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Type | Label | Description |
---|---|---|
Statement | ||
Theorem | bezoutlemmo 10901* | Lemma for Bézout's identity. There is at most one nonnegative integer meeting the greatest common divisor condition. (Contributed by Mario Carneiro and Jim Kingdon, 9-Jan-2022.) |
⊢ (𝜑 → 𝐴 ∈ ℤ) & ⊢ (𝜑 → 𝐵 ∈ ℤ) & ⊢ (𝜑 → 𝐷 ∈ ℕ_{0}) & ⊢ (𝜑 → ∀𝑧 ∈ ℤ (𝑧 ∥ 𝐷 ↔ (𝑧 ∥ 𝐴 ∧ 𝑧 ∥ 𝐵))) & ⊢ (𝜑 → 𝐸 ∈ ℕ_{0}) & ⊢ (𝜑 → ∀𝑧 ∈ ℤ (𝑧 ∥ 𝐸 ↔ (𝑧 ∥ 𝐴 ∧ 𝑧 ∥ 𝐵))) ⇒ ⊢ (𝜑 → 𝐷 = 𝐸) | ||
Theorem | bezoutlemeu 10902* | Lemma for Bézout's identity. There is exactly one nonnegative integer meeting the greatest common divisor condition. (Contributed by Mario Carneiro and Jim Kingdon, 9-Jan-2022.) |
⊢ (𝜑 → 𝐴 ∈ ℤ) & ⊢ (𝜑 → 𝐵 ∈ ℤ) & ⊢ (𝜑 → 𝐷 ∈ ℕ_{0}) & ⊢ (𝜑 → ∀𝑧 ∈ ℤ (𝑧 ∥ 𝐷 ↔ (𝑧 ∥ 𝐴 ∧ 𝑧 ∥ 𝐵))) ⇒ ⊢ (𝜑 → ∃!𝑑 ∈ ℕ_{0} ∀𝑧 ∈ ℤ (𝑧 ∥ 𝑑 ↔ (𝑧 ∥ 𝐴 ∧ 𝑧 ∥ 𝐵))) | ||
Theorem | bezoutlemle 10903* | Lemma for Bézout's identity. The number satisfying the greatest common divisor condition is the largest number which divides both 𝐴 and 𝐵. (Contributed by Mario Carneiro and Jim Kingdon, 9-Jan-2022.) |
⊢ (𝜑 → 𝐴 ∈ ℤ) & ⊢ (𝜑 → 𝐵 ∈ ℤ) & ⊢ (𝜑 → 𝐷 ∈ ℕ_{0}) & ⊢ (𝜑 → ∀𝑧 ∈ ℤ (𝑧 ∥ 𝐷 ↔ (𝑧 ∥ 𝐴 ∧ 𝑧 ∥ 𝐵))) & ⊢ (𝜑 → ¬ (𝐴 = 0 ∧ 𝐵 = 0)) ⇒ ⊢ (𝜑 → ∀𝑧 ∈ ℤ ((𝑧 ∥ 𝐴 ∧ 𝑧 ∥ 𝐵) → 𝑧 ≤ 𝐷)) | ||
Theorem | bezoutlemsup 10904* | Lemma for Bézout's identity. The number satisfying the greatest common divisor condition is the supremum of divisors of both 𝐴 and 𝐵. (Contributed by Mario Carneiro and Jim Kingdon, 9-Jan-2022.) |
⊢ (𝜑 → 𝐴 ∈ ℤ) & ⊢ (𝜑 → 𝐵 ∈ ℤ) & ⊢ (𝜑 → 𝐷 ∈ ℕ_{0}) & ⊢ (𝜑 → ∀𝑧 ∈ ℤ (𝑧 ∥ 𝐷 ↔ (𝑧 ∥ 𝐴 ∧ 𝑧 ∥ 𝐵))) & ⊢ (𝜑 → ¬ (𝐴 = 0 ∧ 𝐵 = 0)) ⇒ ⊢ (𝜑 → 𝐷 = sup({𝑧 ∈ ℤ ∣ (𝑧 ∥ 𝐴 ∧ 𝑧 ∥ 𝐵)}, ℝ, < )) | ||
Theorem | dfgcd3 10905* | Alternate definition of the gcd operator. (Contributed by Jim Kingdon, 31-Dec-2021.) |
⊢ ((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → (𝑀 gcd 𝑁) = (℩𝑑 ∈ ℕ_{0} ∀𝑧 ∈ ℤ (𝑧 ∥ 𝑑 ↔ (𝑧 ∥ 𝑀 ∧ 𝑧 ∥ 𝑁)))) | ||
Theorem | bezout 10906* |
Bézout's identity: For any integers 𝐴 and 𝐵, there are
integers 𝑥, 𝑦 such that (𝐴 gcd 𝐵) = 𝐴 · 𝑥 + 𝐵 · 𝑦. This
is Metamath 100 proof #60.
The proof is constructive, in the sense that it applies the Extended Euclidian Algorithm to constuct a number which can be shown to be (𝐴 gcd 𝐵) and which satisfies the rest of the theorem. In the presence of excluded middle, it is common to prove Bézout's identity by taking the smallest number which satisfies the Bézout condition, and showing it is the greatest common divisor. But we do not have the ability to show that number exists other than by providing a way to determine it. (Contributed by Mario Carneiro, 22-Feb-2014.) |
⊢ ((𝐴 ∈ ℤ ∧ 𝐵 ∈ ℤ) → ∃𝑥 ∈ ℤ ∃𝑦 ∈ ℤ (𝐴 gcd 𝐵) = ((𝐴 · 𝑥) + (𝐵 · 𝑦))) | ||
Theorem | dvdsgcd 10907 | An integer which divides each of two others also divides their gcd. (Contributed by Paul Chapman, 22-Jun-2011.) (Revised by Mario Carneiro, 30-May-2014.) |
⊢ ((𝐾 ∈ ℤ ∧ 𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → ((𝐾 ∥ 𝑀 ∧ 𝐾 ∥ 𝑁) → 𝐾 ∥ (𝑀 gcd 𝑁))) | ||
Theorem | dvdsgcdb 10908 | Biconditional form of dvdsgcd 10907. (Contributed by Scott Fenton, 2-Apr-2014.) (Revised by Mario Carneiro, 19-Apr-2014.) |
⊢ ((𝐾 ∈ ℤ ∧ 𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → ((𝐾 ∥ 𝑀 ∧ 𝐾 ∥ 𝑁) ↔ 𝐾 ∥ (𝑀 gcd 𝑁))) | ||
Theorem | dfgcd2 10909* | Alternate definition of the gcd operator, see definition in [ApostolNT] p. 15. (Contributed by AV, 8-Aug-2021.) |
⊢ ((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → (𝐷 = (𝑀 gcd 𝑁) ↔ (0 ≤ 𝐷 ∧ (𝐷 ∥ 𝑀 ∧ 𝐷 ∥ 𝑁) ∧ ∀𝑒 ∈ ℤ ((𝑒 ∥ 𝑀 ∧ 𝑒 ∥ 𝑁) → 𝑒 ∥ 𝐷)))) | ||
Theorem | gcdass 10910 | Associative law for gcd operator. Theorem 1.4(b) in [ApostolNT] p. 16. (Contributed by Scott Fenton, 2-Apr-2014.) (Revised by Mario Carneiro, 19-Apr-2014.) |
