is not prime. 184 | \begin{proof} 185 | Put 186 | \[ 187 | N = 10^{10^{10^n}} + 10^{10^n} + 10^n - 1. 188 | \] 189 | Write $n = 2^m k$ with $m$ a nonnegative integer and $k$ a positive odd integer. 190 | For any nonnegative integer $j$, 191 | \[ 192 | 10^{2^m j} \equiv (-1)^j \pmod{10^{2^m} + 1}. 193 | \] 194 | Since $10^n \geq n \geq 2^m \geq m+1$, $10^n$ is divisible by $2^n$ and hence by $2^{m+1}$, 195 | and similarly $10^{10^n}$ is divisible by $2^{10^n}$ and hence by $2^{m+1}$. It follows that 196 | \[ 197 | N \equiv 1 + 1 + (-1) + (-1) \equiv 0 \pmod{10^{2^m} + 1}. 198 | \] 199 | Since $N \geq 10^{10^n} > 10^n + 1 \geq 10^{2^m} + 1$, it follows that $N$ is composite. 200 | \end{proof} 201 | 202 | 203 | 204 | \paragraph{Exercise 2001.a5} Prove that there are unique positive integers $a, n$ such that $a^{n+1}-(a+1)^n=2001$. 205 | \begin{proof} 206 | Suppose $a^{n+1} - (a+1)^n = 2001$. 207 | Notice that $a^{n+1} + [(a+1)^n - 1]$ is a multiple of $a$; thus 208 | $a$ divides $2002 = 2 \times 7 \times 11 \times 13$. 209 | 210 | Since $2001$ is divisible by 3, we must have $a \equiv 1 \pmod{3}$, 211 | otherwise one of $a^{n+1}$ and $(a+1)^n$ is a multiple of 3 and the 212 | other is not, so their difference cannot be divisible by 3. Now 213 | $a^{n+1} \equiv 1 \pmod{3}$, so we must have $(a+1)^n \equiv 1 214 | \pmod{3}$, which forces $n$ to be even, and in particular at least 2. 215 | 216 | If $a$ is even, then $a^{n+1} - (a+1)^n \equiv -(a+1)^n \pmod{4}$. 217 | Since $n$ is even, $-(a+1)^n \equiv -1 \pmod{4}$. Since $2001 \equiv 1 218 | \pmod{4}$, this is impossible. Thus $a$ is odd, and so must divide 219 | $1001 = 7 \times 11 \times 13$. Moreover, $a^{n+1} - (a+1)^n \equiv a 220 | \pmod{4}$, so $a \equiv 1 \pmod{4}$. 221 | 222 | Of the divisors of $7 \times 11 \times 13$, those congruent to 1 mod 3 223 | are precisely those not divisible by 11 (since 7 and 13 are both 224 | congruent to 1 mod 3). Thus $a$ divides $7 \times 13$. Now 225 | $a \equiv 1 \pmod{4}$ is only possible if $a$ divides $13$. 226 | 227 | We cannot have $a=1$, since $1 - 2^n \neq 2001$ for any $n$. Thus 228 | the only possibility is $a = 13$. One easily checks that $a=13, n=2$ is a 229 | solution; all that remains is to check that no other $n$ works. In fact, 230 | if $n > 2$, then $13^{n+1} \equiv 2001 \equiv 1 \pmod{8}$. 231 | But $13^{n+1} \equiv 13 \pmod{8}$ since $n$ is even, contradiction. 232 | Thus $a=13, n=2$ is the unique solution. 233 | 234 | Note: once one has that $n$ is even, one can use that $2002 235 | =a^{n+1} + 1 - (a+1)^n$ is divisible by $a+1$ to rule out cases. 236 | \end{proof} 237 | 238 | 239 | 240 | \paragraph{Exercise 2000.a2} Prove that there exist infinitely many integers $n$ such that $n, n+1, n+2$ are each the sum of the squares of two integers. 241 | \begin{proof} 242 | It is well-known that the equation $x^2-2y^2=1$ has infinitely 243 | many solutions (the so-called ``Pell'' equation). Thus setting 244 | $n=2y^2$ (so that $n=y^2+y^2$, $n+1=x^2+0^2$, $n+2=x^2+1^2$) 245 | yields infinitely many $n$ with the desired property. 246 | \end{proof} 247 | 248 | 249 | 250 | \paragraph{Exercise 1999.b4} Let $f$ be a real function with a continuous third derivative such that $f(x), f^{\prime}(x), f^{\prime \prime}(x), f^{\prime \prime \prime}(x)$ are positive for all $x$. Suppose that $f^{\prime \prime \prime}(x) \leq f(x)$ for all $x$. Show that $f^{\prime}(x)<2 f(x)$ for all $x$. 