HomeHome Intuitionistic Logic Explorer
Theorem List (p. 130 of 169)
< Previous  Next >
Bad symbols? Try the
GIF version.

Mirrors  >  Metamath Home Page  >  ILE Home Page  >  Theorem List Contents  >  Recent Proofs       This page: Page List

Theorem List for Intuitionistic Logic Explorer - 12901-13000   *Has distinct variable group(s)
TypeLabelDescription
Statement
 
Definitiondf-phi 12901* Define the Euler phi function (also called "Euler totient function"), which counts the number of integers less than 𝑛 and coprime to it, see definition in [ApostolNT] p. 25. (Contributed by Mario Carneiro, 23-Feb-2014.)
ϕ = (𝑛 ∈ ℕ ↦ (♯‘{𝑥 ∈ (1...𝑛) ∣ (𝑥 gcd 𝑛) = 1}))
 
Theoremphivalfi 12902* Finiteness of an expression used to define the Euler ϕ function. (Contributed by Jim Kingon, 28-May-2022.)
(𝑁 ∈ ℕ → {𝑥 ∈ (1...𝑁) ∣ (𝑥 gcd 𝑁) = 1} ∈ Fin)
 
Theoremphival 12903* Value of the Euler ϕ function. (Contributed by Mario Carneiro, 23-Feb-2014.)
(𝑁 ∈ ℕ → (ϕ‘𝑁) = (♯‘{𝑥 ∈ (1...𝑁) ∣ (𝑥 gcd 𝑁) = 1}))
 
Theoremphicl2 12904 Bounds and closure for the value of the Euler ϕ function. (Contributed by Mario Carneiro, 23-Feb-2014.)
(𝑁 ∈ ℕ → (ϕ‘𝑁) ∈ (1...𝑁))
 
Theoremphicl 12905 Closure for the value of the Euler ϕ function. (Contributed by Mario Carneiro, 28-Feb-2014.)
(𝑁 ∈ ℕ → (ϕ‘𝑁) ∈ ℕ)
 
Theoremphibndlem 12906* Lemma for phibnd 12907. (Contributed by Mario Carneiro, 23-Feb-2014.)
(𝑁 ∈ (ℤ‘2) → {𝑥 ∈ (1...𝑁) ∣ (𝑥 gcd 𝑁) = 1} ⊆ (1...(𝑁 − 1)))
 
Theoremphibnd 12907 A slightly tighter bound on the value of the Euler ϕ function. (Contributed by Mario Carneiro, 23-Feb-2014.)
(𝑁 ∈ (ℤ‘2) → (ϕ‘𝑁) ≤ (𝑁 − 1))
 
Theoremphicld 12908 Closure for the value of the Euler ϕ function. (Contributed by Mario Carneiro, 29-May-2016.)
(𝜑𝑁 ∈ ℕ)       (𝜑 → (ϕ‘𝑁) ∈ ℕ)
 
Theoremphi1 12909 Value of the Euler ϕ function at 1. (Contributed by Mario Carneiro, 23-Feb-2014.)
(ϕ‘1) = 1
 
Theoremdfphi2 12910* 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}))
 
Theoremhashdvds 12911* 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.)
(𝜑𝑁 ∈ ℕ)    &   (𝜑𝐴 ∈ ℤ)    &   (𝜑𝐵 ∈ (ℤ‘(𝐴 − 1)))    &   (𝜑𝐶 ∈ ℤ)       (𝜑 → (♯‘{𝑥 ∈ (𝐴...𝐵) ∣ 𝑁 ∥ (𝑥𝐶)}) = ((⌊‘((𝐵𝐶) / 𝑁)) − (⌊‘(((𝐴 − 1) − 𝐶) / 𝑁))))
 
Theoremphiprmpw 12912 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‘(𝑃𝐾)) = ((𝑃↑(𝐾 − 1)) · (𝑃 − 1)))
 
Theoremphiprm 12913 Value of the Euler ϕ function at a prime. (Contributed by Mario Carneiro, 28-Feb-2014.)
(𝑃 ∈ ℙ → (ϕ‘𝑃) = (𝑃 − 1))
 
Theoremcrth 12914* 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 = (𝑥𝑆 ↦ ⟨(𝑥 mod 𝑀), (𝑥 mod 𝑁)⟩)    &   (𝜑 → (𝑀 ∈ ℕ ∧ 𝑁 ∈ ℕ ∧ (𝑀 gcd 𝑁) = 1))       (𝜑𝐹:𝑆1-1-onto𝑇)
 