⊢ ((𝑁 ∈ ℤ ∧ 𝑀 ∈ ℤ ∧ 𝑃 ∈ ℤ) → ((𝑁 gcd 𝑀) gcd 𝑃) = (𝑁 gcd (𝑀 gcd 𝑃))) | ||
Theorem | mulgcd 10911 | Distribute multiplication by a nonnegative integer over gcd. (Contributed by Paul Chapman, 22-Jun-2011.) (Proof shortened by Mario Carneiro, 30-May-2014.) |
⊢ ((𝐾 ∈ ℕ_{0} ∧ 𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → ((𝐾 · 𝑀) gcd (𝐾 · 𝑁)) = (𝐾 · (𝑀 gcd 𝑁))) | ||
Theorem | absmulgcd 10912 | Distribute absolute value of multiplication over gcd. Theorem 1.4(c) in [ApostolNT] p. 16. (Contributed by Paul Chapman, 22-Jun-2011.) |
⊢ ((𝐾 ∈ ℤ ∧ 𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → ((𝐾 · 𝑀) gcd (𝐾 · 𝑁)) = (abs‘(𝐾 · (𝑀 gcd 𝑁)))) | ||
Theorem | mulgcdr 10913 | Reverse distribution law for the gcd operator. (Contributed by Scott Fenton, 2-Apr-2014.) (Revised by Mario Carneiro, 19-Apr-2014.) |
⊢ ((𝐴 ∈ ℤ ∧ 𝐵 ∈ ℤ ∧ 𝐶 ∈ ℕ_{0}) → ((𝐴 · 𝐶) gcd (𝐵 · 𝐶)) = ((𝐴 gcd 𝐵) · 𝐶)) | ||
Theorem | gcddiv 10914 | Division law for GCD. (Contributed by Scott Fenton, 18-Apr-2014.) (Revised by Mario Carneiro, 19-Apr-2014.) |
⊢ (((𝐴 ∈ ℤ ∧ 𝐵 ∈ ℤ ∧ 𝐶 ∈ ℕ) ∧ (𝐶 ∥ 𝐴 ∧ 𝐶 ∥ 𝐵)) → ((𝐴 gcd 𝐵) / 𝐶) = ((𝐴 / 𝐶) gcd (𝐵 / 𝐶))) | ||
Theorem | gcdmultiple 10915 | The GCD of a multiple of a number is the number itself. (Contributed by Scott Fenton, 12-Apr-2014.) (Revised by Mario Carneiro, 19-Apr-2014.) |
⊢ ((𝑀 ∈ ℕ ∧ 𝑁 ∈ ℕ) → (𝑀 gcd (𝑀 · 𝑁)) = 𝑀) | ||
Theorem | gcdmultiplez 10916 | Extend gcdmultiple 10915 so 𝑁 can be an integer. (Contributed by Scott Fenton, 18-Apr-2014.) (Revised by Mario Carneiro, 19-Apr-2014.) |
⊢ ((𝑀 ∈ ℕ ∧ 𝑁 ∈ ℤ) → (𝑀 gcd (𝑀 · 𝑁)) = 𝑀) | ||
Theorem | gcdzeq 10917 | A positive integer 𝐴 is equal to its gcd with an integer 𝐵 if and only if 𝐴 divides 𝐵. Generalization of gcdeq 10918. (Contributed by AV, 1-Jul-2020.) |
⊢ ((𝐴 ∈ ℕ ∧ 𝐵 ∈ ℤ) → ((𝐴 gcd 𝐵) = 𝐴 ↔ 𝐴 ∥ 𝐵)) | ||
Theorem | gcdeq 10918 | 𝐴 is equal to its gcd with 𝐵 if and only if 𝐴 divides 𝐵. (Contributed by Mario Carneiro, 23-Feb-2014.) (Proof shortened by AV, 8-Aug-2021.) |
⊢ ((𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ) → ((𝐴 gcd 𝐵) = 𝐴 ↔ 𝐴 ∥ 𝐵)) | ||
Theorem | dvdssqim 10919 | Unidirectional form of dvdssq 10926. (Contributed by Scott Fenton, 19-Apr-2014.) |
⊢ ((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → (𝑀 ∥ 𝑁 → (𝑀↑2) ∥ (𝑁↑2))) | ||
Theorem | dvdsmulgcd 10920 | Relationship between the order of an element and that of a multiple. (a divisibility equivalent). (Contributed by Stefan O'Rear, 6-Sep-2015.) |
⊢ ((𝐵 ∈ ℤ ∧ 𝐶 ∈ ℤ) → (𝐴 ∥ (𝐵 · 𝐶) ↔ 𝐴 ∥ (𝐵 · (𝐶 gcd 𝐴)))) | ||
Theorem | rpmulgcd 10921 | If 𝐾 and 𝑀 are relatively prime, then the GCD of 𝐾 and 𝑀 · 𝑁 is the GCD of 𝐾 and 𝑁. (Contributed by Scott Fenton, 12-Apr-2014.) (Revised by Mario Carneiro, 19-Apr-2014.) |
⊢ (((𝐾 ∈ ℕ ∧ 𝑀 ∈ ℕ ∧ 𝑁 ∈ ℕ) ∧ (𝐾 gcd 𝑀) = 1) → (𝐾 gcd (𝑀 · 𝑁)) = (𝐾 gcd 𝑁)) | ||
Theorem | rplpwr 10922 | If 𝐴 and 𝐵 are relatively prime, then so are 𝐴↑𝑁 and 𝐵. (Contributed by Scott Fenton, 12-Apr-2014.) (Revised by Mario Carneiro, 19-Apr-2014.) |
⊢ ((𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝑁 ∈ ℕ) → ((𝐴 gcd 𝐵) = 1 → ((𝐴↑𝑁) gcd 𝐵) = 1)) | ||
Theorem | rppwr 10923 | If 𝐴 and 𝐵 are relatively prime, then so are 𝐴↑𝑁 and 𝐵↑𝑁. (Contributed by Scott Fenton, 12-Apr-2014.) (Revised by Mario Carneiro, 19-Apr-2014.) |
⊢ ((𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝑁 ∈ ℕ) → ((𝐴 gcd 𝐵) = 1 → ((𝐴↑𝑁) gcd (𝐵↑𝑁)) = 1)) | ||
Theorem | sqgcd 10924 | Square distributes over GCD. (Contributed by Scott Fenton, 18-Apr-2014.) (Revised by Mario Carneiro, 19-Apr-2014.) |
⊢ ((𝑀 ∈ ℕ ∧ 𝑁 ∈ ℕ) → ((𝑀 gcd 𝑁)↑2) = ((𝑀↑2) gcd (𝑁↑2))) | ||
Theorem | dvdssqlem 10925 | Lemma for dvdssq 10926. (Contributed by Scott Fenton, 18-Apr-2014.) (Revised by Mario Carneiro, 19-Apr-2014.) |