251 | \begin{proof} 252 | \setcounter{equation}{0} 253 | We make repeated use of the following fact: if $f$ is a differentiable function on all of 254 | $\mathbb{R}$, $\lim_{x \to -\infty} f(x) \geq 0$, and $f'(x) > 0$ for all $x \in \mathbb{R}$, then 255 | $f(x) > 0$ for all $x \in \mathbb{R}$. (Proof: if $f(y) < 0$ for some $x$, then $f(x)< f(y)$ for all 256 | $x0$, but then $\lim_{x \to -\infty} f(x) \leq f(y) < 0$.) 257 | 258 | From the inequality $f'''(x) \leq f(x)$ we obtain 259 | \[ 260 | f'' f'''(x) \leq f''(x) f(x) < f''(x) f(x) + f'(x)^2 261 | \] 262 | since $f'(x)$ is positive. Applying the fact to the difference between the right and left sides, 263 | we get 264 | \begin{equation} 265 | \frac{1}{2} (f''(x))^2 < f(x) f'(x). 266 | \end{equation} 267 | 268 | On the other hand, since $f(x)$ and $f'''(x)$ are both positive for all $x$, 269 | we have 270 | \[ 271 | 2f'(x) f''(x) < 2f'(x)f''(x) + 2f(x) f'''(x). 272 | \] 273 | Applying the fact to the difference between the sides yields 274 | \begin{equation} 275 | f'(x)^2 \leq 2f(x) f''(x). 276 | \end{equation} 277 | Combining (1) and (2), we obtain 278 | \begin{align*} 279 | \frac{1}{2} \left( \frac{f'(x)^2}{2f(x)} \right)^2 280 | &< \frac{1}{2} (f''(x))^2 \\ 281 | &< f(x) f'(x), 282 | \end{align*} 283 | or $(f'(x))^3 < 8 f(x)^3$. We conclude $f'(x) < 2f(x)$, as desired. 284 | \end{proof} 285 | 286 | 287 | 288 | \paragraph{Exercise 1998.a3} Let $f$ be a real function on the real line with continuous third derivative. Prove that there exists a point $a$ such that 289 | $f(a) \cdot f^{\prime}(a) \cdot f^{\prime \prime}(a) \cdot f^{\prime \prime \prime}(a) \geq 0$. 290 | \begin{proof} 291 | If at least one of $f(a)$, $f'(a)$, $f''(a)$, or $f'''(a)$ vanishes 292 | at some point $a$, then we are done. Hence we may assume each of 293 | $f(x)$, $f'(x)$, $f''(x)$, and $f'''(x)$ is either strictly positive 294 | or strictly negative on the real line. By replacing $f(x)$ by $-f(x)$ 295 | if necessary, we may assume $f''(x)>0$; by replacing $f(x)$ 296 | by $f(-x)$ if necessary, we may assume $f'''(x)>0$. (Notice that these 297 | substitutions do not change the sign of $f(x) f'(x) f''(x) f'''(x)$.) 298 | Now $f''(x)>0$ implies that $f'(x)$ is increasing, and $f'''(x)>0$ 299 | implies that $f'(x)$ is convex, so that $f'(x+a)>f'(x)+a f''(x)$ 300 | for all $x$ and $a$. By 301 | letting $a$ increase in the latter inequality, we see that $f'(x+a)$ 302 | must be positive for sufficiently large $a$; it follows that 303 | $f'(x)>0$ 304 | for all $x$. Similarly, $f'(x)>0$ and $f''(x)>0$ imply 305 | that $f(x)>0$ for all $x$. Therefore $f(x) f'(x) f''(x) f'''(x)>0$ for 306 | all $x$, and we are done. 307 | \end{proof} 308 | 309 | 310 | 311 | \paragraph{Exercise 1998.b6} Prove that, for any integers $a, b, c$, there exists a positive integer $n$ such that $\sqrt{n^3+a n^2+b n+c}$ is not an integer. 312 | \begin{proof} 313 | We prove more generally that for any polynomial $P(z)$ with integer 314 | coefficients which is not a perfect square, there exists a positive 315 | integer $n$ such that $P(n)$ is not a perfect square. Of course it 316 | suffices to assume $P(z)$ has no repeated factors, which is to say $P(z)$ 317 | and its derivative $P'(z)$ are relatively prime. 