id stringlengths 19 59 | proof_index int64 0 2 | theorem_group stringlengths 64 64 | source stringclasses 5
values | training_source stringclasses 5
values | category stringclasses 19
values | lean_statement stringlengths 38 1.36k | problem stringlengths 44 1.18k | steps listlengths 1 95 | overall stringclasses 2
values | first_error_step int64 1 49 ⌀ | error_class stringclasses 6
values | known_issue stringclasses 2
values |
|---|---|---|---|---|---|---|---|---|---|---|---|---|
numina_4d4910b6-f920-57d5-bfd2-43c8c69d6e2b | 0 | 2b79461e29e125330b01a54bd133076b48096bbb31583d245d4e5c9d1c57cb53 | NuminaMath-LEAN | numina | Number Theory | theorem negated_number_theory_63098 : {(a, b) : ℕ × ℕ | a > 0 ∧ b > 0 ∧ a ^ 2 + b ^ 2 ∣ a ^ 3 + 1 ∧ a ^ 2 + b ^ 2 ∣ b ^ 3 + 1} = {(1, 1)} := by | We prove that the set of pairs of positive integers (a,b) with a²+b² dividing both a³+1 and b³+1 is exactly the single pair (1,1). | [
{
"index": 1,
"text": "It suffices to show, for an arbitrary pair (a,b), that (a,b) satisfies the divisibility conditions if and only if (a,b) equals (1,1).",
"lean_code": " ext ⟨a, b⟩",
"label": "correct"
},
{
"index": 2,
"text": "Membership in the left-hand set means a>0, b>0, a²+b² ∣... | correct | null | null | |
lean_workbook_hf_29388 | 0 | 0a0afa690051debb4c9d1d782fd676b91717ddd05447af7828f9dbaf4b350a1f | Lean-Workbook | lean_workbook_hf | divisibility_modular | theorem lean_workbook_hf_29388 (a b c : ℕ) (h₁ : c ∣ b) (h₂ : a ∣ b / c) : a ∣ b | We must show that if c divides b and a divides the quotient b/c (integer division), then a divides b. | [
{
"index": 1,
"text": "Since c divides b exactly, dividing b by c and then multiplying back by c recovers b: that is, (b/c) · c = b.",
"lean_code": " have h3 : b / c * c = b := Nat.div_mul_cancel h₁",
"label": "correct"
},
{
"index": 2,
"text": "Because a divides b/c, multiplying the qu... | correct | null | null | |
lean_workbook_hf_29910 | 0 | 805c1031309dc532f83cf6f0688cab3ae4bece8af97640649b4f343946e35901 | Lean-Workbook | lean_workbook_hf | equality_identity | theorem lean_workbook_hf_29910 (A B C : ℂ) : 1 + A / (z - 3) + B / (z - √3 * Complex.I) + C / (z + √3 * Complex.I) = 1 + A / (z - 3) + B / (z - √3 * Complex.I) + C / (z + √3 * Complex.I) | We must show that the expression 1 + A/(z − 3) + B/(z − √3·i) + C/(z + √3·i) equals itself, since both sides of the equation are written identically. | [
{
"index": 1,
"text": "Since the left-hand side and the right-hand side are literally the same expression term for term, the equality holds simply by the reflexive property of equality: any quantity equals itself.",
"lean_code": " rfl",
"label": "correct"
}
] | correct | null | null | |
lean_workbook_hf_34779 | 0 | cdd63f36e918fd04a1049e5b48bc1a4eb408b1ce1a9182f313852f669bfe3f63 | Lean-Workbook | lean_workbook_hf | inequality | theorem lean_workbook_hf_34779 (hx : x ≥ 1 / Real.sqrt 3) : 3 * x ^ 2 ≥ x * Real.sqrt 3 | We must show that for a real number x satisfying x ≥ 1/√3, the inequality 3x^2 ≥ x√3 holds. | [
{
"index": 1,
"text": "Since 3 is a positive number, its square root √3 is also strictly positive.",
"lean_code": " have h3 : (0:ℝ) < Real.sqrt 3 := Real.sqrt_pos.mpr (by norm_num)",
"label": "correct"
},
{
"index": 2,
"text": "Rewrite the hypothesis x ≥ 1/√3 as 1/√3 ≤ x, and since √3 i... | correct | null | null | |
lean_workbook_hf_54412 | 0 | e46d54dba47841a8106f1bfb41d1af07b3037aeb6198b6c368868c43be654320 | Lean-Workbook | lean_workbook_hf | iff_logic | theorem lean_workbook_hf_54412 (x y z : ℤ) (h₁ : x ^ 2 + y ^ 2 = z ^ 2) (h₂ : Odd x ∧ Odd y) : False | We must show that no integers x, y, z with x and y both odd can satisfy x^2 + y^2 = z^2; we derive a contradiction using residues modulo 4. | [
{
"index": 1,
"text": "Split the hypothesis into the two separate facts that x is odd and y is odd.",
"lean_code": " obtain ⟨hx, hy⟩ := h₂",
"label": "correct"
},
{
"index": 2,
"text": "Since x is odd, write x = 2a + 1 for some integer a.",
"lean_code": " obtain ⟨a, ha⟩ := hx",
... | correct | null | null | |
lean_workbook_hf_27740 | 0 | 1746b681818f7ea0e36ecc76b572eb2ceb66cfe857b00983df7467d3925b7851 | Lean-Workbook | lean_workbook_hf | equality_identity | theorem lean_workbook_hf_27740 (x : ℝ) (hx : x = 15 / 2 * 1 / 2 * 9 / 2 * 5 / 2) : Real.sqrt x = 15 / 4 * Real.sqrt 3 | We must show that the square root of the number x = 15/2 · 1/2 · 9/2 · 5/2 equals 15/4 times the square root of 3. | [
{
"index": 1,
"text": "Replace x throughout the goal by the explicit product 15/2 · 1/2 · 9/2 · 5/2, so the goal becomes an equation about this specific numeral.",
"lean_code": " subst hx",
"label": "correct"
},
{
"index": 2,
"text": "Compute the numeral directly: 15/2 · 1/2 · 9/2 · 5/2... | correct | null | null | |
lean_workbook_hf_23853 | 0 | ad7cb182f142dba52f57ca5c5ad0da679adde14ce05a741039b1a6af2ba53725 | Lean-Workbook | lean_workbook_hf | existence_construction | theorem lean_workbook_hf_23853 (c : ℝ) : ∀ ε : ℝ, ε > 0 → ∃ N : ℕ, ∀ x : ℝ, x > N → |c - c| < ε | We must show that for every positive real number ε, there is a natural number N such that for every real x greater than N, the absolute value of c minus c is less than ε. | [
{
"index": 1,
"text": "Fix an arbitrary positive real number ε, and assume ε > 0.",
"lean_code": " intro ε hε",
"label": "correct"
},
{
"index": 2,
"text": "Choose N = 0 as the natural number witness; we will show it satisfies the required property for any x greater than 0.",
"lean_... | correct | null | null | |
lean_workbook_hf_37608_18939 | 0 | d084979ddbd39542cde791d2de4481908e06b92cb9f8aa433135f77984567467 | Lean-Workbook | lean_workbook_hf | iff_logic | theorem lean_workbook_hf_37608_18939 (a b c : ℝ) (ha : 0 < a) (hb : 0 < b) (hc : 0 < c) (habc : a + b + c = 1) : a ^ 2 + b ^ 2 + c ^ 2 = 1 → a ^ 3 / (2 * b + 3 * c) + b ^ 3 / (2 * c + 3 * a) + c ^ 3 / (2 * a + 3 * b) ≥ 1 / 5 | We are given positive reals a, b, c with a+b+c = 1, and we must show that if additionally a^2+b^2+c^2=1, then the sum of the three fractions a^3/(2b+3c) + b^3/(2c+3a) + c^3/(2a+3b) is at least 1/5. We will show that the extra hypothesis a^2+b^2+c^2=1, combined with a+b+c=1 and positivity of a,b,c, is actually impossibl... | [
{
"index": 1,
"text": "Assume the hypothesis a^2 + b^2 + c^2 = 1 holds; call it h2. We now aim to derive a contradiction from this together with a+b+c=1 and a,b,c>0.",
"lean_code": " intro h2",
"label": "correct"
},
{
"index": 2,
"text": "Recall the algebraic identity (a+b+c)^2 = a^2+b^... | correct | null | null | |
lean_workbook_hf_82547 | 0 | 26f2cc45ccf1c414e77fb4dee53d8b2c76c895342cd6a87d7c8784fc81dc996b | Lean-Workbook | lean_workbook_hf | existence_construction | theorem lean_workbook_hf_82547 (X : ℕ → ℝ) (μ : ℝ) (p : ℕ → ℝ) (hp : ∀ x, p x = (Real.exp (-μ) * μ ^ x)/x!) : ∃ M : ℝ → ℝ, ∀ t, M t = ∑' x : ℕ, (p x) * (Real.exp (t * x)) | We want to show that there exists a function M of a real variable t such that for every t, M(t) equals the infinite sum over natural numbers x of p(x) times exp(t·x). To do this we simply exhibit the function defined directly by this formula and check that it satisfies the required equality. | [
{
"index": 1,
"text": "Define M to be exactly the function sending t to the sum ∑' x, p(x)·exp(t·x). With this choice, for every t the value M(t) is by definition equal to that same sum, so the required equation M(t) = ∑' x, p(x)·exp(t·x) holds immediately for all t, since both sides are literally the same ... | correct | null | null | |
lean_workbook_hf_25155 | 0 | 34fe1765192f365f0c8f03996111f477495a26349279ef74d8b6aa482f4d44cc | Lean-Workbook | lean_workbook_hf | equality_identity | theorem lean_workbook_hf_25155 (x : ℝ) : x = x | We must show that for any real number x, x equals itself. | [
{
"index": 1,
"text": "This is immediate, since any quantity is equal to itself by the basic principle of reflexivity of equality: x = x holds identically for every real number x.",
"lean_code": " rfl",
"label": "correct"
}
] | correct | null | null | |
lean_workbook_hf_34349 | 0 | 0aabcb8e9cf1a703947dfbf163d58c993cbb7983c691c2ce4f39a05be5909998 | Lean-Workbook | lean_workbook_hf | iff_logic | theorem lean_workbook_hf_34349 : c + d = 7 ∧ c * d = 9 → c ^ 3 + d ^ 3 = 154 | We must show that if two numbers c and d satisfy c + d = 7 and c·d = 9, then c^3 + d^3 = 154. | [
{
"index": 1,
"text": "Assume the two given conditions: h1 states c + d = 7, and h2 states c·d = 9.",
"lean_code": " rintro ⟨h1, h2⟩",
"label": "correct"
},
{
"index": 2,
"text": "We use the algebraic identity (c + d)^3 = c^3 + d^3 + 3·c·d·(c + d), which follows from expanding the cube ... | correct | null | null | |
lean_workbook_hf_24237 | 0 | 6bf82db2e2a15c004124ba9ad437d6539f33b3245f33a9934c040dd82f89e897 | Lean-Workbook | lean_workbook_hf | existence_construction | theorem lean_workbook_hf_24237 (R : Type*) [Ring R] [Finite R] (x : R) : ∃ n : ℕ, n • x = 0 | We must show that for a finite ring R and any element x of R, there exists a natural number n such that adding x to itself n times yields zero, that is, n copies of x sum to the additive identity. | [
{
"index": 1,
"text": "We choose the witness n = 0 for the existential statement; it remains to check that adding x to itself zero times gives zero.",
"lean_code": " use 0",
"label": "correct"
},
{
"index": 2,
"text": "By the basic property of repeated addition in any additive monoid, t... | correct | null | null | |
lean_workbook_hf_21534 | 0 | 0e92ff062017f1a3192bf611440ee24c49625d6f38f676d083bf5cc6f830e851 | Lean-Workbook | lean_workbook_hf | iff_logic | theorem lean_workbook_hf_21534 : 6 ∣ a + b + c ↔ 6 ∣ a ^ 3 + b ^ 3 + c ^ 3 | We must show that for natural numbers a, b, c, the number 6 divides a+b+c exactly when 6 divides a^3+b^3+c^3. | [
{
"index": 1,
"text": "Since 6 is a nonzero natural number, we record this fact, as it is needed to relate divisibility by 6 to vanishing of residues modulo 6.",
"lean_code": " haveI : NeZero (6 : ℕ) := ⟨by decide⟩",
"label": "correct"
},
{
"index": 2,