Theoremphimullem 12915* Lemma for phimul 12916. (Contributed by Mario Carneiro, 24-Feb-2014.)
𝑆 = (0..^(𝑀 · 𝑁))    &   𝑇 = ((0..^𝑀) × (0..^𝑁))    &   𝐹 = (𝑥𝑆 ↦ ⟨(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 = {𝑦𝑆 ∣ (𝑦 gcd (𝑀 · 𝑁)) = 1}       (𝜑 → (ϕ‘(𝑀 · 𝑁)) = ((ϕ‘𝑀) · (ϕ‘𝑁)))
 
Theoremphimul 12916 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)))
 
Theoremeulerthlem1 12917* Lemma for eulerth 12923. (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))    &   (𝜑𝐹:𝑇1-1-onto𝑆)    &   𝐺 = (𝑥𝑇 ↦ ((𝐴 · (𝐹𝑥)) mod 𝑁))       (𝜑𝐺:𝑇𝑆)
 
Theoremeulerthlemfi 12918* Lemma for eulerth 12923. The set 𝑆 is finite. (Contributed by Mario Carneiro, 28-Feb-2014.) (Revised by Jim Kingdon, 7-Sep-2024.)
(𝜑 → (𝑁 ∈ ℕ ∧ 𝐴 ∈ ℤ ∧ (𝐴 gcd 𝑁) = 1))    &   𝑆 = {𝑦 ∈ (0..^𝑁) ∣ (𝑦 gcd 𝑁) = 1}       (𝜑𝑆 ∈ Fin)
 
Theoremeulerthlemrprm 12919* Lemma for eulerth 12923. 𝑁 and 𝑥 ∈ (1...(ϕ‘𝑁))(𝐹𝑥) are relatively prime. (Contributed by Mario Carneiro, 28-Feb-2014.) (Revised by Jim Kingdon, 2-Sep-2024.)
(𝜑 → (𝑁 ∈ ℕ ∧ 𝐴 ∈ ℤ ∧ (𝐴 gcd 𝑁) = 1))    &   𝑆 = {𝑦 ∈ (0..^𝑁) ∣ (𝑦 gcd 𝑁) = 1}    &   (𝜑𝐹:(1...(ϕ‘𝑁))–1-1-onto𝑆)       (𝜑 → (𝑁 gcd ∏𝑥 ∈ (1...(ϕ‘𝑁))(𝐹𝑥)) = 1)
 
Theoremeulerthlema 12920* Lemma for eulerth 12923. (Contributed by Mario Carneiro, 28-Feb-2014.) (Revised by Jim Kingdon, 2-Sep-2024.)
(𝜑 → (𝑁 ∈ ℕ ∧ 𝐴 ∈ ℤ ∧ (𝐴 gcd 𝑁) = 1))    &   𝑆 = {𝑦 ∈ (0..^𝑁) ∣ (𝑦 gcd 𝑁) = 1}    &   (𝜑𝐹:(1...(ϕ‘𝑁))–1-1-onto𝑆)       (𝜑 → (((𝐴↑(ϕ‘𝑁)) · ∏𝑥 ∈ (1...(ϕ‘𝑁))(𝐹𝑥)) mod 𝑁) = (∏𝑥 ∈ (1...(ϕ‘𝑁))((𝐴 · (𝐹𝑥)) mod 𝑁) mod 𝑁))
 
Theoremeulerthlemh 12921* Lemma for eulerth 12923. A permutation of (1...(ϕ‘𝑁)). (Contributed by Mario Carneiro, 28-Feb-2014.) (Revised by Jim Kingdon, 5-Sep-2024.)
(𝜑 → (𝑁 ∈ ℕ ∧ 𝐴 ∈ ℤ ∧ (𝐴 gcd 𝑁) = 1))    &   𝑆 = {𝑦 ∈ (0..^𝑁) ∣ (𝑦 gcd 𝑁) = 1}    &   (𝜑𝐹:(1...(ϕ‘𝑁))–1-1-onto𝑆)    &   𝐻 = (𝐹 ∘ (𝑦 ∈ (1...(ϕ‘𝑁)) ↦ ((𝐴 · (𝐹𝑦)) mod 𝑁)))       (𝜑𝐻:(1...(ϕ‘𝑁))–1-1-onto→(1...(ϕ‘𝑁)))
 
