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Theorem List for Metamath Proof Explorer - 15001-15100   *Has distinct variable group(s)
TypeLabelDescription
Statement
 
Theoremalgrp1 15001 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 = (ℤ𝑀)    &   𝑅 = seq𝑀((𝐹 ∘ 1st ), (𝑍 × {𝐴}))    &   (𝜑𝑀 ∈ ℤ)    &   (𝜑𝐴𝑆)    &   (𝜑𝐹:𝑆𝑆)       ((𝜑𝐾𝑍) → (𝑅‘(𝐾 + 1)) = (𝐹‘(𝑅𝐾)))
 
Theoremalginv 15002* 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 ∘ 1st ), (ℕ0 × {𝐴}))    &   𝐹:𝑆𝑆    &   𝐼 Fn 𝑆    &   (𝑥𝑆 → (𝐼‘(𝐹𝑥)) = (𝐼𝑥))       ((𝐴𝑆𝐾 ∈ ℕ0) → (𝐼‘(𝑅𝐾)) = (𝐼‘(𝑅‘0)))
 
Theoremalgcvg 15003* One way to prove that an algorithm halts is to construct a countdown function 𝐶:𝑆⟶ℕ0 whose value is guaranteed to decrease for each iteration of 𝐹 until it reaches 0. That is, if 𝑋𝑆 is not a fixed point of 𝐹, then (𝐶‘(𝐹𝑋)) < (𝐶𝑋).

If 𝐶 is a countdown function for algorithm 𝐹, the sequence (𝐶‘(𝑅𝑘)) 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:𝑆𝑆    &   𝑅 = seq0((𝐹 ∘ 1st ), (ℕ0 × {𝐴}))    &   𝐶:𝑆⟶ℕ0    &   (𝑧𝑆 → ((𝐶‘(𝐹𝑧)) ≠ 0 → (𝐶‘(𝐹𝑧)) < (𝐶𝑧)))    &   𝑁 = (𝐶𝐴)       (𝐴𝑆 → (𝐶‘(𝑅𝑁)) = 0)
 
Theoremalgcvgblem 15004 Lemma for algcvgb 15005. (Contributed by Paul Chapman, 31-Mar-2011.)
((𝑀 ∈ ℕ0𝑁 ∈ ℕ0) → ((𝑁 ≠ 0 → 𝑁 < 𝑀) ↔ ((𝑀 ≠ 0 → 𝑁 < 𝑀) ∧ (𝑀 = 0 → 𝑁 = 0))))
 
Theoremalgcvgb 15005 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:𝑆𝑆    &   𝐶:𝑆⟶ℕ0       (𝑋𝑆 → (((𝐶‘(𝐹𝑋)) ≠ 0 → (𝐶‘(𝐹𝑋)) < (𝐶𝑋)) ↔ (((𝐶𝑋) ≠ 0 → (𝐶‘(𝐹𝑋)) < (𝐶𝑋)) ∧ ((𝐶𝑋) = 0 → (𝐶‘(𝐹𝑋)) = 0))))
 
Theoremalgcvga 15006* The countdown function 𝐶 remains 0 after 𝑁 steps. (Contributed by Paul Chapman, 22-Jun-2011.)
𝐹:𝑆𝑆    &   𝑅 = seq0((𝐹 ∘ 1st ), (ℕ0 × {𝐴}))    &   𝐶:𝑆⟶ℕ0    &   (𝑧𝑆 → ((𝐶‘(𝐹𝑧)) ≠ 0 → (𝐶‘(𝐹𝑧)) < (𝐶𝑧)))    &   𝑁 = (𝐶𝐴)       (𝐴𝑆 → (𝐾 ∈ (ℤ𝑁) → (𝐶‘(𝑅𝐾)) = 0))
 
Theoremalgfx 15007* 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:𝑆𝑆    &   𝑅 = seq0((𝐹 ∘ 1st ), (ℕ0 × {𝐴}))    &   𝐶:𝑆⟶ℕ0    &   (𝑧𝑆 → ((𝐶‘(𝐹𝑧)) ≠ 0 → (𝐶‘(𝐹𝑧)) < (𝐶𝑧)))    &   𝑁 = (𝐶𝐴)    &   (𝑧𝑆 → ((𝐶𝑧) = 0 → (𝐹𝑧) = 𝑧))       (𝐴𝑆 → (𝐾 ∈ (ℤ𝑁) → (𝑅𝐾) = (𝑅𝑁)))
 