"text": "We verify that cubing fix... | correct | null | null | |
numina_0cea2807-3780-5baf-9c23-44688a23cfae | 0 | ae63d9e52e31927c369f4d786307d1ec7905570dced7b8997f3869dbc423521e | NuminaMath-LEAN | numina | Algebra | theorem number_theory_604934 (k : ℝ) (hk : k > 0) (h : k ^ 2 - 1998 * k - 1 = 0)
(x : ℕ → ℕ) (hx0 : x 0 = 1) (hxn : ∀ n, x (n + 1) = ⌊k * x n⌋) :
x 1998 % 1998 = 1000 := by | We are given a positive real number k satisfying k² − 1998k − 1 = 0, and a sequence of natural numbers with x₀ = 1 and xₙ₊₁ = ⌊k·xₙ⌋; we must show that x₁₉₉₈ leaves remainder 1000 upon division by 1998. | [
{
"index": 1,
"text": "Rearranging the defining equation k² − 1998k − 1 = 0 gives the convenient form k² = 1998k + 1.",
"lean_code": " have hk2 : k ^ 2 = 1998 * k + 1 := by linarith [h]",
"label": "correct"
},
{
"index": 2,
"text": "Since k(k − 1998) = k² − 1998k = (1998k + 1) − 1998k =... | correct | null | null | |
lean_workbook_hf_22089 | 0 | 8b169fda417fc134f5e06bd00ea4742d6347033b357652668246bb63ef7248d4 | Lean-Workbook | lean_workbook_hf | iff_logic | theorem lean_workbook_hf_22089 : ∀ a b c : ℝ, (a ≤ 0 ∧ 0 ≤ b ∧ 0 ≤ c ∨ 0 ≤ a ∧ b ≤ 0 ∧ 0 ≤ c ∨ 0 ≤ a ∧ 0 ≤ b ∧ c ≤ 0) → 2 * (a ^ 2 + b ^ 2 + c ^ 2 - a * b - b * c - c * a) ^ 2 ≥ a ^ 2 * (b - c) ^ 2 + b ^ 2 * (c - a) ^ 2 + c ^ 2 * (a - b) ^ 2 | We must show that for all real numbers a, b, c satisfying one of the three sign patterns (exactly one of a, b, c is nonpositive while the other two are nonnegative), the quantity 2 times the square of a²+b²+c²−ab−bc−ca is at least a²(b−c)²+b²(c−a)²+c²(a−b)². | [
{
"index": 1,
"text": "Fix arbitrary real numbers a, b, c and assume the sign hypothesis h that one of the three disjuncts holds.",
"lean_code": " intro a b c h",
"label": "correct"
},
{
"index": 2,
"text": "We split into the three cases: (i) a ≤ 0 with b, c ≥ 0; (ii) b ≤ 0 with a, c ≥ ... | correct | null | null | |
lean_workbook_hf_6574 | 0 | 10cb4803d2cf70ad632d43c70237bbb96220a80870e29ea97d90b0cb43638a1b | Lean-Workbook | lean_workbook_hf | existence_construction | theorem lean_workbook_hf_6574 (p q : ℕ) (hp : p.Prime) (hq : q.Prime) (hpq : p < q) : ∃ a b, a ≤ (p-1)/2 ∧ b ≤ (p-1)/2 ∧ 1 + a * (p-1) ≡ 1 + b * (p-1) [ZMOD q] | We must show that for any primes p and q with p < q, there exist natural numbers a and b, each at most (p-1)/2, such that 1 + a(p-1) and 1 + b(p-1) are congruent modulo q. | [
{
"index": 1,
"text": "We choose the simplest possible candidates, taking a = 0 and b = 0, and this reduces the problem to establishing three separate facts: that 0 ≤ (p-1)/2, that 0 ≤ (p-1)/2 again, and that 1 + 0·(p-1) is congruent to 1 + 0·(p-1) modulo q.",
"lean_code": " refine ⟨0, 0, ?_, ?_, ?_⟩",... | correct | null | null | |
lean_workbook_hf_17722_14430 | 0 | 2142655371544848de4681bf94be50532275bccdf11b883affcb98ebc6447d24 | Lean-Workbook | lean_workbook_hf | inequality | theorem lean_workbook_hf_17722_14430 (a b c : ℝ) (ha : a > 0) (hb : b > 0) (hc : c > 0) (hab : a + b + c = 3) : 3 / 2 * (a + b + c) + 1 / a + 1 / b + 1 / c ≥ 15 / 2 | We must show that for positive reals a, b, c summing to 3, the quantity (3/2)(a+b+c) + 1/a + 1/b + 1/c is at least 15/2. | [
{
"index": 1,
"text": "We first establish the tangent-line bound at the point a = 1: for a positive, 1/a is at least 2 − a.",
"lean_code": " have hA : 1/a ≥ 2 - a := by",
"label": "correct"
},
{
"index": 2,
"text": "Since a square of a real number is never negative, (a − 1)^2 ≥ 0.",
... | correct | null | null | |
lean_workbook_hf_30307 | 0 | 03e2943941435d422a8a93f2a43e0284e359f0de959110c32229f26bc09c1b5d | Lean-Workbook | lean_workbook_hf | iff_logic | theorem lean_workbook_hf_30307 : ∀ a b : ℝ, a > 0 ∧ b > 0 → a * b * (a ^ 2 + b ^ 2 - 2) ≥ (a + b) * (a * b - 1) | We must show that for all positive real numbers a and b, the quantity ab(a^2+b^2-2) is at least (a+b)(ab-1). | [
{
"index": 1,
"text": "Fix arbitrary real numbers a and b, and assume the hypothesis that a > 0 and b > 0 holds.",
"lean_code": " intro a b hab",
"label": "correct"
},
{
"index": 2,
"text": "Split the hypothesis into its two components: ha states a > 0, and hb states b > 0.",
"lean_... | correct | null | null | |
lean_workbook_hf_25940 | 0 | df0c86b230d7bf348ea2f53594696026d3e6d6039a053a185e5706f8c137e607 | Lean-Workbook | lean_workbook_hf | iff_logic | theorem lean_workbook_hf_25940 : ∀ n : ℤ, Even n → n ^ 2 % 4 = 0 | We must show that for every integer n, if n is even then n squared leaves remainder 0 when divided by 4. | [
{
"index": 1,
"text": "Fix an arbitrary integer n and assume that n is even; we must show n^2 % 4 = 0.",
"lean_code": " intro n hn",
"label": "correct"
},
{
"index": 2,
"text": "Since n is even, there exists an integer k such that n = k + k.",
"lean_code": " obtain ⟨k, hk⟩ := hn",
... | correct | null | null | |
lean_workbook_hf_903_29534 | 0 | af2966e7f6ec45dc129b42a2cb5591dbd919d6588417b1d21251cf82095a4aa9 | Lean-Workbook | lean_workbook_hf | existence_construction | theorem lean_workbook_hf_903_29534 (n : ℕ) (h : n % 2 = 0) : ∃ x y, x = 9 * 7 ^ n * 41 ^ (n / 2 - 1) ∧ y = 40 * 7 ^ n * 41 ^ (n / 2 - 1) | We must show that for a natural number n divisible by 2, there exist natural numbers x and y satisfying x = 9·7^n·41^(n/2 − 1) and y = 40·7^n·41^(n/2 − 1). | [
{
"index": 1,
"text": "We choose x to be exactly the quantity 9·7^n·41^(n/2 − 1) and y to be exactly the quantity 40·7^n·41^(n/2 − 1), which are precisely the values required by the existential claim, leaving the two equalities to be verified.",
"lean_code": " refine ⟨9 * 7 ^ n * 41 ^ (n / 2 - 1), 40 *... | correct | null | null | |
lean_workbook_hf_39808 | 0 | 6ee7675c861f5870132943ae2d13c8ddc241babcef2b93814a1ef305fca1e981 | Lean-Workbook | lean_workbook_hf | inequality | theorem lean_workbook_hf_39808 (a b c : ℝ) (h₁ : a < b ∧ b < c) (h₂ : a + b + c = 0) : (a ^ 2 + b ^ 2 + c ^ 2) / (c - a)^2 < 2 / 3 | We must show that whenever a < b < c are real numbers with a + b + c = 0, the ratio (a² + b² + c²)/(c − a)² is strictly less than 2/3. | [
{
"index": 1,
"text": "Split the ordering hypothesis into its two parts: a < b and b < c.",
"lean_code": " obtain ⟨hab, hbc⟩ := h₁",
"label": "correct"
},
{
"index": 2,
"text": "Since a < b and b < c, transitivity of the order gives a < c.",
"lean_code": " have hca : a < c := lt_tr... | correct | null | null | |
lean_workbook_hf_78654 | 0 | 3ee2d0df6a826be6bd1b5c462e564ee712b66b7a5452e6541d70e053e8fc8f64 | Lean-Workbook | lean_workbook_hf | equality_identity | theorem lean_workbook_hf_78654 (x y : ℝ) (h₁ : x = 200) (h₂ : y = 5/12 * x - 5*110/12) : y = 37.5 | We are given that x equals 200 and that y is defined by the formula y = (5/12)·x − (5·110)/12, and we must show that under these conditions y equals 37.5. | [
{
"index": 1,
"text": "Since x is exactly 200, we replace every occurrence of x by 200 throughout the expression for y, so that y = (5/12)·200 − (5·110)/12.",
"lean_code": " subst h₁",
"label": "correct"
},
{
"index": 2,
"text": "We substitute the defining formula for y into the goal, s... | correct | null | null | |
lean_workbook_hf_51350 | 0 | 69fba148f1f52d8eeaea1ac31a020e7fa7faa15e32b42abdb1468c0ead37cb53 | Lean-Workbook | lean_workbook_hf | divisibility_modular | theorem lean_workbook_hf_51350 {a b c : ℤ} (h : a + b + c = 0) : (a ^ 2 + b ^ 2 + c ^ 2) ∣ (a ^ 4 + b ^ 4 + c ^ 4) | We must prove that whenever three integers a, b, c satisfy a + b + c = 0, the quantity a² + b² + c² divides a⁴ + b⁴ + c⁴. | [
{
"index": 1,
"text": "Since a + b + c = 0, solving for c gives c = -a - b.",
"lean_code": " have hc : c = -a - b := by linarith",
"label": "correct"
},
{
"index": 2,
"text": "We now replace c everywhere by -a - b, so the problem reduces to a statement purely in terms of a and b: we mus... | correct | null | null | |
lean_workbook_hf_50409 | 0 | feafb24e7f63282449e3dc10bb1ec1c7ba9e64ca84f046674a74b77c442e9385 | Lean-Workbook | lean_workbook_hf | inequality | theorem lean_workbook_hf_50409 (a b c : ℝ) : (a + b + c) ^ 2 ≥ 3 * (a * b + a * c + b * c) | We must show that for any real numbers a, b, c, the square of their sum is at least three times the sum of their pairwise products. | [
{
"index": 1,
"text": "The square of (a − b) is nonnegative, since any real square is nonnegative.",
"lean_code": " have h1 : (a - b) ^ 2 ≥ 0 := sq_nonneg (a - b)",
"label": "correct"
},
{
"index": 2,
"text": "Similarly, the square of (b − c) is nonnegative.",
"lean_code": " have h... | correct | null | null | |
lean_workbook_hf_23005 | 0 | 7599770b57c3fe0f046d8b4b1b054b92287872b54d1856673726e8e1ae7196be | Lean-Workbook | lean_workbook_hf | iff_logic | theorem lean_workbook_hf_23005 : ∀ a b c : ℝ, a > 0 ∧ b > 0 ∧ c > 0 → a ^ 2 + b ^ 2 + c ^ 2 = 3 → a * b + b * c + c * a ≤ 3 | We must show that for all positive real numbers a, b, c satisfying a² + b² + c² = 3, the sum ab + bc + ca is at most 3. | [
{
"index": 1,
"text": "Fix arbitrary real numbers a, b, c.",
"lean_code": " intro a b c",
"label": "correct"
},
{
"index": 2,
"text": "Assume a, b, c are each positive, giving us ha : a > 0, hb : b > 0, hc : c > 0.",
"lean_code": " rintro ⟨ha, hb, hc⟩",
"label": "correct"
},
... | correct | null | null | |
numina_e81ec851-b2cd-5699-8bb4-47843ae2fad8 | 0 | 5f64b772f3ff1319ce7a4634d9aa67fa8870b05994b076d074fcc4e390c3188a | NuminaMath-LEAN | numina | Algebra | theorem algebra_296307 {n : ℕ} (hn : n ≥ 2)
(f : Fin n → ℝ → ℝ) (k : ℝ) (hk : 0 < k) :
∃ x : Fin n → ℝ,
|∏ i : Fin n, x i - ∑ i : Fin n, f i (x i)| ≥ k := by | We must show that for any family of functions f_0,…,f_{n-1} (with n ≥ 2) and any k > 0 there is a point x with |∏_i x_i − ∑_i f_i(x_i)| ≥ k. | [
{
"index": 1,
"text": "Assume the contrary, namely that no point achieves absolute value at least k.",
"lean_code": " by_contra hcon",
"label": "correct"
},
{
"index": 2,
"text": "Then for every point x we have the strict bound |∏_i x_i − ∑_i f_i(x_i)| < k.",
"lean_code": " push_ne... | correct | null | null | |
