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Type | Label | Description |
---|---|---|
Statement | ||
Theorem | dfphi2 16101* | Alternate definition of the Euler ϕ function. (Contributed by Mario Carneiro, 23-Feb-2014.) (Revised by Mario Carneiro, 2-May-2016.) |
⊢ (𝑁 ∈ ℕ → (ϕ‘𝑁) = (♯‘{𝑥 ∈ (0..^𝑁) ∣ (𝑥 gcd 𝑁) = 1})) | ||
Theorem | hashdvds 16102* | The number of numbers in a given residue class in a finite set of integers. (Contributed by Mario Carneiro, 12-Mar-2014.) (Proof shortened by Mario Carneiro, 7-Jun-2016.) |
⊢ (𝜑 → 𝑁 ∈ ℕ) & ⊢ (𝜑 → 𝐴 ∈ ℤ) & ⊢ (𝜑 → 𝐵 ∈ (ℤ≥‘(𝐴 − 1))) & ⊢ (𝜑 → 𝐶 ∈ ℤ) ⇒ ⊢ (𝜑 → (♯‘{𝑥 ∈ (𝐴...𝐵) ∣ 𝑁 ∥ (𝑥 − 𝐶)}) = ((⌊‘((𝐵 − 𝐶) / 𝑁)) − (⌊‘(((𝐴 − 1) − 𝐶) / 𝑁)))) | ||
Theorem | phiprmpw 16103 | Value of the Euler ϕ function at a prime power. Theorem 2.5(a) in [ApostolNT] p. 28. (Contributed by Mario Carneiro, 24-Feb-2014.) |
⊢ ((𝑃 ∈ ℙ ∧ 𝐾 ∈ ℕ) → (ϕ‘(𝑃↑𝐾)) = ((𝑃↑(𝐾 − 1)) · (𝑃 − 1))) | ||
Theorem | phiprm 16104 | Value of the Euler ϕ function at a prime. (Contributed by Mario Carneiro, 28-Feb-2014.) |
⊢ (𝑃 ∈ ℙ → (ϕ‘𝑃) = (𝑃 − 1)) | ||
Theorem | crth 16105* | The Chinese Remainder Theorem: the function that maps 𝑥 to its remainder classes mod 𝑀 and mod 𝑁 is 1-1 and onto when 𝑀 and 𝑁 are coprime. (Contributed by Mario Carneiro, 24-Feb-2014.) (Proof shortened by Mario Carneiro, 2-May-2016.) |
⊢ 𝑆 = (0..^(𝑀 · 𝑁)) & ⊢ 𝑇 = ((0..^𝑀) × (0..^𝑁)) & ⊢ 𝐹 = (𝑥 ∈ 𝑆 ↦ 〈(𝑥 mod 𝑀), (𝑥 mod 𝑁)〉) & ⊢ (𝜑 → (𝑀 ∈ ℕ ∧ 𝑁 ∈ ℕ ∧ (𝑀 gcd 𝑁) = 1)) ⇒ ⊢ (𝜑 → 𝐹:𝑆–1-1-onto→𝑇) | ||
Theorem | phimullem 16106* | Lemma for phimul 16107. (Contributed by Mario Carneiro, 24-Feb-2014.) |
⊢ 𝑆 = (0..^(𝑀 · 𝑁)) & ⊢ 𝑇 = ((0..^𝑀) × (0..^𝑁)) & ⊢ 𝐹 = (𝑥 ∈ 𝑆 ↦ 〈(𝑥 mod 𝑀), (𝑥 mod 𝑁)〉) & ⊢ (𝜑 → (𝑀 ∈ ℕ ∧ 𝑁 ∈ ℕ ∧ (𝑀 gcd 𝑁) = 1)) & ⊢ 𝑈 = {𝑦 ∈ (0..^𝑀) ∣ (𝑦 gcd 𝑀) = 1} & ⊢ 𝑉 = {𝑦 ∈ (0..^𝑁) ∣ (𝑦 gcd 𝑁) = 1} & ⊢ 𝑊 = {𝑦 ∈ 𝑆 ∣ (𝑦 gcd (𝑀 · 𝑁)) = 1} ⇒ ⊢ (𝜑 → (ϕ‘(𝑀 · 𝑁)) = ((ϕ‘𝑀) · (ϕ‘𝑁))) | ||
Theorem | phimul 16107 | The Euler ϕ function is a multiplicative function, meaning that it distributes over multiplication at relatively prime arguments. Theorem 2.5(c) in [ApostolNT] p. 28. (Contributed by Mario Carneiro, 24-Feb-2014.) |
⊢ ((𝑀 ∈ ℕ ∧ 𝑁 ∈ ℕ ∧ (𝑀 gcd 𝑁) = 1) → (ϕ‘(𝑀 · 𝑁)) = ((ϕ‘𝑀) · (ϕ‘𝑁))) | ||
Theorem | eulerthlem1 16108* | Lemma for eulerth 16110. (Contributed by Mario Carneiro, 8-May-2015.) |
⊢ (𝜑 → (𝑁 ∈ ℕ ∧ 𝐴 ∈ ℤ ∧ (𝐴 gcd 𝑁) = 1)) & ⊢ 𝑆 = {𝑦 ∈ (0..^𝑁) ∣ (𝑦 gcd 𝑁) = 1} & ⊢ 𝑇 = (1...(ϕ‘𝑁)) & ⊢ (𝜑 → 𝐹:𝑇–1-1-onto→𝑆) & ⊢ 𝐺 = (𝑥 ∈ 𝑇 ↦ ((𝐴 · (𝐹‘𝑥)) mod 𝑁)) ⇒ ⊢ (𝜑 → 𝐺:𝑇⟶𝑆) | ||
Theorem | eulerthlem2 16109* | Lemma for eulerth 16110. (Contributed by Mario Carneiro, 28-Feb-2014.) |
⊢ (𝜑 → (𝑁 ∈ ℕ ∧ 𝐴 ∈ ℤ ∧ (𝐴 gcd 𝑁) = 1)) & ⊢ 𝑆 = {𝑦 ∈ (0..^𝑁) ∣ (𝑦 gcd 𝑁) = 1} & ⊢ 𝑇 = (1...(ϕ‘𝑁)) & ⊢ (𝜑 → 𝐹:𝑇–1-1-onto→𝑆) & ⊢ 𝐺 = (𝑥 ∈ 𝑇 ↦ ((𝐴 · (𝐹‘𝑥)) mod 𝑁)) ⇒ ⊢ (𝜑 → ((𝐴↑(ϕ‘𝑁)) mod 𝑁) = (1 mod 𝑁)) | ||
Theorem | eulerth 16110 | Euler's theorem, a generalization of Fermat's little theorem. If 𝐴 and 𝑁 are coprime, then 𝐴↑ϕ(𝑁)≡1 (mod 𝑁). This is Metamath 100 proof #10. Also called Euler-Fermat theorem, see theorem 5.17 in [ApostolNT] p. 113. (Contributed by Mario Carneiro, 28-Feb-2014.) |
⊢ ((𝑁 ∈ ℕ ∧ 𝐴 ∈ ℤ ∧ (𝐴 gcd 𝑁) = 1) → ((𝐴↑(ϕ‘𝑁)) mod 𝑁) = (1 mod 𝑁)) | ||
Theorem | fermltl 16111 | Fermat's little theorem. When 𝑃 is prime, 𝐴↑𝑃≡𝐴 (mod 𝑃) for any 𝐴, see theorem 5.19 in [ApostolNT] p. 114. (Contributed by Mario Carneiro, 28-Feb-2014.) (Proof shortened by AV, 19-Mar-2022.) |
⊢ ((𝑃 ∈ ℙ ∧ 𝐴 ∈ ℤ) → ((𝐴↑𝑃) mod 𝑃) = (𝐴 mod 𝑃)) | ||
Theorem | prmdiv 16112 | Show an explicit expression for the modular inverse of 𝐴 mod 𝑃. (Contributed by Mario Carneiro, 24-Jan-2015.) |