6.1.10  Euclid's Algorithm
 
Theoremeucalgval2 15008* 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 ∈ ℕ0, 𝑦 ∈ ℕ0 ↦ if(𝑦 = 0, ⟨𝑥, 𝑦⟩, ⟨𝑦, (𝑥 mod 𝑦)⟩))       ((𝑀 ∈ ℕ0𝑁 ∈ ℕ0) → (𝑀𝐸𝑁) = if(𝑁 = 0, ⟨𝑀, 𝑁⟩, ⟨𝑁, (𝑀 mod 𝑁)⟩))
 
Theoremeucalgval 15009* Euclid's Algorithm eucalg 15014 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 ∈ ℕ0, 𝑦 ∈ ℕ0 ↦ if(𝑦 = 0, ⟨𝑥, 𝑦⟩, ⟨𝑦, (𝑥 mod 𝑦)⟩))       (𝑋 ∈ (ℕ0 × ℕ0) → (𝐸𝑋) = if((2nd𝑋) = 0, 𝑋, ⟨(2nd𝑋), ( mod ‘𝑋)⟩))
 
Theoremeucalgf 15010* 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 ∈ ℕ0, 𝑦 ∈ ℕ0 ↦ if(𝑦 = 0, ⟨𝑥, 𝑦⟩, ⟨𝑦, (𝑥 mod 𝑦)⟩))       𝐸:(ℕ0 × ℕ0)⟶(ℕ0 × ℕ0)
 
Theoremeucalginv 15011* 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 ∈ ℕ0, 𝑦 ∈ ℕ0 ↦ if(𝑦 = 0, ⟨𝑥, 𝑦⟩, ⟨𝑦, (𝑥 mod 𝑦)⟩))       (𝑋 ∈ (ℕ0 × ℕ0) → ( gcd ‘(𝐸𝑋)) = ( gcd ‘𝑋))
 
Theoremeucalglt 15012* 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 ∈ ℕ0, 𝑦 ∈ ℕ0 ↦ if(𝑦 = 0, ⟨𝑥, 𝑦⟩, ⟨𝑦, (𝑥 mod 𝑦)⟩))       (𝑋 ∈ (ℕ0 × ℕ0) → ((2nd ‘(𝐸𝑋)) ≠ 0 → (2nd ‘(𝐸𝑋)) < (2nd𝑋)))
 
Theoremeucalgcvga 15013* Once Euclid's Algorithm halts after 𝑁 steps, the second element of the state remains 0 . (Contributed by Paul Chapman, 22-Jun-2011.) (Revised by Mario Carneiro, 29-May-2014.)
𝐸 = (𝑥 ∈ ℕ0, 𝑦 ∈ ℕ0 ↦ if(𝑦 = 0, ⟨𝑥, 𝑦⟩, ⟨𝑦, (𝑥 mod 𝑦)⟩))    &   𝑅 = seq0((𝐸 ∘ 1st ), (ℕ0 × {𝐴}))    &   𝑁 = (2nd𝐴)       (𝐴 ∈ (ℕ0 × ℕ0) → (𝐾 ∈ (ℤ𝑁) → (2nd ‘(𝑅𝐾)) = 0))
 
Theoremeucalg 15014* 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 (𝑅𝑁) 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 Paul Chapman, 31-Mar-2011.) (Proof shortened by Mario Carneiro, 29-May-2014.)

𝐸 = (𝑥 ∈ ℕ0, 𝑦 ∈ ℕ0 ↦ if(𝑦 = 0, ⟨𝑥, 𝑦⟩, ⟨𝑦, (𝑥 mod 𝑦)⟩))    &   𝑅 = seq0((𝐸 ∘ 1st ), (ℕ0 × {𝐴}))    &   𝐴 = ⟨𝑀, 𝑁       ((𝑀 ∈ ℕ0𝑁 ∈ ℕ0) → (1st ‘(𝑅𝑁)) = (𝑀 gcd 𝑁))
 
6.1.11  The least common multiple

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. ... The lcm of more than two integers is also well-defined: it is the smallest positive integer that is divisible by each of them."

In this section, an operation calculating the least common multiple of two integers (df-lcm 15017) as well as a function mapping a set of integers to their least common multiple (df-lcmf 15018) are provided. Both definitions are valid for all integers, including negative integers and 0, obeying the above mentioned convention. It is shown by lcmfpr 15054 that the two definitions are compatible.