318 | 319 | In particular, if we carry out the Euclidean algorithm on $P(z)$ and $P'(z)$ 320 | without dividing, we get an integer $D$ (the discriminant of $P$) such that 321 | the greatest common divisor of $P(n)$ and $P'(n)$ divides $D$ for any $n$. 322 | Now there exist infinitely many primes $p$ such that $p$ divides $P(n)$ for 323 | some $n$: if there were only finitely many, say, $p_1, \dots, p_k$, then 324 | for any $n$ divisible by $m = P(0) p_1 p_2 \cdots p_k$, we have $P(n) 325 | \equiv P(0) \pmod{m}$, that is, $P(n)/P(0)$ is not divisible by $p_1, 326 | \dots, p_k$, so must be $\pm 1$, but then $P$ takes some value infinitely 327 | many times, contradiction. In particular, we can choose some such $p$ not 328 | dividing $D$, and choose $n$ such that $p$ divides $P(n)$. Then $P(n+kp) 329 | \equiv P(n) + kp P'(n) (\mathrm{mod}\,p)$ 330 | (write out the Taylor series of the left side); 331 | in particular, since $p$ does not divide $P'(n)$, we can find some $k$ 332 | such that $P(n+kp)$ is divisible by $p$ but not by $p^2$, and so 333 | is not a perfect square. 334 | \end{proof} 335 | 336 | 337 | 338 | 339 | \end{document} 340 | -------------------------------------------------------------------------------- /informal/Rudin.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/rahul3613/ProofNet-lean4/1893b2f054fddb1c4c90b91732173528c16f3a37/informal/Rudin.pdf -------------------------------------------------------------------------------- /informal/Shakarchi.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/rahul3613/ProofNet-lean4/1893b2f054fddb1c4c90b91732173528c16f3a37/informal/Shakarchi.pdf -------------------------------------------------------------------------------- /informal/Shakarchi.tex: -------------------------------------------------------------------------------- 1 | \documentclass{article} 2 | 3 | \title{\textbf{ 4 | Exercises from \\ 5 | \textit{Complex analysis} \\ 6 | by Elias M. Stein and Rami Shakarchi 7 | }} 8 | 9 | \date{} 10 | 11 | \usepackage{amsmath} 12 | \usepackage{amssymb} 13 | \usepackage{amsthm} 14 | 15 | \begin{document} 16 | \maketitle 17 | 18 | 19 | \paragraph{Exercise 1.13a} Suppose that $f$ is holomorphic in an open set $\Omega$. Prove that if $\text{Re}(f)$ is constant, then $f$ is constant. 20 | \begin{proof} 21 | Let $f(z)=f(x, y)=u(x, y)+i v(x, y)$, where $z=x+i y$. 22 | Since $\operatorname{Re}(f)=$ constant, 23 | $$ 24 | \frac{\partial u}{\partial x}=0, \frac{\partial u}{\partial y}=0 . 25 | $$ 26 | By the Cauchy-Riemann equations, 27 | $$ 28 | \frac{\partial v}{\partial x}=-\frac{\partial u}{\partial y}=0 . 29 | $$ 30 | Thus, in $\Omega$, 31 | $$ 32 | f^{\prime}(z)=\frac{\partial f}{\partial x}=\frac{\partial u}{\partial x}+i \frac{\partial v}{\partial x}=0+0=0 . 33 | $$ 34 | 3 35 | Thus $f(z)$ is constant. 36 | \end{proof} 37 | 38 | 39 | 40 | \paragraph{Exercise 1.13b} Suppose that $f$ is holomorphic in an open set $\Omega$. Prove that if $\text{Im}(f)$ is constant, then $f$ is constant. 41 | \begin{proof} 42 | Let $f(z)=f(x, y)=u(x, y)+i v(x, y)$, where $z=x+i y$. 43 | Since $\operatorname{Im}(f)=$ constant, 44 | $$ 45 | \frac{\partial v}{\partial x}=0, \frac{\partial v}{\partial y}=0 . 46 | $$ 47 | By the Cauchy-Riemann equations, 48 | $$ 49 | \frac{\partial u}{\partial x}=\frac{\partial v}{\partial y}=0 . 