Theorem List for Intuitionistic Logic Explorer - 12201-12300 *Has distinct variable
group(s)
Type | Label | Description |
Statement |
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5.2.6 Arithmetic modulo a prime
number
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Theorem | modprm1div 12201 |
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.)
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⊢ ((𝑃 ∈ ℙ ∧ 𝐴 ∈ ℤ) → ((𝐴 mod 𝑃) = 1 ↔ 𝑃 ∥ (𝐴 − 1))) |
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Theorem | m1dvdsndvds 12202 |
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.)
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⊢ ((𝑃 ∈ ℙ ∧ 𝐴 ∈ ℤ) → (𝑃 ∥ (𝐴 − 1) → ¬ 𝑃 ∥ 𝐴)) |
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Theorem | modprminv 12203 |
Show an explicit expression for the modular inverse of 𝐴 mod 𝑃.
This is an application of prmdiv 12189. (Contributed by Alexander van der
Vekens, 15-May-2018.)
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⊢ 𝑅 = ((𝐴↑(𝑃 − 2)) mod 𝑃) ⇒ ⊢ ((𝑃 ∈ ℙ ∧ 𝐴 ∈ ℤ ∧ ¬ 𝑃 ∥ 𝐴) → (𝑅 ∈ (1...(𝑃 − 1)) ∧ ((𝐴 · 𝑅) mod 𝑃) = 1)) |
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Theorem | modprminveq 12204 |
The modular inverse of 𝐴 mod 𝑃 is unique. (Contributed by
Alexander
van der Vekens, 17-May-2018.)
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⊢ 𝑅 = ((𝐴↑(𝑃 − 2)) mod 𝑃) ⇒ ⊢ ((𝑃 ∈ ℙ ∧ 𝐴 ∈ ℤ ∧ ¬ 𝑃 ∥ 𝐴) → ((𝑆 ∈ (0...(𝑃 − 1)) ∧ ((𝐴 · 𝑆) mod 𝑃) = 1) ↔ 𝑆 = 𝑅)) |
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Theorem | vfermltl 12205 |
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.)
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⊢ ((𝑃 ∈ ℙ ∧ 𝐴 ∈ ℤ ∧ ¬ 𝑃 ∥ 𝐴) → ((𝐴↑(𝑃 − 1)) mod 𝑃) = 1) |
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Theorem | powm2modprm 12206 |
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.)
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⊢ ((𝑃 ∈ ℙ ∧ 𝐴 ∈ ℤ) → (𝑃 ∥ (𝐴 − 1) → ((𝐴↑(𝑃 − 2)) mod 𝑃) = 1)) |
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Theorem | reumodprminv 12207* |
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.)
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⊢ ((𝑃 ∈ ℙ ∧ 𝑁 ∈ (1..^𝑃)) → ∃!𝑖 ∈ (1...(𝑃 − 1))((𝑁 · 𝑖) mod 𝑃) = 1) |
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Theorem | modprm0 12208* |
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.)
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⊢ ((𝑃 ∈ ℙ ∧ 𝑁 ∈ (1..^𝑃) ∧ 𝐼 ∈ (1..^𝑃)) → ∃𝑗 ∈ (0..^𝑃)((𝐼 + (𝑗 · 𝑁)) mod 𝑃) = 0) |
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Theorem | nnnn0modprm0 12209* |
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.)
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⊢ ((𝑃 ∈ ℙ ∧ 𝑁 ∈ (1..^𝑃) ∧ 𝐼 ∈ (0..^𝑃)) → ∃𝑗 ∈ (0..^𝑃)((𝐼 + (𝑗 · 𝑁)) mod 𝑃) = 0) |
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Theorem | modprmn0modprm0 12210* |
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.)
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⊢ ((𝑃 ∈ ℙ ∧ 𝑁 ∈ ℤ ∧ (𝑁 mod 𝑃) ≠ 0) → (𝐼 ∈ (0..^𝑃) → ∃𝑗 ∈ (0..^𝑃)((𝐼 + (𝑗 · 𝑁)) mod 𝑃) = 0)) |
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5.2.7 Pythagorean Triples
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Theorem | coprimeprodsq 12211 |
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.)