It seems a little confusing at the first glance that the least common multiple is defined by a supremum. Actually, it is an infimum, because the inverse of "less than" < is used to turn supremum into infimum - currently we don't have infimum defined separately...

 
Syntaxclcm 15015 Extend the definition of a class to include the least common multiple operator.
class lcm
 
Syntaxclcmf 15016 Extend the definition of a class to include the least common multiple function.
class lcm
 
Definitiondf-lcm 15017* Define the lcm operator. For example, (6 lcm 9) = 18 (ex-lcm 26445). (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 ∣ (𝑥𝑛𝑦𝑛)}, ℝ, < )))
 
Definitiondf-lcmf 15018* Define the lcm function on a set of integers. (Contributed by AV, 21-Aug-2020.) (Revised by AV, 16-Sep-2020.)
lcm = (𝑧 ∈ 𝒫 ℤ ↦ if(0 ∈ 𝑧, 0, inf({𝑛 ∈ ℕ ∣ ∀𝑚𝑧 𝑚𝑛}, ℝ, < )))
 
Theoremlcmval 15019* 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 14928 and gcdval 14929. (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 ∣ (𝑀𝑛𝑁𝑛)}, ℝ, < )))
 
Theoremlcmcom 15020 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))
 
Theoremlcm0val 15021 The value, by convention, of the lcm operator when either operand is 0. (Use lcmcom 15020 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)
 
Theoremlcmn0val 15022* 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 ∣ (𝑀𝑛𝑁𝑛)}, ℝ, < ))
 
Theoremlcmcllem 15023* Lemma for lcmn0cl 15024 and dvdslcm 15025. (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 ∣ (𝑀𝑛𝑁𝑛)})
 
Theoremlcmn0cl 15024 Closure of the lcm operator. (Contributed by Steve Rodriguez, 20-Jan-2020.)
(((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) ∧ ¬ (𝑀 = 0 ∨ 𝑁 = 0)) → (𝑀 lcm 𝑁) ∈ ℕ)
 
Theoremdvdslcm 15025 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)))
 
Theoremlcmledvds 15026 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)) → ((𝑀𝐾𝑁𝐾) → (𝑀 lcm 𝑁) ≤ 𝐾))
 
Theoremlcmeq0 15027 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)))
 
Theoremlcmcl 15028 Closure of the lcm operator. (Contributed by Steve Rodriguez, 20-Jan-2020.)
((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → (𝑀 lcm 𝑁) ∈ ℕ0)
 
Theoremgcddvdslcm 15029 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))
 
Theoremlcmneg 15030 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))
 
Theoremneglcm 15031 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))
 
Theoremlcmabs 15032 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))
 
Theoremlcmgcdlem 15033 Lemma for lcmgcd 15034 and lcmdvds 15035. 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 ∧ (𝑀𝐾𝑁𝐾)) → (𝑀 lcm 𝑁) ∥ 𝐾)))
 
Theoremlcmgcd 15034 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 1arith 15353 or of Bézout's identity bezout 14974; 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 integer 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 15021 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)))
 
Theoremlcmdvds 15035 The lcm of two integers divides any integer the two divide. (Contributed by Steve Rodriguez, 20-Jan-2020.)
((𝐾 ∈ ℤ ∧ 𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → ((𝑀𝐾𝑁𝐾) → (𝑀 lcm 𝑁) ∥ 𝐾))
 
Theoremlcmid 15036 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))
 
Theoremlcm1 15037 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))
 
Theoremlcmgcdnn 15038 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))
 
Theoremlcmgcdeq 15039 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)))
 
Theoremlcmdvdsb 15040 Biconditional form of lcmdvds 15035. (Contributed by Steve Rodriguez, 20-Jan-2020.)
((𝐾 ∈ ℤ ∧ 𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → ((𝑀𝐾𝑁𝐾) ↔ (𝑀 lcm 𝑁) ∥ 𝐾))
 
Theoremlcmass 15041 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)))
 
Theorem3lcm2e6woprm 15042 The least common multiple of three and two is six. In contrast to 3lcm2e6 15154, this proof does not use the property of 2 and 3 being prime, therefore it is much longer. (Contributed by Steve Rodriguez, 20-Jan-2020.) (Revised by AV, 27-Aug-2020.) (Proof modification is discouraged.) (New usage is discouraged.)
(3 lcm 2) = 6
 