50 | $$ 51 | Thus in $\Omega$, 52 | $$ 53 | f^{\prime}(z)=\frac{\partial f}{\partial x}=\frac{\partial u}{\partial x}+i \frac{\partial v}{\partial x}=0+0=0 . 54 | $$ 55 | Thus $f$ is constant. 56 | \end{proof} 57 | 58 | 59 | \paragraph{Exercise 1.13c} Suppose that $f$ is holomorphic in an open set $\Omega$. Prove that if $|f|$ is constant, then $f$ is constant. 60 | \begin{proof} 61 | Let $f(z)=f(x, y)=u(x, y)+i v(x, y)$, where $z=x+i y$. 62 | We first give a mostly correct argument; the reader should pay attention to find the difficulty. Since $|f|=\sqrt{u^2+v^2}$ is constant, 63 | $$ 64 | \left\{\begin{array}{l} 65 | 0=\frac{\partial\left(u^2+v^2\right)}{\partial x}=2 u \frac{\partial u}{\partial x}+2 v \frac{\partial v}{\partial x} . \\ 66 | 0=\frac{\partial\left(u^2+v^2\right)}{\partial y}=2 u \frac{\partial u}{\partial y}+2 v \frac{\partial v}{\partial y} . 67 | \end{array}\right. 68 | $$ 69 | Plug in the Cauchy-Riemann equations and we get 70 | $$ 71 | \begin{gathered} 72 | u \frac{\partial v}{\partial y}+v \frac{\partial v}{\partial x}=0 \\ 73 | -u \frac{\partial v}{\partial x}+v \frac{\partial v}{\partial y}=0 \\ 74 | (1.14) \Rightarrow \frac{\partial v}{\partial x}=\frac{v}{u} \frac{\partial v}{\partial y} 75 | \end{gathered} 76 | $$ 77 | Plug (1.15) into (1.13) and we get 78 | $$ 79 | \frac{u^2+v^2}{u} \frac{\partial v}{\partial y}=0 . 80 | $$ 81 | So $u^2+v^2=0$ or $\frac{\partial v}{\partial y}=0$. 82 | 83 | If $u^2+v^2=0$, then, since $u, v$ are real, $u=v=0$, and thus $f=0$ which is constant. 84 | 85 | Thus we may assume $u^2+v^2$ equals a non-zero constant, and we may divide by it. We multiply both sides by $u$ and find $\frac{\partial v}{\partial y}=0$, then by (1.15), $\frac{\partial v}{\partial x}=0$, and by Cauchy-Riemann, $\frac{\partial u}{\partial x}=0$. 86 | $$ 87 | f^{\prime}=\frac{\partial f}{\partial x}=\frac{\partial u}{\partial x}+i \frac{\partial v}{\partial x}=0 . 88 | $$ 89 | Thus $f$ is constant. 90 | Why is the above only mostly a proof? The problem is we have a division by $u$, and need to make sure everything is well-defined. Specifically, we need to know that $u$ is never zero. We do have $f^{\prime}=0$ except at points where $u=0$, but we would need to investigate that a bit more. 91 | Let's return to 92 | $$ 93 | \left\{\begin{array}{l} 94 | 0=\frac{\partial\left(u^2+v^2\right)}{\partial x}=2 u \frac{\partial u}{\partial x}+2 v \frac{\partial v}{\partial x} . \\ 95 | 0=\frac{\partial\left(u^2+v^2\right)}{\partial y}=2 u \frac{\partial u}{\partial y}+2 v \frac{\partial v}{\partial y} . 96 | \end{array}\right. 97 | $$ 98 | Plug in the Cauchy-Riemann equations and we get 99 | $$ 100 | \begin{array}{r} 101 | u \frac{\partial v}{\partial y}+v \frac{\partial v}{\partial x}=0 \\ 102 | -u \frac{\partial v}{\partial x}+v \frac{\partial v}{\partial y}=0 . 103 | \end{array} 104 | $$ 105 | We multiply the first equation $u$ and the second by $v$, and obtain 106 | $$ 107 | \begin{aligned} 108 | u^2 \frac{\partial v}{\partial y}+u v \frac{\partial v}{\partial x} & =0 \\ 109 | -u v \frac{\partial v}{\partial x}+v^2 \frac{\partial v}{\partial y} & =0 . 110 | \end{aligned} 111 | $$ 112 | Adding the two yields 113 | $$ 114 | u^2 \frac{\partial v}{\partial y}+v^2 \frac{\partial v}{\partial y}=0, 115 | $$ 116 | or equivalently 117 | $$ 118 | \left(u^2+v^2\right) \frac{\partial v}{\partial y}=0 . 