Theoremeulerthlemth 12922* Lemma for eulerth 12923. The result. (Contributed by Mario Carneiro, 28-Feb-2014.) (Revised by Jim Kingdon, 2-Sep-2024.)
(𝜑 → (𝑁 ∈ ℕ ∧ 𝐴 ∈ ℤ ∧ (𝐴 gcd 𝑁) = 1))    &   𝑆 = {𝑦 ∈ (0..^𝑁) ∣ (𝑦 gcd 𝑁) = 1}    &   (𝜑𝐹:(1...(ϕ‘𝑁))–1-1-onto𝑆)       (𝜑 → ((𝐴↑(ϕ‘𝑁)) mod 𝑁) = (1 mod 𝑁))
 
Theoremeulerth 12923 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))
 
Theoremfermltl 12924 Fermat's little theorem. When 𝑃 is prime, 𝐴𝑃𝐴 (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) → ((𝐴𝑃) mod 𝑃) = (𝐴 mod 𝑃))
 
Theoremprmdiv 12925 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 ∧ ¬ 𝑃𝐴) → (𝑅 ∈ (1...(𝑃 − 1)) ∧ 𝑃 ∥ ((𝐴 · 𝑅) − 1)))
 
Theoremprmdiveq 12926 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 ∧ ¬ 𝑃𝐴) → ((𝑆 ∈ (0...(𝑃 − 1)) ∧ 𝑃 ∥ ((𝐴 · 𝑆) − 1)) ↔ 𝑆 = 𝑅))
 
Theoremprmdivdiv 12927 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))
 
Theoremhashgcdlem 12928* 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 = (𝑥𝐴 ↦ (𝑥 · 𝑁))       ((𝑀 ∈ ℕ ∧ 𝑁 ∈ ℕ ∧ 𝑁𝑀) → 𝐹:𝐴1-1-onto𝐵)
 
Theoremdvdsfi 12929* A natural number has finitely many divisors. (Contributed by Jim Kingdon, 9-Oct-2025.)
(𝑁 ∈ ℕ → {𝑥 ∈ ℕ ∣ 𝑥𝑁} ∈ Fin)
 
Theoremhashgcdeq 12930* 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(𝑁𝑀, (ϕ‘(𝑀 / 𝑁)), 0))
 
Theoremphisum 12931* 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 ∣ 𝑥𝑁} (ϕ‘𝑑) = 𝑁)
 
Theoremodzval 12932* 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) → ((od𝑁)‘𝐴) = inf({𝑛 ∈ ℕ ∣ 𝑁 ∥ ((𝐴𝑛) − 1)}, ℝ, < ))
 
Theoremodzcllem 12933 - Lemma for odzcl 12934, 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) → (((od𝑁)‘𝐴) ∈ ℕ ∧ 𝑁 ∥ ((𝐴↑((od𝑁)‘𝐴)) − 1)))
 
Theoremodzcl 12934 The order of a group element is an integer. (Contributed by Mario Carneiro, 28-Feb-2014.)
((𝑁 ∈ ℕ ∧ 𝐴 ∈ ℤ ∧ (𝐴 gcd 𝑁) = 1) → ((od𝑁)‘𝐴) ∈ ℕ)
 
Theoremodzid 12935 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↑((od𝑁)‘𝐴)) − 1))
 
Theoremodzdvds 12936 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 ∥ ((𝐴𝐾) − 1) ↔ ((od𝑁)‘𝐴) ∥ 𝐾))
 
Theoremodzphi 12937 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) → ((od𝑁)‘𝐴) ∥ (ϕ‘𝑁))
 
5.2.6  Arithmetic modulo a prime number
 
Theoremmodprm1div 12938 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)))
 
Theoremm1dvdsndvds 12939 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) → ¬ 𝑃𝐴))
 
Theoremmodprminv 12940 Show an explicit expression for the modular inverse of 𝐴 mod 𝑃. This is an application of prmdiv 12925. (Contributed by Alexander van der Vekens, 15-May-2018.)
𝑅 = ((𝐴↑(𝑃 − 2)) mod 𝑃)       ((𝑃 ∈ ℙ ∧ 𝐴 ∈ ℤ ∧ ¬ 𝑃𝐴) → (𝑅 ∈ (1...(𝑃 − 1)) ∧ ((𝐴 · 𝑅) mod 𝑃) = 1))
 