Theorem6lcm4e12 15043 The least common multiple of six and four is twelve. (Contributed by AV, 27-Aug-2020.)
(6 lcm 4) = 12
 
Theoremabsproddvds 15044* The absolute value of the product of the elements of a finite subset of the integers is divisible by each element of this subset. (Contributed by AV, 21-Aug-2020.)
(𝜑𝑍 ⊆ ℤ)    &   (𝜑𝑍 ∈ Fin)    &   𝑃 = (abs‘∏𝑧𝑍 𝑧)       (𝜑 → ∀𝑚𝑍 𝑚𝑃)
 
Theoremabsprodnn 15045* The absolute value of the product of the elements of a finite subset of the integers not containing 0 is a poitive integer. (Contributed by AV, 21-Aug-2020.)
(𝜑𝑍 ⊆ ℤ)    &   (𝜑𝑍 ∈ Fin)    &   𝑃 = (abs‘∏𝑧𝑍 𝑧)    &   (𝜑 → 0 ∉ 𝑍)       (𝜑𝑃 ∈ ℕ)
 
Theoremfissn0dvds 15046* For each finite subset of the integers not containing 0 there is a positive integer which is divisible by each element of this subset. (Contributed by AV, 21-Aug-2020.)
((𝑍 ⊆ ℤ ∧ 𝑍 ∈ Fin ∧ 0 ∉ 𝑍) → ∃𝑛 ∈ ℕ ∀𝑚𝑍 𝑚𝑛)
 
Theoremfissn0dvdsn0 15047* For each finite subset of the integers not containing 0 there is a positive integer which is divisible by each element of this subset. (Contributed by AV, 21-Aug-2020.)
((𝑍 ⊆ ℤ ∧ 𝑍 ∈ Fin ∧ 0 ∉ 𝑍) → {𝑛 ∈ ℕ ∣ ∀𝑚𝑍 𝑚𝑛} ≠ ∅)
 
Theoremlcmfval 15048* Value of the lcm function. (lcm𝑍) is the least common multiple of the integers contained in the finite subset of integers 𝑍. If at least one of the elements of 𝑍 is 0, the result is defined conventionally as 0. (Contributed by AV, 21-Apr-2020.) (Revised by AV, 16-Sep-2020.)
((𝑍 ⊆ ℤ ∧ 𝑍 ∈ Fin) → (lcm𝑍) = if(0 ∈ 𝑍, 0, inf({𝑛 ∈ ℕ ∣ ∀𝑚𝑍 𝑚𝑛}, ℝ, < )))
 
Theoremlcmf0val 15049 The value, by convention, of the least common multiple for a set containing 0 is 0. (Contributed by AV, 21-Apr-2020.) (Proof shortened by AV, 16-Sep-2020.)
((𝑍 ⊆ ℤ ∧ 0 ∈ 𝑍) → (lcm𝑍) = 0)
 
Theoremlcmfn0val 15050* The value of the lcm function for a set without 0. (Contributed by AV, 21-Aug-2020.) (Revised by AV, 16-Sep-2020.)
((𝑍 ⊆ ℤ ∧ 𝑍 ∈ Fin ∧ 0 ∉ 𝑍) → (lcm𝑍) = inf({𝑛 ∈ ℕ ∣ ∀𝑚𝑍 𝑚𝑛}, ℝ, < ))
 
Theoremlcmfnnval 15051* The value of the lcm function for a subset of the positive integers. (Contributed by AV, 21-Aug-2020.) (Revised by AV, 16-Sep-2020.)
((𝑍 ⊆ ℕ ∧ 𝑍 ∈ Fin) → (lcm𝑍) = inf({𝑛 ∈ ℕ ∣ ∀𝑚𝑍 𝑚𝑛}, ℝ, < ))
 
Theoremlcmfcllem 15052* Lemma for lcmfn0cl 15053 and dvdslcmf 15058. (Contributed by AV, 21-Aug-2020.) (Proof shortened by AV, 16-Sep-2020.)
((𝑍 ⊆ ℤ ∧ 𝑍 ∈ Fin ∧ 0 ∉ 𝑍) → (lcm𝑍) ∈ {𝑛 ∈ ℕ ∣ ∀𝑚𝑍 𝑚𝑛})
 
Theoremlcmfn0cl 15053 Closure of the lcm function. (Contributed by AV, 21-Aug-2020.)
((𝑍 ⊆ ℤ ∧ 𝑍 ∈ Fin ∧ 0 ∉ 𝑍) → (lcm𝑍) ∈ ℕ)
 