⊢ ((𝑀 ∈ ℕ ∧ 𝑁 ∈ ℕ) → (𝑀 ∥ 𝑁 ↔ (𝑀↑2) ∥ (𝑁↑2))) | ||
Theorem | dvdssq 10926 | Two numbers are divisible iff their squares are. (Contributed by Scott Fenton, 18-Apr-2014.) (Revised by Mario Carneiro, 19-Apr-2014.) |
⊢ ((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → (𝑀 ∥ 𝑁 ↔ (𝑀↑2) ∥ (𝑁↑2))) | ||
Theorem | bezoutr 10927 | Partial converse to bezout 10906. Existence of a linear combination does not set the GCD, but it does upper bound it. (Contributed by Stefan O'Rear, 23-Sep-2014.) |
⊢ (((𝐴 ∈ ℤ ∧ 𝐵 ∈ ℤ) ∧ (𝑋 ∈ ℤ ∧ 𝑌 ∈ ℤ)) → (𝐴 gcd 𝐵) ∥ ((𝐴 · 𝑋) + (𝐵 · 𝑌))) | ||
Theorem | bezoutr1 10928 | Converse of bezout 10906 for when the greater common divisor is one (sufficient condition for relative primality). (Contributed by Stefan O'Rear, 23-Sep-2014.) |
⊢ (((𝐴 ∈ ℤ ∧ 𝐵 ∈ ℤ) ∧ (𝑋 ∈ ℤ ∧ 𝑌 ∈ ℤ)) → (((𝐴 · 𝑋) + (𝐵 · 𝑌)) = 1 → (𝐴 gcd 𝐵) = 1)) | ||
Theorem | nn0seqcvgd 10929* | A strictly-decreasing nonnegative integer sequence with initial term 𝑁 reaches zero by the 𝑁 th term. Deduction version. (Contributed by Paul Chapman, 31-Mar-2011.) |
⊢ (𝜑 → 𝐹:ℕ_{0}⟶ℕ_{0}) & ⊢ (𝜑 → 𝑁 = (𝐹‘0)) & ⊢ ((𝜑 ∧ 𝑘 ∈ ℕ_{0}) → ((𝐹‘(𝑘 + 1)) ≠ 0 → (𝐹‘(𝑘 + 1)) < (𝐹‘𝑘))) ⇒ ⊢ (𝜑 → (𝐹‘𝑁) = 0) | ||
Theorem | ialgrlem1st 10930 | Lemma for ialgr0 10932. Expressing algrflemg 5954 in a form suitable for theorems such as iseq1 9794 or iseqfcl 9796. (Contributed by Jim Kingdon, 22-Jul-2021.) |
⊢ (𝜑 → 𝐹:𝑆⟶𝑆) ⇒ ⊢ ((𝜑 ∧ (𝑥 ∈ 𝑆 ∧ 𝑦 ∈ 𝑆)) → (𝑥(𝐹 ∘ 1^{st} )𝑦) ∈ 𝑆) | ||
Theorem | ialgrlemconst 10931 | Lemma for ialgr0 10932. Closure of a constant function, in a form suitable for theorems such as iseq1 9794 or iseqfcl 9796. (Contributed by Jim Kingdon, 22-Jul-2021.) |
⊢ 𝑍 = (ℤ_{≥}‘𝑀) & ⊢ (𝜑 → 𝐴 ∈ 𝑆) ⇒ ⊢ ((𝜑 ∧ 𝑥 ∈ (ℤ_{≥}‘𝑀)) → ((𝑍 × {𝐴})‘𝑥) ∈ 𝑆) | ||
Theorem | ialgr0 10932 | The value of the algorithm iterator 𝑅 at 0 is the initial state 𝐴. (Contributed by Paul Chapman, 31-Mar-2011.) (Revised by Mario Carneiro, 28-May-2014.) |
⊢ 𝑍 = (ℤ_{≥}‘𝑀) & ⊢ 𝑅 = seq𝑀((𝐹 ∘ 1^{st} ), (𝑍 × {𝐴}), 𝑆) & ⊢ (𝜑 → 𝑀 ∈ ℤ) & ⊢ (𝜑 → 𝐴 ∈ 𝑆) & ⊢ (𝜑 → 𝐹:𝑆⟶𝑆) & ⊢ (𝜑 → 𝑆 ∈ 𝑉) ⇒ ⊢ (𝜑 → (𝑅‘𝑀) = 𝐴) | ||
Theorem | ialgrf 10933 |
An algorithm is a step function 𝐹:𝑆⟶𝑆 on a state space 𝑆.
An algorithm acts on an initial state 𝐴 ∈ 𝑆 by iteratively applying
𝐹 to give 𝐴, (𝐹‘𝐴), (𝐹‘(𝐹‘𝐴)) and so
on. An algorithm is said to halt if a fixed point of 𝐹 is
reached
after a finite number of iterations.
The algorithm iterator 𝑅:ℕ_{0}⟶𝑆 "runs" the algorithm 𝐹 so that (𝑅‘𝑘) is the state after 𝑘 iterations of 𝐹 on the initial state 𝐴. Domain and codomain of the algorithm iterator 𝑅. (Contributed by Paul Chapman, 31-Mar-2011.) (Revised by Mario Carneiro, 28-May-2014.) |
⊢ 𝑍 = (ℤ_{≥}‘𝑀) & ⊢ 𝑅 = seq𝑀((𝐹 ∘ 1^{st} ), (𝑍 × {𝐴}), 𝑆) & ⊢ (𝜑 → 𝑀 ∈ ℤ) & ⊢ (𝜑 → 𝐴 ∈ 𝑆) & ⊢ (𝜑 → 𝐹:𝑆⟶𝑆) & ⊢ (𝜑 → 𝑆 ∈ 𝑉) ⇒ ⊢ (𝜑 → 𝑅:𝑍⟶𝑆) | ||
Theorem | ialgrp1 10934 | The value of the algorithm iterator 𝑅 at (𝐾 + 1). (Contributed by Paul Chapman, 31-Mar-2011.) (Revised by Mario Carneiro, 27-Dec-2014.) |
⊢ 𝑍 = (ℤ_{≥}‘𝑀) & ⊢ 𝑅 = seq𝑀((𝐹 ∘ 1^{st} ), (𝑍 × {𝐴}), 𝑆) & ⊢ (𝜑 → 𝑀 ∈ ℤ) & ⊢ (𝜑 → 𝐴 ∈ 𝑆) & ⊢ (𝜑 → 𝐹:𝑆⟶𝑆) & ⊢ (𝜑 → 𝑆 ∈ 𝑉) ⇒ ⊢ ((𝜑 ∧ 𝐾 ∈ 𝑍) → (𝑅‘(𝐾 + 1)) = (𝐹‘(𝑅‘𝐾))) | ||
Theorem | ialginv 10935* | If 𝐼 is an invariant of 𝐹, its value is unchanged after any number of iterations of 𝐹. (Contributed by Paul Chapman, 31-Mar-2011.) |
⊢ 𝑅 = seq0((𝐹 ∘ 1^{st} ), (ℕ_{0} × {𝐴}), 𝑆) & ⊢ 𝐹:𝑆⟶𝑆 & ⊢ 𝐼 Fn 𝑆 & ⊢ (𝑥 ∈ 𝑆 → (𝐼‘(𝐹‘𝑥)) = (𝐼‘𝑥)) & ⊢ 𝑆 ∈ 𝑉 ⇒ ⊢ ((𝐴 ∈ 𝑆 ∧ 𝐾 ∈ ℕ_{0}) → (𝐼‘(𝑅‘𝐾)) = (𝐼‘(𝑅‘0))) | ||
Theorem | ialgcvg 10936* |
One way to prove that an algorithm halts is to construct a countdown
function 𝐶:𝑆⟶ℕ_{0} whose
value is guaranteed to decrease for
each iteration of 𝐹 until it reaches 0. That is, if 𝑋 ∈ 𝑆
is not a fixed point of 𝐹, then
(𝐶‘(𝐹‘𝑋)) < (𝐶‘𝑋).