⊢ 𝑅 = ((𝐴↑(𝑃 − 2)) mod 𝑃) ⇒ ⊢ ((𝑃 ∈ ℙ ∧ 𝐴 ∈ ℤ ∧ ¬ 𝑃 ∥ 𝐴) → (𝑅 ∈ (1...(𝑃 − 1)) ∧ 𝑃 ∥ ((𝐴 · 𝑅) − 1))) | ||
Theorem | prmdiveq 16113 | The modular inverse of 𝐴 mod 𝑃 is unique. (Contributed by Mario Carneiro, 24-Jan-2015.) |
⊢ 𝑅 = ((𝐴↑(𝑃 − 2)) mod 𝑃) ⇒ ⊢ ((𝑃 ∈ ℙ ∧ 𝐴 ∈ ℤ ∧ ¬ 𝑃 ∥ 𝐴) → ((𝑆 ∈ (0...(𝑃 − 1)) ∧ 𝑃 ∥ ((𝐴 · 𝑆) − 1)) ↔ 𝑆 = 𝑅)) | ||
Theorem | prmdivdiv 16114 | The (modular) inverse of the inverse of a number is itself. (Contributed by Mario Carneiro, 24-Jan-2015.) |
⊢ 𝑅 = ((𝐴↑(𝑃 − 2)) mod 𝑃) ⇒ ⊢ ((𝑃 ∈ ℙ ∧ 𝐴 ∈ (1...(𝑃 − 1))) → 𝐴 = ((𝑅↑(𝑃 − 2)) mod 𝑃)) | ||
Theorem | hashgcdlem 16115* | A correspondence between elements of specific GCD and relative primes in a smaller ring. (Contributed by Stefan O'Rear, 12-Sep-2015.) |
⊢ 𝐴 = {𝑦 ∈ (0..^(𝑀 / 𝑁)) ∣ (𝑦 gcd (𝑀 / 𝑁)) = 1} & ⊢ 𝐵 = {𝑧 ∈ (0..^𝑀) ∣ (𝑧 gcd 𝑀) = 𝑁} & ⊢ 𝐹 = (𝑥 ∈ 𝐴 ↦ (𝑥 · 𝑁)) ⇒ ⊢ ((𝑀 ∈ ℕ ∧ 𝑁 ∈ ℕ ∧ 𝑁 ∥ 𝑀) → 𝐹:𝐴–1-1-onto→𝐵) | ||
Theorem | hashgcdeq 16116* | Number of initial positive integers with specified divisors. (Contributed by Stefan O'Rear, 12-Sep-2015.) |
⊢ ((𝑀 ∈ ℕ ∧ 𝑁 ∈ ℕ) → (♯‘{𝑥 ∈ (0..^𝑀) ∣ (𝑥 gcd 𝑀) = 𝑁}) = if(𝑁 ∥ 𝑀, (ϕ‘(𝑀 / 𝑁)), 0)) | ||
Theorem | phisum 16117* | The divisor sum identity of the totient function. Theorem 2.2 in [ApostolNT] p. 26. (Contributed by Stefan O'Rear, 12-Sep-2015.) |
⊢ (𝑁 ∈ ℕ → Σ𝑑 ∈ {𝑥 ∈ ℕ ∣ 𝑥 ∥ 𝑁} (ϕ‘𝑑) = 𝑁) | ||
Theorem | odzval 16118* | Value of the order function. This is a function of functions; the inner argument selects the base (i.e., mod 𝑁 for some 𝑁, often prime) and the outer argument selects the integer or equivalence class (if you want to think about it that way) from the integers mod 𝑁. In order to ensure the supremum is well-defined, we only define the expression when 𝐴 and 𝑁 are coprime. (Contributed by Mario Carneiro, 23-Feb-2014.) (Revised by AV, 26-Sep-2020.) |
⊢ ((𝑁 ∈ ℕ ∧ 𝐴 ∈ ℤ ∧ (𝐴 gcd 𝑁) = 1) → ((odℤ‘𝑁)‘𝐴) = inf({𝑛 ∈ ℕ ∣ 𝑁 ∥ ((𝐴↑𝑛) − 1)}, ℝ, < )) | ||
Theorem | odzcllem 16119 | - Lemma for odzcl 16120, showing existence of a recurrent point for the exponential. (Contributed by Mario Carneiro, 28-Feb-2014.) (Proof shortened by AV, 26-Sep-2020.) |
⊢ ((𝑁 ∈ ℕ ∧ 𝐴 ∈ ℤ ∧ (𝐴 gcd 𝑁) = 1) → (((odℤ‘𝑁)‘𝐴) ∈ ℕ ∧ 𝑁 ∥ ((𝐴↑((odℤ‘𝑁)‘𝐴)) − 1))) | ||
Theorem | odzcl 16120 | The order of a group element is an integer. (Contributed by Mario Carneiro, 28-Feb-2014.) |
⊢ ((𝑁 ∈ ℕ ∧ 𝐴 ∈ ℤ ∧ (𝐴 gcd 𝑁) = 1) → ((odℤ‘𝑁)‘𝐴) ∈ ℕ) | ||
Theorem | odzid 16121 | Any element raised to the power of its order is 1. (Contributed by Mario Carneiro, 28-Feb-2014.) |
⊢ ((𝑁 ∈ ℕ ∧ 𝐴 ∈ ℤ ∧ (𝐴 gcd 𝑁) = 1) → 𝑁 ∥ ((𝐴↑((odℤ‘𝑁)‘𝐴)) − 1)) | ||
Theorem | odzdvds 16122 | The only powers of 𝐴 that are congruent to 1 are the multiples of the order of 𝐴. (Contributed by Mario Carneiro, 28-Feb-2014.) (Proof shortened by AV, 26-Sep-2020.) |
⊢ (((𝑁 ∈ ℕ ∧ 𝐴 ∈ ℤ ∧ (𝐴 gcd 𝑁) = 1) ∧ 𝐾 ∈ ℕ0) → (𝑁 ∥ ((𝐴↑𝐾) − 1) ↔ ((odℤ‘𝑁)‘𝐴) ∥ 𝐾)) | ||
Theorem | odzphi 16123 | The order of any group element is a divisor of the Euler ϕ function. (Contributed by Mario Carneiro, 28-Feb-2014.) |
⊢ ((𝑁 ∈ ℕ ∧ 𝐴 ∈ ℤ ∧ (𝐴 gcd 𝑁) = 1) → ((odℤ‘𝑁)‘𝐴) ∥ (ϕ‘𝑁)) | ||
Theorem | modprm1div 16124 | A prime number divides an integer minus 1 iff the integer modulo the prime number is 1. (Contributed by Alexander van der Vekens, 17-May-2018.) (Proof shortened by AV, 30-May-2023.) |
⊢ ((𝑃 ∈ ℙ ∧ 𝐴 ∈ ℤ) → ((𝐴 mod 𝑃) = 1 ↔ 𝑃 ∥ (𝐴 − 1))) | ||
Theorem | m1dvdsndvds 16125 | If an integer minus 1 is divisible by a prime number, the integer itself is not divisible by this prime number. (Contributed by Alexander van der Vekens, 30-Aug-2018.) |