Theoremlcmfpr 15054 The value of the lcm function for an unordered pair is the value of the lcm operator for both elements. (Contributed by AV, 22-Aug-2020.) (Proof shortened by AV, 16-Sep-2020.)
((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → (lcm‘{𝑀, 𝑁}) = (𝑀 lcm 𝑁))
 
Theoremlcmfcl 15055 Closure of the lcm function. (Contributed by AV, 21-Aug-2020.)
((𝑍 ⊆ ℤ ∧ 𝑍 ∈ Fin) → (lcm𝑍) ∈ ℕ0)
 
Theoremlcmfnncl 15056 Closure of the lcm function. (Contributed by AV, 20-Apr-2020.)
((𝑍 ⊆ ℕ ∧ 𝑍 ∈ Fin) → (lcm𝑍) ∈ ℕ)
 
Theoremlcmfeq0b 15057 The least common multiple of a set of integers is 0 iff at least one of its element is 0. (Contributed by AV, 21-Aug-2020.)
((𝑍 ⊆ ℤ ∧ 𝑍 ∈ Fin) → ((lcm𝑍) = 0 ↔ 0 ∈ 𝑍))
 
Theoremdvdslcmf 15058* The least common multiple of a set of integers is divisible by each of its elements. (Contributed by AV, 22-Aug-2020.)
((𝑍 ⊆ ℤ ∧ 𝑍 ∈ Fin) → ∀𝑥𝑍 𝑥 ∥ (lcm𝑍))
 
Theoremlcmfledvds 15059* A positive integer which is divisible by all elements of a set of integers bounds the least common multiple of the set. (Contributed by AV, 22-Aug-2020.) (Proof shortened by AV, 16-Sep-2020.)
((𝑍 ⊆ ℤ ∧ 𝑍 ∈ Fin ∧ 0 ∉ 𝑍) → ((𝐾 ∈ ℕ ∧ ∀𝑚𝑍 𝑚𝐾) → (lcm𝑍) ≤ 𝐾))
 
Theoremlcmf 15060* Characterization of the least common multiple of a set of integers (without 0): A positiven integer is the least common multiple of a set of integers iff it divides each of the elements of the set and every integer which divides each of the elements of the set is greater than or equal to this integer. (Contributed by AV, 22-Aug-2020.)
((𝐾 ∈ ℕ ∧ (𝑍 ⊆ ℤ ∧ 𝑍 ∈ Fin ∧ 0 ∉ 𝑍)) → (𝐾 = (lcm𝑍) ↔ (∀𝑚𝑍 𝑚𝐾 ∧ ∀𝑘 ∈ ℕ (∀𝑚𝑍 𝑚𝑘𝐾𝑘))))
 
Theoremlcmf0 15061 The least common multiple of the empty set is 1. (Contributed by AV, 22-Aug-2020.) (Proof shortened by AV, 16-Sep-2020.)
(lcm‘∅) = 1
 
Theoremlcmfsn 15062 The least common multiple of a singleton is its absolute value. (Contributed by AV, 22-Aug-2020.)
(𝑀 ∈ ℤ → (lcm‘{𝑀}) = (abs‘𝑀))
 
Theoremlcmftp 15063 The least common multiple of a triple of integers is the least common multiple of the third integer and the the least common multiple of the first two integers. Although there would be a shorter proof using lcmfunsn 15071, this explicit proof (not based on induction) should be kept. (Proof modification is discouraged.) (Contributed by AV, 23-Aug-2020.)
((𝐴 ∈ ℤ ∧ 𝐵 ∈ ℤ ∧ 𝐶 ∈ ℤ) → (lcm‘{𝐴, 𝐵, 𝐶}) = ((𝐴 lcm 𝐵) lcm 𝐶))
 
Theoremlcmfunsnlem1 15064* Lemma for lcmfdvds 15069 and lcmfunsnlem 15068 (Induction step part 1). (Contributed by AV, 25-Aug-2020.)
(((𝑧 ∈ ℤ ∧ 𝑦 ⊆ ℤ ∧ 𝑦 ∈ Fin) ∧ (∀𝑘 ∈ ℤ (∀𝑚𝑦 𝑚𝑘 → (lcm𝑦) ∥ 𝑘) ∧ ∀𝑛 ∈ ℤ (lcm‘(𝑦 ∪ {𝑛})) = ((lcm𝑦) lcm 𝑛))) → ∀𝑘 ∈ ℤ (∀𝑚 ∈ (𝑦 ∪ {𝑧})𝑚𝑘 → (lcm‘(𝑦 ∪ {𝑧})) ∥ 𝑘))
 