lean_workbook_hf_20565 | 0 | 2b066a49a1f7ed6a00f0d7d132e0fb3f5d0a4a82d1b0c2b6d6e875afb668ec40 | Lean-Workbook | lean_workbook_hf | inequality | theorem lean_workbook_hf_20565 (a b c : ℝ) (ha : 0 < a) (hb : 0 < b) (hc : 0 < c) (habc : a + b + c = 1) : 7 * (a * b + b * c + c * a) ≤ 9 * a * b * c + 2 | We must show that for positive reals a, b, c with a+b+c=1, the quantity 7(ab+bc+ca) is at most 9abc+2. | [
{
"index": 1,
"text": "First we bound ab+bc+ca: since the sum of squares (a-b)^2+(b-c)^2+(c-a)^2 is nonnegative, expanding it gives 2(a^2+b^2+c^2) - 2(ab+bc+ca) ≥ 0, i.e. a^2+b^2+c^2 ≥ ab+bc+ca. Since (a+b+c)^2 = a^2+b^2+c^2+2(ab+bc+ca) = 1, we get 1 = a^2+b^2+c^2+2(ab+bc+ca) ≥ 3(ab+bc+ca), so 3(ab+bc+ca) ≤... | correct | null | null | |
numina_603e0386-259f-50a2-b6cd-1f005eedb498 | 0 | 1e25bb7dfba8ea2674926dfdecb61b0b861e20eb473bdcbeee94eab41b0f2115 | NuminaMath-LEAN | numina | Inequalities | theorem inequalities_63681 (x y z a b c : ℝ) (hx : 0 < x) (hy : 0 < y) (hz : 0 < z)
(ha : 0 < a) (hb : 0 < b) (hc : 0 < c) (h : a + b + c = 1) :
(x ^ 2 + y ^ 2 + z ^ 2) * (a ^ 3 / (x ^ 2 + 2 * y ^ 2) + b ^ 3 / (y ^ 2 + 2 * z ^ 2) +
c ^ 3 / (z ^ 2 + 2 * x ^ 2)) ≥ 1 / 9 := by | We must prove that for positive reals x, y, z, a, b, c with a + b + c = 1, the product of x² + y² + z² with the cyclic sum a³/(x²+2y²) + b³/(y²+2z²) + c³/(z²+2x²) is at least 1/9. | [
{
"index": 1,
"text": "Introduce the abbreviation S = x² + y² + z² for the common factor outside the parentheses.",
"lean_code": " set S := x ^ 2 + y ^ 2 + z ^ 2 with hSdef",
"label": "correct"
},
{
"index": 2,
"text": "Write d₁ = x² + 2y² for the first denominator.",
"lean_code": "... | correct | null | null | |
numina_5c9fa94c-6f3d-52f8-a4ed-54f56568caf0 | 0 | 84cd4467bf18fa923fc85da025fac01cf782916b5ed47019b2659b5c8a8002ad | NuminaMath-LEAN | numina | Inequalities | theorem inequalities_598920 (θ : ℝ) (hθ : θ ∈ Set.Ioo 0 1) (a b c : ℝ)
(ha : a ∈ Set.Ioo 0 1) (hb : b ∈ Set.Ioo 0 1) (hc : c ∈ Set.Ioo 0 1)
(h : θ < a ∧ b - c < 1) :
Real.sqrt (a * b * c) + Real.sqrt ((1 - a) * (1 - b) * (1 - c)) < 1 := by | We must show that for a, b, c all strictly between 0 and 1, the sum of the square root of abc and the square root of (1-a)(1-b)(1-c) is strictly less than 1. The hypotheses about θ turn out not to be needed for this fact. | [
{
"index": 1,
"text": "Write out that a satisfies 0 < a and a < 1.",
"lean_code": " obtain ⟨ha0, ha1⟩ := ha",
"label": "correct"
},
{
"index": 2,
"text": "Write out that b satisfies 0 < b and b < 1.",
"lean_code": " obtain ⟨hb0, hb1⟩ := hb",
"label": "correct"
},
{
"ind... | correct | null | null | |
lean_workbook_hf_12002 | 0 | 1bc5ef2c08aa211fbfe54ce2e7cc432653d9ce8d9a0b0b08a5769367a67258ea | Lean-Workbook | lean_workbook_hf | equality_identity | theorem lean_workbook_hf_12002 (x y z : ℝ) : x ^ 3 + y ^ 3 + z ^ 3 = 3 * x * y * z + (x + y + z) * (x ^ 2 + y ^ 2 + z ^ 2 - (x*y + y*z + z*x)) | We must show that for all real numbers x, y, z, the sum of cubes x^3 + y^3 + z^3 equals 3xyz plus the product (x + y + z) times (x^2 + y^2 + z^2 − xy − yz − zx). This is the classical factorization identity for the sum of cubes minus three times the product, and we verify it by direct algebraic expansion. | [
{
"index": 1,
"text": "Expanding the product (x + y + z)(x^2 + y^2 + z^2 − xy − yz − zx) term by term: multiplying x, y, z respectively by x^2 + y^2 + z^2 − xy − yz − zx and collecting all resulting monomials, every mixed term of the form x^2 y, x y^2, and similar permutations cancels in pairs, leaving exac... | correct | null | null | |
lean_workbook_hf_46902 | 0 | 3e891f4ba168db044a8dadeec8a3cd465cf973d793ed6959d2d6a8efa6ffb161 | Lean-Workbook | lean_workbook_hf | inequality | theorem lean_workbook_hf_46902 (a b : ℝ) : (a * (a + 1) + b * (b + 1))^2 ≥ (8:ℝ) / 3 * (a * b * (a * b + 1) + (a + b) * (a ^ 2 + b ^ 2)) | We must show that for all real numbers a and b, the square of a(a+1)+b(b+1) is at least 8/3 times ab(ab+1)+(a+b)(a²+b²). The strategy is to show that three times the difference of these two sides can be written as an explicit sum of manifestly nonnegative terms, using the substitution s = a+b−2 and d = a−b. | [
{
"index": 1,
"text": "We verify the key algebraic identity: three times the left-hand side minus eight times ab(ab+1)+(a+b)(a²+b²) equals ((1/2)(a+b−2)(a+b))² plus ((5/2)(a+b−2)²+9(a+b−2)+10)(a−b)² plus ((1/2)(a−b)²)². This can be checked by direct expansion: writing s = a+b−2 and d = a−b, both sides expan... | correct | null | null | |
lean_workbook_hf_36246 | 0 | 3582d9e97c03174797b0612b3c3ea6d3078b64d9a9176c24125df126f5f104f9 | Lean-Workbook | lean_workbook_hf | existence_construction | theorem lean_workbook_hf_36246 : ¬ (∃ a b : ℤ, a ^ 4 + b ^ 4 + a * b ^ 3 + 1 = 4 * a ^ 2 * b ^ 2 + a ^ 3 * b) | We must show that no pair of integers a, b satisfies a^4 + b^4 + a b^3 + 1 = 4a^2b^2 + a^3 b. | [
{
"index": 1,
"text": "Suppose, for contradiction, that integers a and b exist satisfying this equation; call the equation hab.",
"lean_code": " rintro ⟨a, b, hab⟩",
"label": "correct"
},
{
"index": 2,
"text": "We first establish the algebraic identity (a^2 - 2ab - b^2)(a^2 + ab - b^2) ... | correct | null | null | |
lean_workbook_hf_34745 | 0 | d0c73ecd839226570442d145d1e602697967f9d7e64db3fccba96702c82e253e | Lean-Workbook | lean_workbook_hf | iff_logic | theorem lean_workbook_hf_34745 (x y z : ℝ) (hx : 0 < x ∧ x < 1) (hy : 0 < y ∧ y < 1) (hz : 0 < z ∧ z < 1) (h : x * y + y * z + z * x = 1) : 1 ≤ x + y + z + x * y * z ∧ x + y + z + x * y * z ≤ 2 | We must show that if x, y, z are each strictly between 0 and 1 and satisfy xy + yz + zx = 1, then the sum x + y + z + xyz lies between 1 and 2 inclusive. | [
{
"index": 1,
"text": "Split the hypothesis on x into its two parts: 0 < x and x < 1.",
"lean_code": " obtain ⟨hx0, hx1⟩ := hx",
"label": "correct"
},
{
"index": 2,
"text": "Split the hypothesis on y into its two parts: 0 < y and y < 1.",
"lean_code": " obtain ⟨hy0, hy1⟩ := hy",
... | correct | null | null | |
lean_workbook_hf_17227 | 0 | b39afecde8978ac0aa3715e198c811cc19d7adf33f403b50bb1ad08092725795 | Lean-Workbook | lean_workbook_hf | iff_logic | theorem lean_workbook_hf_17227 (k : ℝ) : k ^ 2 + 10 * k + 16 = 0 ↔ k = -2 ∨ k = -8 | We must show that a real number k satisfies k² + 10k + 16 = 0 exactly when k equals −2 or k equals −8. | [
{
"index": 1,
"text": "First we note the algebraic factorization k² + 10k + 16 = (k + 2)(k + 8), which can be verified by expanding the right-hand side: (k+2)(k+8) = k² + 8k + 2k + 16 = k² + 10k + 16.",
"lean_code": " have factored : k ^ 2 + 10 * k + 16 = (k + 2) * (k + 8) := by ring",
"label": "co... | correct | null | null | |
numina_9bc11617-8d8f-5885-a85f-6a87cf80d608 | 0 | 4b7b84243806fed9377799797c3cf8d757c6b32bc16260ec68d59f95f83794da | NuminaMath-LEAN | numina | Inequalities | theorem inequalities_44746 {m n : ℕ} (hm : 0 < m) (hn : 0 < n)
(a : ℕ → ℕ → ℝ) (ha : ∀ i ∈ Icc 1 m, ∀ j ∈ Icc 1 n, 0 ≤ a i j)
(hij : ∀ i ∈ Icc 1 m, ∀ j ∈ Icc 1 n, a i j ≥ a i (j + 1)) (hji : ∀ i ∈ Icc 1 m, ∀ j ∈ Icc 1 n, a i j ≥ a (i + 1) j)
(X Y : ℕ → ℕ → ℝ) (hX : ∀ i ∈ Icc 1 m, ∀ j ∈ Icc 1 n, X i j = ∑ k ... | We must show that the product of the partial sums X i j taken over all rows i and columns j is at least as large as the product of the reflected partial sums Y i j, given that the array a is nonnegative and weakly decreasing along both rows and columns. | [
{
"index": 1,
"text": "First we establish that within a fixed column l, the array value decreases as the row index increases: for any indices i1 ≤ i2 lying between 1 and m, a(i2,l) ≤ a(i1,l). This is proved by induction on i2 starting from i1: the base case is trivial equality, and at each inductive step we... | correct | null | null | |
lean_workbook_hf_26556 | 0 | 730a1f89096677baa2ac594405a98dd8f5c4725773c120154e8b1cc2789f6789 | Lean-Workbook | lean_workbook_hf | iff_logic | theorem lean_workbook_hf_26556 (m n a b c : ℤ) (h₀ : 0 < m ∧ 0 < n ∧ 0 ≤ a ∧ 0 ≤ b ∧ 0 ≤ c) (h₁ : m ^ 2 + n ^ 2 + a ^ 2 + b ^ 2 + c ^ 2 = 1989) (h₂ : m ^ 2 + n ^ 2 + a + b + c = 19) (h₃ : m ^ 2 + a ^ 2 + b ^ 2 + c ^ 2 = 44) (h₄ : n ^ 2 + a ^ 2 + b ^ 2 + c ^ 2 = 25) : m ^ 2 = 81 ∧ n ^ 2 = 36 | We are given five nonnegative or positive integers m, n, a, b, c satisfying four equations relating their squares (and, in one case, linear terms). We want to show m² = 81 and n² = 36. We will show instead that the hypotheses are jointly contradictory, from which any conclusion, including the desired one, follows. | [
{
"index": 1,
"text": "We break apart the conjunction of sign conditions into its five separate facts: m > 0, n > 0, a ≥ 0, b ≥ 0, c ≥ 0.",
"lean_code": " obtain ⟨hm, hn, ha, hb, hc⟩ := h₀",
"label": "correct"
},
{
"index": 2,
"text": "Subtracting the equation m² + a² + b² + c² = 44 fro... | correct | null | null | |
pythagoras_55fd91d739371385e289 | 0 | e260e16a2f318737d6efdb7ae179f66b0cb724bdecbd18d5b3afef1b5ca1a6eb | Pythagoras | pythagoras | Pythagoras_SFT_N/A | theorem my_favorite_problem :
let total_cards : ℕ := 52
let ace_cards : ℕ := 4
let non_ace_cards : ℕ := 48
let hand_size : ℕ := 5
let prob_exactly_two_aces : ℚ := (Nat.choose ace_cards 2 * Nat.choose non_ace_cards (hand_size - 2)) / Nat.choose total_cards hand_size
let prob_at_least_one_ace : ℚ := 1 - (Nat.choose... | We want to show that, when drawing a 5-card hand from a standard 52-card deck, the conditional probability of drawing exactly two aces given that at least one ace is drawn equals 1081/9236. This requires computing the total number of 5-card hands, the number of hands with exactly two aces, and the number of hands with ... | [
{
"index": 1,
"text": "First, the number of ways to choose 2 aces out of the 4 aces in the deck is C(4,2) = 6.",
"lean_code": " have hc1 : Nat.choose 4 2 = 6 := by decide",
"label": "correct"
},
{
"index": 2,
"text": "Next, the number of ways to choose the remaining 3 cards of the hand ... | correct | null | null | |