⊢ ((𝑃 ∈ ℙ ∧ 𝐴 ∈ ℤ) → (𝑃 ∥ (𝐴 − 1) → ¬ 𝑃 ∥ 𝐴)) | ||
Theorem | modprminv 16126 | Show an explicit expression for the modular inverse of 𝐴 mod 𝑃. This is an application of prmdiv 16112. (Contributed by Alexander van der Vekens, 15-May-2018.) |
⊢ 𝑅 = ((𝐴↑(𝑃 − 2)) mod 𝑃) ⇒ ⊢ ((𝑃 ∈ ℙ ∧ 𝐴 ∈ ℤ ∧ ¬ 𝑃 ∥ 𝐴) → (𝑅 ∈ (1...(𝑃 − 1)) ∧ ((𝐴 · 𝑅) mod 𝑃) = 1)) | ||
Theorem | modprminveq 16127 | The modular inverse of 𝐴 mod 𝑃 is unique. (Contributed by Alexander van der Vekens, 17-May-2018.) |
⊢ 𝑅 = ((𝐴↑(𝑃 − 2)) mod 𝑃) ⇒ ⊢ ((𝑃 ∈ ℙ ∧ 𝐴 ∈ ℤ ∧ ¬ 𝑃 ∥ 𝐴) → ((𝑆 ∈ (0...(𝑃 − 1)) ∧ ((𝐴 · 𝑆) mod 𝑃) = 1) ↔ 𝑆 = 𝑅)) | ||
Theorem | vfermltl 16128 | Variant of Fermat's little theorem if 𝐴 is not a multiple of 𝑃, see theorem 5.18 in [ApostolNT] p. 113. (Contributed by AV, 21-Aug-2020.) (Proof shortened by AV, 5-Sep-2020.) |
⊢ ((𝑃 ∈ ℙ ∧ 𝐴 ∈ ℤ ∧ ¬ 𝑃 ∥ 𝐴) → ((𝐴↑(𝑃 − 1)) mod 𝑃) = 1) | ||
Theorem | vfermltlALT 16129 | Alternate proof of vfermltl 16128, not using Euler's theorem. (Contributed by AV, 21-Aug-2020.) (New usage is discouraged.) (Proof modification is discouraged.) |
⊢ ((𝑃 ∈ ℙ ∧ 𝐴 ∈ ℤ ∧ ¬ 𝑃 ∥ 𝐴) → ((𝐴↑(𝑃 − 1)) mod 𝑃) = 1) | ||
Theorem | powm2modprm 16130 | If an integer minus 1 is divisible by a prime number, then the integer to the power of the prime number minus 2 is 1 modulo the prime number. (Contributed by Alexander van der Vekens, 30-Aug-2018.) |
⊢ ((𝑃 ∈ ℙ ∧ 𝐴 ∈ ℤ) → (𝑃 ∥ (𝐴 − 1) → ((𝐴↑(𝑃 − 2)) mod 𝑃) = 1)) | ||
Theorem | reumodprminv 16131* | For any prime number and for any positive integer less than this prime number, there is a unique modular inverse of this positive integer. (Contributed by Alexander van der Vekens, 12-May-2018.) |
⊢ ((𝑃 ∈ ℙ ∧ 𝑁 ∈ (1..^𝑃)) → ∃!𝑖 ∈ (1...(𝑃 − 1))((𝑁 · 𝑖) mod 𝑃) = 1) | ||
Theorem | modprm0 16132* | For two positive integers less than a given prime number there is always a nonnegative integer (less than the given prime number) so that the sum of one of the two positive integers and the other of the positive integers multiplied by the nonnegative integer is 0 ( modulo the given prime number). (Contributed by Alexander van der Vekens, 17-May-2018.) |
⊢ ((𝑃 ∈ ℙ ∧ 𝑁 ∈ (1..^𝑃) ∧ 𝐼 ∈ (1..^𝑃)) → ∃𝑗 ∈ (0..^𝑃)((𝐼 + (𝑗 · 𝑁)) mod 𝑃) = 0) | ||
Theorem | nnnn0modprm0 16133* | For a positive integer and a nonnegative integer both less than a given prime number there is always a second nonnegative integer (less than the given prime number) so that the sum of this second nonnegative integer multiplied with the positive integer and the first nonnegative integer is 0 ( modulo the given prime number). (Contributed by Alexander van der Vekens, 8-Nov-2018.) |
⊢ ((𝑃 ∈ ℙ ∧ 𝑁 ∈ (1..^𝑃) ∧ 𝐼 ∈ (0..^𝑃)) → ∃𝑗 ∈ (0..^𝑃)((𝐼 + (𝑗 · 𝑁)) mod 𝑃) = 0) | ||
Theorem | modprmn0modprm0 16134* | For an integer not being 0 modulo a given prime number and a nonnegative integer less than the prime number, there is always a second nonnegative integer (less than the given prime number) so that the sum of this second nonnegative integer multiplied with the integer and the first nonnegative integer is 0 ( modulo the given prime number). (Contributed by Alexander van der Vekens, 10-Nov-2018.) |
⊢ ((𝑃 ∈ ℙ ∧ 𝑁 ∈ ℤ ∧ (𝑁 mod 𝑃) ≠ 0) → (𝐼 ∈ (0..^𝑃) → ∃𝑗 ∈ (0..^𝑃)((𝐼 + (𝑗 · 𝑁)) mod 𝑃) = 0)) | ||
Theorem | coprimeprodsq 16135 | If three numbers are coprime, and the square of one is the product of the other two, then there is a formula for the other two in terms of gcd and square. (Contributed by Scott Fenton, 2-Apr-2014.) (Revised by Mario Carneiro, 19-Apr-2014.) |