If 𝐶 is a countdown function for algorithm 𝐹, the sequence (𝐶‘(𝑅‘𝑘)) reaches 0 after at most 𝑁 steps, where 𝑁 is the value of 𝐶 for the initial state 𝐴. (Contributed by Paul Chapman, 22-Jun-2011.) |
⊢ 𝐹:𝑆⟶𝑆 & ⊢ 𝑅 = seq0((𝐹 ∘ 1^{st} ), (ℕ_{0} × {𝐴}), 𝑆) & ⊢ 𝐶:𝑆⟶ℕ_{0} & ⊢ (𝑧 ∈ 𝑆 → ((𝐶‘(𝐹‘𝑧)) ≠ 0 → (𝐶‘(𝐹‘𝑧)) < (𝐶‘𝑧))) & ⊢ 𝑁 = (𝐶‘𝐴) & ⊢ 𝑆 ∈ 𝑉 ⇒ ⊢ (𝐴 ∈ 𝑆 → (𝐶‘(𝑅‘𝑁)) = 0) | ||
Theorem | algcvgblem 10937 | Lemma for algcvgb 10938. (Contributed by Paul Chapman, 31-Mar-2011.) |
⊢ ((𝑀 ∈ ℕ_{0} ∧ 𝑁 ∈ ℕ_{0}) → ((𝑁 ≠ 0 → 𝑁 < 𝑀) ↔ ((𝑀 ≠ 0 → 𝑁 < 𝑀) ∧ (𝑀 = 0 → 𝑁 = 0)))) | ||
Theorem | algcvgb 10938 | Two ways of expressing that 𝐶 is a countdown function for algorithm 𝐹. The first is used in these theorems. The second states the condition more intuitively as a conjunction: if the countdown function's value is currently nonzero, it must decrease at the next step; if it has reached zero, it must remain zero at the next step. (Contributed by Paul Chapman, 31-Mar-2011.) |
⊢ 𝐹:𝑆⟶𝑆 & ⊢ 𝐶:𝑆⟶ℕ_{0} ⇒ ⊢ (𝑋 ∈ 𝑆 → (((𝐶‘(𝐹‘𝑋)) ≠ 0 → (𝐶‘(𝐹‘𝑋)) < (𝐶‘𝑋)) ↔ (((𝐶‘𝑋) ≠ 0 → (𝐶‘(𝐹‘𝑋)) < (𝐶‘𝑋)) ∧ ((𝐶‘𝑋) = 0 → (𝐶‘(𝐹‘𝑋)) = 0)))) | ||
Theorem | ialgcvga 10939* | The countdown function 𝐶 remains 0 after 𝑁 steps. (Contributed by Paul Chapman, 22-Jun-2011.) |
⊢ 𝐹:𝑆⟶𝑆 & ⊢ 𝑅 = seq0((𝐹 ∘ 1^{st} ), (ℕ_{0} × {𝐴}), 𝑆) & ⊢ 𝐶:𝑆⟶ℕ_{0} & ⊢ (𝑧 ∈ 𝑆 → ((𝐶‘(𝐹‘𝑧)) ≠ 0 → (𝐶‘(𝐹‘𝑧)) < (𝐶‘𝑧))) & ⊢ 𝑁 = (𝐶‘𝐴) & ⊢ 𝑆 ∈ 𝑉 ⇒ ⊢ (𝐴 ∈ 𝑆 → (𝐾 ∈ (ℤ_{≥}‘𝑁) → (𝐶‘(𝑅‘𝐾)) = 0)) | ||
Theorem | ialgfx 10940* | If 𝐹 reaches a fixed point when the countdown function 𝐶 reaches 0, 𝐹 remains fixed after 𝑁 steps. (Contributed by Paul Chapman, 22-Jun-2011.) |
⊢ 𝐹:𝑆⟶𝑆 & ⊢ 𝑅 = seq0((𝐹 ∘ 1^{st} ), (ℕ_{0} × {𝐴}), 𝑆) & ⊢ 𝐶:𝑆⟶ℕ_{0} & ⊢ (𝑧 ∈ 𝑆 → ((𝐶‘(𝐹‘𝑧)) ≠ 0 → (𝐶‘(𝐹‘𝑧)) < (𝐶‘𝑧))) & ⊢ 𝑁 = (𝐶‘𝐴) & ⊢ 𝑆 ∈ 𝑉 & ⊢ (𝑧 ∈ 𝑆 → ((𝐶‘𝑧) = 0 → (𝐹‘𝑧) = 𝑧)) ⇒ ⊢ (𝐴 ∈ 𝑆 → (𝐾 ∈ (ℤ_{≥}‘𝑁) → (𝑅‘𝐾) = (𝑅‘𝑁))) | ||
Theorem | eucalgval2 10941* | The value of the step function 𝐸 for Euclid's Algorithm on an ordered pair. (Contributed by Paul Chapman, 31-Mar-2011.) (Revised by Mario Carneiro, 28-May-2014.) |
⊢ 𝐸 = (𝑥 ∈ ℕ_{0}, 𝑦 ∈ ℕ_{0} ↦ if(𝑦 = 0, ⟨𝑥, 𝑦⟩, ⟨𝑦, (𝑥 mod 𝑦)⟩)) ⇒ ⊢ ((𝑀 ∈ ℕ_{0} ∧ 𝑁 ∈ ℕ_{0}) → (𝑀𝐸𝑁) = if(𝑁 = 0, ⟨𝑀, 𝑁⟩, ⟨𝑁, (𝑀 mod 𝑁)⟩)) | ||
Theorem | eucalgval 10942* |
Euclid's Algorithm eucialg 10947 computes the greatest common divisor of two
nonnegative integers by repeatedly replacing the larger of them with its
remainder modulo the smaller until the remainder is 0.