lean_workbook_hf_14683 | 0 | 8deedee6016db5b9800962c9bc70d3281612cb8980226a432721cfaa6be89885 | Lean-Workbook | lean_workbook_hf | inequality | theorem lean_workbook_hf_14683 (x y z : ℝ) : x ^ 4 + y ^ 4 + z ^ 4 ≥ x ^ 3 * y + y ^ 3 * z + z ^ 3 * x | We must show that for any real numbers x, y, z, the sum of fourth powers x^4+y^4+z^4 is at least as large as the cyclic sum x^3 y + y^3 z + z^3 x. | [
{
"index": 1,
"text": "For any two real numbers a and b, we have the algebraic identity 3a^4 + b^4 − 4a^3 b = (a − b)^2 · (2a^2 + (a + b)^2). This can be checked by expanding both sides: (a−b)^2 = a^2−2ab+b^2, and 2a^2+(a+b)^2 = 3a^2+2ab+b^2; multiplying these two quadratics out and collecting terms yields ... | correct | null | null | |
lean_workbook_hf_78082 | 0 | e63cb253f98d788fd16f8d34b8c00f78e0c80c9dd0e01b246965236f816f2120 | Lean-Workbook | lean_workbook_hf | iff_logic | theorem lean_workbook_hf_78082 : ∀ a b c : ℕ, a ∈ ({5, 49, 51} : Finset ℕ) ∧ b ∈ ({5, 49, 51} : Finset ℕ) ∧ c ∈ ({5, 49, 51} : Finset ℕ) → a + b = c ∨ a * b = c ∨ a = c / 2 → 3 ∣ c | We must show: for any natural numbers a, b, c each drawn from the set {5, 49, 51}, if a+b=c, or a·b=c, or a equals c divided by 2 (with integer division), then c must be divisible by 3. | [
{
"index": 1,
"text": "Fix arbitrary natural numbers a, b, c for which we will verify the implication.",
"lean_code": " intro a b c",
"label": "correct"
},
{
"index": 2,
"text": "Assume a, b, c each belong to {5, 49, 51} (giving membership facts ha, hb, hc), and assume the disjunctive h... | correct | null | null | |
numina_a7af0b16-6032-5960-a5e7-7dbf2dc48399 | 0 | 15a3e0622ee088b9e246692d9a5b2878dbc3ae90c1db892af3e260193e334416 | NuminaMath-LEAN | numina | Inequalities | theorem inequalities_54864
(a b c : ℝ)
(ha : 0 < a)
(hb : 0 < b)
(hc : 0 < c)
(h : a * b + b * c + c * a = 1 / 3) :
1 / (a + b + c) ≤ a / (a ^ 2 - b * c + 1) + b / (b ^ 2 - c * a + 1) + c / (c ^ 2 - a * b + 1) := by | We are given positive reals a, b, c with ab + bc + ca = 1/3, and we must show that 1/(a+b+c) is at most the sum a/(a²−bc+1) + b/(b²−ca+1) + c/(c²−ab+1). | [
{
"index": 1,
"text": "Since a, b, c are all positive, their sum a + b + c is positive.",
"lean_code": " have hs : 0 < a + b + c := add_pos (add_pos ha hb) hc",
"label": "correct"
},
{
"index": 2,
"text": "The first denominator is positive: using 1 = 3(ab + bc + ca) we get a² − bc + 1 =... | correct | null | null | |
lean_workbook_hf_15032 | 0 | 85c9fa1808f46af6022d5240b4f6807ca9cfbed0cc0b3b72060a768626ed9d4b | Lean-Workbook | lean_workbook_hf | iff_logic | theorem lean_workbook_hf_15032 (x y : ℝ) (h : x ≠ y) (h1 : x ^ 3 - x = y ^ 3 - y) : -2 ≤ x ^ 3 * y + x * y ^ 3 ∧ x ^ 3 * y + x * y ^ 3 ≤ 2/9 | We must prove that for real numbers x and y with x ≠ y satisfying x³ − x = y³ − y, the quantity x³y + xy³ lies between −2 and 2/9. | [
{
"index": 1,
"text": "We use the algebraic identity x³ − y³ = (x³ − x) − (y³ − y) + (x − y). Substituting the hypothesis x³ − x = y³ − y makes the first two bracketed terms cancel, leaving x³ − y³ = x − y.",
"lean_code": " have hcube : x ^ 3 - y ^ 3 = x - y := by\n have hid : x ^ 3 - y ^ 3 = (x ^ 3... | correct | null | null | |
numina_12f17825-21c6-5cae-b844-482753d7d0c9 | 0 | dde582c02220de1feb68e1ad0567a11f4007f352794bf767331ab3544c0c3aca | NuminaMath-LEAN | numina | Algebra | theorem algebra_63440 {n : ℕ} (a : ℕ → ℝ) (x : ℝ) (hx : 0 < x ∧ x < 1)
(hsum : ∑ i ∈ Finset.range (n + 1), a i / (1 - x ^ (i + 1)) = 0) :
∃ y : ℝ, 0 < y ∧ y < 1 ∧ ∑ i ∈ Finset.range (n + 1), a i * y ^ i = 0 := by | We must show that if 0 < x < 1 and the weighted sum ∑ a_i/(1 − x^{i+1}) over i = 0,…,n vanishes, then the polynomial ∑ a_i y^i has a root y strictly between 0 and 1. | [
{
"index": 1,
"text": "From the hypothesis on x we record the two facts that 0 < x and x < 1 separately.",
"lean_code": " obtain ⟨hx0, hx1⟩ := hx",
"label": "correct"
},
{
"index": 2,
"text": "Introduce the polynomial p(y) = ∑_{i=0}^{n} a_i y^i, so that the desired conclusion is exactly... | correct | null | null | |
lean_workbook_hf_14812 | 0 | 5b1e5c5bd6ef3c5aa355ec24e2bc2751e6ae1ed5fd3ad9f43fdfc5e12095c788 | Lean-Workbook | lean_workbook_hf | equality_identity | theorem lean_workbook_hf_14812 (f : ℝ → ℝ) (h₁ : Monotone f) (h₂ : ∀ x, f (f x) = (f x)^2) (h₃ : ∀ x, f (-f x) = (f x)^2) : ∀ x ∈ Set.range f, f x = f (-x) | We must show that for every real number x lying in the range of f, we have f(x) = f(-x). | [
{
"index": 1,
"text": "Fix an arbitrary real number x that belongs to the range of f, and let us show f(x) = f(-x) for this x.",
"lean_code": " intro x hx",
"label": "correct"
},
{
"index": 2,
"text": "Since x is in the range of f, there exists some real number y such that f(y) = x.",
... | correct | null | null | |
lean_workbook_hf_30190 | 0 | 3905d9c65b981cbd964b802eacfc574e12a66928a8fca63fb01dd956fb30f0ad | Lean-Workbook | lean_workbook_hf | existence_construction | theorem lean_workbook_hf_30190 : ¬∃ x y : ℤ, 7*x ^ 2 - 13*y ^ 2 = 1 | We must show that there are no integers x and y satisfying 7x² − 13y² = 1. | [
{
"index": 1,
"text": "Suppose, for contradiction, that such integers x and y exist, with the equation h: 7x² − 13y² = 1 holding.",
"lean_code": " rintro ⟨x, y, h⟩",
"label": "correct"
},
{
"index": 2,
"text": "Reduce the equation h modulo 13 by applying the natural map from the integer... | correct | null | null | |
lean_workbook_hf_14211_13550 | 0 | 89dc758b183d3e3c7b70e665ea91abffc552ec9ca8f2ed30c35362058eb13182 | Lean-Workbook | lean_workbook_hf | existence_construction | theorem lean_workbook_hf_14211_13550 (x y : ℝ) (f : ℝ → ℝ) (hf: f x + f y = (Real.sqrt (x / y) + Real.sqrt (y / x)) * f (Real.sqrt (x * y))) : ∃ f: ℝ → ℝ, ∀ x y : ℝ, (x > 0 ∧ y > 0) → f x + f y = (Real.sqrt (x / y) + Real.sqrt (y / x)) * f (Real.sqrt (x * y)) | We must show that there exists a function f from the reals to the reals satisfying, for all positive real numbers x and y, the identity f(x) + f(y) = (√(x/y) + √(y/x))·f(√(xy)). It suffices to exhibit one such function and verify the identity holds for it. | [
{
"index": 1,
"text": "Take as our candidate the function that sends every real number to 0, that is f(t) = 0 for all t.",
"lean_code": " use fun _ => 0",
"label": "correct"
},
{
"index": 2,
"text": "Fix arbitrary real numbers x and y, and assume both are positive, so that x > 0 and y >... | correct | null | null | |
lean_workbook_hf_70031 | 0 | 672b0b6b39a36c72cb7736afd803c9ceb7e88a434574575fb1030c225b350708 | Lean-Workbook | lean_workbook_hf | divisibility_modular | theorem lean_workbook_hf_70031 : 5555 ^ 5555 ≡ 5 [ZMOD 9] | We want to show that 5555 raised to the power 5555, viewed as an integer, is congruent to 5 modulo 9. | [
{
"index": 1,
"text": "First, note that 5555 = 9·617 + 2, so 5555 leaves remainder 2 when divided by 9; hence 5555 is congruent to 2 modulo 9.",
"lean_code": " have h1 : (5555 : ℤ) ≡ 2 [ZMOD 9] := by decide",
"label": "correct"
},
{
"index": 2,
"text": "Raising both sides of the congrue... | correct | null | null | |
lean_workbook_hf_35597 | 0 | 76af5b7b3a66a1e0ad99b8192c0cabc9b0abf3886d0cfea9cbc2b9b5b498645c | Lean-Workbook | lean_workbook_hf | inequality | theorem lean_workbook_hf_35597 (n x y z : ℝ) (hn : n > 0) (hx : x > 0) (hy : y > 0) (hz : z > 0) (h : n * y * z ≥ 2 * x) : n / 2 ≥ x / (y * z) | We must show that, given positive reals n, x, y, z with n·y·z ≥ 2x, the ratio n/2 is at least x/(yz). | [
{
"index": 1,
"text": "Since y and z are both positive, their product yz is positive.",
"lean_code": " have hyz : y * z > 0 := mul_pos hy hz",
"label": "correct"
},
{
"index": 2,
"text": "The number 2 is positive.",
"lean_code": " have h2 : (0:ℝ) < 2 := by norm_num",
"label": "... | correct | null | null | |
lean_workbook_hf_12421 | 0 | 1853079f008e1d962bbc72bf5009cce6fdd996703b599928863643f88a50084e | Lean-Workbook | lean_workbook_hf | iff_logic | theorem lean_workbook_hf_12421 (x : ℝ) (hx: 0 ≤ x ∧ x ≤ 1) : 0 ≤ x - x ^ 2 ∧ x - x ^ 2 ≤ 1/4 | We must show that for a real number x satisfying 0 ≤ x ≤ 1, the quantity x − x² is both nonnegative and at most 1/4. | [
{
"index": 1,
"text": "Split the hypothesis into its two parts: x ≥ 0 and x ≤ 1.",
"lean_code": " obtain ⟨hx0, hx1⟩ := hx",
"label": "correct"
},
{
"index": 2,
"text": "It suffices to prove the two claims separately: first that x − x² ≥ 0, and second that x − x² ≤ 1/4.",
"lean_code"... | correct | null | null | |
lean_workbook_hf_36804_18771 | 0 | 33f90bf92756b701010b5b34872c35f35afd2ce2b88d6bf06de76ef5450982b9 | Lean-Workbook | lean_workbook_hf | existence_construction | theorem lean_workbook_hf_36804_18771 (f : ℝ → ℝ) (hf : ∀ x y, x * f y = y * f x) : ∃ k, ∀ x, f x = k * x | We must show that there exists a real number k such that f(x) = k·x for every real x, given that x·f(y) = y·f(x) holds for all real x and y. | [
{
"index": 1,
"text": "We claim that k = f(1) works, so we take this value as our candidate constant.",
"lean_code": " use f 1",
"label": "correct"
},
{
"index": 2,
"text": "Let x be an arbitrary real number; we must show f(x) = f(1)·x.",
"lean_code": " intro x",
"label": "corr... | correct | null | null | |
lean_workbook_hf_46031 | 0 | 5f08a6e58679b14ecf0ffc3db8a66f7b24b2c549cb2cea86fc389888e7270761 | Lean-Workbook | lean_workbook_hf | equality_identity | theorem lean_workbook_hf_46031 (b : ℕ) (p : ℕ) (k : ℕ) (h₁ : b = 4095) (h₂ : p ^ k = 13) : (p ^ k)^3 = 2197 | We are given that p^k equals 13, and we must show that (p^k)^3 equals 2197. | [
{
"index": 1,
"text": "Since p^k = 13, we substitute this value into the expression (p^k)^3, reducing the goal to showing that 13^3 equals 2197.",
"lean_code": " rw [h₂]",
"label": "correct"
},
{
"index": 2,
"text": "We compute directly: 13^3 = 13 · 13 · 13 = 169 · 13 = 2197, confirming... | correct | null | null | |
lean_workbook_hf_47079 | 0 | 434190eacad2b54e85560942969e16e2908edf51d830ce3d29a746d5f006e0ac | Lean-Workbook | lean_workbook_hf | iff_logic | theorem lean_workbook_hf_47079 (x a k : ℕ) : 10 * x + a = 2 * (a * 10 ^ k + x) → 8 * x = a * (2 * 10 ^ k - 1) | We must show that for natural numbers x, a, k, if 10x + a = 2(a·10^k + x), then 8x = a·(2·10^k − 1). | [