Theoremmodprminveq 12941 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 ∧ ¬ 𝑃𝐴) → ((𝑆 ∈ (0...(𝑃 − 1)) ∧ ((𝐴 · 𝑆) mod 𝑃) = 1) ↔ 𝑆 = 𝑅))
 
Theoremvfermltl 12942 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 ∧ ¬ 𝑃𝐴) → ((𝐴↑(𝑃 − 1)) mod 𝑃) = 1)
 
Theorempowm2modprm 12943 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))
 
Theoremreumodprminv 12944* 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)
 
Theoremmodprm0 12945* 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)
 
Theoremnnnn0modprm0 12946* 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)
 
Theoremmodprmn0modprm0 12947* 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))
 
5.2.7  Pythagorean Triples
 
Theoremcoprimeprodsq 12948 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 ∈ ℕ0𝐵 ∈ ℤ ∧ 𝐶 ∈ ℕ0) ∧ ((𝐴 gcd 𝐵) gcd 𝐶) = 1) → ((𝐶↑2) = (𝐴 · 𝐵) → 𝐴 = ((𝐴 gcd 𝐶)↑2)))
 
Theoremcoprimeprodsq2 12949 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 ∈ ℕ0𝐶 ∈ ℕ0) ∧ ((𝐴 gcd 𝐵) gcd 𝐶) = 1) → ((𝐶↑2) = (𝐴 · 𝐵) → 𝐵 = ((𝐵 gcd 𝐶)↑2)))
 
Theoremoddprm 12950 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)
 
Theoremnnoddn2prm 12951 A prime not equal to 2 is an odd positive integer. (Contributed by AV, 28-Jun-2021.)
(𝑁 ∈ (ℙ ∖ {2}) → (𝑁 ∈ ℕ ∧ ¬ 2 ∥ 𝑁))
 
Theoremoddn2prm 12952 A prime not equal to 2 is odd. (Contributed by AV, 28-Jun-2021.)
(𝑁 ∈ (ℙ ∖ {2}) → ¬ 2 ∥ 𝑁)
 
Theoremnnoddn2prmb 12953 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))
 
Theoremprm23lt5 12954 A prime less than 5 is either 2 or 3. (Contributed by AV, 5-Jul-2021.)
((𝑃 ∈ ℙ ∧ 𝑃 < 5) → (𝑃 = 2 ∨ 𝑃 = 3))
 
Theoremprm23ge5 12955 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)))
 
Theorempythagtriplem1 12956* Lemma for pythagtrip 12974. 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))
 
Theorempythagtriplem2 12957* Lemma for pythagtrip 12974. 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)))
 
Theorempythagtriplem3 12958 Lemma for pythagtrip 12974. 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)
 
Theorempythagtriplem4 12959 Lemma for pythagtrip 12974. Show that 𝐶𝐵 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)) → ((𝐶𝐵) gcd (𝐶 + 𝐵)) = 1)
 
Theorempythagtriplem10 12960 Lemma for pythagtrip 12974. Show that 𝐶𝐵 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 < (𝐶𝐵))
 
Theorempythagtriplem6 12961 Lemma for pythagtrip 12974. Calculate (√‘(𝐶𝐵)). (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)) → (√‘(𝐶𝐵)) = ((𝐶𝐵) gcd 𝐴))
 
Theorempythagtriplem7 12962 Lemma for pythagtrip 12974. 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))
 
Theorempythagtriplem8 12963 Lemma for pythagtrip 12974. Show that (√‘(𝐶𝐵)) 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)) → (√‘(𝐶𝐵)) ∈ ℕ)
 
Theorempythagtriplem9 12964 Lemma for pythagtrip 12974. 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)
 
Theorempythagtriplem11 12965 Lemma for pythagtrip 12974. 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)) + (√‘(𝐶𝐵))) / 2)       (((𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝐶 ∈ ℕ) ∧ ((𝐴↑2) + (𝐵↑2)) = (𝐶↑2) ∧ ((𝐴 gcd 𝐵) = 1 ∧ ¬ 2 ∥ 𝐴)) → 𝑀 ∈ ℕ)
 