119 | $$ 120 | We now argue in a similar manner as before, except now we don't have the annoying $u$ in the denominator. If $u^2+v^2=0$ then $u=v=0$, else we can divide by $u^2+v^2$ and find $\partial v / \partial y=0$. Arguing along these lines finishes the proof. 121 | \end{proof} 122 | 123 | 124 | 125 | \paragraph{Exercise 1.19a} Prove that the power series $\sum nz^n$ does not converge on any point of the unit circle. 126 | \begin{proof} 127 | For $z \in S:=\{z \in \mathbb{C}:|z|=1\}$ it also holds $z^n \in S$ for all $n \in \mathbb{N}$ (since in this case $\left.\left|z^n\right|=|z|^n=1^n=1\right)$ 128 | Thus, the sequence $\left(a_n\right)_{n \in \mathbb{N}}$ with $a_n=n z^n$ does not converge to zero which is necessary for the corresponding sum $\sum_{n \in \mathbb{N}} a_n$ to be convergent. Hence this sum does not converge. 129 | \end{proof} 130 | 131 | 132 | 133 | \paragraph{Exercise 1.19b} Prove that the power series $\sum zn/n^2$ converges at every point of the unit circle. 134 | \begin{proof} 135 | Since $\left|z^n / n^2\right|=1 / n^2$ for all $|z|=1$, then $\sum z^n / n^2$ converges at every point in the unit circle as $\sum 1 / n^2$ does ( $p$-series $p=2$.) 136 | \end{proof} 137 | 138 | 139 | 140 | \paragraph{Exercise 1.19c} Prove that the power series $\sum zn/n$ converges at every point of the unit circle except $z = 1$. 141 | \begin{proof} 142 | If $z=1$ then $\sum z^n / n=\sum 1 / n$ is divergent (harmonic series). If $|z|=1$ and $z \neq 1$, write $z=e^{2 \pi i t}$ with $t \in(0,1)$ and apply Dirichlet's test: if $\left\{a_n\right\}$ is a sequence of real numbers and $\left\{b_n\right\}$ a sequence of complex numbers satisfying 143 | - $a_{n+1} \leq a_n$ 144 | - $\lim _{n \rightarrow \infty} a_n=0$ 145 | - $\left|\sum_{n=1}^N b_n\right| \leq M$ for every positive integer $N$ and some $M>0$, 146 | then $\sum a_n b_n$ converges. Let $a_n=1 / n$, so $a_n$ satisfies $a_{n+1} \leq a_n$ and $\lim _{n \rightarrow \infty} a_n=0$. Let $b_n=e^{2 \pi i n t}$, then 147 | $$ 148 | \left|\sum_{n=1}^N b_n\right|=\left|\sum_{n=1}^N e^{2 \pi i n t}\right|=\left|\frac{e^{2 \pi i t}-e^{2 \pi i(N+1) t}}{1-e^{2 \pi i t}}\right| \leq \frac{2}{\left|1-e^{2 \pi i t}\right|}=M \text { for all } N 149 | $$ 150 | Thus $\sum a_n b_n=\sum z^n / n$ converges for every point in the unit circle except $z=1$. 151 | \end{proof} 152 | 153 | 154 | \paragraph{Exercise 1.26} Suppose $f$ is continuous in a region $\Omega$. Prove that any two primitives of $f$ (if they exist) differ by a constant. 155 | \begin{proof} 156 | Suppose $F_1$ and $F_2$ are primitives of $F$. Then $(F_1-F_2)^\prime = f - f = 0$, therefore $F_1$ and $F_2$ differ by a constant. 157 | \end{proof} 158 | 159 | 160 | 161 | \paragraph{Exercise 2.2} Show that $\int_{0}^{\infty} \frac{\sin x}{x} d x=\frac{\pi}{2}$. 162 | \begin{proof} 163 | We have $\int_0^{\infty} \frac{\sin x}{x} d x=\frac{1}{2 * i} \int_0^{\infty} \frac{e^{i * x}-e^{-i * x}}{x} d x=\frac{1}{2 * i}\left(\int_0^{\infty} \frac{e^{i * x}-1}{x} d x-\int_0^{\infty} \frac{e^{-i * x}-1}{x} d x=\right.$ $\frac{1}{2 * i} \int_{-\infty}^{\infty} \frac{e^{i * x}-1}{x} d x$. Now integrate along the big and small semicircles $C_0$ and $C_1$ shown below. For $C_0$ : we have that $\int_{C_0} \frac{1}{x} d x=\pi * i$ and $\left|\int_{C_0} \frac{e^{i * x}}{x} d x\right| \leq$ $2 *\left|\int_{C_{00}} \frac{e^{i * x}}{x} d x\right|+\left|\int_{C_{01}} \frac{e^{i * x}}{x} d x\right|$ where $C_{00}$ and $C_{01}$ are shown below $\left(C_{01}\right.