⊢ (((𝐴 ∈ ℕ0 ∧ 𝐵 ∈ ℤ ∧ 𝐶 ∈ ℕ0) ∧ ((𝐴 gcd 𝐵) gcd 𝐶) = 1) → ((𝐶↑2) = (𝐴 · 𝐵) → 𝐴 = ((𝐴 gcd 𝐶)↑2))) | ||
Theorem | coprimeprodsq2 16136 | If three numbers are coprime, and the square of one is the product of the other two, then there is a formula for the other two in terms of gcd and square. (Contributed by Scott Fenton, 17-Apr-2014.) (Revised by Mario Carneiro, 19-Apr-2014.) |
⊢ (((𝐴 ∈ ℤ ∧ 𝐵 ∈ ℕ0 ∧ 𝐶 ∈ ℕ0) ∧ ((𝐴 gcd 𝐵) gcd 𝐶) = 1) → ((𝐶↑2) = (𝐴 · 𝐵) → 𝐵 = ((𝐵 gcd 𝐶)↑2))) | ||
Theorem | oddprm 16137 | A prime not equal to 2 is odd. (Contributed by Mario Carneiro, 4-Feb-2015.) (Proof shortened by AV, 10-Jul-2022.) |
⊢ (𝑁 ∈ (ℙ ∖ {2}) → ((𝑁 − 1) / 2) ∈ ℕ) | ||
Theorem | nnoddn2prm 16138 | A prime not equal to 2 is an odd positive integer. (Contributed by AV, 28-Jun-2021.) |
⊢ (𝑁 ∈ (ℙ ∖ {2}) → (𝑁 ∈ ℕ ∧ ¬ 2 ∥ 𝑁)) | ||
Theorem | oddn2prm 16139 | A prime not equal to 2 is odd. (Contributed by AV, 28-Jun-2021.) |
⊢ (𝑁 ∈ (ℙ ∖ {2}) → ¬ 2 ∥ 𝑁) | ||
Theorem | nnoddn2prmb 16140 | A number is a prime number not equal to 2 iff it is an odd prime number. Conversion theorem for two representations of odd primes. (Contributed by AV, 14-Jul-2021.) |
⊢ (𝑁 ∈ (ℙ ∖ {2}) ↔ (𝑁 ∈ ℙ ∧ ¬ 2 ∥ 𝑁)) | ||
Theorem | prm23lt5 16141 | A prime less than 5 is either 2 or 3. (Contributed by AV, 5-Jul-2021.) |
⊢ ((𝑃 ∈ ℙ ∧ 𝑃 < 5) → (𝑃 = 2 ∨ 𝑃 = 3)) | ||
Theorem | prm23ge5 16142 | A prime is either 2 or 3 or greater than or equal to 5. (Contributed by AV, 5-Jul-2021.) |
⊢ (𝑃 ∈ ℙ → (𝑃 = 2 ∨ 𝑃 = 3 ∨ 𝑃 ∈ (ℤ≥‘5))) | ||
Theorem | pythagtriplem1 16143* | Lemma for pythagtrip 16161. Prove a weaker version of one direction of the theorem. (Contributed by Scott Fenton, 28-Mar-2014.) (Revised by Mario Carneiro, 19-Apr-2014.) |
⊢ (∃𝑛 ∈ ℕ ∃𝑚 ∈ ℕ ∃𝑘 ∈ ℕ (𝐴 = (𝑘 · ((𝑚↑2) − (𝑛↑2))) ∧ 𝐵 = (𝑘 · (2 · (𝑚 · 𝑛))) ∧ 𝐶 = (𝑘 · ((𝑚↑2) + (𝑛↑2)))) → ((𝐴↑2) + (𝐵↑2)) = (𝐶↑2)) | ||
Theorem | pythagtriplem2 16144* | Lemma for pythagtrip 16161. Prove the full version of one direction of the theorem. (Contributed by Scott Fenton, 28-Mar-2014.) (Revised by Mario Carneiro, 19-Apr-2014.) |
⊢ ((𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ) → (∃𝑛 ∈ ℕ ∃𝑚 ∈ ℕ ∃𝑘 ∈ ℕ ({𝐴, 𝐵} = {(𝑘 · ((𝑚↑2) − (𝑛↑2))), (𝑘 · (2 · (𝑚 · 𝑛)))} ∧ 𝐶 = (𝑘 · ((𝑚↑2) + (𝑛↑2)))) → ((𝐴↑2) + (𝐵↑2)) = (𝐶↑2))) | ||
Theorem | pythagtriplem3 16145 | Lemma for pythagtrip 16161. Show that 𝐶 and 𝐵 are relatively prime under some conditions. (Contributed by Scott Fenton, 8-Apr-2014.) (Revised by Mario Carneiro, 19-Apr-2014.) |
⊢ (((𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝐶 ∈ ℕ) ∧ ((𝐴↑2) + (𝐵↑2)) = (𝐶↑2) ∧ ((𝐴 gcd 𝐵) = 1 ∧ ¬ 2 ∥ 𝐴)) → (𝐵 gcd 𝐶) = 1) | ||
Theorem | pythagtriplem4 16146 | Lemma for pythagtrip 16161. Show that 𝐶 − 𝐵 and 𝐶 + 𝐵 are relatively prime. (Contributed by Scott Fenton, 12-Apr-2014.) (Revised by Mario Carneiro, 19-Apr-2014.) |
⊢ (((𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝐶 ∈ ℕ) ∧ ((𝐴↑2) + (𝐵↑2)) = (𝐶↑2) ∧ ((𝐴 gcd 𝐵) = 1 ∧ ¬ 2 ∥ 𝐴)) → ((𝐶 − 𝐵) gcd (𝐶 + 𝐵)) = 1) | ||
Theorem | pythagtriplem10 16147 | Lemma for pythagtrip 16161. Show that 𝐶 − 𝐵 is positive. (Contributed by Scott Fenton, 17-Apr-2014.) (Revised by Mario Carneiro, 19-Apr-2014.) |
⊢ (((𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝐶 ∈ ℕ) ∧ ((𝐴↑2) + (𝐵↑2)) = (𝐶↑2)) → 0 < (𝐶 − 𝐵)) | ||
Theorem | pythagtriplem6 16148 | Lemma for pythagtrip 16161. Calculate (√‘(𝐶 − 𝐵)). (Contributed by Scott Fenton, 18-Apr-2014.) (Revised by Mario Carneiro, 19-Apr-2014.) |
⊢ (((𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝐶 ∈ ℕ) ∧ ((𝐴↑2) + (𝐵↑2)) = (𝐶↑2) ∧ ((𝐴 gcd 𝐵) = 1 ∧ ¬ 2 ∥ 𝐴)) → (√‘(𝐶 − 𝐵)) = ((𝐶 − 𝐵) gcd 𝐴)) | ||