Theorempythagtriplem12 12966 Lemma for pythagtrip 12974. Calculate the square of 𝑀. (Contributed by Scott Fenton, 17-Apr-2014.) (Revised by Mario Carneiro, 19-Apr-2014.)
𝑀 = (((√‘(𝐶 + 𝐵)) + (√‘(𝐶𝐵))) / 2)       (((𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝐶 ∈ ℕ) ∧ ((𝐴↑2) + (𝐵↑2)) = (𝐶↑2) ∧ ((𝐴 gcd 𝐵) = 1 ∧ ¬ 2 ∥ 𝐴)) → (𝑀↑2) = ((𝐶 + 𝐴) / 2))
 
Theorempythagtriplem13 12967 Lemma for pythagtrip 12974. 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)) − (√‘(𝐶𝐵))) / 2)       (((𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝐶 ∈ ℕ) ∧ ((𝐴↑2) + (𝐵↑2)) = (𝐶↑2) ∧ ((𝐴 gcd 𝐵) = 1 ∧ ¬ 2 ∥ 𝐴)) → 𝑁 ∈ ℕ)
 
Theorempythagtriplem14 12968 Lemma for pythagtrip 12974. Calculate the square of 𝑁. (Contributed by Scott Fenton, 17-Apr-2014.) (Revised by Mario Carneiro, 19-Apr-2014.)
𝑁 = (((√‘(𝐶 + 𝐵)) − (√‘(𝐶𝐵))) / 2)       (((𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝐶 ∈ ℕ) ∧ ((𝐴↑2) + (𝐵↑2)) = (𝐶↑2) ∧ ((𝐴 gcd 𝐵) = 1 ∧ ¬ 2 ∥ 𝐴)) → (𝑁↑2) = ((𝐶𝐴) / 2))
 
Theorempythagtriplem15 12969 Lemma for pythagtrip 12974. 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)) + (√‘(𝐶𝐵))) / 2)    &   𝑁 = (((√‘(𝐶 + 𝐵)) − (√‘(𝐶𝐵))) / 2)       (((𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝐶 ∈ ℕ) ∧ ((𝐴↑2) + (𝐵↑2)) = (𝐶↑2) ∧ ((𝐴 gcd 𝐵) = 1 ∧ ¬ 2 ∥ 𝐴)) → 𝐴 = ((𝑀↑2) − (𝑁↑2)))
 
Theorempythagtriplem16 12970 Lemma for pythagtrip 12974. 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)) + (√‘(𝐶𝐵))) / 2)    &   𝑁 = (((√‘(𝐶 + 𝐵)) − (√‘(𝐶𝐵))) / 2)       (((𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝐶 ∈ ℕ) ∧ ((𝐴↑2) + (𝐵↑2)) = (𝐶↑2) ∧ ((𝐴 gcd 𝐵) = 1 ∧ ¬ 2 ∥ 𝐴)) → 𝐵 = (2 · (𝑀 · 𝑁)))
 
Theorempythagtriplem17 12971 Lemma for pythagtrip 12974. 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)) + (√‘(𝐶𝐵))) / 2)    &   𝑁 = (((√‘(𝐶 + 𝐵)) − (√‘(𝐶𝐵))) / 2)       (((𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝐶 ∈ ℕ) ∧ ((𝐴↑2) + (𝐵↑2)) = (𝐶↑2) ∧ ((𝐴 gcd 𝐵) = 1 ∧ ¬ 2 ∥ 𝐴)) → 𝐶 = ((𝑀↑2) + (𝑁↑2)))
 
Theorempythagtriplem18 12972* Lemma for pythagtrip 12974. 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))))
 
Theorempythagtriplem19 12973* Lemma for pythagtrip 12974. 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)))))
 
Theorempythagtrip 12974* 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))))))
 
5.2.8  The prime count function
 
Syntaxcpc 12975 Extend class notation with the prime count function.
class pCnt
 
Definitiondf-pc 12976* 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, +∞, (℩𝑧𝑥 ∈ ℤ ∃𝑦 ∈ ℕ (𝑟 = (𝑥 / 𝑦) ∧ 𝑧 = (sup({𝑛 ∈ ℕ0 ∣ (𝑝𝑛) ∥ 𝑥}, ℝ, < ) − sup({𝑛 ∈ ℕ0 ∣ (𝑝𝑛) ∥ 𝑦}, ℝ, < ))))))
 