The value of the step function 𝐸 for Euclid's Algorithm. (Contributed by Paul Chapman, 31-Mar-2011.) (Revised by Mario Carneiro, 28-May-2014.) |
⊢ 𝐸 = (𝑥 ∈ ℕ_{0}, 𝑦 ∈ ℕ_{0} ↦ if(𝑦 = 0, ⟨𝑥, 𝑦⟩, ⟨𝑦, (𝑥 mod 𝑦)⟩)) ⇒ ⊢ (𝑋 ∈ (ℕ_{0} × ℕ_{0}) → (𝐸‘𝑋) = if((2^{nd} ‘𝑋) = 0, 𝑋, ⟨(2^{nd} ‘𝑋), ( mod ‘𝑋)⟩)) | ||
Theorem | eucalgf 10943* | Domain and codomain of the step function 𝐸 for Euclid's Algorithm. (Contributed by Paul Chapman, 31-Mar-2011.) (Revised by Mario Carneiro, 28-May-2014.) |
⊢ 𝐸 = (𝑥 ∈ ℕ_{0}, 𝑦 ∈ ℕ_{0} ↦ if(𝑦 = 0, ⟨𝑥, 𝑦⟩, ⟨𝑦, (𝑥 mod 𝑦)⟩)) ⇒ ⊢ 𝐸:(ℕ_{0} × ℕ_{0})⟶(ℕ_{0} × ℕ_{0}) | ||
Theorem | eucalginv 10944* | The invariant of the step function 𝐸 for Euclid's Algorithm is the gcd operator applied to the state. (Contributed by Paul Chapman, 31-Mar-2011.) (Revised by Mario Carneiro, 29-May-2014.) |
⊢ 𝐸 = (𝑥 ∈ ℕ_{0}, 𝑦 ∈ ℕ_{0} ↦ if(𝑦 = 0, ⟨𝑥, 𝑦⟩, ⟨𝑦, (𝑥 mod 𝑦)⟩)) ⇒ ⊢ (𝑋 ∈ (ℕ_{0} × ℕ_{0}) → ( gcd ‘(𝐸‘𝑋)) = ( gcd ‘𝑋)) | ||
Theorem | eucalglt 10945* | The second member of the state decreases with each iteration of the step function 𝐸 for Euclid's Algorithm. (Contributed by Paul Chapman, 31-Mar-2011.) (Revised by Mario Carneiro, 29-May-2014.) |
⊢ 𝐸 = (𝑥 ∈ ℕ_{0}, 𝑦 ∈ ℕ_{0} ↦ if(𝑦 = 0, ⟨𝑥, 𝑦⟩, ⟨𝑦, (𝑥 mod 𝑦)⟩)) ⇒ ⊢ (𝑋 ∈ (ℕ_{0} × ℕ_{0}) → ((2^{nd} ‘(𝐸‘𝑋)) ≠ 0 → (2^{nd} ‘(𝐸‘𝑋)) < (2^{nd} ‘𝑋))) | ||
Theorem | eucialgcvga 10946* | Once Euclid's Algorithm halts after 𝑁 steps, the second element of the state remains 0 . (Contributed by Jim Kingdon, 11-Jan-2022.) |
⊢ 𝐸 = (𝑥 ∈ ℕ_{0}, 𝑦 ∈ ℕ_{0} ↦ if(𝑦 = 0, ⟨𝑥, 𝑦⟩, ⟨𝑦, (𝑥 mod 𝑦)⟩)) & ⊢ 𝑅 = seq0((𝐸 ∘ 1^{st} ), (ℕ_{0} × {𝐴}), (ℕ_{0} × ℕ_{0})) & ⊢ 𝑁 = (2^{nd} ‘𝐴) ⇒ ⊢ (𝐴 ∈ (ℕ_{0} × ℕ_{0}) → (𝐾 ∈ (ℤ_{≥}‘𝑁) → (2^{nd} ‘(𝑅‘𝐾)) = 0)) | ||
Theorem | eucialg 10947* |
Euclid's Algorithm computes the greatest common divisor of two
nonnegative integers by repeatedly replacing the larger of them with
its remainder modulo the smaller until the remainder is 0. Theorem
1.15 in [ApostolNT] p. 20.
Upon halting, the 1st member of the final state (𝑅‘𝑁) is equal to the gcd of the values comprising the input state ⟨𝑀, 𝑁⟩. This is Metamath 100 proof #69 (greatest common divisor algorithm). (Contributed by Jim Kingdon, 11-Jan-2022.) |
⊢ 𝐸 = (𝑥 ∈ ℕ_{0}, 𝑦 ∈ ℕ_{0} ↦ if(𝑦 = 0, ⟨𝑥, 𝑦⟩, ⟨𝑦, (𝑥 mod 𝑦)⟩)) & ⊢ 𝑅 = seq0((𝐸 ∘ 1^{st} ), (ℕ_{0} × {𝐴}), (ℕ_{0} × ℕ_{0})) & ⊢ 𝑁 = (2^{nd} ‘𝐴) & ⊢ 𝐴 = ⟨𝑀, 𝑁⟩ ⇒ ⊢ ((𝑀 ∈ ℕ_{0} ∧ 𝑁 ∈ ℕ_{0}) → (1^{st} ‘(𝑅‘𝑁)) = (𝑀 gcd 𝑁)) | ||
According to Wikipedia ("Least common multiple", 27-Aug-2020, https://en.wikipedia.org/wiki/Least_common_multiple): "In arithmetic and number theory, the least common multiple, lowest common multiple, or smallest common multiple of two integers a and b, usually denoted by lcm(a, b), is the smallest positive integer that is divisible by both a and b. Since division of integers by zero is undefined, this definition has meaning only if a and b are both different from zero. However, some authors define lcm(a,0) as 0 for all a, which is the result of taking the lcm to be the least upper bound in the lattice of divisibility." In this section, an operation calculating the least common multiple of two integers (df-lcm 10949). The definition is valid for all integers, including negative integers and 0, obeying the above mentioned convention. | ||
Syntax | clcm 10948 | Extend the definition of a class to include the least common multiple operator. |
class lcm | ||
Definition | df-lcm 10949* | Define the lcm operator. For example, (6 lcm 9) = 18. (Contributed by Steve Rodriguez, 20-Jan-2020.) (Revised by AV, 16-Sep-2020.) |
⊢ lcm = (𝑥 ∈ ℤ, 𝑦 ∈ ℤ ↦ if((𝑥 = 0 ∨ 𝑦 = 0), 0, inf({𝑛 ∈ ℕ ∣ (𝑥 ∥ 𝑛 ∧ 𝑦 ∥ 𝑛)}, ℝ, < ))) | ||
Theorem | lcmmndc 10950 | Decidablity lemma used in various proofs related to lcm. (Contributed by Jim Kingdon, 21-Jan-2022.) |
⊢ ((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → DECID (𝑀 = 0 ∨ 𝑁 = 0)) | ||
Theorem | lcmval 10951* | Value of the lcm operator. (𝑀 lcm 𝑁) is the least common multiple of 𝑀 and 𝑁. If either 𝑀 or 𝑁 is 0, the result is defined conventionally as 0. Contrast with df-gcd 10845 and gcdval 10857. (Contributed by Steve Rodriguez, 20-Jan-2020.) (Revised by AV, 16-Sep-2020.) |
⊢ ((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → (𝑀 lcm 𝑁) = if((𝑀 = 0 ∨ 𝑁 = 0), 0, inf({𝑛 ∈ ℕ ∣ (𝑀 ∥ 𝑛 ∧ 𝑁 ∥ 𝑛)}, ℝ, < ))) | ||
Theorem | lcmcom 10952 | The lcm operator is commutative. (Contributed by Steve Rodriguez, 20-Jan-2020.) (Proof shortened by AV, 16-Sep-2020.) |
⊢ ((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → (𝑀 lcm 𝑁) = (𝑁 lcm 𝑀)) | ||
Theorem | lcm0val 10953 | The value, by convention, of the lcm operator when either operand is 0. (Use lcmcom 10952 for a left-hand 0.) (Contributed by Steve Rodriguez, 20-Jan-2020.) (Proof shortened by AV, 16-Sep-2020.) |
⊢ (𝑀 ∈ ℤ → (𝑀 lcm 0) = 0) | ||
Theorem | lcmn0val 10954* | The value of the lcm operator when both operands are nonzero. (Contributed by Steve Rodriguez, 20-Jan-2020.) (Revised by AV, 16-Sep-2020.) |