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⊢ (((𝐴 ∈ ℕ0 ∧ 𝐵 ∈ ℤ ∧ 𝐶 ∈ ℕ0)
∧ ((𝐴 gcd 𝐵) gcd 𝐶) = 1) → ((𝐶↑2) = (𝐴 · 𝐵) → 𝐴 = ((𝐴 gcd 𝐶)↑2))) |
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Theorem | coprimeprodsq2 12212 |
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.)
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⊢ (((𝐴 ∈ ℤ ∧ 𝐵 ∈ ℕ0 ∧ 𝐶 ∈ ℕ0)
∧ ((𝐴 gcd 𝐵) gcd 𝐶) = 1) → ((𝐶↑2) = (𝐴 · 𝐵) → 𝐵 = ((𝐵 gcd 𝐶)↑2))) |
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Theorem | oddprm 12213 |
A prime not equal to 2 is odd. (Contributed by Mario
Carneiro,
4-Feb-2015.) (Proof shortened by AV, 10-Jul-2022.)
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⊢ (𝑁 ∈ (ℙ ∖ {2}) →
((𝑁 − 1) / 2) ∈
ℕ) |
|
Theorem | nnoddn2prm 12214 |
A prime not equal to 2 is an odd positive integer.
(Contributed by
AV, 28-Jun-2021.)
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⊢ (𝑁 ∈ (ℙ ∖ {2}) → (𝑁 ∈ ℕ ∧ ¬ 2
∥ 𝑁)) |
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Theorem | oddn2prm 12215 |
A prime not equal to 2 is odd. (Contributed by AV,
28-Jun-2021.)
|
⊢ (𝑁 ∈ (ℙ ∖ {2}) → ¬
2 ∥ 𝑁) |
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Theorem | nnoddn2prmb 12216 |
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.)
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⊢ (𝑁 ∈ (ℙ ∖ {2}) ↔ (𝑁 ∈ ℙ ∧ ¬ 2
∥ 𝑁)) |
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Theorem | prm23lt5 12217 |
A prime less than 5 is either 2 or 3. (Contributed by AV, 5-Jul-2021.)
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⊢ ((𝑃 ∈ ℙ ∧ 𝑃 < 5) → (𝑃 = 2 ∨ 𝑃 = 3)) |
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Theorem | prm23ge5 12218 |
A prime is either 2 or 3 or greater than or equal to 5. (Contributed by
AV, 5-Jul-2021.)
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⊢ (𝑃 ∈ ℙ → (𝑃 = 2 ∨ 𝑃 = 3 ∨ 𝑃 ∈
(ℤ≥‘5))) |
|
Theorem | pythagtriplem1 12219* |
Lemma for pythagtrip 12237. 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)) |
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Theorem | pythagtriplem2 12220* |
Lemma for pythagtrip 12237. Prove the full version of one direction of
the
theorem. (Contributed by Scott Fenton, 28-Mar-2014.) (Revised by Mario
Carneiro, 19-Apr-2014.)
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⊢ ((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))) |
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Theorem | pythagtriplem3 12221 |
Lemma for pythagtrip 12237. 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.)
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⊢ (((𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝐶 ∈ ℕ) ∧ ((𝐴↑2) + (𝐵↑2)) = (𝐶↑2) ∧ ((𝐴 gcd 𝐵) = 1 ∧ ¬ 2 ∥ 𝐴)) → (𝐵 gcd 𝐶) = 1) |
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Theorem | pythagtriplem4 12222 |
Lemma for pythagtrip 12237. Show that 𝐶 − 𝐵 and 𝐶 + 𝐵 are relatively
prime. (Contributed by Scott Fenton, 12-Apr-2014.) (Revised by Mario
Carneiro, 19-Apr-2014.)
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⊢ (((𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝐶 ∈ ℕ) ∧ ((𝐴↑2) + (𝐵↑2)) = (𝐶↑2) ∧ ((𝐴 gcd 𝐵) = 1 ∧ ¬ 2 ∥ 𝐴)) → ((𝐶 − 𝐵) gcd (𝐶 + 𝐵)) = 1) |
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Theorem | pythagtriplem10 12223 |
Lemma for pythagtrip 12237. Show that 𝐶 − 𝐵 is positive. (Contributed
by Scott Fenton, 17-Apr-2014.) (Revised by Mario Carneiro,
19-Apr-2014.)