$ contains the part of $C_0$ that has points with imaginary parts more than $a$ and $C_{00}$ is one of the other 2 components). We have $\left|\int_{C_{00}} \frac{e^{i * x}}{x} d x\right| \leq$ $\sup _{x \in C_{00}}\left(e^{i * x}\right) / R * \int_{C_{00}}|d x| \leq e^{-a} * \pi$ and $\left|\int_{C_{01}} \frac{e^{i * x}}{x} d x\right| \leq\left|\int_{C_{01}} \frac{1}{x} d x\right| \leq$ $\frac{1}{R} * C * a$ for some constant $C$ (the constant $C$ exists because the length of the curve approaches $a$ as $a / R \rightarrow 0)$. Thus, the integral of $e^{i * x} / x$ over $C_0$ is bounded by $A * e^{-a}+B * a / R$ for some constants $A$ and $B$. Pick $R$ large and $a=\sqrt{R}$ and note that the above tends to 0 . About the integral over $C_1$ : We have $e^{i * x}-1=1+O(x)$ for $x \rightarrow 0$ (this is again from $\sin (x) / x \rightarrow 1$ ), 164 | 4 165 | so $\left|\int_{C_1} \frac{e^{i * x}-1}{x} d x\right| \leq O(1) *\left|\int_{C_1} d x\right| \rightarrow 0$ as $x \rightarrow 0$. Thus, we only care about the integral over $C_{00}$ which is $-\pi * i$. Using Cauchy's theorem we get that our integral equals $\frac{1}{2 * i}(-(\pi * i))=\pi / 2$. 166 | \end{proof} 167 | 168 | 169 | 170 | \paragraph{Exercise 2.9} Let $\Omega$ be a bounded open subset of $\mathbb{C}$, and $\varphi: \Omega \rightarrow \Omega$ a holomorphic function. Prove that if there exists a point $z_{0} \in \Omega$ such that $\varphi\left(z_{0}\right)=z_{0} \quad \text { and } \quad \varphi^{\prime}\left(z_{0}\right)=1$ then $\varphi$ is linear. 171 | \begin{proof} 172 | When $\Omega$ is connected, if $\varphi$ is not linear, then there exists $n \geq 2$ and $a_n \neq 0$, such that 173 | $$ 174 | \varphi(z)=z+a_n\left(z-z_0\right)^n+O\left(\left(z-z_0\right)^{n+1}\right) . 175 | $$ 176 | As you have noticed, by induction, it follows that for every $k \geq 1$, 177 | $$ 178 | \varphi^k(z)=z+k a_n\left(z-z_0\right)^n+O\left(\left(z-z_0\right)^{n+1}\right) . 179 | $$ 180 | Let $r>0$ be such that when $\left|z-z_0\right| \leq r$, then $z \in \Omega$. Then by (1), 181 | $$ 182 | k a_n=\frac{1}{2 \pi i} \int_{\left|z-z_0\right|=r} \frac{\varphi^k(z)}{\left(z-z_0\right)^{n+1}} d z . 183 | $$ 184 | Since $\varphi^k(\Omega) \subset \Omega$ and since $\Omega$ is bounded, there exists $M>0$, independent of $k$, such that $\left|\varphi^k\right| \leq M$ on $\Omega$. Then by (2), 185 | $$ 186 | k\left|a_n\right| \leq M r^{-n} . 187 | $$ 188 | Since $k$ is arbitrary, $a_n=0$, a contradiction. 189 | \end{proof} 190 | 191 | 192 | 193 | \paragraph{Exercise 2.13} Suppose $f$ is an analytic function defined everywhere in $\mathbb{C}$ and such that for each $z_0 \in \mathbb{C}$ at least one coefficient in the expansion $f(z) = \sum_{n=0}^\infty c_n(z - z_0)^n$ is equal to 0. Prove that $f$ is a polynomial. 194 | \begin{proof} 195 | Say that at least one of the coefficients of the Taylor series vanishes is the same as saying that for every $a \in \mathbb{C}$ there is $m \in \mathbb{N}$ such that $f^{(m)}(a)=0$. 196 | Consider $A_n:=\left\{z \in \mathbb{C}: f^{(n)}(z)=0\right\}$ for each $n \in \mathbb{N}$. Note that: 197 | $f$ is polynomial iff $A_n$ is not countable for some $n \in \mathbb{N}$. 