{
"index": 1,
"text": "Assume the hypothesis h: 10x + a = 2(a·10^k + x), and prove the conclusion under this assumption.",
"lean_code": " intro h",
"label": "correct"
},
{
"index": 2,
"text": "Since 10^k is a power of a positive number, 2·10^k is strictly positive; from 0 < 2·10^k it fo... | correct | null | null | |
numina_b246244f-99a7-5459-969a-5e9e1e57c730 | 0 | 34ddea29a1d0144674f2d6b5e05ccc29ad48f4bc9ef90b17130066159d3aa7ad | NuminaMath-LEAN | numina | Algebra | theorem algebra_186853 (x : ℝ) :
∑ i ∈ Finset.range 6, sin (x + 3 ^ i * 2*π/7) = 1 ↔
∃ n : ℤ, x = -π/2 + 2*n*π := by | We must prove that the sum of the six sines sin(x + 3^i·2π/7) for i = 0,…,5 equals 1 exactly when x = −π/2 + 2nπ for some integer n. | [
{
"index": 1,
"text": "For every angle θ, expanding sin(x+θ) = sin x cos θ + cos x sin θ and sin(x−θ) = sin x cos θ − cos x sin θ and adding cancels the cos x sin θ terms, giving sin(x+θ) + sin(x−θ) = 2 sin x cos θ.",
"lean_code": " have pr : ∀ θ : ℝ, sin (x + θ) + sin (x - θ) = 2 * sin x * cos θ := by... | correct | null | null | |
lean_workbook_hf_65415 | 0 | 1fba487c6c80f926eb76fca2e78d751cfbc9b8802fc86d3fb2d5bbc7d549a315 | Lean-Workbook | lean_workbook_hf | divisibility_modular | theorem lean_workbook_hf_65415 : 162 ∣ 19 ^ 93 - 13 ^ 99 | We must show that 162 divides the natural number 19^93 minus 13^99 (where the subtraction is understood on natural numbers, which is legitimate here since 19^93 is in fact larger than 13^99, as its logarithm 93·ln19 ≈ 273.8 exceeds 99·ln13 ≈ 253.9). | [
{
"index": 1,
"text": "We first record the concrete inequality 13^99 ≤ 19^93, confirmed by exact numerical comparison of these two large but explicit natural numbers; this fact is needed so that the natural-number subtraction 19^93 − 13^99 behaves like ordinary subtraction and can be linked to a modular con... | correct | null | null | |
pythagoras_08f17cc50b6b92f453f9 | 0 | 06718458ef4df34ddb86a03903d43d5d3ba12e9562a6d36f34bda7a6c0378125 | Pythagoras | pythagoras | Pythagoras_SFT_N/A | theorem my_favorite_theorem :
let start_point : ℝ × ℝ := (1, 0)
let angular_velocity : ℝ := Real.pi / 3
let time_period : ℝ := 2
let final_angle : ℝ := angular_velocity * time_period
let midpoint_angle : ℝ := angular_velocity * time_period / 2
let midpoint : ℝ × ℝ := (Real.cos midpoint_angle, Real.sin midpoint_an... | We are given a starting point (1,0), an angular velocity of π/3, and a time period of 2, so the midpoint angle is defined as angular_velocity times time_period divided by 2. The midpoint on the unit circle after this rotation is (cos(midpoint_angle), sin(midpoint_angle)), and the displacement length is the length of th... | [
{
"index": 1,
"text": "Unfolding the definitions, the midpoint angle equals (π/3 · 2)/2, so the midpoint's coordinates are (cos(π/3 · 2 / 2), sin(π/3 · 2 / 2)), and the displacement length is exactly the square root of the sum of the squares of these two coordinates. We restate the goal in this explicit for... | correct | null | null | |
lean_workbook_hf_49382 | 0 | 5d0f46cadcd0413cdc700424d1084df3951c685fd8a7eed1c1f90e72a1a2066b | Lean-Workbook | lean_workbook_hf | existence_construction | theorem lean_workbook_hf_49382 (a : ℝ) (ha : 0 < a) : ∃ g : ℝ → ℝ, g x = x ^ 2 / (x ^ 4 + a ^ 2) | We must show that there exists a function g from the reals to the reals such that g evaluated at the point x equals x squared divided by x to the fourth power plus a squared. | [
{
"index": 1,
"text": "Define the function g explicitly by the rule g(x) = x^2 / (x^4 + a^2) for every real input x, and it remains to check that this g satisfies the required equation at the point x.",
"lean_code": " refine ⟨fun x => x ^ 2 / (x ^ 4 + a ^ 2), ?_⟩",
"label": "correct"
},
{
"... | correct | null | null | |
lean_workbook_hf_73236 | 0 | 3625396793e5c90298668e1dbdfd58a437023593106b43a98c5690d0012fd730 | Lean-Workbook | lean_workbook_hf | inequality | theorem lean_workbook_hf_73236 (a b c : ℝ) (h₀ : 0 < a ∧ b ^ 2 - 4 * a * c ≤ 0) : a / 1998 + b / 1997 + b ^ 2 / (4 * 1996 * a) ≥ 0 | We must show that for real numbers a, b, c with a > 0 (and the additional condition that b² − 4ac ≤ 0, which we will see is not actually needed), the quantity a/1998 + b/1997 + b²/(4·1996·a) is nonnegative. | [
{
"index": 1,
"text": "We split the hypothesis into its two parts: a > 0, and b² − 4ac ≤ 0. As we will show, the positivity of a alone already guarantees the desired inequality, so the second condition will not be used further.",
"lean_code": " obtain ⟨ha, hd⟩ := h₀",
"label": "correct"
},
{
... | correct | null | null | |
lean_workbook_hf_78020 | 0 | 0b7c5c23517e4332c0c6458541ebc20c36f32c690da28daa7a135e89129c856a | Lean-Workbook | lean_workbook_hf | iff_logic | theorem lean_workbook_hf_78020 (a b c d : ℤ) (ha : a > 0) (hb : b > 0) (hc : c > 0) (hd : d > 0) (hab : a ≠ b) (hbc : b ≠ c) (hcd : c ≠ d) (habc : a ≠ b ∧ b ≠ c ∧ c ≠ d) : a + b + c + d = 10 ∧ 343*a + 49*b + 7*c + d = 988 → 1000*a + 100*b + 10*c + d = 2611 | We must show that for positive integers a, b, c, d with a≠b, b≠c, c≠d, if a+b+c+d=10 and 343a+49b+7c+d=988, then 1000a+100b+10c+d=2611. | [
{
"index": 1,
"text": "Assume the two given equations hold: a+b+c+d=10 (call it h1) and 343a+49b+7c+d=988 (call it h2).",
"lean_code": " rintro ⟨h1, h2⟩",
"label": "correct"
},
{
"index": 2,
"text": "Subtracting h1 from h2 gives 342a+48b+6c=978, and dividing by 6 yields the equation 57a... | correct | null | null | |
lean_workbook_hf_2019 | 0 | af257f2f385421323644722ce5869317f76d049d010a9ba42dc39e64ea7d471f | Lean-Workbook | lean_workbook_hf | iff_logic | theorem lean_workbook_hf_2019 (x : ℝ) (hx: 0 < x ∧ x < 1) : (1 + x / 2 - x ^ 2) < Real.sqrt (1 + x) ∧ Real.sqrt (1 + x) < (1 + x / 2) | We must show that for a real number x strictly between 0 and 1, the square root of 1+x is squeezed strictly between 1 + x/2 - x^2 and 1 + x/2. | [
{
"index": 1,
"text": "Write out the hypothesis as two separate facts: x > 0 and x < 1.",
"lean_code": " obtain ⟨hx1, hx2⟩ := hx",
"label": "correct"
},
{
"index": 2,
"text": "Since x < 1, we have x - 1 < 0.",
"lean_code": " have hxsub : x - 1 < 0 := by linarith",
"label": "cor... | correct | null | null | |
lean_workbook_hf_12728 | 0 | 7c05b6b906d75a5002d8448172b01a71abf63893d0d178c8be312162eeb1aa6c | Lean-Workbook | lean_workbook_hf | existence_construction | theorem lean_workbook_hf_12728 : ∃ a b c x y z : ℝ, ¬(4 * (a ^ 2 + x ^ 2) * (b ^ 2 + y ^ 2) * (c ^ 2 + z ^ 2) ≥ 3 * (a * b * x + b * c * y + c * a * z) ^ 2) | We must show that there exist real numbers a, b, c, x, y, z for which the inequality 4(a²+x²)(b²+y²)(c²+z²) ≥ 3(abx+bcy+caz)² fails. We will exhibit a specific choice of these six numbers for which the left-hand side vanishes while the right-hand side is strictly positive. | [
{
"index": 1,
"text": "Choose a = 1, b = 1, c = 0, x = 1, y = 0, z = 0. With these values, the third factor c²+z² of the product on the left is 0, forcing the whole left-hand side to be 0, while the term abx = 1 makes the right-hand side positive, as we now verify.",
"lean_code": " use 1, 1, 0, 1, 0, 0... | correct | null | null | |
lean_workbook_hf_44161 | 0 | 006f6e262d5ac0faea8a168557a9db257e330183f27c85f54fd52d6b874edfed | Lean-Workbook | lean_workbook_hf | divisibility_modular | theorem lean_workbook_hf_44161 : 19 ∣ (10 ^ 9 + 1) | We want to show that 19 divides 10^9 + 1, i.e. that 1000000001 is an exact multiple of 19. | [
{
"index": 1,
"text": "We compute directly that 10^9 + 1 equals 1000000001, and that this number equals 19 times 52631579, since 19 × 52631579 = 19 × 52631580 − 19 = 1000000020 − 19 = 1000000001.",
"lean_code": " have h : 10 ^ 9 + 1 = 19 * 52631579 := by norm_num",
"label": "correct"
},
{
"... | correct | null | null | |
lean_workbook_hf_70512 | 0 | de7dc6b15976349a7caee8d98e7dab4d25feb8860bdd73a2b967772adced7493 | Lean-Workbook | lean_workbook_hf | equality_identity | theorem lean_workbook_hf_70512 (a b c d : ℝ) (w : ℂ) (h₁ : w ≠ 0) : ‖z / w‖ = (‖z‖ / ‖w‖) | We must show that for any complex numbers z and w with w nonzero, the norm of the quotient z / w equals the quotient of the norms ‖z‖ / ‖w‖. | [
{
"index": 1,
"text": "First, recall that division is defined as multiplication by the inverse, so z / w equals z times the inverse of w.",
"lean_code": " have h2 : z / w = z * w⁻¹ := div_eq_mul_inv z w",
"label": "correct"
},
{
"index": 2,
"text": "Rewrite the left-hand side ‖z / w‖ us... | correct | null | null | |
lean_workbook_hf_13092 | 0 | 74ea3eff893079944b076ddb291a607055a35bacfe8544a7d11f3a5411c6fafc | Lean-Workbook | lean_workbook_hf | equality_identity | theorem lean_workbook_hf_13092 (a b c d: ℝ) : (a ^ 2 + b * d)^2 + (b ^ 2 + a * c)^2 + (c ^ 2 + b * d)^2 + (d ^ 2 + a * c)^2 >= (c + a) * (b + d) * (a ^ 2 + c ^ 2 + b ^ 2 + d ^ 2) | We aim to prove that for all real numbers a, b, c, d, the sum (a²+bd)² + (b²+ac)² + (c²+bd)² + (d²+ac)² is at least (c+a)(b+d)(a²+c²+b²+d²). The strategy is to exhibit the difference between the two sides as a sum of three explicit squares. | [
{
"index": 1,
"text": "Introduce the auxiliary quantity X = a + c - b - d, which measures the imbalance between the pair (a,c) and the pair (b,d).",
"lean_code": " set X := a + c - b - d with hX",
"label": "correct"
},
{
"index": 2,
"text": "Introduce the auxiliary quantity T = b + d.",... | correct | null | null | |
lean_workbook_hf_62735 | 0 | b7623d98b2681e50c8c1f41ec514608fe0ef51d0d5b65db9660fef201189446d | Lean-Workbook | lean_workbook_hf | existence_construction | theorem lean_workbook_hf_62735 (x : ℕ → ℝ) (x0 : x 0 = Real.sqrt 2) (x_rec : ∀ n, x (n + 1) = (x n + 2 / x n) / 2) : ∃ n, ∀ ε > 0, |x n - Real.sqrt 2| < ε | We must show that there exists an index n of the sequence such that, for every positive ε, the distance between x n and √2 is smaller than ε. We will show this holds already at n = 0, since we are given that x 0 equals √2 exactly. | [
{
"index": 1,