Theoremlcmfunsnlem2lem1 15065* Lemma 1 for lcmfunsnlem2 15067. (Contributed by AV, 26-Aug-2020.)
(((0 ∉ 𝑦𝑧 ≠ 0 ∧ 𝑛 ≠ 0) ∧ (𝑛 ∈ ℤ ∧ ((𝑧 ∈ ℤ ∧ 𝑦 ⊆ ℤ ∧ 𝑦 ∈ Fin) ∧ (∀𝑘 ∈ ℤ (∀𝑚𝑦 𝑚𝑘 → (lcm𝑦) ∥ 𝑘) ∧ ∀𝑛 ∈ ℤ (lcm‘(𝑦 ∪ {𝑛})) = ((lcm𝑦) lcm 𝑛))))) → ∀𝑘 ∈ ℕ (∀𝑖 ∈ ((𝑦 ∪ {𝑧}) ∪ {𝑛})𝑖𝑘 → ((lcm‘(𝑦 ∪ {𝑧})) lcm 𝑛) ≤ 𝑘))
 
Theoremlcmfunsnlem2lem2 15066* Lemma 2 for lcmfunsnlem2 15067. (Contributed by AV, 26-Aug-2020.)
(((0 ∉ 𝑦𝑧 ≠ 0 ∧ 𝑛 ≠ 0) ∧ (𝑛 ∈ ℤ ∧ ((𝑧 ∈ ℤ ∧ 𝑦 ⊆ ℤ ∧ 𝑦 ∈ Fin) ∧ (∀𝑘 ∈ ℤ (∀𝑚𝑦 𝑚𝑘 → (lcm𝑦) ∥ 𝑘) ∧ ∀𝑛 ∈ ℤ (lcm‘(𝑦 ∪ {𝑛})) = ((lcm𝑦) lcm 𝑛))))) → (lcm‘((𝑦 ∪ {𝑧}) ∪ {𝑛})) = ((lcm‘(𝑦 ∪ {𝑧})) lcm 𝑛))
 
Theoremlcmfunsnlem2 15067* Lemma for lcmfunsn 15071 and lcmfunsnlem 15068 (Induction step part 2). (Contributed by AV, 26-Aug-2020.)
(((𝑧 ∈ ℤ ∧ 𝑦 ⊆ ℤ ∧ 𝑦 ∈ Fin) ∧ (∀𝑘 ∈ ℤ (∀𝑚𝑦 𝑚𝑘 → (lcm𝑦) ∥ 𝑘) ∧ ∀𝑛 ∈ ℤ (lcm‘(𝑦 ∪ {𝑛})) = ((lcm𝑦) lcm 𝑛))) → ∀𝑛 ∈ ℤ (lcm‘((𝑦 ∪ {𝑧}) ∪ {𝑛})) = ((lcm‘(𝑦 ∪ {𝑧})) lcm 𝑛))
 
Theoremlcmfunsnlem 15068* Lemma for lcmfdvds 15069 and lcmfunsn 15071. These two theorems must be proven simultaneously by induction on the cardinality of a finite set 𝑌, because they depend on each other. This can be seen by the two parts lcmfunsnlem1 15064 and lcmfunsnlem2 15067 of the induction step, each of them using both induction hypotheses. (Contributed by AV, 26-Aug-2020.)
((𝑌 ⊆ ℤ ∧ 𝑌 ∈ Fin) → (∀𝑘 ∈ ℤ (∀𝑚𝑌 𝑚𝑘 → (lcm𝑌) ∥ 𝑘) ∧ ∀𝑛 ∈ ℤ (lcm‘(𝑌 ∪ {𝑛})) = ((lcm𝑌) lcm 𝑛)))
 
Theoremlcmfdvds 15069* The least common multiple of a set of integers divides any integer which is divisible by all elements of the set. (Contributed by AV, 26-Aug-2020.)
((𝐾 ∈ ℤ ∧ 𝑍 ⊆ ℤ ∧ 𝑍 ∈ Fin) → (∀𝑚𝑍 𝑚𝐾 → (lcm𝑍) ∥ 𝐾))
 