198 | Indeed, if $f$ is polynomial of degree $n$, then $f^{(n+1)}(z)=0$ for all $z \in \mathbb{C}$, then $A_{n+1}=\mathbb{C}$, so, $A_{n+1}$ is not countable. Conversely, if there is $n \in \mathbb{C}$ such that $A_n$ is not countable, then $A_n$ has a limit point, then by Identity principle we have $f^{(n)}(z)=0$ for all $z \in \mathbb{C}$, so, $f$ is a polynomial of degree at most $n-1$. 199 | 200 | Therefore, tt suffices to show that there is $n \in \mathbb{N}$ such that $A_n$ is not countable. Indeed, consider $\bigcup_{n \in \mathbb{N}} A_n$, by hypothesis for each $a \in \mathbb{C}$ there is $m \in \mathbb{N}$ such that $f^{(m)}(a)=0$, then $\mathbb{C} \subseteq \bigcup_{n \in \mathbb{N}} A_n$. Therefore, $\bigcup_{n \in \mathbb{N}} A_n$ is not countable, then there is $n \in \mathbb{N}$ such that $A_n$ is not countable. 201 | \end{proof} 202 | 203 | 204 | 205 | \paragraph{Exercise 3.3} Show that $ \int_{-\infty}^{\infty} \frac{\cos x}{x^2 + a^2} dx = \pi \frac{e^{-a}}{a}$ for $a > 0$. 206 | \begin{proof} 207 | $\cos x=\frac{e^{i x}+e^{-i x}}{2}$. changing $x \rightarrow-x$ we see that we can just integrate $e^{i x} /\left(x^2+a^2\right)$ and we'll get the same answer. Again, we use the same semicircle and part of the real line. The only pole is $x=i a$, it has order 1 and the residue at it is $\lim _{x \rightarrow i a} \frac{e^{i x}}{x^2+a^2}(x-i a)=\frac{e^{-a}}{2 i a}$, which multiplied by $2 \pi i$ gives the answer. 208 | \end{proof} 209 | 210 | 211 | 212 | \paragraph{Exercise 3.4} Show that $ \int_{-\infty}^{\infty} \frac{x \sin x}{x^2 + a^2} dx = \pi e^{-a}$ for $a > 0$. 213 | \begin{proof} 214 | $$ 215 | x /\left(x^2+a^2\right)=x / 2 i a(1 /(x-i a)-1 /(x+i a))=1 / 2 i a(i a /(x-i a)+i a /(x+ 216 | $$ 217 | $i a))=(1 /(x-i a)+1 /(x+i a)) / 2$. So we care about $\sin (x)(1 /(x-i a)+$ $1 /(x+i a)) / 2$. Its residue at $x=i a$ is $\sin (i a) / 2=\left(e^{-a}-e^a\right) / 4 i$.? 218 | \end{proof} 219 | 220 | 221 | 222 | \paragraph{Exercise 3.9} Show that $\int_0^1 \log(\sin \pi x) dx = - \log 2$. 223 | \begin{proof} 224 | Consider 225 | $$ 226 | \begin{gathered} 227 | f(z)=\log \left(1-e^{2 \pi z i}\right)=\log \left(e^{\pi z i}\left(e^{-\pi z i}-e^{\pi z i}\right)\right)=\log (-2 i)+\pi z i+\log \\ 228 | (\sin (\pi z)) 229 | \end{gathered} 230 | $$ 231 | Then we have 232 | $$ 233 | \begin{aligned} 234 | \int_0^1 f(z) d z & =\log (-2 i)+\frac{i \pi}{2}+\int_0^1 \log (\sin (\pi z)) d z \\ 235 | & =\int_0^1 \log (\sin (\pi z)) d z+\log (-2 i)+\log (i) \\ 236 | & =\log (2)+\int_0^1 \log (\sin (\pi z)) d z 237 | \end{aligned} 238 | $$ 239 | Now it suffices to show that $\int_0^1 f(z) d z=0$. Consider the contour $C(\epsilon, R)$ (which is the contour given in your question) given by the following. 240 | 1. $C_1(\epsilon, R)$ : The vertical line along the imaginary axis from $i R$ to $i \epsilon$. 241 | 2. $C_2(\epsilon)$ : The quarter turn of radius $\epsilon$ about 0 . 242 | 3. $C_3(\epsilon)$ : Along the real axis from $(\epsilon, 1-\epsilon)$. 243 | 4. $C_4(\epsilon)$ : The quarter turn of radius $\epsilon$ about 1 . 244 | 5. $C_5(\epsilon, R)$ : The vertical line from $1+i \epsilon$ to $1+i R$. 245 | 6. $C_6(R)$ : The horizontal line from $1+i R$ to $i R$. 