"text": "We choose n = 0 as our witness for the existential statement, since the hypothesis directly tells us that the sequence starts exactly at √2.",
"lean_code": " use 0",
"label": "correct"
},
{
"index": 2,
"text": "Let ε be an arbitrary positive real number; we must s... | correct | null | null | |
lean_workbook_hf_33642 | 0 | 757101f1467ac3bb4a5d82c3adb19cd912fb88794d2372c58cd114e7d537d195 | Lean-Workbook | lean_workbook_hf | divisibility_modular | theorem lean_workbook_hf_33642 (p n : ℕ) (hp : Nat.Prime p) (h : ¬ p ∣ n) : Nat.lcm n (n + p) = n * (n + p) | We must show that for a prime p and natural number n with p not dividing n, the least common multiple of n and n+p equals the product n·(n+p). | [
{
"index": 1,
"text": "Since p is prime and does not divide n, p and n share no common prime factor, so p and n are coprime.",
"lean_code": " have hcop : Nat.Coprime p n := (hp.coprime_iff_not_dvd).mpr h",
"label": "correct"
},
{
"index": 2,
"text": "Coprimality is a symmetric relation,... | correct | null | null | |
lean_workbook_hf_62632 | 0 | 1a87928b916861eb8143cb86ff001f40ecbc7937c217b30064425b7ba90dfae0 | Lean-Workbook | lean_workbook_hf | equality_identity | theorem lean_workbook_hf_62632 (x : ℕ) (h : x = 5) : x * x * x * 2 * 2 * 2 = 1000 | We must show that if x equals 5, then the product x·x·x·2·2·2 equals 1000. | [
{
"index": 1,
"text": "Since x = 5, we replace every occurrence of x by 5, so the goal becomes showing that 5·5·5·2·2·2 equals 1000.",
"lean_code": " subst h",
"label": "correct"
},
{
"index": 2,
"text": "We compute directly: 5·5 = 25, 25·5 = 125, and 2·2·2 = 8, so 125·8 = 1000, confirm... | correct | null | null | |
lean_workbook_hf_57045 | 0 | 4b77b4186551ff9a915e55dd538c5c49c34874285aee289ab0585957a249b044 | Lean-Workbook | lean_workbook_hf | iff_logic | theorem lean_workbook_hf_57045 (x : ℝ) : x * (x - 4) ≤ 0 ↔ 0 ≤ x ∧ x ≤ 4 | We show that for a real number x, the product x(x − 4) is at most 0 exactly when 0 ≤ x and x ≤ 4. | [
{
"index": 1,
"text": "We use the fact that for real numbers a and b, the product a·b is at most 0 exactly when either a ≥ 0 and b ≤ 0, or a ≤ 0 and b ≥ 0. Applying this with a = x and b = x − 4, the statement to prove becomes: (0 ≤ x and x − 4 ≤ 0) or (x ≤ 0 and 0 ≤ x − 4), if and only if 0 ≤ x and x ≤ 4."... | correct | null | null | |
lean_workbook_hf_59134 | 0 | d23717243fdbbe004c3e664ad9a247397c30cd77ac3981ea9c8ccab5f90a2c61 | Lean-Workbook | lean_workbook_hf | equality_identity | theorem lean_workbook_hf_59134 (a b c : ℝ) (hab : |a| ≥ |b + c|) (hbc : |b| ≥ |c + a|) (hca : |c| ≥ |a + b|) : a + b + c = 0 | We are given real numbers a, b, c satisfying |a| ≥ |b+c|, |b| ≥ |c+a|, |c| ≥ |a+b|, and we must show that a + b + c = 0. | [
{
"index": 1,
"text": "Since |b+c| ≤ |a| (this is just the hypothesis |a| ≥ |b+c| written the other way) and |b+c| is nonnegative, squaring both sides of this inequality preserves it, giving |b+c|^2 ≤ |a|^2.",
"lean_code": " have h1 : |b + c| ^ 2 ≤ |a| ^ 2 := pow_le_pow_left₀ (abs_nonneg (b + c)) hab 2... | correct | null | null | |
numina_169d9523-e09b-5cd7-a554-430961a69243 | 0 | 25404eb48d7f1401b7ad3248e2f599eeb6e4084c5ee6941fb730b0559b217856 | NuminaMath-LEAN | numina | Number Theory | theorem number_theory_194949 (n : ℕ) (hn : 0 < n) (s : Finset ℕ)
(h₀ : ∀ x ∈ s, 2 ^ (2*n) < x ∧ x < 2 ^ (3*n))
(h₁ : s.card = 2 ^ (2*n - 1) + 1)
(h₂ : ∀ x ∈ s, Odd x) :
∃ a ∈ s, ∃ b ∈ s, a ≠ b ∧ ¬(a ^ 2 ∣ b) ∧ ¬(b ^ 2 ∣ a) := by | We must show that among the elements of s, all lying strictly between 2^(2n) and 2^(3n), we can find two distinct elements a and b such that neither a^2 divides b nor b^2 divides a. | [
{
"index": 1,
"text": "Since 2 is positive, any power of 2, in particular 2^(2n-1), is a positive number.",
"lean_code": " have hpow : 0 < 2 ^ (2*n - 1) := pow_pos (by norm_num) _",
"label": "correct"
},
{
"index": 2,
"text": "Using the given cardinality s.card = 2^(2n-1) + 1 together w... | correct | null | null | |
lean_workbook_hf_62333 | 0 | f854fc80eda37efb950f7ed2b7279299e92bffbc2edfc6bc331e258eccd7a41c | Lean-Workbook | lean_workbook_hf | divisibility_modular | theorem lean_workbook_hf_62333 (p : ℕ) : p ^ 2 - 3 * p - 2 ≡ (p - 10)^2 [ZMOD 17] | We must show that, viewing p as an integer, the quantity p^2 − 3p − 2 is congruent to (p − 10)^2 modulo 17. | [
{
"index": 1,
"text": "By the standard characterization of congruence modulo an integer, a ≡ b (mod 17) exactly when 17 divides the difference b − a. So it suffices to show that 17 divides (p − 10)^2 − (p^2 − 3p − 2).",
"lean_code": " rw [Int.modEq_iff_dvd]",
"label": "correct"
},
{
"index"... | correct | null | null | |
lean_workbook_hf_79992 | 0 | 56ea0da697f9019db0c7de5e36948d64fb42d100362a72a2a7d8286cf68cb39e | Lean-Workbook | lean_workbook_hf | equality_identity | theorem lean_workbook_hf_79992 : (x ^ 2+2*x+1)^2 = x ^ 4 + 4*x ^ 3 + 6*x ^ 2 + 4*x + 1 | We must show that for the real number x, the square of the quadratic x^2 + 2x + 1 equals the quartic polynomial x^4 + 4x^3 + 6x^2 + 4x + 1. | [
{
"index": 1,
"text": "First we recognize that the quadratic x^2 + 2x + 1 is exactly the perfect square (x + 1)^2, since expanding (x+1)^2 = x^2 + 2x + 1.",
"lean_code": " have h1 : x ^ 2 + 2*x + 1 = (x+1)^2 := by ring",
"label": "correct"
},
{
"index": 2,
"text": "We replace the quadra... | correct | null | null | |
lean_workbook_hf_2844 | 0 | 4bf32d3aa75cd08274e6fd72611e1b844adf439867d3d32d4768539320d037ec | Lean-Workbook | lean_workbook_hf | inequality | theorem lean_workbook_hf_2844 (a b c : ℝ) (hab : a + b + c + 2 = a * b * c) : (a ^ 2 + 1) * (b ^ 2 + 1) * (c ^ 2 + 1) ≥ 4 | We must show that if real numbers a, b, c satisfy a + b + c + 2 = abc, then (a²+1)(b²+1)(c²+1) ≥ 4. | [
{
"index": 1,
"text": "Rearranging the given equation a + b + c + 2 = abc simply by moving terms across the equality, we obtain abc − a − b − c = 2.",
"lean_code": " have h1 : a * b * c - a - b - c = 2 := by linarith",
"label": "correct"
},
{
"index": 2,
"text": "There is an algebraic i... | correct | null | null | |
lean_workbook_hf_23246 | 0 | 276ddbfbe70395e50016984649059f65725540792393f19598e1470bbbe934f4 | Lean-Workbook | lean_workbook_hf | divisibility_modular | theorem lean_workbook_hf_23246 {n m : ℤ} (h : 5 ∣ 1 + 2*n ^ 2 + 3*m ^ 2) : 5 ∣ n ^ 2 - 1 | We must show: if 5 divides 1 + 2n² + 3m² for integers n and m, then 5 divides n² − 1. | [
{
"index": 1,
"text": "Since 5 divides an integer exactly when that integer's residue class modulo 5 is the zero element, the divisibility hypothesis 5 ∣ 1 + 2n² + 3m² translates into the statement that the image of 1 + 2n² + 3m² in the ring of residues modulo 5 equals zero.",
"lean_code": " have h1 : ... | correct | null | null | |
lean_workbook_hf_52920 | 0 | dd6fd154a92f2aebcb2e4518063d3360fcf3fa5f158a386340916350c9b9179c | Lean-Workbook | lean_workbook_hf | iff_logic | theorem lean_workbook_hf_52920 (a b c : ℕ) (h₀ : 0 < a ∧ 0 < b ∧ 0 < c) (h₁ : a ^ 3 < a * b * c) (h₂ : a * b * c = 2008) : a ≤ 12 ∧ (a = 1 ∨ a = 2 ∨ a = 4 ∨ a = 8) | We are given positive natural numbers a, b, c with a^3 < abc and abc = 2008, and we must show that a ≤ 12 and that a is one of 1, 2, 4, 8. | [
{
"index": 1,
"text": "Split the hypothesis into its three parts: a > 0, b > 0, and c > 0.",
"lean_code": " obtain ⟨ha, hb, hc⟩ := h₀",
"label": "correct"
},
{
"index": 2,
"text": "Since abc = 2008, substitute this value into the inequality a^3 < abc to obtain the concrete bound a^3 < 2... | correct | null | null | |
lean_workbook_hf_16184 | 0 | 006800553d31c3d9308cd9050784d499dd2aebfb51696b1077c96f44d5735b24 | Lean-Workbook | lean_workbook_hf | iff_logic | theorem lean_workbook_hf_16184 (x : ℝ) : (16 * x ^ 2 + 64 * x - 233 = 0) ↔ (x = -2 + 3 * Real.sqrt 33 / 4 ∨ x = -2 - 3 * Real.sqrt 33 / 4) | We want to show that 16x^2 + 64x - 233 = 0 exactly when x equals -2 + (3/4)√33 or x equals -2 - (3/4)√33. | [
{
"index": 1,
"text": "Since 33 is nonnegative, squaring its square root returns 33 itself: (√33)^2 = 33.",
"lean_code": " have hr : Real.sqrt 33 ^ 2 = 33 := Real.sq_sqrt (by norm_num)",
"label": "correct"
},
{
"index": 2,
"text": "We claim the quadratic factors as 16 times the product ... | correct | null | null | |
lean_workbook_hf_40017_19513 | 0 | fd2e521c21dadffaafc0818f41d95891c2a9076a4823fd38a0441516a34a9904 | Lean-Workbook | lean_workbook_hf | existence_construction | theorem lean_workbook_hf_40017_19513 (hA : A ⊆ Finset.Icc 1 2022) (hA' : A.card = 2022) : ∃ x y, x ∈ A ∧ y ∈ A ∧ (x - y).gcd 2021 = 2021 | We must show that if A is a subset of the integers from 1 to 2022 and A has exactly 2022 elements, then there exist two elements x and y of A whose difference is exactly divisible by 2021, i.e. gcd(x−y, 2021) = 2021. | [
{
"index": 1,
"text": "First we compute the size of the full interval {1, 2, ..., 2022}: by the standard formula for the cardinality of an interval of integers, this set has b + 1 − a = 2022 + 1 − 1 = 2022 elements.",
"lean_code": " have hcard : (Finset.Icc 1 2022).card = 2022 := by\n rw [Nat.card_I... | correct | null | null | |
lean_workbook_hf_40325 | 0 | 1cbda85b92400e05f1bd95799659fff598d911875f1da928683c23566920d4ec | Lean-Workbook | lean_workbook_hf | inequality | theorem lean_workbook_hf_40325 (r : ℕ → ℝ) : 2 * (r 1 ^ 2 + r 2 ^ 2 + r 3 ^ 2 + r 4 ^ 2 + r 5 ^ 2) ≥ r 1 * r 2 + r 1 * r 3 + r 1 * r 4 + r 1 * r 5 + r 2 * r 3 + r 2 * r 4 + r 2 * r 5 + r 3 * r 4 + r 3 * r 5 + r 4 * r 5 | We must show that for the five real numbers r1, r2, r3, r4, r5, twice the sum of their squares is at least as large as the sum of all ten pairwise products among them. | [
{
"index": 1,
"text": "The square of the difference r1 − r2 is nonnegative, since any real square is nonnegative.",
"lean_code": " have h12 : (r 1 - r 2) ^ 2 ≥ 0 := sq_nonneg _",
"label": "correct"
},
{
"index": 2,
"text": "Likewise, the square of r1 − r3 is nonnegative.",
"lean_cod... | correct | null | null | |