Theorem | pythagtriplem7 16149 | Lemma for pythagtrip 16161. Calculate (√‘(𝐶 + 𝐵)). (Contributed by Scott Fenton, 18-Apr-2014.) (Revised by Mario Carneiro, 19-Apr-2014.) |
⊢ (((𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝐶 ∈ ℕ) ∧ ((𝐴↑2) + (𝐵↑2)) = (𝐶↑2) ∧ ((𝐴 gcd 𝐵) = 1 ∧ ¬ 2 ∥ 𝐴)) → (√‘(𝐶 + 𝐵)) = ((𝐶 + 𝐵) gcd 𝐴)) | ||
Theorem | pythagtriplem8 16150 | Lemma for pythagtrip 16161. Show that (√‘(𝐶 − 𝐵)) is a positive integer. (Contributed by Scott Fenton, 17-Apr-2014.) (Revised by Mario Carneiro, 19-Apr-2014.) |
⊢ (((𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝐶 ∈ ℕ) ∧ ((𝐴↑2) + (𝐵↑2)) = (𝐶↑2) ∧ ((𝐴 gcd 𝐵) = 1 ∧ ¬ 2 ∥ 𝐴)) → (√‘(𝐶 − 𝐵)) ∈ ℕ) | ||
Theorem | pythagtriplem9 16151 | Lemma for pythagtrip 16161. Show that (√‘(𝐶 + 𝐵)) is a positive integer. (Contributed by Scott Fenton, 17-Apr-2014.) (Revised by Mario Carneiro, 19-Apr-2014.) |
⊢ (((𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝐶 ∈ ℕ) ∧ ((𝐴↑2) + (𝐵↑2)) = (𝐶↑2) ∧ ((𝐴 gcd 𝐵) = 1 ∧ ¬ 2 ∥ 𝐴)) → (√‘(𝐶 + 𝐵)) ∈ ℕ) | ||
Theorem | pythagtriplem11 16152 | Lemma for pythagtrip 16161. Show that 𝑀 (which will eventually be closely related to the 𝑚 in the final statement) is a natural. (Contributed by Scott Fenton, 17-Apr-2014.) (Revised by Mario Carneiro, 19-Apr-2014.) |
⊢ 𝑀 = (((√‘(𝐶 + 𝐵)) + (√‘(𝐶 − 𝐵))) / 2) ⇒ ⊢ (((𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝐶 ∈ ℕ) ∧ ((𝐴↑2) + (𝐵↑2)) = (𝐶↑2) ∧ ((𝐴 gcd 𝐵) = 1 ∧ ¬ 2 ∥ 𝐴)) → 𝑀 ∈ ℕ) | ||
Theorem | pythagtriplem12 16153 | Lemma for pythagtrip 16161. Calculate the square of 𝑀. (Contributed by Scott Fenton, 17-Apr-2014.) (Revised by Mario Carneiro, 19-Apr-2014.) |
⊢ 𝑀 = (((√‘(𝐶 + 𝐵)) + (√‘(𝐶 − 𝐵))) / 2) ⇒ ⊢ (((𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝐶 ∈ ℕ) ∧ ((𝐴↑2) + (𝐵↑2)) = (𝐶↑2) ∧ ((𝐴 gcd 𝐵) = 1 ∧ ¬ 2 ∥ 𝐴)) → (𝑀↑2) = ((𝐶 + 𝐴) / 2)) | ||
Theorem | pythagtriplem13 16154 | Lemma for pythagtrip 16161. Show that 𝑁 (which will eventually be closely related to the 𝑛 in the final statement) is a natural. (Contributed by Scott Fenton, 17-Apr-2014.) (Revised by Mario Carneiro, 19-Apr-2014.) |
⊢ 𝑁 = (((√‘(𝐶 + 𝐵)) − (√‘(𝐶 − 𝐵))) / 2) ⇒ ⊢ (((𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝐶 ∈ ℕ) ∧ ((𝐴↑2) + (𝐵↑2)) = (𝐶↑2) ∧ ((𝐴 gcd 𝐵) = 1 ∧ ¬ 2 ∥ 𝐴)) → 𝑁 ∈ ℕ) | ||
Theorem | pythagtriplem14 16155 | Lemma for pythagtrip 16161. Calculate the square of 𝑁. (Contributed by Scott Fenton, 17-Apr-2014.) (Revised by Mario Carneiro, 19-Apr-2014.) |
⊢ 𝑁 = (((√‘(𝐶 + 𝐵)) − (√‘(𝐶 − 𝐵))) / 2) ⇒ ⊢ (((𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝐶 ∈ ℕ) ∧ ((𝐴↑2) + (𝐵↑2)) = (𝐶↑2) ∧ ((𝐴 gcd 𝐵) = 1 ∧ ¬ 2 ∥ 𝐴)) → (𝑁↑2) = ((𝐶 − 𝐴) / 2)) | ||
Theorem | pythagtriplem15 16156 | Lemma for pythagtrip 16161. Show the relationship between 𝑀, 𝑁, and 𝐴. (Contributed by Scott Fenton, 17-Apr-2014.) (Revised by Mario Carneiro, 19-Apr-2014.) |
⊢ 𝑀 = (((√‘(𝐶 + 𝐵)) + (√‘(𝐶 − 𝐵))) / 2) & ⊢ 𝑁 = (((√‘(𝐶 + 𝐵)) − (√‘(𝐶 − 𝐵))) / 2) ⇒ ⊢ (((𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝐶 ∈ ℕ) ∧ ((𝐴↑2) + (𝐵↑2)) = (𝐶↑2) ∧ ((𝐴 gcd 𝐵) = 1 ∧ ¬ 2 ∥ 𝐴)) → 𝐴 = ((𝑀↑2) − (𝑁↑2))) | ||
Theorem | pythagtriplem16 16157 | Lemma for pythagtrip 16161. Show the relationship between 𝑀, 𝑁, and 𝐵. (Contributed by Scott Fenton, 17-Apr-2014.) (Revised by Mario Carneiro, 19-Apr-2014.) |
⊢ 𝑀 = (((√‘(𝐶 + 𝐵)) + (√‘(𝐶 − 𝐵))) / 2) & ⊢ 𝑁 = (((√‘(𝐶 + 𝐵)) − (√‘(𝐶 − 𝐵))) / 2) ⇒ ⊢ (((𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝐶 ∈ ℕ) ∧ ((𝐴↑2) + (𝐵↑2)) = (𝐶↑2) ∧ ((𝐴 gcd 𝐵) = 1 ∧ ¬ 2 ∥ 𝐴)) → 𝐵 = (2 · (𝑀 · 𝑁))) | ||