246 | $f(z)$ is analytic inside the contour $C$ and hence $\oint_C f(z)=0$. This gives us 247 | $$ 248 | \begin{aligned} 249 | \int_{C_1(\epsilon, R)} f d z+\int_{C_2(\epsilon)} f d z+\int_{C_3(\epsilon)} f d z+\int_{C_4(\epsilon)} f d z+\int_{C_5(\epsilon, R)} f d z+\int_{C_6(R)} f d z \\ 250 | =0 251 | \end{aligned} 252 | $$ 253 | Now the integral along 1 cancels with the integral along 5 due to symmetry. Integrals along 2 and 4 scale as $\epsilon \log (\epsilon)$. Integral along 6 goes to 0 as $R \rightarrow \infty$. This gives us 254 | $$ 255 | \lim _{\epsilon \rightarrow 0} \int_{C_3(\epsilon)} f d z=0 256 | $$ 257 | which is what we need. 258 | \end{proof} 259 | 260 | 261 | 262 | \paragraph{Exercise 3.14} Prove that all entire functions that are also injective take the form $f(z) = az + b$, $a, b \in \mathbb{C}$ and $a \neq 0$. 263 | \begin{proof} 264 | Look at $f(1 / z)$. If it has an essential singularity at 0 , then pick any $z_0 \neq 0$. Now we know that the range of $f$ is dense as $z \rightarrow 0$. We also know that the image of $f$ in some small ball around $z_0$ contains a ball around $f\left(z_0\right)$. But this means that the image of $f$ around this ball intersects the image of $f$ in any arbitrarily small ball around 0 (because of the denseness). Thus, $f$ cannot be injective. So the singularity at 0 is not essential, so $f(1 / z)$ is some polynomial of $1 / z$, so $f$ is some polynomial of $z$. If its degree is more than 1 it is not injective (fundamental theorem of algebra), so the degree of $f$ is 1 . 265 | \end{proof} 266 | 267 | 268 | 269 | \paragraph{Exercise 3.22} Show that there is no holomorphic function $f$ in the unit disc $D$ that extends continuously to $\partial D$ such that $f(z) = 1/z$ for $z \in \partial D$. 270 | \begin{proof} 271 | Consider $g(r)=\int_{|z|=r} f(z) d z$. Cauchy theorem implies that $g(r)=0$ for all $r<1$. Now since $\left.f\right|_{\partial D}=1 / z$ we have $\lim _{r \rightarrow 1} \int_{|z|=r} f(z) d z=\int_{|z|=1} \frac{1}{z} d z=\frac{2}{\pi i} \neq 0$. Contradiction. 272 | \end{proof} 273 | 274 | 275 | 276 | \paragraph{Exercise 5.1} Prove that if $f$ is holomorphic in the unit disc, bounded and not identically zero, and $z_{1}, z_{2}, \ldots, z_{n}, \ldots$ are its zeros $\left(\left|z_{k}\right|<1\right)$, then $\sum_{n}\left(1-\left|z_{n}\right|\right)<\infty$. 277 | \begin{proof} 278 | Fix $\mathrm{N}$ and let $D(0, R)$ contains the first $\mathrm{N}$ zeroes of f. Let $S_N=\sum_{k=1}^N\left(1-\left|z_k\right|\right)=$ $\sum_{k=1}^N \int_{\left|z_k\right|}^1 1 d r$. Let $\eta_k$ be the characteristic function of the interval $\left.\| z_k \mid, 1\right]$. We have $S_N=\sum_{k=1}^N \int_0^1 \eta(r) d r=\int_0^1\left(\sum_{k=1}^N \eta_k(r)\right) d r \leq \int_0^1 n(r) d r$, where $n(r)$ is the number of zeroes of $f$ at the disk $D(0, r)$. For $r \leq 1$ we have $n(r) \leq \frac{n(r)}{r}$. This means that $S_N \leq \int_0^1 n(r) \frac{d r}{r}$. If $f(0)=0$ then we have $f(z)=z^m g(z)$ for some integer $\mathrm{m}$ and some holomorphic $g$ with $g(0) \neq 0$. The other zeroes of $\mathrm{f}$ are precisely the zeroes of $g$. Thus we have reduced the problem to $f(0) \neq 0$. By the Corollary of the Jensen's equality we get $S_N \leq \int_0^1 n(r) \frac{d r}{r}=\frac{1}{2 \pi} \int_0^{2 \pi} \log \left|f\left(R e^{i \pi}\right)\right| d \phi-\log |f(0)|