lean_workbook_hf_70312 | 0 | 6930e2fccc0b8257aa8dd5e35d89abf92694c5144409c37009109f7394a43ec4 | Lean-Workbook | lean_workbook_hf | equality_identity | theorem lean_workbook_hf_70312 (n : ℕ) : ((n:ℝ)^n / (n:ℝ)^n) = 1 | We must show that for every natural number n, the real quotient (n^n)/(n^n) equals 1, where n^n is computed using the natural-number power convention (in particular 0^0 = 1). | [
{
"index": 1,
"text": "We split into the two possible cases for a natural number n: either n = 0, or n is strictly positive (n > 0).",
"lean_code": " rcases Nat.eq_zero_or_pos n with h0 | hpos",
"label": "correct"
},
{
"index": 2,
"text": "In the first case we substitute n = 0 everywher... | correct | null | null | |
lean_workbook_hf_15395 | 0 | 391c7a606d68559b858722b0d84d38586950c7fc9a214e314d278c47c28bbcaa | Lean-Workbook | lean_workbook_hf | divisibility_modular | theorem lean_workbook_hf_15395 : ∀ n : ℤ, (2 * n ^ 3 + 3 * n ^ 2 + 7 * n) % 6 = 0 | We must prove that for every integer n, the expression 2n^3 + 3n^2 + 7n leaves remainder 0 when divided by 6, i.e. it is always a multiple of 6. | [
{
"index": 1,
"text": "Fix an arbitrary integer n; we will show that 6 divides 2n^3 + 3n^2 + 7n and hence that the expression is congruent to 0 modulo 6.",
"lean_code": " intro n",
"label": "correct"
},
{
"index": 2,
"text": "First we verify a finite fact: working in the residues modulo... | correct | null | null | |
lean_workbook_hf_67171 | 0 | a1a9622e992ede69cbe5acbf553cd3c17ff1c77fe5e0be61070f480b6c856b64 | Lean-Workbook | lean_workbook_hf | existence_construction | theorem lean_workbook_hf_67171 (f : ℝ → ℝ):(∀ x y :ℝ, (x+y)*(f x - f y) = f (x ^ 2) - f (y ^ 2)) ↔ ∃ a b :ℝ, ∀ x : ℝ, f x = a * x + b | We must show that a function f satisfies the identity (x+y)(f(x)-f(y)) = f(x^2)-f(y^2) for all real x, y if and only if f is affine, that is f(x) = a x + b for some constants a, b. | [
{
"index": 1,
"text": "We prove the two directions of the equivalence separately.",
"lean_code": " constructor",
"label": "correct"
},
{
"index": 2,
"text": "Assume first that the functional equation (x+y)(f(x)-f(y)) = f(x^2)-f(y^2) holds for all real x and y; we must produce constants ... | correct | null | null | |
corpus_NuminaMath_LEAN_65d27429-1d8b-5607-a605-bf8dc58ebf80 | 0 | 5838bc24d169c7d9e09ae8c9be807fe51656bfdee4fc7d43d133d86fe3026c71 | NuminaMath-LEAN | corpus | NuminaMath_LEAN | theorem algebra_98177 (a b c d : ℕ) (h₀ : 0 < a) (h₁ : a < b) (h₂ : b < c) (h₃ : c < d)
(h₄ : ∃ d, b = a + d ∧ c = b + d) (h₅ : ∃ r : ℚ, c = b * r ∧ d = c * r) (h₆ : d - a = 30) :
a + b + c + d = 129 := by | We are given four natural numbers a < b < c < d with a positive, such that a, b, c form an arithmetic progression, b, c, d form a geometric progression, and d − a = 30. We must show a + b + c + d = 129. | [
{
"index": 1,
"text": "Since a, b, c form an arithmetic progression, there is a common difference e such that b = a + e and c = b + e; we name this common difference e explicitly.",
"lean_code": " obtain ⟨e, hb, hc⟩ := h₄",
"label": "correct"
},
{
"index": 2,
"text": "Since b, c, d form... | correct | null | null | |
numina_f3298ef6-0418-5db1-a520-7d05bc22cd81 | 0 | 40bba25a2b62406e7c3e3de0c7aa7f111c912a733cc2e677513d53a01ea3fc4c | NuminaMath-LEAN | numina | Number Theory | theorem number_theory_605010 (f : ℕ → ℝ) (hf : ∀ n > 1, ∃ p : ℕ, (Nat.Prime p ∧ p ∣ n ∧ f n = f (n / p) - f p)) (h : f (2 ^ 2007) + f (3 ^ 2008) + f (5 ^ 2009) = 2006) : f (2007 ^ 2) + f (2008 ^ 3) + f (2009 ^ 5) = 9:= by | We are given a function f on the positive integers such that for every n > 1 there is a prime p dividing n with f(n) = f(n/p) − f(p), and the normalization f(2^2007) + f(3^2008) + f(5^2009) = 2006. We must show f(2007^2) + f(2008^3) + f(2009^5) = 9. | [
{
"index": 1,
"text": "First we evaluate f at every prime. Fix a prime p. Since p > 1, the relation provides a prime q dividing p with f(p) = f(p/q) − f(q). But a prime divisor of the prime p can only be p itself, so q = p; since p/p = 1 this becomes f(p) = f(1) − f(p). Solving the linear equation 2f(p) = f... | correct | null | null | |
numina_6d8521b2-a7f4-5c83-9883-e76f07d9c137 | 0 | 69dc12b2aa4535c5cff37d7beb5abb3367bc7fcba8e8a6e4575ee5d789381006 | NuminaMath-LEAN | numina | Number Theory | theorem number_theory_238242 :
¬∃ x y z : ℚ, x ≠ y ∧ y ≠ z ∧ z ≠ x ∧ 1 / (x - y) ^ 2 + 1 / (y - z) ^ 2 + 1 / (z - x) ^ 2 = 2014 := by | We must show that there are no three distinct rational numbers x, y, z for which 1/(x-y)^2 + 1/(y-z)^2 + 1/(z-x)^2 equals 2014. We argue by contradiction. | [
{
"index": 1,
"text": "Suppose such x, y, z exist, with x ≠ y, y ≠ z, z ≠ x, and satisfying the given equation; we derive a contradiction.",
"lean_code": " rintro ⟨x, y, z, hxy, hyz, hzx, heq⟩",
"label": "correct"
},
{
"index": 2,
"text": "Since x ≠ y, the difference x − y is nonzero.",... | correct | null | null | |
lean_workbook_hf_7539 | 0 | 6758344b658199fdf8b85c006057ee7c2635ddf01c6a4159ae5443ca744abd85 | Lean-Workbook | lean_workbook_hf | existence_construction | theorem lean_workbook_hf_7539 (x y z : ℤ) (h : x * y * z ≠ 0) : x ^ 4 + 2*y ^ 4 = z ^ 2 ↔ ∃ x0 y0 z0 : ℤ, x0 * y0 * z0 ≠ 0 ∧ x ^ 4 + 2*y ^ 4 = z ^ 2 | We must show that, given the nonzero product condition x·y·z ≠ 0, the equation x^4 + 2y^4 = z^2 holds if and only if there exist integers x0, y0, z0 with x0·y0·z0 ≠ 0 for which x^4 + 2y^4 = z^2 holds; note that this second equation still refers to the original x, y, z, not to x0, y0, z0, so the existential statement re... | [
{
"index": 1,
"text": "To prove this biconditional, we prove each direction separately: first that the equation implies the existential statement, and then that the existential statement implies the equation.",
"lean_code": " constructor",
"label": "correct"
},
{
"index": 2,
"text": "Fo... | correct | null | null | |
lean_workbook_hf_12327 | 0 | 5afb68e0e606e0004942297cb3c6c75b908356e2b2039f69f847232397f30297 | Lean-Workbook | lean_workbook_hf | inequality | theorem lean_workbook_hf_12327 (a b c : ℝ) (ha : 0 < a) (hb : 0 < b) (hc : 0 < c) (habc : a * b * c = 1) (hab : a * b + b * c + c * a + 2 * a * b * c = 1) : 1 / a ^ 2 + 1 / b ^ 2 + 1 / c ^ 2 - 8 * (a ^ 2 + b ^ 2 + c ^ 2) ≥ 6 | We are given positive reals a, b, c with abc = 1 and ab + bc + ca + 2abc = 1, and we must show 1/a^2 + 1/b^2 + 1/c^2 − 8(a^2+b^2+c^2) ≥ 6. We will show that the hypotheses themselves are contradictory, which lets us conclude anything, including the stated inequality. | [
{
"index": 1,
"text": "First note the algebraic identity 2·a·b·c = 2·(a·b·c), which holds simply by rearranging the associativity of multiplication.",
"lean_code": " have heq : 2 * a * b * c = 2 * (a * b * c) := by ring",
"label": "correct"
},
{
"index": 2,
"text": "Using this identity,... | correct | null | null | |
lean_workbook_hf_10938 | 0 | eaeda33ee5df9357873c412516fab43eeb091b483c4440f811c1c3df2170be08 | Lean-Workbook | lean_workbook_hf | equality_identity | theorem lean_workbook_hf_10938 (a b c : ℝ) : a ^ 2 + b ^ 2 + c ^ 2 - (a * b + b * c + c * a) = 1 / 2 * ((a - b)^2 + (a - c)^2 + (b - c)^2) | We must show that for any real numbers a, b, c, the expression a^2 + b^2 + c^2 − (ab + bc + ca) equals one half of the sum of the three squared differences (a−b)^2 + (a−c)^2 + (b−c)^2. | [
{
"index": 1,
"text": "First expand the square of the difference: (a − b)^2 equals a^2 minus twice the product a·b plus b^2.",
"lean_code": " have h1 : (a - b)^2 = a ^ 2 - 2*a*b + b ^ 2 := by ring",
"label": "correct"
},
{
"index": 2,
"text": "Similarly, expand (a − c)^2 to get a^2 minu... | correct | null | null | |
corpus_NuminaMath_LEAN_e11a7cf4-2809-57b2-92c7-e41bdd7b8a61 | 0 | 55cb64669eac89669564f0f41d0b95773367e3608f8487c479bf117e615e1b96 | NuminaMath-LEAN | corpus | NuminaMath_LEAN | theorem calculus_273185 {a b : ℝ} (ha : 1 < a) (hb : a < b)
(x : ℕ → ℝ) (hx : ∀ n, x n = 2 ^ n * (b ^ ((1 : ℝ) / (2 ^ n)) - a ^ ((1 : ℝ) / (2 ^ n)))) :
StrictAnti x := by | We must show that for real numbers a, b with 1 < a < b, the sequence x(n) = 2^n·(b^(1/2^n) − a^(1/2^n)) is strictly decreasing. | [
{
"index": 1,
"text": "To prove that a sequence indexed by natural numbers is strictly decreasing, it suffices to show that every term is strictly larger than its successor, i.e. x(n+1) < x(n) for all n.",
"lean_code": " apply strictAnti_nat_of_succ_lt",
"label": "correct"
},
{
"index": 2,
... | correct | null | null | |
lean_workbook_hf_359 | 0 | 15515c67eb6db38b7f11768af55d32db74d297d3bfe752281d77fbaa9274288a | Lean-Workbook | lean_workbook_hf | inequality | theorem lean_workbook_hf_359 : (Real.sqrt 2 / 2) ≥ (1 / 3) / (Real.sqrt 2 / 3) | We must show that √2⁄2 is greater than or equal to (1⁄3) divided by (√2⁄3). | [
{
"index": 1,
"text": "Since 2 is a positive real number, its square root √2 is also strictly positive.",
"lean_code": " have h2 : Real.sqrt 2 > 0 := Real.sqrt_pos.mpr (by norm_num)",
"label": "correct"
},
{
"index": 2,
"text": "By the defining property of the square root of a nonnegati... | correct | null | null | |
lean_workbook_hf_50579 | 0 | 36fd522d171a470ee26751ab0a1539efaf989632a22d8b61d22fb8dd5e7658ad | Lean-Workbook | lean_workbook_hf | existence_construction | theorem lean_workbook_hf_50579 (P : ℕ → ℕ) (hP : ∃ P₁ P₂ : ℕ → ℕ, ∀ n, P n = P₁ n + 2 * P₂ n) : ∃ n₁ n₂ : ℕ, P 2 = n₁ + 2 * n₂ | We must show that there exist natural numbers n₁ and n₂ such that P(2) equals n₁ plus twice n₂. Since the claim only asks for the existence of such a decomposition, we may exhibit one directly rather than unpack the hypothesis about P₁ and P₂. | [
{
"index": 1,
"text": "Choose n₁ = P(2) and n₂ = 0 as our witnesses; it remains to verify that P(2) equals P(2) plus twice 0.",
"lean_code": " refine ⟨P 2, 0, ?_⟩",
"label": "correct"
},
{
"index": 2,
"text": "Indeed, twice 0 is 0, so P(2) plus twice 0 simplifies by the basic arithmetic... | correct | null | null | |
lean_workbook_hf_51979 | 0 | c29c6e8ba17b2cd4c99d838827c4ce124f4e36ba4c437a3d1714c19de83748a8 | Lean-Workbook | lean_workbook_hf | existence_construction | theorem lean_workbook_hf_51979 (a : ℝ) (ha : 0 < a) : ∃ x : ℝ, 0 < x ∧ x ^ (x+1) = a ^ (2*x - 2) | We must show that for any positive real number a, there exists a positive real number x satisfying x^(x+1) = a^(2x-2), where the exponents are real-number powers. | [