Theorem | pythagtriplem17 16158 | Lemma for pythagtrip 16161. Show the relationship between 𝑀, 𝑁, and 𝐶. (Contributed by Scott Fenton, 17-Apr-2014.) (Revised by Mario Carneiro, 19-Apr-2014.) |
⊢ 𝑀 = (((√‘(𝐶 + 𝐵)) + (√‘(𝐶 − 𝐵))) / 2) & ⊢ 𝑁 = (((√‘(𝐶 + 𝐵)) − (√‘(𝐶 − 𝐵))) / 2) ⇒ ⊢ (((𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝐶 ∈ ℕ) ∧ ((𝐴↑2) + (𝐵↑2)) = (𝐶↑2) ∧ ((𝐴 gcd 𝐵) = 1 ∧ ¬ 2 ∥ 𝐴)) → 𝐶 = ((𝑀↑2) + (𝑁↑2))) | ||
Theorem | pythagtriplem18 16159* | Lemma for pythagtrip 16161. Wrap the previous 𝑀 and 𝑁 up in quantifiers. (Contributed by Scott Fenton, 18-Apr-2014.) (Revised by Mario Carneiro, 19-Apr-2014.) |
⊢ (((𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝐶 ∈ ℕ) ∧ ((𝐴↑2) + (𝐵↑2)) = (𝐶↑2) ∧ ((𝐴 gcd 𝐵) = 1 ∧ ¬ 2 ∥ 𝐴)) → ∃𝑛 ∈ ℕ ∃𝑚 ∈ ℕ (𝐴 = ((𝑚↑2) − (𝑛↑2)) ∧ 𝐵 = (2 · (𝑚 · 𝑛)) ∧ 𝐶 = ((𝑚↑2) + (𝑛↑2)))) | ||
Theorem | pythagtriplem19 16160* | Lemma for pythagtrip 16161. Introduce 𝑘 and remove the relative primality requirement. (Contributed by Scott Fenton, 18-Apr-2014.) (Revised by Mario Carneiro, 19-Apr-2014.) |
⊢ (((𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝐶 ∈ ℕ) ∧ ((𝐴↑2) + (𝐵↑2)) = (𝐶↑2) ∧ ¬ 2 ∥ (𝐴 / (𝐴 gcd 𝐵))) → ∃𝑛 ∈ ℕ ∃𝑚 ∈ ℕ ∃𝑘 ∈ ℕ (𝐴 = (𝑘 · ((𝑚↑2) − (𝑛↑2))) ∧ 𝐵 = (𝑘 · (2 · (𝑚 · 𝑛))) ∧ 𝐶 = (𝑘 · ((𝑚↑2) + (𝑛↑2))))) | ||
Theorem | pythagtrip 16161* | Parameterize the Pythagorean triples. If 𝐴, 𝐵, and 𝐶 are naturals, then they obey the Pythagorean triple formula iff they are parameterized by three naturals. This proof follows the Isabelle proof at http://afp.sourceforge.net/entries/Fermat3_4.shtml. This is Metamath 100 proof #23. (Contributed by Scott Fenton, 19-Apr-2014.) |
⊢ ((𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝐶 ∈ ℕ) → (((𝐴↑2) + (𝐵↑2)) = (𝐶↑2) ↔ ∃𝑛 ∈ ℕ ∃𝑚 ∈ ℕ ∃𝑘 ∈ ℕ ({𝐴, 𝐵} = {(𝑘 · ((𝑚↑2) − (𝑛↑2))), (𝑘 · (2 · (𝑚 · 𝑛)))} ∧ 𝐶 = (𝑘 · ((𝑚↑2) + (𝑛↑2)))))) | ||
Theorem | iserodd 16162* | Collect the odd terms in a sequence. (Contributed by Mario Carneiro, 7-Apr-2015.) (Proof shortened by AV, 10-Jul-2022.) |
⊢ ((𝜑 ∧ 𝑘 ∈ ℕ0) → 𝐶 ∈ ℂ) & ⊢ (𝑛 = ((2 · 𝑘) + 1) → 𝐵 = 𝐶) ⇒ ⊢ (𝜑 → (seq0( + , (𝑘 ∈ ℕ0 ↦ 𝐶)) ⇝ 𝐴 ↔ seq1( + , (𝑛 ∈ ℕ ↦ if(2 ∥ 𝑛, 0, 𝐵))) ⇝ 𝐴)) | ||
Syntax | cpc 16163 | Extend class notation with the prime count function. |
class pCnt | ||
Definition | df-pc 16164* | Define the prime count function, which returns the largest exponent of a given prime (or other positive integer) that divides the number. For rational numbers, it returns negative values according to the power of a prime in the denominator. (Contributed by Mario Carneiro, 23-Feb-2014.) |
⊢ pCnt = (𝑝 ∈ ℙ, 𝑟 ∈ ℚ ↦ if(𝑟 = 0, +∞, (℩𝑧∃𝑥 ∈ ℤ ∃𝑦 ∈ ℕ (𝑟 = (𝑥 / 𝑦) ∧ 𝑧 = (sup({𝑛 ∈ ℕ0 ∣ (𝑝↑𝑛) ∥ 𝑥}, ℝ, < ) − sup({𝑛 ∈ ℕ0 ∣ (𝑝↑𝑛) ∥ 𝑦}, ℝ, < )))))) | ||
Theorem | pclem 16165* | - Lemma for the prime power pre-function's properties. (Contributed by Mario Carneiro, 23-Feb-2014.) |
⊢ 𝐴 = {𝑛 ∈ ℕ0 ∣ (𝑃↑𝑛) ∥ 𝑁} ⇒ ⊢ ((𝑃 ∈ (ℤ≥‘2) ∧ (𝑁 ∈ ℤ ∧ 𝑁 ≠ 0)) → (𝐴 ⊆ ℤ ∧ 𝐴 ≠ ∅ ∧ ∃𝑥 ∈ ℤ ∀𝑦 ∈ 𝐴 𝑦 ≤ 𝑥)) | ||
Theorem | pcprecl 16166* | Closure of the prime power pre-function. (Contributed by Mario Carneiro, 23-Feb-2014.) |
⊢ 𝐴 = {𝑛 ∈ ℕ0 ∣ (𝑃↑𝑛) ∥ 𝑁} & ⊢ 𝑆 = sup(𝐴, ℝ, < ) ⇒ ⊢ ((𝑃 ∈ (ℤ≥‘2) ∧ (𝑁 ∈ ℤ ∧ 𝑁 ≠ 0)) → (𝑆 ∈ ℕ0 ∧ (𝑃↑𝑆) ∥ 𝑁)) | ||
Theorem | pcprendvds 16167* | Non-divisibility property of the prime power pre-function. (Contributed by Mario Carneiro, 23-Feb-2014.) |
⊢ 𝐴 = {𝑛 ∈ ℕ0 ∣ (𝑃↑𝑛) ∥ 𝑁} & ⊢ 𝑆 = sup(𝐴, ℝ, < ) ⇒ ⊢ ((𝑃 ∈ (ℤ≥‘2) ∧ (𝑁 ∈ ℤ ∧ 𝑁 ≠ 0)) → ¬ (𝑃↑(𝑆 + 1)) ∥ 𝑁) | ||
Theorem | pcprendvds2 16168* | Non-divisibility property of the prime power pre-function. (Contributed by Mario Carneiro, 23-Feb-2014.) |
