Theorem List for Intuitionistic Logic Explorer - 12601-12700 *Has distinct variable
group(s)
| Type | Label | Description |
| Statement |
| |
| Theorem | prmdiv 12601 |
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 12602 |
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 12603 |
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 12604* |
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 | dvdsfi 12605* |
A natural number has finitely many divisors. (Contributed by Jim
Kingdon, 9-Oct-2025.)
|
| ⊢ (𝑁 ∈ ℕ → {𝑥 ∈ ℕ ∣ 𝑥 ∥ 𝑁} ∈ Fin) |
| |
| Theorem | hashgcdeq 12606* |
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 12607* |
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 12608* |
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 12609 |
- Lemma for odzcl 12610, 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 12610 |
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 12611 |
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 12612 |
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 12613 |
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)) |
| |
| 5.2.6 Arithmetic modulo a prime
number
|
| |
| Theorem | modprm1div 12614 |
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 12615 |
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 12616 |
Show an explicit expression for the modular inverse of 𝐴 mod 𝑃.
This is an application of prmdiv 12601. (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 12617 |
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 12618 |
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 | powm2modprm 12619 |
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 12620* |
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 12621* |
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 12622* |
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 12623* |
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
|
| |
| Theorem | coprimeprodsq 12624 |
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 12625 |
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 12626 |
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 12627 |
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 12628 |
A prime not equal to 2 is odd. (Contributed by AV,
28-Jun-2021.)
|
| ⊢ (𝑁 ∈ (ℙ ∖ {2}) → ¬
2 ∥ 𝑁) |
| |
| Theorem | nnoddn2prmb 12629 |
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 12630 |
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 12631 |
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 12632* |
Lemma for pythagtrip 12650. 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 12633* |
Lemma for pythagtrip 12650. 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 12634 |
Lemma for pythagtrip 12650. 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 12635 |
Lemma for pythagtrip 12650. 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 12636 |
Lemma for pythagtrip 12650. 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 12637 |
Lemma for pythagtrip 12650. 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 12638 |
Lemma for pythagtrip 12650. 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 12639 |
Lemma for pythagtrip 12650. 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 12640 |
Lemma for pythagtrip 12650. 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 12641 |
Lemma for pythagtrip 12650. 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 12642 |
Lemma for pythagtrip 12650. 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 12643 |
Lemma for pythagtrip 12650. 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 12644 |
Lemma for pythagtrip 12650. 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 12645 |
Lemma for pythagtrip 12650. 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 12646 |
Lemma for pythagtrip 12650. 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 12647 |
Lemma for pythagtrip 12650. 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 12648* |
Lemma for pythagtrip 12650. 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 12649* |
Lemma for pythagtrip 12650. 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 12650* |
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
|
| |
| Syntax | cpc 12651 |
Extend class notation with the prime count function.
|
| class pCnt |
| |
| Definition | df-pc 12652* |
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 | pclem0 12653* |
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 ∣ (𝑃↑𝑛) ∥ 𝑁} ⇒ ⊢ ((𝑃 ∈ (ℤ≥‘2)
∧ (𝑁 ∈ ℤ
∧ 𝑁 ≠ 0)) → 0
∈ 𝐴) |
| |
| Theorem | pclemub 12654* |
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 ∣ (𝑃↑𝑛) ∥ 𝑁} ⇒ ⊢ ((𝑃 ∈ (ℤ≥‘2)
∧ (𝑁 ∈ ℤ
∧ 𝑁 ≠ 0)) →
∃𝑥 ∈ ℤ
∀𝑦 ∈ 𝐴 𝑦 ≤ 𝑥) |
| |
| Theorem | pclemdc 12655* |
Lemma for the prime power pre-function's properties. (Contributed by
Jim Kingdon, 8-Oct-2024.)
|
| ⊢ 𝐴 = {𝑛 ∈ ℕ0 ∣ (𝑃↑𝑛) ∥ 𝑁} ⇒ ⊢ ((𝑃 ∈ (ℤ≥‘2)
∧ (𝑁 ∈ ℤ
∧ 𝑁 ≠ 0)) →
∀𝑥 ∈ ℤ
DECID 𝑥
∈ 𝐴) |
| |
| Theorem | pcprecl 12656* |
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 12657* |
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 12658* |
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 12659* |
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 12660* |
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 | pceulem 12661* |
Lemma for pceu 12662. (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 12662* |
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 | pcval 12663* |
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 | pczpre 12664* |
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 12665 |
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 12666 |
Closure of the prime power function. (Contributed by Mario Carneiro,
23-Feb-2014.)
|
| ⊢ ((𝑃 ∈ ℙ ∧ 𝑁 ∈ ℕ) → (𝑃 pCnt 𝑁) ∈
ℕ0) |
| |
| Theorem | pccld 12667 |
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 12668 |
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 12669 |
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 12670 |
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 12671 |
The value of the prime power function at zero. (Contributed by Mario
Carneiro, 3-Oct-2014.)
|
| ⊢ (𝑃 ∈ ℙ → (𝑃 pCnt 0) = +∞) |
| |
| Theorem | pc1 12672 |
Value of the prime count function at 1. (Contributed by Mario Carneiro,
23-Feb-2014.)
|
| ⊢ (𝑃 ∈ ℙ → (𝑃 pCnt 1) = 0) |
| |
| Theorem | pcqcl 12673 |
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 12674 |
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 12675 |
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 12676 |
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 | pcxnn0cl 12677 |
Extended nonnegative integer closure of the general prime count
function. (Contributed by Jim Kingdon, 13-Oct-2024.)
|
| ⊢ ((𝑃 ∈ ℙ ∧ 𝑁 ∈ ℤ) → (𝑃 pCnt 𝑁) ∈
ℕ0*) |
| |
| Theorem | pcxcl 12678 |
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 | pcxqcl 12679 |
The general prime count function is an integer or infinite.