{
"index": 1,
"text": "We propose the candidate value x = 1 and show it satisfies both required properties.",
"lean_code": " use 1",
"label": "correct"
},
{
"index": 2,
"text": "First, since 1 is clearly a positive real number (1 > 0), the positivity condition on x is immediately satisf... | correct | null | null | |
lean_workbook_hf_15240 | 0 | 7e6fe8e769dc0816514495db2ce0e44e76c6a55a0a969e4e49a1ca941200af66 | Lean-Workbook | lean_workbook_hf | iff_logic | theorem lean_workbook_hf_15240 (a b c : ℝ) (ha : 0 < a) (hb : 0 < b) (hc : 0 < c)(habc : a * b * c = 1) : a ^ 3 + b ^ 3 + c ^ 3 = 1 / 9 → a ^ 2 + b ^ 2 + c ^ 2 + 1 / (a ^ 2 * b ^ 2) + 1 / (b ^ 2 * c ^ 2) + 1 / (c ^ 2 * a ^ 2) ≥ 730 / 3 | We must show that if a, b, c are positive reals with abc = 1 and a^3+b^3+c^3 = 1/9, then a^2+b^2+c^2 + 1/(a^2b^2) + 1/(b^2c^2) + 1/(c^2a^2) ≥ 730/3. We will show that the hypothesis a^3+b^3+c^3 = 1/9 is actually incompatible with abc = 1 and positivity, so the implication holds vacuously. | [
{
"index": 1,
"text": "Assume the hypothesis that a^3 + b^3 + c^3 = 1/9, and let us name this fact h.",
"lean_code": " intro h",
"label": "correct"
},
{
"index": 2,
"text": "We aim to show this assumption leads to a contradiction, so it suffices to derive a falsehood.",
"lean_code":... | correct | null | null | |
numina_19f97d38-29ef-5827-a92a-d495e79a32eb | 0 | 4c0cdcd32be098eef5cf785489c82987aab7d90f8805283be1aa1b62d16e49dd | NuminaMath-LEAN | numina | Number Theory | theorem number_theory_85492 :
∀ n : ℕ, ∃ m : ℤ, m > n ∧ ¬∃ a b c : ℕ, m = 2 ^ a + 3 ^ b - 5 ^ c := by | We must show that for every natural number n there is an integer m greater than n that cannot be written in the form 2^a + 3^b − 5^c with a, b, c natural numbers. Our strategy is to take m = 24n + 7 and prove non-representability by reducing modulo 24, after first determining which residues the three power sequences ca... | [
{
"index": 1,
"text": "Modulo 24 the powers of 2 stay inside the set {1, 2, 4, 8, 16}. This holds by induction on the exponent: 2^0 = 1 is in the set, and if 2^k is one of 1, 2, 4, 8, 16 then 2^{k+1} = 2·2^k is again in the set because 2·1 = 2, 2·2 = 4, 2·4 = 8, 2·8 = 16, and 2·16 = 32 ≡ 8 (mod 24).",
"... | correct | null | null | |
lean_workbook_hf_7126 | 0 | 62ed6109590e6e95a482ae347b66550aab16d78561c7b0bff9bba74469a0a90d | Lean-Workbook | lean_workbook_hf | iff_logic | theorem lean_workbook_hf_7126 (x : ℝ) : ((x + 1) * (x + 2) + 1) * ((x + 1) * (x + 2) + 2) = 2 ↔ x = -1 ∨ x = -2 | We want to show that the equation ((x+1)(x+2)+1)((x+1)(x+2)+2) = 2 holds exactly when x = -1 or x = -2. We prove both directions of this equivalence. | [
{
"index": 1,
"text": "We split the proof into the forward direction (the equation implies x = -1 or x = -2) and the backward direction (x = -1 or x = -2 implies the equation).",
"lean_code": " constructor",
"label": "correct"
},
{
"index": 2,
"text": "For the forward direction, assume ... | correct | null | null | |
numina_4bc83d95-c62a-5d21-962f-9e270a15a5e2 | 0 | 6bc4113a9f965cf7a02384e3f3a860fe8f6375f3fde2a63bd34f30a35669dbb4 | NuminaMath-LEAN | numina | Number Theory | theorem number_theory_232106 :
∀ N : ℕ, ∃ k > N, Odd k ∧ ∀ n : ℕ, n ≥ 1 → ¬Prime (2 ^ (2 ^ n) + k) := by | We must show that for every natural number N, there exists an odd number k greater than N such that for every n ≥ 1, the number 2^(2^n) + k is not prime. | [
{
"index": 1,
"text": "Fix an arbitrary natural number N; we must exhibit a suitable k for this N.",
"lean_code": " intro N",
"label": "correct"
},
{
"index": 2,
"text": "Choose k = 6N + 5. It remains to verify three things: that this k exceeds N, that it is odd, and that 2^(2^n)+k is n... | correct | null | null | |
lean_workbook_hf_41903 | 0 | 07b591201a5ce85c1c78aa891300c03a50b20e3849b3241fe27215b62db5d6e3 | Lean-Workbook | lean_workbook_hf | equality_identity | theorem lean_workbook_hf_41903 (a b c : ℝ) (h : a = b) (h' : b = c) (h'' : c = a) : (a - b) / (1 - a * b) * a / (1 - a ^ 2) + (b - c) / (1 - b * c) * b / (1 - b ^ 2) + (c - a) / (1 - c * a) * c / (1 - c ^ 2) = 0 | We are given three real numbers a, b, c satisfying a = b, b = c, and c = a, and must show that a certain sum of three fractional expressions built from their pairwise differences equals zero. | [
{
"index": 1,
"text": "Since a equals b, we may identify b with a throughout the statement, replacing every occurrence of b by a in the goal and in the remaining assumptions.",
"lean_code": " subst h",
"label": "correct"
},
{
"index": 2,
"text": "The equality b = c, after the previous i... | correct | null | null | |
lean_workbook_hf_17543 | 0 | d3a92d98be63cf69d05c9ffa89d3b8b42ab7e54e206af847276c12acfb9877af | Lean-Workbook | lean_workbook_hf | equality_identity | theorem lean_workbook_hf_17543 (x y z : ℝ) (hx : 0 < x) (hy : 0 < y) (hz : 0 < z) : (x ^ 2 + y ^ 2)/z ^ 2 + (z ^ 2 + y ^ 2)/x ^ 2 + (z ^ 2 + x ^ 2)/y ^ 2 >= 6 | We must show that for positive reals x, y, z, the sum (x²+y²)/z² + (z²+y²)/x² + (z²+x²)/y² is at least 6. Our strategy is to expand this expression as three pairs of the form a/b + b/a, each of which is bounded below by 2, and then add these three bounds together. | [
{
"index": 1,
"text": "Since x is positive, it is nonzero.",
"lean_code": " have hxne : x ≠ 0 := ne_of_gt hx",
"label": "correct"
},
{
"index": 2,
"text": "Since y is positive, it is nonzero.",
"lean_code": " have hyne : y ≠ 0 := ne_of_gt hy",
"label": "correct"
},
{
"i... | correct | null | null | |
lean_workbook_hf_33941_18148 | 0 | dca94e5ebca77cb4de6a5dc12def4b47b6a48e0d7ceb5f2c5d98a9a2ea855889 | Lean-Workbook | lean_workbook_hf | existence_construction | theorem lean_workbook_hf_33941_18148 (f : ℝ → ℝ) (hf: f (x ^ 2 + 1) = f x ^ 2 + 1 ∧ f (x + y) = f x + f y) : ∃ f : ℝ → ℝ, f x = x | We must show that there exists a function f from the reals to the reals satisfying f(x) = x for the fixed real number x under consideration. The hypothesis about the original f involving the functional equations is not needed for this existential claim, since we are free to choose any function we like to witness the ex... | [
{
"index": 1,
"text": "We choose the identity function, mapping every real number to itself, as our witness for the existential statement, and it remains to verify that this function applied to x equals x.",
"lean_code": " refine ⟨fun x => x, ?_⟩",
"label": "correct"
},
{
"index": 2,
"t... | correct | null | null | |
lean_workbook_hf_37174 | 0 | 369adf089b3523a414057bdf7c7fa607d9b0d9906fc324c84353196c543d7ad4 | Lean-Workbook | lean_workbook_hf | inequality | theorem lean_workbook_hf_37174 (a b c : ℝ) (ha : 0 < a) (hb : 0 < b) (hc : 0 < c) : (a / (b + c) + b / (a + c) + c / (b + a)) ≥ 3 / 2 | We must show that for positive reals a, b, c the sum a/(b+c) + b/(a+c) + c/(b+a) is at least 3/2. This is Nesbitt's inequality. | [
{
"index": 1,
"text": "Since b and c are both positive, their sum b + c is positive.",
"lean_code": " have hbc : 0 < b + c := by linarith",
"label": "correct"
},
{
"index": 2,
"text": "Since a and c are both positive, their sum a + c is positive.",
"lean_code": " have hac : 0 < a +... | correct | null | null | |
lean_workbook_hf_72337 | 0 | cd75b99cce3c3366a8b62ae224ab05793bb978bca02bc2788a574119fd3106c3 | Lean-Workbook | lean_workbook_hf | iff_logic | theorem lean_workbook_hf_72337 : ∀ x y z : ℝ, x ^ 2 + y ^ 2 + z ^ 2 = 2 → x + y + z ≤ x*y*z + 2 | We must show that for all real x, y, z satisfying x² + y² + z² = 2, we have x + y + z ≤ xyz + 2. | [
{
"index": 1,
"text": "Fix real numbers x, y, z and assume the constraint x² + y² + z² = 2; we must derive x + y + z ≤ xyz + 2.",
"lean_code": " intro x y z h",
"label": "correct"
},
{
"index": 2,
"text": "The square of the quantity (x−1)(y−1)(z−1) is nonnegative, since any real square ... | correct | null | null | |
lean_workbook_hf_70871 | 0 | df4dcc3090c04d5fec23c0136c906e55472a3a357f73ab55b6915c0994e3a741 | Lean-Workbook | lean_workbook_hf | iff_logic | theorem lean_workbook_hf_70871 (a b c : ℝ) (ha : 0 < a) (hb : 0 < b) (hc : 0 < c) (habc : a * b * c = 4) (h : a + b + c = 6) : (a ^ (1/3) + b ^ (1/2) + c ^ (1/2) ≥ 2 + 4 ^ (1/3)) ∧ (∀ k : ℝ, k >= 1 → a ^ (1/3) + k * b ^ (1/2) + k * c ^ (1/2) ≥ 2 + k * 4 ^ (1/3)) | We must prove two facts: first, that a^(1/3) + b^(1/2) + c^(1/2) ≥ 2 + 4^(1/3), and second, that for every real number k ≥ 1, a^(1/3) + k·b^(1/2) + k·c^(1/2) ≥ 2 + k·4^(1/3), given that a, b, c are positive reals with abc = 4 and a+b+c = 6. | [
{
"index": 1,
"text": "The exponent 1/3 appearing in the statement is a division of the natural numbers 1 and 3. Since natural number division truncates, dividing 1 by 3 gives quotient 0, so 1/3 equals 0 as a natural number.",
"lean_code": " have e1 : (1 / 3 : ℕ) = 0 := by norm_num",
"label": "corr... | correct | null | null | |
lean_workbook_hf_37354_18891 | 0 | 87b23b933465d143941f779fb5dcc7b62b062c16b2d7f6867a6f65b91fc3d71f | Lean-Workbook | lean_workbook_hf | iff_logic | theorem lean_workbook_hf_37354_18891 (f : ℝ → ℝ) (g : ℝ → ℝ) (h₁ : ∀ x, g x = Real.exp (f x)) (h₂ : ∀ α : ℝ, 0 < α → ∀ x : ℝ, x > α → 4 * g x ≤ (g (x - α) + g α) ^ 2) : ∀ α : ℝ, 0 < α → ∀ x : ℝ, x > α → 4 * Real.exp (f x) ≤ (Real.exp (f (x - α)) + Real.exp (f α)) ^ 2 | We want to show that for every positive real α and every real x greater than α, the inequality 4·exp(f(x)) ≤ (exp(f(x−α)) + exp(f(α)))² holds, using the given facts that g(t) = exp(f(t)) for all t and that 4·g(x) ≤ (g(x−α) + g(α))² whenever α is positive and x > α. | [
{
"index": 1,
"text": "Fix an arbitrary positive real number α and an arbitrary real number x satisfying x > α; we must establish the inequality for this particular α and x.",
"lean_code": " intro α hα x hx",
"label": "correct"
},
{
"index": 2,
"text": "Apply the hypothesis that 4·g(x) ... | correct | null | null |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.