⊢ (((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) ∧ ¬ (𝑀 = 0 ∨ 𝑁 = 0)) → (𝑀 lcm 𝑁) = inf({𝑛 ∈ ℕ ∣ (𝑀 ∥ 𝑛 ∧ 𝑁 ∥ 𝑛)}, ℝ, < )) | ||
Theorem | lcmcllem 10955* | Lemma for lcmn0cl 10956 and dvdslcm 10957. (Contributed by Steve Rodriguez, 20-Jan-2020.) (Proof shortened by AV, 16-Sep-2020.) |
⊢ (((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) ∧ ¬ (𝑀 = 0 ∨ 𝑁 = 0)) → (𝑀 lcm 𝑁) ∈ {𝑛 ∈ ℕ ∣ (𝑀 ∥ 𝑛 ∧ 𝑁 ∥ 𝑛)}) | ||
Theorem | lcmn0cl 10956 | Closure of the lcm operator. (Contributed by Steve Rodriguez, 20-Jan-2020.) |
⊢ (((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) ∧ ¬ (𝑀 = 0 ∨ 𝑁 = 0)) → (𝑀 lcm 𝑁) ∈ ℕ) | ||
Theorem | dvdslcm 10957 | The lcm of two integers is divisible by each of them. (Contributed by Steve Rodriguez, 20-Jan-2020.) |
⊢ ((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → (𝑀 ∥ (𝑀 lcm 𝑁) ∧ 𝑁 ∥ (𝑀 lcm 𝑁))) | ||
Theorem | lcmledvds 10958 | A positive integer which both operands of the lcm operator divide bounds it. (Contributed by Steve Rodriguez, 20-Jan-2020.) (Proof shortened by AV, 16-Sep-2020.) |
⊢ (((𝐾 ∈ ℕ ∧ 𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) ∧ ¬ (𝑀 = 0 ∨ 𝑁 = 0)) → ((𝑀 ∥ 𝐾 ∧ 𝑁 ∥ 𝐾) → (𝑀 lcm 𝑁) ≤ 𝐾)) | ||
Theorem | lcmeq0 10959 | The lcm of two integers is zero iff either is zero. (Contributed by Steve Rodriguez, 20-Jan-2020.) |
⊢ ((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → ((𝑀 lcm 𝑁) = 0 ↔ (𝑀 = 0 ∨ 𝑁 = 0))) | ||
Theorem | lcmcl 10960 | Closure of the lcm operator. (Contributed by Steve Rodriguez, 20-Jan-2020.) |
⊢ ((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → (𝑀 lcm 𝑁) ∈ ℕ_{0}) | ||
Theorem | gcddvdslcm 10961 | The greatest common divisor of two numbers divides their least common multiple. (Contributed by Steve Rodriguez, 20-Jan-2020.) |
⊢ ((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → (𝑀 gcd 𝑁) ∥ (𝑀 lcm 𝑁)) | ||
Theorem | lcmneg 10962 | Negating one operand of the lcm operator does not alter the result. (Contributed by Steve Rodriguez, 20-Jan-2020.) |
⊢ ((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → (𝑀 lcm -𝑁) = (𝑀 lcm 𝑁)) | ||
Theorem | neglcm 10963 | Negating one operand of the lcm operator does not alter the result. (Contributed by Steve Rodriguez, 20-Jan-2020.) |
⊢ ((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → (-𝑀 lcm 𝑁) = (𝑀 lcm 𝑁)) | ||
Theorem | lcmabs 10964 | The lcm of two integers is the same as that of their absolute values. (Contributed by Steve Rodriguez, 20-Jan-2020.) |
⊢ ((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → ((abs‘𝑀) lcm (abs‘𝑁)) = (𝑀 lcm 𝑁)) | ||
Theorem | lcmgcdlem 10965 | Lemma for lcmgcd 10966 and lcmdvds 10967. Prove them for positive 𝑀, 𝑁, and 𝐾. (Contributed by Steve Rodriguez, 20-Jan-2020.) (Proof shortened by AV, 16-Sep-2020.) |
⊢ ((𝑀 ∈ ℕ ∧ 𝑁 ∈ ℕ) → (((𝑀 lcm 𝑁) · (𝑀 gcd 𝑁)) = (abs‘(𝑀 · 𝑁)) ∧ ((𝐾 ∈ ℕ ∧ (𝑀 ∥ 𝐾 ∧ 𝑁 ∥ 𝐾)) → (𝑀 lcm 𝑁) ∥ 𝐾))) | ||
Theorem | lcmgcd 10966 |
The product of two numbers' least common multiple and greatest common
divisor is the absolute value of the product of the two numbers. In
particular, that absolute value is the least common multiple of two
coprime numbers, for which (𝑀 gcd 𝑁) = 1.
Multiple methods exist for proving this, and it is often proven either as a consequence of the fundamental theorem of arithmetic or of Bézout's identity bezout 10906; see e.g. https://proofwiki.org/wiki/Product_of_GCD_and_LCM and https://math.stackexchange.com/a/470827. This proof uses the latter to first confirm it for positive integers 𝑀 and 𝑁 (the "Second Proof" in the above Stack Exchange page), then shows that implies it for all nonzero integer inputs, then finally uses lcm0val 10953 to show it applies when either or both inputs are zero. (Contributed by Steve Rodriguez, 20-Jan-2020.) |
⊢ ((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → ((𝑀 lcm 𝑁) · (𝑀 gcd 𝑁)) = (abs‘(𝑀 · 𝑁))) | ||
Theorem | lcmdvds 10967 | The lcm of two integers divides any integer the two divide. (Contributed by Steve Rodriguez, 20-Jan-2020.) |
⊢ ((𝐾 ∈ ℤ ∧ 𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → ((𝑀 ∥ 𝐾 ∧ 𝑁 ∥ 𝐾) → (𝑀 lcm 𝑁) ∥ 𝐾)) | ||
Theorem | lcmid 10968 | The lcm of an integer and itself is its absolute value. (Contributed by Steve Rodriguez, 20-Jan-2020.) |
⊢ (𝑀 ∈ ℤ → (𝑀 lcm 𝑀) = (abs‘𝑀)) | ||
Theorem | lcm1 10969 | The lcm of an integer and 1 is the absolute value of the integer. (Contributed by AV, 23-Aug-2020.) |
⊢ (𝑀 ∈ ℤ → (𝑀 lcm 1) = (abs‘𝑀)) | ||
Theorem | lcmgcdnn 10970 | The product of two positive integers' least common multiple and greatest common divisor is the product of the two integers. (Contributed by AV, 27-Aug-2020.) |
⊢ ((𝑀 ∈ ℕ ∧ 𝑁 ∈ ℕ) → ((𝑀 lcm 𝑁) · (𝑀 gcd 𝑁)) = (𝑀 · 𝑁)) | ||
Theorem | lcmgcdeq 10971 | Two integers' absolute values are equal iff their least common multiple and greatest common divisor are equal. (Contributed by Steve Rodriguez, 20-Jan-2020.) |