Theoremlcmfdvdsb 15070* Biconditional form of lcmfdvds 15069. (Contributed by AV, 26-Aug-2020.)
((𝐾 ∈ ℤ ∧ 𝑍 ⊆ ℤ ∧ 𝑍 ∈ Fin) → (∀𝑚𝑍 𝑚𝐾 ↔ (lcm𝑍) ∥ 𝐾))
 
Theoremlcmfunsn 15071 The lcm function for a union of a set of integer and a singleton. (Contributed by AV, 26-Aug-2020.)
((𝑌 ⊆ ℤ ∧ 𝑌 ∈ Fin ∧ 𝑁 ∈ ℤ) → (lcm‘(𝑌 ∪ {𝑁})) = ((lcm𝑌) lcm 𝑁))
 
Theoremlcmfun 15072 The lcm function for a union of sets of integers. (Contributed by AV, 27-Aug-2020.)
(((𝑌 ⊆ ℤ ∧ 𝑌 ∈ Fin) ∧ (𝑍 ⊆ ℤ ∧ 𝑍 ∈ Fin)) → (lcm‘(𝑌𝑍)) = ((lcm𝑌) lcm (lcm𝑍)))
 
Theoremlcmfass 15073 Associative law for the lcm function. (Contributed by AV, 27-Aug-2020.)
(((𝑌 ⊆ ℤ ∧ 𝑌 ∈ Fin) ∧ (𝑍 ⊆ ℤ ∧ 𝑍 ∈ Fin)) → (lcm‘({(lcm𝑌)} ∪ 𝑍)) = (lcm‘(𝑌 ∪ {(lcm𝑍)})))
 
Theoremlcmf2a3a4e12 15074 The least common multiple of 2 , 3 and 4 is 12. (Contributed by AV, 27-Aug-2020.)
(lcm‘{2, 3, 4}) = 12
 
Theoremlcmflefac 15075 The least common multiple of all positive integers less than or equal to an integer is less than or equal to the factorial of the integer. (Contributed by AV, 16-Aug-2020.) (Revised by AV, 27-Aug-2020.)
(𝑁 ∈ ℕ → (lcm‘(1...𝑁)) ≤ (!‘𝑁))
 
6.1.12  Coprimality and Euclid's lemma

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 15076. The negation, i.e. two integers are not coprime, can be expressed either by (𝐴 gcd 𝐵) ≠ 1, see ncoprmgcdne1b 15077, or equivalently by 1 < (𝐴 gcd 𝐵), see ncoprmgcdgt1b 15078.

A proof of Euclid's lemma based on coprimality is provided in coprmdvds 15080 (see euclemma 15139 for a version of Euclid's lemma for primes).

 
Theoremcoprmgcdb 15076* 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 ((𝑖𝐴𝑖𝐵) → 𝑖 = 1) ↔ (𝐴 gcd 𝐵) = 1))
 
Theoremncoprmgcdne1b 15077* 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)(𝑖𝐴𝑖𝐵) ↔ (𝐴 gcd 𝐵) ≠ 1))
 
Theoremncoprmgcdgt1b 15078* 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)(𝑖𝐴𝑖𝐵) ↔ 1 < (𝐴 gcd 𝐵)))
 
Theoremcoprmdvds1 15079 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 ∧ 𝐼𝐹𝐼𝐺) → 𝐼 = 1))
 
Theoremcoprmdvds 15080 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. Generalization of euclemma 15139. (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) → 𝐾𝑁))
 
TheoremcoprmdvdsOLD 15081 If an integer divides the product of two integers and is coprime to one of them, then it divides the other. (Contributed by Paul Chapman, 22-Jun-2011.) Obsolete version of coprmdvds 15080 as of 10-Jul-2021. (New usage is discouraged.) (Proof modification is discouraged.)
((𝐾 ∈ ℤ ∧ 𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → ((𝐾 ∥ (𝑀 · 𝑁) ∧ (𝐾 gcd 𝑀) = 1) → 𝐾𝑁))
 
Theoremcoprmdvds2 15082 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) → ((𝑀𝐾𝑁𝐾) → (𝑀 · 𝑁) ∥ 𝐾))
 
Theoremmulgcddvds 15083 One half of rpmulgcd2 15084, 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)))
 
Theoremrpmulgcd2 15084 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)))
 
Theoremqredeq 15085 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 = 𝑃𝑁 = 𝑄))
 