⊢ 𝐴 = {𝑛 ∈ ℕ0 ∣ (𝑃↑𝑛) ∥ 𝑁} & ⊢ 𝑆 = sup(𝐴, ℝ, < ) ⇒ ⊢ ((𝑃 ∈ (ℤ≥‘2) ∧ (𝑁 ∈ ℤ ∧ 𝑁 ≠ 0)) → ¬ 𝑃 ∥ (𝑁 / (𝑃↑𝑆))) | ||
Theorem | pcpre1 16169* | Value of the prime power pre-function at 1. (Contributed by Mario Carneiro, 23-Feb-2014.) (Revised by Mario Carneiro, 26-Apr-2016.) |
⊢ 𝐴 = {𝑛 ∈ ℕ0 ∣ (𝑃↑𝑛) ∥ 𝑁} & ⊢ 𝑆 = sup(𝐴, ℝ, < ) ⇒ ⊢ ((𝑃 ∈ (ℤ≥‘2) ∧ 𝑁 = 1) → 𝑆 = 0) | ||
Theorem | pcpremul 16170* | Multiplicative property of the prime count pre-function. Note that the primality of 𝑃 is essential for this property; (4 pCnt 2) = 0 but (4 pCnt (2 · 2)) = 1 ≠ 2 · (4 pCnt 2) = 0. Since this is needed to show uniqueness for the real prime count function (over ℚ), we don't bother to define it off the primes. (Contributed by Mario Carneiro, 23-Feb-2014.) |
⊢ 𝑆 = sup({𝑛 ∈ ℕ0 ∣ (𝑃↑𝑛) ∥ 𝑀}, ℝ, < ) & ⊢ 𝑇 = sup({𝑛 ∈ ℕ0 ∣ (𝑃↑𝑛) ∥ 𝑁}, ℝ, < ) & ⊢ 𝑈 = sup({𝑛 ∈ ℕ0 ∣ (𝑃↑𝑛) ∥ (𝑀 · 𝑁)}, ℝ, < ) ⇒ ⊢ ((𝑃 ∈ ℙ ∧ (𝑀 ∈ ℤ ∧ 𝑀 ≠ 0) ∧ (𝑁 ∈ ℤ ∧ 𝑁 ≠ 0)) → (𝑆 + 𝑇) = 𝑈) | ||
Theorem | pcval 16171* | The value of the prime power function. (Contributed by Mario Carneiro, 23-Feb-2014.) (Revised by Mario Carneiro, 3-Oct-2014.) |
⊢ 𝑆 = sup({𝑛 ∈ ℕ0 ∣ (𝑃↑𝑛) ∥ 𝑥}, ℝ, < ) & ⊢ 𝑇 = sup({𝑛 ∈ ℕ0 ∣ (𝑃↑𝑛) ∥ 𝑦}, ℝ, < ) ⇒ ⊢ ((𝑃 ∈ ℙ ∧ (𝑁 ∈ ℚ ∧ 𝑁 ≠ 0)) → (𝑃 pCnt 𝑁) = (℩𝑧∃𝑥 ∈ ℤ ∃𝑦 ∈ ℕ (𝑁 = (𝑥 / 𝑦) ∧ 𝑧 = (𝑆 − 𝑇)))) | ||
Theorem | pceulem 16172* | Lemma for pceu 16173. (Contributed by Mario Carneiro, 23-Feb-2014.) |
⊢ 𝑆 = sup({𝑛 ∈ ℕ0 ∣ (𝑃↑𝑛) ∥ 𝑥}, ℝ, < ) & ⊢ 𝑇 = sup({𝑛 ∈ ℕ0 ∣ (𝑃↑𝑛) ∥ 𝑦}, ℝ, < ) & ⊢ 𝑈 = sup({𝑛 ∈ ℕ0 ∣ (𝑃↑𝑛) ∥ 𝑠}, ℝ, < ) & ⊢ 𝑉 = sup({𝑛 ∈ ℕ0 ∣ (𝑃↑𝑛) ∥ 𝑡}, ℝ, < ) & ⊢ (𝜑 → 𝑃 ∈ ℙ) & ⊢ (𝜑 → 𝑁 ≠ 0) & ⊢ (𝜑 → (𝑥 ∈ ℤ ∧ 𝑦 ∈ ℕ)) & ⊢ (𝜑 → 𝑁 = (𝑥 / 𝑦)) & ⊢ (𝜑 → (𝑠 ∈ ℤ ∧ 𝑡 ∈ ℕ)) & ⊢ (𝜑 → 𝑁 = (𝑠 / 𝑡)) ⇒ ⊢ (𝜑 → (𝑆 − 𝑇) = (𝑈 − 𝑉)) | ||
Theorem | pceu 16173* | Uniqueness for the prime power function. (Contributed by Mario Carneiro, 23-Feb-2014.) |
⊢ 𝑆 = sup({𝑛 ∈ ℕ0 ∣ (𝑃↑𝑛) ∥ 𝑥}, ℝ, < ) & ⊢ 𝑇 = sup({𝑛 ∈ ℕ0 ∣ (𝑃↑𝑛) ∥ 𝑦}, ℝ, < ) ⇒ ⊢ ((𝑃 ∈ ℙ ∧ (𝑁 ∈ ℚ ∧ 𝑁 ≠ 0)) → ∃!𝑧∃𝑥 ∈ ℤ ∃𝑦 ∈ ℕ (𝑁 = (𝑥 / 𝑦) ∧ 𝑧 = (𝑆 − 𝑇))) | ||
Theorem | pczpre 16174* | Connect the prime count pre-function to the actual prime count function, when restricted to the integers. (Contributed by Mario Carneiro, 23-Feb-2014.) (Proof shortened by Mario Carneiro, 24-Dec-2016.) |
⊢ 𝑆 = sup({𝑛 ∈ ℕ0 ∣ (𝑃↑𝑛) ∥ 𝑁}, ℝ, < ) ⇒ ⊢ ((𝑃 ∈ ℙ ∧ (𝑁 ∈ ℤ ∧ 𝑁 ≠ 0)) → (𝑃 pCnt 𝑁) = 𝑆) | ||
Theorem | pczcl 16175 | Closure of the prime power function. (Contributed by Mario Carneiro, 23-Feb-2014.) |
⊢ ((𝑃 ∈ ℙ ∧ (𝑁 ∈ ℤ ∧ 𝑁 ≠ 0)) → (𝑃 pCnt 𝑁) ∈ ℕ0) | ||
Theorem | pccl 16176 | Closure of the prime power function. (Contributed by Mario Carneiro, 23-Feb-2014.) |
⊢ ((𝑃 ∈ ℙ ∧ 𝑁 ∈ ℕ) → (𝑃 pCnt 𝑁) ∈ ℕ0) | ||
Theorem | pccld 16177 | Closure of the prime power function. (Contributed by Mario Carneiro, 29-May-2016.) |
⊢ (𝜑 → 𝑃 ∈ ℙ) & ⊢ (𝜑 → 𝑁 ∈ ℕ) ⇒ ⊢ (𝜑 → (𝑃 pCnt 𝑁) ∈ ℕ0) | ||
Theorem | pcmul 16178 | Multiplication property of the prime power function. (Contributed by Mario Carneiro, 23-Feb-2014.) |
⊢ ((𝑃 ∈ ℙ ∧ (𝐴 ∈ ℤ ∧ 𝐴 ≠ 0) ∧ (𝐵 ∈ ℤ ∧ 𝐵 ≠ 0)) → (𝑃 pCnt (𝐴 · 𝐵)) = ((𝑃 pCnt 𝐴) + (𝑃 pCnt 𝐵))) | ||
Theorem | pcdiv 16179 | Division property of the prime power function. (Contributed by Mario Carneiro, 1-Mar-2014.) |
⊢ ((𝑃 ∈ ℙ ∧ (𝐴 ∈ ℤ ∧ 𝐴 ≠ 0) ∧ 𝐵 ∈ ℕ) → (𝑃 pCnt (𝐴 / 𝐵)) = ((𝑃 pCnt 𝐴) − (𝑃 pCnt 𝐵))) | ||
Theorem | pcqmul 16180 | Multiplication property of the prime power function. (Contributed by Mario Carneiro, 9-Sep-2014.) |
⊢ ((𝑃 ∈ ℙ ∧ (𝐴 ∈ ℚ ∧ 𝐴 ≠ 0) ∧ (𝐵 ∈ ℚ ∧ 𝐵 ≠ 0)) → (𝑃 pCnt (𝐴 · 𝐵)) = ((𝑃 pCnt 𝐴) + (𝑃 pCnt 𝐵))) | ||
Theorem | pc0 16181 | The value of the prime power function at zero. (Contributed by Mario Carneiro, 3-Oct-2014.) |