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⊢ (((𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝐶 ∈ ℕ) ∧ ((𝐴↑2) + (𝐵↑2)) = (𝐶↑2)) → 0 < (𝐶 − 𝐵)) |
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Theorem | pythagtriplem6 12224 |
Lemma for pythagtrip 12237. Calculate (√‘(𝐶 − 𝐵)).
(Contributed by Scott Fenton, 18-Apr-2014.) (Revised by Mario Carneiro,
19-Apr-2014.)
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⊢ (((𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝐶 ∈ ℕ) ∧ ((𝐴↑2) + (𝐵↑2)) = (𝐶↑2) ∧ ((𝐴 gcd 𝐵) = 1 ∧ ¬ 2 ∥ 𝐴)) → (√‘(𝐶 − 𝐵)) = ((𝐶 − 𝐵) gcd 𝐴)) |
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Theorem | pythagtriplem7 12225 |
Lemma for pythagtrip 12237. Calculate (√‘(𝐶 + 𝐵)).
(Contributed by Scott Fenton, 18-Apr-2014.) (Revised by Mario Carneiro,
19-Apr-2014.)
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⊢ (((𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝐶 ∈ ℕ) ∧ ((𝐴↑2) + (𝐵↑2)) = (𝐶↑2) ∧ ((𝐴 gcd 𝐵) = 1 ∧ ¬ 2 ∥ 𝐴)) → (√‘(𝐶 + 𝐵)) = ((𝐶 + 𝐵) gcd 𝐴)) |
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Theorem | pythagtriplem8 12226 |
Lemma for pythagtrip 12237. 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 12227 |
Lemma for pythagtrip 12237. 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 12228 |
Lemma for pythagtrip 12237. 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 12229 |
Lemma for pythagtrip 12237. 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 12230 |
Lemma for pythagtrip 12237. 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 12231 |
Lemma for pythagtrip 12237. 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 12232 |
Lemma for pythagtrip 12237. 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 12233 |
Lemma for pythagtrip 12237. 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 12234 |
Lemma for pythagtrip 12237. 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 12235* |
Lemma for pythagtrip 12237. 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 12236* |
Lemma for pythagtrip 12237. 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 12237* |
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 12238 |
Extend class notation with the prime count function.
|
class pCnt |
|
Definition | df-pc 12239* |
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 12240* |
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 12241* |
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 12242* |
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 12243* |
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 12244* |
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 12245* |
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 12246* |
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 12247* |
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 12248* |
Lemma for pceu 12249. (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 12249* |
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 12250* |
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 12251* |
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 12252 |
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 12253 |
Closure of the prime power function. (Contributed by Mario Carneiro,
23-Feb-2014.)
|
⊢ ((𝑃 ∈ ℙ ∧ 𝑁 ∈ ℕ) → (𝑃 pCnt 𝑁) ∈
ℕ0) |
|
Theorem | pccld 12254 |
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 12255 |
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 12256 |
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 12257 |
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 12258 |
The value of the prime power function at zero. (Contributed by Mario
Carneiro, 3-Oct-2014.)
|
⊢ (𝑃 ∈ ℙ → (𝑃 pCnt 0) = +∞) |
|
Theorem | pc1 12259 |
Value of the prime count function at 1. (Contributed by Mario Carneiro,
23-Feb-2014.)
|
⊢ (𝑃 ∈ ℙ → (𝑃 pCnt 1) = 0) |
|
Theorem | pcqcl 12260 |
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 12261 |
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 12262 |
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 12263 |
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 12264 |
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 12265 |
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 12266 |
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 12267 |
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 12268 |
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 12269 |
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 12270 |
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 12271 |
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 12272 |
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 12273 |
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 12274 |
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 12275 |
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 12276 |
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 12277 |
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 12278 |
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 12279 |
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 12280 |
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 12281 |
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 12282 |
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 12283* |
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 12284* |
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 12285* |
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 12286* |
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)))) |
|
Theorem | pcprmpw 12287* |
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)))) |
|
Theorem | dvdsprmpweq 12288* |
If a positive integer divides a prime power, it is a prime power.