Theoremqredeu 15086* Every rational number has a unique reduced form. (Contributed by Jeff Hankins, 29-Sep-2013.)
(𝐴 ∈ ℚ → ∃!𝑥 ∈ (ℤ × ℕ)(((1st𝑥) gcd (2nd𝑥)) = 1 ∧ 𝐴 = ((1st𝑥) / (2nd𝑥))))
 
Theoremrpmul 15087 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))
 
Theoremrpdvds 15088 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 ∧ 𝑀𝑁)) → (𝐾 gcd 𝑀) = 1)
 
Theoremcoprmprod 15089* The product of the elements of a sequence of pairwise coprime positive integers is coprime to a positive integer which is coprime to all integers of the sequence. (Contributed by AV, 18-Aug-2020.)
(((𝑀 ∈ Fin ∧ 𝑀 ⊆ ℕ ∧ 𝑁 ∈ ℕ) ∧ 𝐹:ℕ⟶ℕ ∧ ∀𝑚𝑀 ((𝐹𝑚) gcd 𝑁) = 1) → (∀𝑚𝑀𝑛 ∈ (𝑀 ∖ {𝑚})((𝐹𝑚) gcd (𝐹𝑛)) = 1 → (∏𝑚𝑀 (𝐹𝑚) gcd 𝑁) = 1))
 
Theoremcoprmproddvdslem 15090* Lemma for coprmproddvds 15091: Induction step. (Contributed by AV, 19-Aug-2020.)
((𝑦 ∈ Fin ∧ ¬ 𝑧𝑦) → ((((𝑦 ⊆ ℕ ∧ (𝐾 ∈ ℕ ∧ 𝐹:ℕ⟶ℕ)) ∧ (∀𝑚𝑦𝑛 ∈ (𝑦 ∖ {𝑚})((𝐹𝑚) gcd (𝐹𝑛)) = 1 ∧ ∀𝑚𝑦 (𝐹𝑚) ∥ 𝐾)) → ∏𝑚𝑦 (𝐹𝑚) ∥ 𝐾) → ((((𝑦 ∪ {𝑧}) ⊆ ℕ ∧ (𝐾 ∈ ℕ ∧ 𝐹:ℕ⟶ℕ)) ∧ (∀𝑚 ∈ (𝑦 ∪ {𝑧})∀𝑛 ∈ ((𝑦 ∪ {𝑧}) ∖ {𝑚})((𝐹𝑚) gcd (𝐹𝑛)) = 1 ∧ ∀𝑚 ∈ (𝑦 ∪ {𝑧})(𝐹𝑚) ∥ 𝐾)) → ∏𝑚 ∈ (𝑦 ∪ {𝑧})(𝐹𝑚) ∥ 𝐾)))
 
Theoremcoprmproddvds 15091* If a positive integer is divisible by each element of a set of pairwise coprime positive integers, then it is divisible by their product. (Contributed by AV, 19-Aug-2020.)
(((𝑀 ⊆ ℕ ∧ 𝑀 ∈ Fin) ∧ (𝐾 ∈ ℕ ∧ 𝐹:ℕ⟶ℕ) ∧ (∀𝑚𝑀𝑛 ∈ (𝑀 ∖ {𝑚})((𝐹𝑚) gcd (𝐹𝑛)) = 1 ∧ ∀𝑚𝑀 (𝐹𝑚) ∥ 𝐾)) → ∏𝑚𝑀 (𝐹𝑚) ∥ 𝐾)
 
6.1.13  Cancellability of congruences
 
Theoremcongr 15092* 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 𝑎𝑏 (mod 𝑚) if 𝑚 divides the difference 𝑎𝑏", 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) = (𝐴𝐵)))
 
Theoremdivgcdcoprm0 15093 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)
 
Theoremdivgcdcoprmex 15094* 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))
 
Theoremcncongr1 15095 One direction of the bicondition in cncongr 15097. 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)))
 
Theoremcncongr2 15096 The other direction of the bicondition in cncongr 15097. (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)))
 
Theoremcncongr 15097 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)))
 
Theoremcncongrcoprm 15098 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)))
 
6.2  Elementary prime number theory
 
6.2.1  Elementary properties

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 4163.

 
Syntaxcprime 15099 Extend the definition of a class to include the set of prime numbers.
class
 
Definitiondf-prm 15100* Define the set of prime numbers. (Contributed by Paul Chapman, 22-Jun-2011.)
ℙ = {𝑝 ∈ ℕ ∣ {𝑛 ∈ ℕ ∣ 𝑛𝑝} ≈ 2𝑜}
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