(Contributed by Jim Kingdon, 6-Jun-2025.)
|
| ⊢ ((𝑃 ∈ ℙ ∧ 𝑁 ∈ ℚ) → ((𝑃 pCnt 𝑁) ∈ ℤ ∨ (𝑃 pCnt 𝑁) = +∞)) |
| |
| Theorem | pcge0 12680 |
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 12681 |
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 12682 |
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 12683 |
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 12684 |
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 12685 |
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 12686 |
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 12687 |
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 12688 |
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 12689 |
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 12690 |
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 12691 |
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 12692 |
The prime count of a negative number. (Contributed by Mario Carneiro,
13-Mar-2014.)
|
| ⊢ ((𝑃 ∈ ℙ ∧ 𝐴 ∈ ℚ) → (𝑃 pCnt -𝐴) = (𝑃 pCnt 𝐴)) |
| |
| Theorem | pcabs 12693 |
The prime count of an absolute value. (Contributed by Mario Carneiro,
13-Mar-2014.)
|
| ⊢ ((𝑃 ∈ ℙ ∧ 𝐴 ∈ ℚ) → (𝑃 pCnt (abs‘𝐴)) = (𝑃 pCnt 𝐴)) |
| |
| Theorem | pcdvdstr 12694 |
The prime count increases under the divisibility relation. (Contributed
by Mario Carneiro, 13-Mar-2014.)
|
| ⊢ ((𝑃 ∈ ℙ ∧ (𝐴 ∈ ℤ ∧ 𝐵 ∈ ℤ ∧ 𝐴 ∥ 𝐵)) → (𝑃 pCnt 𝐴) ≤ (𝑃 pCnt 𝐵)) |
| |
| Theorem | pcgcd1 12695 |
The prime count of a GCD is the minimum of the prime counts of the
arguments. (Contributed by Mario Carneiro, 3-Oct-2014.)
|
| ⊢ (((𝑃 ∈ ℙ ∧ 𝐴 ∈ ℤ ∧ 𝐵 ∈ ℤ) ∧ (𝑃 pCnt 𝐴) ≤ (𝑃 pCnt 𝐵)) → (𝑃 pCnt (𝐴 gcd 𝐵)) = (𝑃 pCnt 𝐴)) |
| |
| Theorem | pcgcd 12696 |
The prime count of a GCD is the minimum of the prime counts of the
arguments. (Contributed by Mario Carneiro, 3-Oct-2014.)
|
| ⊢ ((𝑃 ∈ ℙ ∧ 𝐴 ∈ ℤ ∧ 𝐵 ∈ ℤ) → (𝑃 pCnt (𝐴 gcd 𝐵)) = if((𝑃 pCnt 𝐴) ≤ (𝑃 pCnt 𝐵), (𝑃 pCnt 𝐴), (𝑃 pCnt 𝐵))) |
| |
| Theorem | pc2dvds 12697* |
A characterization of divisibility in terms of prime count.
(Contributed by Mario Carneiro, 23-Feb-2014.) (Revised by Mario
Carneiro, 3-Oct-2014.)
|
| ⊢ ((𝐴 ∈ ℤ ∧ 𝐵 ∈ ℤ) → (𝐴 ∥ 𝐵 ↔ ∀𝑝 ∈ ℙ (𝑝 pCnt 𝐴) ≤ (𝑝 pCnt 𝐵))) |
| |
| Theorem | pc11 12698* |
The prime count function, viewed as a function from ℕ to
(ℕ ↑𝑚 ℙ), is
one-to-one. (Contributed by Mario Carneiro,
23-Feb-2014.)
|
| ⊢ ((𝐴 ∈ ℕ0 ∧ 𝐵 ∈ ℕ0)
→ (𝐴 = 𝐵 ↔ ∀𝑝 ∈ ℙ (𝑝 pCnt 𝐴) = (𝑝 pCnt 𝐵))) |
| |
| Theorem | pcz 12699* |
The prime count function can be used as an indicator that a given
rational number is an integer. (Contributed by Mario Carneiro,
23-Feb-2014.)
|
| ⊢ (𝐴 ∈ ℚ → (𝐴 ∈ ℤ ↔ ∀𝑝 ∈ ℙ 0 ≤ (𝑝 pCnt 𝐴))) |
| |
| Theorem | pcprmpw2 12700* |
Self-referential expression for a prime power. (Contributed by Mario
Carneiro, 16-Jan-2015.)
|
| ⊢ ((𝑃 ∈ ℙ ∧ 𝐴 ∈ ℕ) → (∃𝑛 ∈ ℕ0
𝐴 ∥ (𝑃↑𝑛) ↔ 𝐴 = (𝑃↑(𝑃 pCnt 𝐴)))) |