(Contributed by AV, 25-Jul-2021.)
|
⊢ ((𝑃 ∈ ℙ ∧ 𝐴 ∈ ℕ ∧ 𝑁 ∈ ℕ0) → (𝐴 ∥ (𝑃↑𝑁) → ∃𝑛 ∈ ℕ0 𝐴 = (𝑃↑𝑛))) |
|
Theorem | dvdsprmpweqnn 12289* |
If an integer greater than 1 divides a prime power, it is a (proper)
prime power. (Contributed by AV, 13-Aug-2021.)
|
⊢ ((𝑃 ∈ ℙ ∧ 𝐴 ∈ (ℤ≥‘2)
∧ 𝑁 ∈
ℕ0) → (𝐴 ∥ (𝑃↑𝑁) → ∃𝑛 ∈ ℕ 𝐴 = (𝑃↑𝑛))) |
|
Theorem | dvdsprmpweqle 12290* |
If a positive integer divides a prime power, it is a prime power with a
smaller exponent. (Contributed by AV, 25-Jul-2021.)
|
⊢ ((𝑃 ∈ ℙ ∧ 𝐴 ∈ ℕ ∧ 𝑁 ∈ ℕ0) → (𝐴 ∥ (𝑃↑𝑁) → ∃𝑛 ∈ ℕ0 (𝑛 ≤ 𝑁 ∧ 𝐴 = (𝑃↑𝑛)))) |
|
Theorem | difsqpwdvds 12291 |
If the difference of two squares is a power of a prime, the prime
divides twice the second squared number. (Contributed by AV,
13-Aug-2021.)
|
⊢ (((𝐴 ∈ ℕ0 ∧ 𝐵 ∈ ℕ0
∧ (𝐵 + 1) < 𝐴) ∧ (𝐶 ∈ ℙ ∧ 𝐷 ∈ ℕ0)) →
((𝐶↑𝐷) = ((𝐴↑2) − (𝐵↑2)) → 𝐶 ∥ (2 · 𝐵))) |
|
Theorem | pcaddlem 12292 |
Lemma for pcadd 12293. The original numbers 𝐴 and
𝐵
have been
decomposed using the prime count function as (𝑃↑𝑀) · (𝑅 / 𝑆)
where 𝑅, 𝑆 are both not divisible by 𝑃 and
𝑀 =
(𝑃 pCnt 𝐴), and similarly for 𝐵.
(Contributed by Mario
Carneiro, 9-Sep-2014.)
|
⊢ (𝜑 → 𝑃 ∈ ℙ) & ⊢ (𝜑 → 𝐴 = ((𝑃↑𝑀) · (𝑅 / 𝑆))) & ⊢ (𝜑 → 𝐵 = ((𝑃↑𝑁) · (𝑇 / 𝑈))) & ⊢ (𝜑 → 𝑁 ∈ (ℤ≥‘𝑀)) & ⊢ (𝜑 → (𝑅 ∈ ℤ ∧ ¬ 𝑃 ∥ 𝑅)) & ⊢ (𝜑 → (𝑆 ∈ ℕ ∧ ¬ 𝑃 ∥ 𝑆)) & ⊢ (𝜑 → (𝑇 ∈ ℤ ∧ ¬ 𝑃 ∥ 𝑇)) & ⊢ (𝜑 → (𝑈 ∈ ℕ ∧ ¬ 𝑃 ∥ 𝑈)) ⇒ ⊢ (𝜑 → 𝑀 ≤ (𝑃 pCnt (𝐴 + 𝐵))) |
|
Theorem | pcadd 12293 |
An inequality for the prime count of a sum. This is the source of the
ultrametric inequality for the p-adic metric. (Contributed by Mario
Carneiro, 9-Sep-2014.)
|
⊢ (𝜑 → 𝑃 ∈ ℙ) & ⊢ (𝜑 → 𝐴 ∈ ℚ) & ⊢ (𝜑 → 𝐵 ∈ ℚ) & ⊢ (𝜑 → (𝑃 pCnt 𝐴) ≤ (𝑃 pCnt 𝐵)) ⇒ ⊢ (𝜑 → (𝑃 pCnt 𝐴) ≤ (𝑃 pCnt (𝐴 + 𝐵))) |
|
Theorem | pcmptcl 12294 |
Closure for the prime power map. (Contributed by Mario Carneiro,
12-Mar-2014.)
|
⊢ 𝐹 = (𝑛 ∈ ℕ ↦ if(𝑛 ∈ ℙ, (𝑛↑𝐴), 1)) & ⊢ (𝜑 → ∀𝑛 ∈ ℙ 𝐴 ∈
ℕ0) ⇒ ⊢ (𝜑 → (𝐹:ℕ⟶ℕ ∧ seq1(
· , 𝐹):ℕ⟶ℕ)) |
|
Theorem | pcmpt 12295* |
Construct a function with given prime count characteristics.