⊢ ((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → ((𝑀 lcm 𝑁) = (𝑀 gcd 𝑁) ↔ (abs‘𝑀) = (abs‘𝑁))) | ||
Theorem | lcmdvdsb 10972 | Biconditional form of lcmdvds 10967. (Contributed by Steve Rodriguez, 20-Jan-2020.) |
⊢ ((𝐾 ∈ ℤ ∧ 𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → ((𝑀 ∥ 𝐾 ∧ 𝑁 ∥ 𝐾) ↔ (𝑀 lcm 𝑁) ∥ 𝐾)) | ||
Theorem | lcmass 10973 | Associative law for lcm operator. (Contributed by Steve Rodriguez, 20-Jan-2020.) (Proof shortened by AV, 16-Sep-2020.) |
⊢ ((𝑁 ∈ ℤ ∧ 𝑀 ∈ ℤ ∧ 𝑃 ∈ ℤ) → ((𝑁 lcm 𝑀) lcm 𝑃) = (𝑁 lcm (𝑀 lcm 𝑃))) | ||
Theorem | 3lcm2e6woprm 10974 | The least common multiple of three and two is six. This proof does not use the property of 2 and 3 being prime. (Contributed by Steve Rodriguez, 20-Jan-2020.) (Revised by AV, 27-Aug-2020.) |
⊢ (3 lcm 2) = 6 | ||
Theorem | 6lcm4e12 10975 | The least common multiple of six and four is twelve. (Contributed by AV, 27-Aug-2020.) |
⊢ (6 lcm 4) = ;12 | ||
According to Wikipedia "Coprime integers", see https://en.wikipedia.org/wiki/Coprime_integers (16-Aug-2020) "[...] two integers a and b are said to be relatively prime, mutually prime, or coprime [...] if the only positive integer (factor) that divides both of them is 1. Consequently, any prime number that divides one does not divide the other. This is equivalent to their greatest common divisor (gcd) being 1.". In the following, we use this equivalent characterization to say that 𝐴 ∈ ℤ and 𝐵 ∈ ℤ are coprime (or relatively prime) if (𝐴 gcd 𝐵) = 1. The equivalence of the definitions is shown by coprmgcdb 10976. The negation, i.e. two integers are not coprime, can be expressed either by (𝐴 gcd 𝐵) ≠ 1, see ncoprmgcdne1b 10977, or equivalently by 1 < (𝐴 gcd 𝐵), see ncoprmgcdgt1b 10978. A proof of Euclid's lemma based on coprimality is provided in coprmdvds 10980 (as opposed to Euclid's lemma for primes). | ||
Theorem | coprmgcdb 10976* | Two positive integers are coprime, i.e. the only positive integer that divides both of them is 1, iff their greatest common divisor is 1. (Contributed by AV, 9-Aug-2020.) |
⊢ ((𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ) → (∀𝑖 ∈ ℕ ((𝑖 ∥ 𝐴 ∧ 𝑖 ∥ 𝐵) → 𝑖 = 1) ↔ (𝐴 gcd 𝐵) = 1)) | ||
Theorem | ncoprmgcdne1b 10977* | Two positive integers are not coprime, i.e. there is an integer greater than 1 which divides both integers, iff their greatest common divisor is not 1. (Contributed by AV, 9-Aug-2020.) |
⊢ ((𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ) → (∃𝑖 ∈ (ℤ_{≥}‘2)(𝑖 ∥ 𝐴 ∧ 𝑖 ∥ 𝐵) ↔ (𝐴 gcd 𝐵) ≠ 1)) | ||
Theorem | ncoprmgcdgt1b 10978* | Two positive integers are not coprime, i.e. there is an integer greater than 1 which divides both integers, iff their greatest common divisor is greater than 1. (Contributed by AV, 9-Aug-2020.) |
⊢ ((𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ) → (∃𝑖 ∈ (ℤ_{≥}‘2)(𝑖 ∥ 𝐴 ∧ 𝑖 ∥ 𝐵) ↔ 1 < (𝐴 gcd 𝐵))) | ||
Theorem | coprmdvds1 10979 | If two positive integers are coprime, i.e. their greatest common divisor is 1, the only positive integer that divides both of them is 1. (Contributed by AV, 4-Aug-2021.) |
⊢ ((𝐹 ∈ ℕ ∧ 𝐺 ∈ ℕ ∧ (𝐹 gcd 𝐺) = 1) → ((𝐼 ∈ ℕ ∧ 𝐼 ∥ 𝐹 ∧ 𝐼 ∥ 𝐺) → 𝐼 = 1)) | ||
Theorem | coprmdvds 10980 | Euclid's Lemma (see ProofWiki "Euclid's Lemma", 10-Jul-2021, https://proofwiki.org/wiki/Euclid's_Lemma): If an integer divides the product of two integers and is coprime to one of them, then it divides the other. See also theorem 1.5 in [ApostolNT] p. 16. (Contributed by Paul Chapman, 22-Jun-2011.) (Proof shortened by AV, 10-Jul-2021.) |
⊢ ((𝐾 ∈ ℤ ∧ 𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → ((𝐾 ∥ (𝑀 · 𝑁) ∧ (𝐾 gcd 𝑀) = 1) → 𝐾 ∥ 𝑁)) | ||
Theorem | coprmdvds2 10981 | If an integer is divisible by two coprime integers, then it is divisible by their product. (Contributed by Mario Carneiro, 24-Feb-2014.) |
⊢ (((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ ∧ 𝐾 ∈ ℤ) ∧ (𝑀 gcd 𝑁) = 1) → ((𝑀 ∥ 𝐾 ∧ 𝑁 ∥ 𝐾) → (𝑀 · 𝑁) ∥ 𝐾)) | ||
Theorem | mulgcddvds 10982 | One half of rpmulgcd2 10983, which does not need the coprimality assumption. (Contributed by Mario Carneiro, 2-Jul-2015.) |
⊢ ((𝐾 ∈ ℤ ∧ 𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → (𝐾 gcd (𝑀 · 𝑁)) ∥ ((𝐾 gcd 𝑀) · (𝐾 gcd 𝑁))) | ||
Theorem | rpmulgcd2 10983 | If 𝑀 is relatively prime to 𝑁, then the GCD of 𝐾 with 𝑀 · 𝑁 is the product of the GCDs with 𝑀 and 𝑁 respectively. (Contributed by Mario Carneiro, 2-Jul-2015.) |
⊢ (((𝐾 ∈ ℤ ∧ 𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) ∧ (𝑀 gcd 𝑁) = 1) → (𝐾 gcd (𝑀 · 𝑁)) = ((𝐾 gcd 𝑀) · (𝐾 gcd 𝑁))) | ||
Theorem | qredeq 10984 | Two equal reduced fractions have the same numerator and denominator. (Contributed by Jeff Hankins, 29-Sep-2013.) |
⊢ (((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℕ ∧ (𝑀 gcd 𝑁) = 1) ∧ (𝑃 ∈ ℤ ∧ 𝑄 ∈ ℕ ∧ (𝑃 gcd 𝑄) = 1) ∧ (𝑀 / 𝑁) = (𝑃 / 𝑄)) → (𝑀 = 𝑃 ∧ 𝑁 = 𝑄)) | ||
Theorem | qredeu 10985* | Every rational number has a unique reduced form. (Contributed by Jeff Hankins, 29-Sep-2013.) |