Theorempclem0 12977* Lemma for the prime power pre-function's properties. (Contributed by Mario Carneiro, 23-Feb-2014.) (Revised by Jim Kingdon, 7-Oct-2024.)
𝐴 = {𝑛 ∈ ℕ0 ∣ (𝑃𝑛) ∥ 𝑁}       ((𝑃 ∈ (ℤ‘2) ∧ (𝑁 ∈ ℤ ∧ 𝑁 ≠ 0)) → 0 ∈ 𝐴)
 
Theorempclemub 12978* Lemma for the prime power pre-function's properties. (Contributed by Mario Carneiro, 23-Feb-2014.) (Revised by Jim Kingdon, 7-Oct-2024.)
𝐴 = {𝑛 ∈ ℕ0 ∣ (𝑃𝑛) ∥ 𝑁}       ((𝑃 ∈ (ℤ‘2) ∧ (𝑁 ∈ ℤ ∧ 𝑁 ≠ 0)) → ∃𝑥 ∈ ℤ ∀𝑦𝐴 𝑦𝑥)
 
Theorempclemdc 12979* Lemma for the prime power pre-function's properties. (Contributed by Jim Kingdon, 8-Oct-2024.)
𝐴 = {𝑛 ∈ ℕ0 ∣ (𝑃𝑛) ∥ 𝑁}       ((𝑃 ∈ (ℤ‘2) ∧ (𝑁 ∈ ℤ ∧ 𝑁 ≠ 0)) → ∀𝑥 ∈ ℤ DECID 𝑥𝐴)
 
Theorempcprecl 12980* Closure of the prime power pre-function. (Contributed by Mario Carneiro, 23-Feb-2014.)
𝐴 = {𝑛 ∈ ℕ0 ∣ (𝑃𝑛) ∥ 𝑁}    &   𝑆 = sup(𝐴, ℝ, < )       ((𝑃 ∈ (ℤ‘2) ∧ (𝑁 ∈ ℤ ∧ 𝑁 ≠ 0)) → (𝑆 ∈ ℕ0 ∧ (𝑃𝑆) ∥ 𝑁))
 
Theorempcprendvds 12981* Non-divisibility property of the prime power pre-function. (Contributed by Mario Carneiro, 23-Feb-2014.)
𝐴 = {𝑛 ∈ ℕ0 ∣ (𝑃𝑛) ∥ 𝑁}    &   𝑆 = sup(𝐴, ℝ, < )       ((𝑃 ∈ (ℤ‘2) ∧ (𝑁 ∈ ℤ ∧ 𝑁 ≠ 0)) → ¬ (𝑃↑(𝑆 + 1)) ∥ 𝑁)
 
Theorempcprendvds2 12982* Non-divisibility property of the prime power pre-function. (Contributed by Mario Carneiro, 23-Feb-2014.)
𝐴 = {𝑛 ∈ ℕ0 ∣ (𝑃𝑛) ∥ 𝑁}    &   𝑆 = sup(𝐴, ℝ, < )       ((𝑃 ∈ (ℤ‘2) ∧ (𝑁 ∈ ℤ ∧ 𝑁 ≠ 0)) → ¬ 𝑃 ∥ (𝑁 / (𝑃𝑆)))
 
Theorempcpre1 12983* 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 ∣ (𝑃𝑛) ∥ 𝑁}    &   𝑆 = sup(𝐴, ℝ, < )       ((𝑃 ∈ (ℤ‘2) ∧ 𝑁 = 1) → 𝑆 = 0)
 
Theorempcpremul 12984* 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 ), we don't bother to define it off the primes. (Contributed by Mario Carneiro, 23-Feb-2014.)
𝑆 = sup({𝑛 ∈ ℕ0 ∣ (𝑃𝑛) ∥ 𝑀}, ℝ, < )    &   𝑇 = sup({𝑛 ∈ ℕ0 ∣ (𝑃𝑛) ∥ 𝑁}, ℝ, < )    &   𝑈 = sup({𝑛 ∈ ℕ0 ∣ (𝑃𝑛) ∥ (𝑀 · 𝑁)}, ℝ, < )       ((𝑃 ∈ ℙ ∧ (𝑀 ∈ ℤ ∧ 𝑀 ≠ 0) ∧ (𝑁 ∈ ℤ ∧ 𝑁 ≠ 0)) → (𝑆 + 𝑇) = 𝑈)
 