(Contributed by Mario Carneiro, 12-Mar-2014.)
|
⊢ 𝐹 = (𝑛 ∈ ℕ ↦ if(𝑛 ∈ ℙ, (𝑛↑𝐴), 1)) & ⊢ (𝜑 → ∀𝑛 ∈ ℙ 𝐴 ∈
ℕ0)
& ⊢ (𝜑 → 𝑁 ∈ ℕ) & ⊢ (𝜑 → 𝑃 ∈ ℙ) & ⊢ (𝑛 = 𝑃 → 𝐴 = 𝐵) ⇒ ⊢ (𝜑 → (𝑃 pCnt (seq1( · , 𝐹)‘𝑁)) = if(𝑃 ≤ 𝑁, 𝐵, 0)) |
|
Theorem | pcmpt2 12296* |
Dividing two prime count maps yields a number with all dividing primes
confined to an interval. (Contributed by Mario Carneiro,
14-Mar-2014.)
|
⊢ 𝐹 = (𝑛 ∈ ℕ ↦ if(𝑛 ∈ ℙ, (𝑛↑𝐴), 1)) & ⊢ (𝜑 → ∀𝑛 ∈ ℙ 𝐴 ∈
ℕ0)
& ⊢ (𝜑 → 𝑁 ∈ ℕ) & ⊢ (𝜑 → 𝑃 ∈ ℙ) & ⊢ (𝑛 = 𝑃 → 𝐴 = 𝐵)
& ⊢ (𝜑 → 𝑀 ∈ (ℤ≥‘𝑁))
⇒ ⊢ (𝜑 → (𝑃 pCnt ((seq1( · , 𝐹)‘𝑀) / (seq1( · , 𝐹)‘𝑁))) = if((𝑃 ≤ 𝑀 ∧ ¬ 𝑃 ≤ 𝑁), 𝐵, 0)) |
|
Theorem | pcmptdvds 12297 |
The partial products of the prime power map form a divisibility chain.
(Contributed by Mario Carneiro, 12-Mar-2014.)
|
⊢ 𝐹 = (𝑛 ∈ ℕ ↦ if(𝑛 ∈ ℙ, (𝑛↑𝐴), 1)) & ⊢ (𝜑 → ∀𝑛 ∈ ℙ 𝐴 ∈
ℕ0)
& ⊢ (𝜑 → 𝑁 ∈ ℕ) & ⊢ (𝜑 → 𝑀 ∈ (ℤ≥‘𝑁))
⇒ ⊢ (𝜑 → (seq1( · , 𝐹)‘𝑁) ∥ (seq1( · , 𝐹)‘𝑀)) |
|
Theorem | pcprod 12298* |
The product of the primes taken to their respective powers reconstructs
the original number. (Contributed by Mario Carneiro, 12-Mar-2014.)
|
⊢ 𝐹 = (𝑛 ∈ ℕ ↦ if(𝑛 ∈ ℙ, (𝑛↑(𝑛 pCnt 𝑁)), 1)) ⇒ ⊢ (𝑁 ∈ ℕ → (seq1( · ,
𝐹)‘𝑁) = 𝑁) |
|
Theorem | sumhashdc 12299* |
The sum of 1 over a set is the size of the set. (Contributed by Mario
Carneiro, 8-Mar-2014.) (Revised by Mario Carneiro, 20-May-2014.)
|
⊢ ((𝐵 ∈ Fin ∧ 𝐴 ⊆ 𝐵 ∧ ∀𝑥 ∈ 𝐵 DECID 𝑥 ∈ 𝐴) → Σ𝑘 ∈ 𝐵 if(𝑘 ∈ 𝐴, 1, 0) = (♯‘𝐴)) |
|
Theorem | fldivp1 12300 |
The difference between the floors of adjacent fractions is either 1 or 0.
(Contributed by Mario Carneiro, 8-Mar-2014.)
|
⊢ ((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℕ) →
((⌊‘((𝑀 + 1) /
𝑁)) −
(⌊‘(𝑀 / 𝑁))) = if(𝑁 ∥ (𝑀 + 1), 1, 0)) |