⊢ (𝑃 ∈ ℙ → (𝑃 pCnt 0) = +∞) | ||
Theorem | pc1 16182 | Value of the prime count function at 1. (Contributed by Mario Carneiro, 23-Feb-2014.) |
⊢ (𝑃 ∈ ℙ → (𝑃 pCnt 1) = 0) | ||
Theorem | pcqcl 16183 | Closure of the general prime count function. (Contributed by Mario Carneiro, 23-Feb-2014.) |
⊢ ((𝑃 ∈ ℙ ∧ (𝑁 ∈ ℚ ∧ 𝑁 ≠ 0)) → (𝑃 pCnt 𝑁) ∈ ℤ) | ||
Theorem | pcqdiv 16184 | Division property of the prime power function. (Contributed by Mario Carneiro, 10-Aug-2015.) |
⊢ ((𝑃 ∈ ℙ ∧ (𝐴 ∈ ℚ ∧ 𝐴 ≠ 0) ∧ (𝐵 ∈ ℚ ∧ 𝐵 ≠ 0)) → (𝑃 pCnt (𝐴 / 𝐵)) = ((𝑃 pCnt 𝐴) − (𝑃 pCnt 𝐵))) | ||
Theorem | pcrec 16185 | Prime power of a reciprocal. (Contributed by Mario Carneiro, 10-Aug-2015.) |
⊢ ((𝑃 ∈ ℙ ∧ (𝐴 ∈ ℚ ∧ 𝐴 ≠ 0)) → (𝑃 pCnt (1 / 𝐴)) = -(𝑃 pCnt 𝐴)) | ||
Theorem | pcexp 16186 | Prime power of an exponential. (Contributed by Mario Carneiro, 10-Aug-2015.) |
⊢ ((𝑃 ∈ ℙ ∧ (𝐴 ∈ ℚ ∧ 𝐴 ≠ 0) ∧ 𝑁 ∈ ℤ) → (𝑃 pCnt (𝐴↑𝑁)) = (𝑁 · (𝑃 pCnt 𝐴))) | ||
Theorem | pcxcl 16187 | Extended real closure of the general prime count function. (Contributed by Mario Carneiro, 3-Oct-2014.) |
⊢ ((𝑃 ∈ ℙ ∧ 𝑁 ∈ ℚ) → (𝑃 pCnt 𝑁) ∈ ℝ*) | ||
Theorem | pcge0 16188 | The prime count of an integer is greater than or equal to zero. (Contributed by Mario Carneiro, 3-Oct-2014.) |
⊢ ((𝑃 ∈ ℙ ∧ 𝑁 ∈ ℤ) → 0 ≤ (𝑃 pCnt 𝑁)) | ||
Theorem | pczdvds 16189 | Defining property of the prime count function. (Contributed by Mario Carneiro, 9-Sep-2014.) |
⊢ ((𝑃 ∈ ℙ ∧ (𝑁 ∈ ℤ ∧ 𝑁 ≠ 0)) → (𝑃↑(𝑃 pCnt 𝑁)) ∥ 𝑁) | ||
Theorem | pcdvds 16190 | Defining property of the prime count function. (Contributed by Mario Carneiro, 23-Feb-2014.) |
⊢ ((𝑃 ∈ ℙ ∧ 𝑁 ∈ ℕ) → (𝑃↑(𝑃 pCnt 𝑁)) ∥ 𝑁) | ||
Theorem | pczndvds 16191 | Defining property of the prime count function. (Contributed by Mario Carneiro, 3-Oct-2014.) |
⊢ ((𝑃 ∈ ℙ ∧ (𝑁 ∈ ℤ ∧ 𝑁 ≠ 0)) → ¬ (𝑃↑((𝑃 pCnt 𝑁) + 1)) ∥ 𝑁) | ||
Theorem | pcndvds 16192 | Defining property of the prime count function. (Contributed by Mario Carneiro, 23-Feb-2014.) |
⊢ ((𝑃 ∈ ℙ ∧ 𝑁 ∈ ℕ) → ¬ (𝑃↑((𝑃 pCnt 𝑁) + 1)) ∥ 𝑁) | ||
Theorem | pczndvds2 16193 | The remainder after dividing out all factors of 𝑃 is not divisible by 𝑃. (Contributed by Mario Carneiro, 9-Sep-2014.) |
⊢ ((𝑃 ∈ ℙ ∧ (𝑁 ∈ ℤ ∧ 𝑁 ≠ 0)) → ¬ 𝑃 ∥ (𝑁 / (𝑃↑(𝑃 pCnt 𝑁)))) | ||
Theorem | pcndvds2 16194 | The remainder after dividing out all factors of 𝑃 is not divisible by 𝑃. (Contributed by Mario Carneiro, 23-Feb-2014.) |
⊢ ((𝑃 ∈ ℙ ∧ 𝑁 ∈ ℕ) → ¬ 𝑃 ∥ (𝑁 / (𝑃↑(𝑃 pCnt 𝑁)))) | ||
Theorem | pcdvdsb 16195 | 𝑃↑𝐴 divides 𝑁 if and only if 𝐴 is at most the count of 𝑃. (Contributed by Mario Carneiro, 3-Oct-2014.) |
⊢ ((𝑃 ∈ ℙ ∧ 𝑁 ∈ ℤ ∧ 𝐴 ∈ ℕ0) → (𝐴 ≤ (𝑃 pCnt 𝑁) ↔ (𝑃↑𝐴) ∥ 𝑁)) | ||
Theorem | pcelnn 16196 | There are a positive number of powers of a prime 𝑃 in 𝑁 iff 𝑃 divides 𝑁. (Contributed by Mario Carneiro, 23-Feb-2014.) |
⊢ ((𝑃 ∈ ℙ ∧ 𝑁 ∈ ℕ) → ((𝑃 pCnt 𝑁) ∈ ℕ ↔ 𝑃 ∥ 𝑁)) | ||
Theorem | pceq0 16197 | There are zero powers of a prime 𝑃 in 𝑁 iff 𝑃 does not divide 𝑁. (Contributed by Mario Carneiro, 23-Feb-2014.) |
⊢ ((𝑃 ∈ ℙ ∧ 𝑁 ∈ ℕ) → ((𝑃 pCnt 𝑁) = 0 ↔ ¬ 𝑃 ∥ 𝑁)) | ||
Theorem | pcidlem 16198 | The prime count of a prime power. (Contributed by Mario Carneiro, 12-Mar-2014.) |
⊢ ((𝑃 ∈ ℙ ∧ 𝐴 ∈ ℕ0) → (𝑃 pCnt (𝑃↑𝐴)) = 𝐴) | ||
Theorem | pcid 16199 | The prime count of a prime power. (Contributed by Mario Carneiro, 9-Sep-2014.) |
⊢ ((𝑃 ∈ ℙ ∧ 𝐴 ∈ ℤ) → (𝑃 pCnt (𝑃↑𝐴)) = 𝐴) | ||
Theorem | pcneg 16200 | The prime count of a negative number. (Contributed by Mario Carneiro, 13-Mar-2014.) |
⊢ ((𝑃 ∈ ℙ ∧ 𝐴 ∈ ℚ) → (𝑃 pCnt -𝐴) = (𝑃 pCnt 𝐴)) |
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