⊢ (𝐴 ∈ ℚ → ∃!𝑥 ∈ (ℤ × ℕ)(((1^{st} ‘𝑥) gcd (2^{nd} ‘𝑥)) = 1 ∧ 𝐴 = ((1^{st} ‘𝑥) / (2^{nd} ‘𝑥)))) | ||
Theorem | rpmul 10986 | If 𝐾 is relatively prime to 𝑀 and to 𝑁, it is also relatively prime to their product. (Contributed by Mario Carneiro, 24-Feb-2014.) (Proof shortened by Mario Carneiro, 2-Jul-2015.) |
⊢ ((𝐾 ∈ ℤ ∧ 𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → (((𝐾 gcd 𝑀) = 1 ∧ (𝐾 gcd 𝑁) = 1) → (𝐾 gcd (𝑀 · 𝑁)) = 1)) | ||
Theorem | rpdvds 10987 | If 𝐾 is relatively prime to 𝑁 then it is also relatively prime to any divisor 𝑀 of 𝑁. (Contributed by Mario Carneiro, 19-Jun-2015.) |
⊢ (((𝐾 ∈ ℤ ∧ 𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) ∧ ((𝐾 gcd 𝑁) = 1 ∧ 𝑀 ∥ 𝑁)) → (𝐾 gcd 𝑀) = 1) | ||
Theorem | congr 10988* | Definition of congruence by integer multiple (see ProofWiki "Congruence (Number Theory)", 11-Jul-2021, https://proofwiki.org/wiki/Definition:Congruence_(Number_Theory)): An integer 𝐴 is congruent to an integer 𝐵 modulo 𝑀 if their difference is a multiple of 𝑀. See also the definition in [ApostolNT] p. 104: "... 𝑎 is congruent to 𝑏 modulo 𝑚, and we write 𝑎≡𝑏 (mod 𝑚) if 𝑚 divides the difference 𝑎 − 𝑏", or Wikipedia "Modular arithmetic - Congruence", https://en.wikipedia.org/wiki/Modular_arithmetic#Congruence, 11-Jul-2021,: "Given an integer n > 1, called a modulus, two integers are said to be congruent modulo n, if n is a divisor of their difference (i.e., if there is an integer k such that a-b = kn)". (Contributed by AV, 11-Jul-2021.) |
⊢ ((𝐴 ∈ ℤ ∧ 𝐵 ∈ ℤ ∧ 𝑀 ∈ ℕ) → ((𝐴 mod 𝑀) = (𝐵 mod 𝑀) ↔ ∃𝑛 ∈ ℤ (𝑛 · 𝑀) = (𝐴 − 𝐵))) | ||
Theorem | divgcdcoprm0 10989 | Integers divided by gcd are coprime. (Contributed by AV, 12-Jul-2021.) |
⊢ ((𝐴 ∈ ℤ ∧ 𝐵 ∈ ℤ ∧ 𝐵 ≠ 0) → ((𝐴 / (𝐴 gcd 𝐵)) gcd (𝐵 / (𝐴 gcd 𝐵))) = 1) | ||
Theorem | divgcdcoprmex 10990* | Integers divided by gcd are coprime (see ProofWiki "Integers Divided by GCD are Coprime", 11-Jul-2021, https://proofwiki.org/wiki/Integers_Divided_by_GCD_are_Coprime): Any pair of integers, not both zero, can be reduced to a pair of coprime ones by dividing them by their gcd. (Contributed by AV, 12-Jul-2021.) |
⊢ ((𝐴 ∈ ℤ ∧ (𝐵 ∈ ℤ ∧ 𝐵 ≠ 0) ∧ 𝑀 = (𝐴 gcd 𝐵)) → ∃𝑎 ∈ ℤ ∃𝑏 ∈ ℤ (𝐴 = (𝑀 · 𝑎) ∧ 𝐵 = (𝑀 · 𝑏) ∧ (𝑎 gcd 𝑏) = 1)) | ||
Theorem | cncongr1 10991 | One direction of the bicondition in cncongr 10993. Theorem 5.4 in [ApostolNT] p. 109. (Contributed by AV, 13-Jul-2021.) |
⊢ (((𝐴 ∈ ℤ ∧ 𝐵 ∈ ℤ ∧ 𝐶 ∈ ℤ) ∧ (𝑁 ∈ ℕ ∧ 𝑀 = (𝑁 / (𝐶 gcd 𝑁)))) → (((𝐴 · 𝐶) mod 𝑁) = ((𝐵 · 𝐶) mod 𝑁) → (𝐴 mod 𝑀) = (𝐵 mod 𝑀))) | ||
Theorem | cncongr2 10992 | The other direction of the bicondition in cncongr 10993. (Contributed by AV, 11-Jul-2021.) |
⊢ (((𝐴 ∈ ℤ ∧ 𝐵 ∈ ℤ ∧ 𝐶 ∈ ℤ) ∧ (𝑁 ∈ ℕ ∧ 𝑀 = (𝑁 / (𝐶 gcd 𝑁)))) → ((𝐴 mod 𝑀) = (𝐵 mod 𝑀) → ((𝐴 · 𝐶) mod 𝑁) = ((𝐵 · 𝐶) mod 𝑁))) | ||
Theorem | cncongr 10993 | Cancellability of Congruences (see ProofWiki "Cancellability of Congruences, https://proofwiki.org/wiki/Cancellability_of_Congruences, 10-Jul-2021): Two products with a common factor are congruent modulo a positive integer iff the other factors are congruent modulo the integer divided by the greates common divisor of the integer and the common factor. See also Theorem 5.4 "Cancellation law" in [ApostolNT] p. 109. (Contributed by AV, 13-Jul-2021.) |
⊢ (((𝐴 ∈ ℤ ∧ 𝐵 ∈ ℤ ∧ 𝐶 ∈ ℤ) ∧ (𝑁 ∈ ℕ ∧ 𝑀 = (𝑁 / (𝐶 gcd 𝑁)))) → (((𝐴 · 𝐶) mod 𝑁) = ((𝐵 · 𝐶) mod 𝑁) ↔ (𝐴 mod 𝑀) = (𝐵 mod 𝑀))) | ||
Theorem | cncongrcoprm 10994 | Corollary 1 of Cancellability of Congruences: Two products with a common factor are congruent modulo an integer being coprime to the common factor iff the other factors are congruent modulo the integer. (Contributed by AV, 13-Jul-2021.) |
⊢ (((𝐴 ∈ ℤ ∧ 𝐵 ∈ ℤ ∧ 𝐶 ∈ ℤ) ∧ (𝑁 ∈ ℕ ∧ (𝐶 gcd 𝑁) = 1)) → (((𝐴 · 𝐶) mod 𝑁) = ((𝐵 · 𝐶) mod 𝑁) ↔ (𝐴 mod 𝑁) = (𝐵 mod 𝑁))) | ||
Remark: to represent odd prime numbers, i.e., all prime numbers except 2, the idiom 𝑃 ∈ (ℙ ∖ {2}) is used. It is a little bit shorter than (𝑃 ∈ ℙ ∧ 𝑃 ≠ 2). Both representations can be converted into each other by eldifsn 3552. | ||
Syntax | cprime 10995 | Extend the definition of a class to include the set of prime numbers. |
class ℙ | ||
Definition | df-prm 10996* | Define the set of prime numbers. (Contributed by Paul Chapman, 22-Jun-2011.) |
⊢ ℙ = {𝑝 ∈ ℕ ∣ {𝑛 ∈ ℕ ∣ 𝑛 ∥ 𝑝} ≈ 2_{𝑜}} | ||
Theorem | isprm 10997* | The predicate "is a prime number". A prime number is a positive integer with exactly two positive divisors. (Contributed by Paul Chapman, 22-Jun-2011.) |
⊢ (𝑃 ∈ ℙ ↔ (𝑃 ∈ ℕ ∧ {𝑛 ∈ ℕ ∣ 𝑛 ∥ 𝑃} ≈ 2_{𝑜})) | ||
Theorem | prmnn 10998 | A prime number is a positive integer. (Contributed by Paul Chapman, 22-Jun-2011.) |
⊢ (𝑃 ∈ ℙ → 𝑃 ∈ ℕ) | ||
Theorem | prmz 10999 | A prime number is an integer. (Contributed by Paul Chapman, 22-Jun-2011.) (Proof shortened by Jonathan Yan, 16-Jul-2017.) |
⊢ (𝑃 ∈ ℙ → 𝑃 ∈ ℤ) | ||
Theorem | prmssnn 11000 | The prime numbers are a subset of the positive integers. (Contributed by AV, 22-Jul-2020.) |
⊢ ℙ ⊆ ℕ |
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