Theorempceulem 12985* Lemma for pceu 12986. (Contributed by Mario Carneiro, 23-Feb-2014.)
𝑆 = sup({𝑛 ∈ ℕ0 ∣ (𝑃𝑛) ∥ 𝑥}, ℝ, < )    &   𝑇 = sup({𝑛 ∈ ℕ0 ∣ (𝑃𝑛) ∥ 𝑦}, ℝ, < )    &   𝑈 = sup({𝑛 ∈ ℕ0 ∣ (𝑃𝑛) ∥ 𝑠}, ℝ, < )    &   𝑉 = sup({𝑛 ∈ ℕ0 ∣ (𝑃𝑛) ∥ 𝑡}, ℝ, < )    &   (𝜑𝑃 ∈ ℙ)    &   (𝜑𝑁 ≠ 0)    &   (𝜑 → (𝑥 ∈ ℤ ∧ 𝑦 ∈ ℕ))    &   (𝜑𝑁 = (𝑥 / 𝑦))    &   (𝜑 → (𝑠 ∈ ℤ ∧ 𝑡 ∈ ℕ))    &   (𝜑𝑁 = (𝑠 / 𝑡))       (𝜑 → (𝑆𝑇) = (𝑈𝑉))
 
Theorempceu 12986* Uniqueness for the prime power function. (Contributed by Mario Carneiro, 23-Feb-2014.)
𝑆 = sup({𝑛 ∈ ℕ0 ∣ (𝑃𝑛) ∥ 𝑥}, ℝ, < )    &   𝑇 = sup({𝑛 ∈ ℕ0 ∣ (𝑃𝑛) ∥ 𝑦}, ℝ, < )       ((𝑃 ∈ ℙ ∧ (𝑁 ∈ ℚ ∧ 𝑁 ≠ 0)) → ∃!𝑧𝑥 ∈ ℤ ∃𝑦 ∈ ℕ (𝑁 = (𝑥 / 𝑦) ∧ 𝑧 = (𝑆𝑇)))
 
Theorempcval 12987* 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 ∣ (𝑃𝑛) ∥ 𝑥}, ℝ, < )    &   𝑇 = sup({𝑛 ∈ ℕ0 ∣ (𝑃𝑛) ∥ 𝑦}, ℝ, < )       ((𝑃 ∈ ℙ ∧ (𝑁 ∈ ℚ ∧ 𝑁 ≠ 0)) → (𝑃 pCnt 𝑁) = (℩𝑧𝑥 ∈ ℤ ∃𝑦 ∈ ℕ (𝑁 = (𝑥 / 𝑦) ∧ 𝑧 = (𝑆𝑇))))
 
Theorempczpre 12988* 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 ∣ (𝑃𝑛) ∥ 𝑁}, ℝ, < )       ((𝑃 ∈ ℙ ∧ (𝑁 ∈ ℤ ∧ 𝑁 ≠ 0)) → (𝑃 pCnt 𝑁) = 𝑆)
 
Theorempczcl 12989 Closure of the prime power function. (Contributed by Mario Carneiro, 23-Feb-2014.)
((𝑃 ∈ ℙ ∧ (𝑁 ∈ ℤ ∧ 𝑁 ≠ 0)) → (𝑃 pCnt 𝑁) ∈ ℕ0)
 
Theorempccl 12990 Closure of the prime power function. (Contributed by Mario Carneiro, 23-Feb-2014.)
((𝑃 ∈ ℙ ∧ 𝑁 ∈ ℕ) → (𝑃 pCnt 𝑁) ∈ ℕ0)
 
Theorempccld 12991 Closure of the prime power function. (Contributed by Mario Carneiro, 29-May-2016.)
(𝜑𝑃 ∈ ℙ)    &   (𝜑𝑁 ∈ ℕ)       (𝜑 → (𝑃 pCnt 𝑁) ∈ ℕ0)
 
Theorempcmul 12992 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)))
 
Theorempcdiv 12993 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)))
 
Theorempcqmul 12994 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)))
 
Theorempc0 12995 The value of the prime power function at zero. (Contributed by Mario Carneiro, 3-Oct-2014.)
(𝑃 ∈ ℙ → (𝑃 pCnt 0) = +∞)
 
Theorempc1 12996 Value of the prime count function at 1. (Contributed by Mario Carneiro, 23-Feb-2014.)
(𝑃 ∈ ℙ → (𝑃 pCnt 1) = 0)
 
Theorempcqcl 12997 Closure of the general prime count function. (Contributed by Mario Carneiro, 23-Feb-2014.)
((𝑃 ∈ ℙ ∧ (𝑁 ∈ ℚ ∧ 𝑁 ≠ 0)) → (𝑃 pCnt 𝑁) ∈ ℤ)
 
Theorempcqdiv 12998 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)))
 
Theorempcrec 12999 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))
 
Theorempcexp 13000 Prime power of an exponential. (Contributed by Mario Carneiro, 10-Aug-2015.)
((𝑃 ∈ ℙ ∧ (𝐴 ∈ ℚ ∧ 𝐴 ≠ 0) ∧ 𝑁 ∈ ℤ) → (𝑃 pCnt (𝐴𝑁)) = (𝑁 · (𝑃 pCnt 𝐴)))
    < Previous  Next >

Page List
Jump to page: Contents  1 1-100 2 101-200 3 201-300 4 301-400 5 401-500 6 501-600 7 601-700 8 701-800 9 801-900 10 901-1000 11 1001-1100 12 1101-1200 13 1201-1300 14 1301-1400 15 1401-1500 16 1501-1600 17 1601-1700 18 1701-1800 19 1801-1900 20 1901-2000 21 2001-2100 22 2101-2200 23 2201-2300 24 2301-2400 25 2401-2500 26 2501-2600 27 2601-2700 28 2701-2800 29 2801-2900 30 2901-3000 31 3001-3100 32 3101-3200 33 3201-3300 34 3301-3400 35 3401-3500 36 3501-3600 37 3601-3700 38 3701-3800 39 3801-3900 40 3901-4000 41 4001-4100 42 4101-4200 43 4201-4300 44 4301-4400 45 4401-4500 46 4501-4600 47 4601-4700 48 4701-4800 49 4801-4900 50 4901-5000 51 5001-5100 52 5101-5200 53 5201-5300 54 5301-5400 55 5401-5500 56 5501-5600 57 5601-5700 58 5701-5800 59 5801-5900 60 5901-6000 61 6001-6100 62 6101-6200 63 6201-6300 64 6301-6400 65 6401-6500 66 6501-6600 67 6601-6700 68 6701-6800 69 6801-6900 70 6901-7000 71 7001-7100 72 7101-7200 73 7201-7300 74 7301-7400 75 7401-7500 76 7501-7600 77 7601-7700 78 7701-7800 79 7801-7900 80 7901-8000 81 8001-8100 82 8101-8200 83 8201-8300 84 8301-8400 85 8401-8500 86 8501-8600 87 8601-8700 88 8701-8800 89 8801-8900 90 8901-9000 91 9001-9100 92 9101-9200 93 9201-9300 94 9301-9400 95 9401-9500 96 9501-9600 97 9601-9700 98 9701-9800 99 9801-9900 100 9901-10000 101 10001-10100 102 10101-10200 103 10201-10300 104 10301-10400 105 10401-10500 106 10501-10600 107 10601-10700 108 10701-10800 109 10801-10900 110 10901-11000 111 11001-11100 112 11101-11200 113 11201-11300 114 11301-11400 115 11401-11500 116 11501-11600 117 11601-11700 118 11701-11800 119 11801-11900 120 11901-12000 121 12001-12100 122 12101-12200 123 12201-12300 124 12301-12400 125 12401-12500 126 12501-12600 127 12601-12700 128 12701-12800 129 12801-12900 130 12901-13000 131 13001-13100 132 13101-13200 133 13201-13300 134 13301-13400 135 13401-13500 136 13501-13600 137 13601-13700 138 13701-13800 139 13801-13900 140 13901-14000 141 14001-14100 142 14101-14200 143 14201-14300 144 14301-14400 145 14401-14500 146 14501-14600 147 14601-14700 148 14701-14800 149 14801-14900 150 14901-15000 151 15001-15100 152 15101-15200 153 15201-15300 154 15301-15400 155 15401-15500 156 15501-15600 157 15601-15700 158 15701-15800 159 15801-15900 160 15901-16000 161 16001-16100 162 16101-16200 163 16201-16300 164 16301-16400 165 16401-16500 166 16501-16600 167 16601-16700 168 16701-16800 169 16801-16867
  Copyright terms: Public domain < Previous  Next >