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Theorem List for Intuitionistic Logic Explorer - 12001-12100   *Has distinct variable group(s)
TypeLabelDescription
Statement
 
Theoremgcddvdslcm 12001 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 12002 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 12003 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 12004 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 12005 Lemma for lcmgcd 12006 and lcmdvds 12007. 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 12006 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 11940; see, e.g., https://proofwiki.org/wiki/Product_of_GCD_and_LCM 11940 and https://math.stackexchange.com/a/470827 11940. 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 11993 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 12007 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 12008 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 12009 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 12010 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 12011 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 12012 Biconditional form of lcmdvds 12007. (Contributed by Steve Rodriguez, 20-Jan-2020.)
((𝐾 ∈ ℤ ∧ 𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → ((𝑀𝐾𝑁𝐾) ↔ (𝑀 lcm 𝑁) ∥ 𝐾))
 
Theoremlcmass 12013 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 12014 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
 
Theorem6lcm4e12 12015 The least common multiple of six and four is twelve. (Contributed by AV, 27-Aug-2020.)
(6 lcm 4) = 12
 
5.1.10  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 12016. The negation, i.e. two integers are not coprime, can be expressed either by (𝐴 gcd 𝐵) ≠ 1, see ncoprmgcdne1b 12017, or equivalently by 1 < (𝐴 gcd 𝐵), see ncoprmgcdgt1b 12018.

A proof of Euclid's lemma based on coprimality is provided in coprmdvds 12020 (as opposed to Euclid's lemma for primes).

 
Theoremcoprmgcdb 12016* 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 12017* 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 12018* 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 12019 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 12020 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) → 𝐾𝑁))
 
Theoremcoprmdvds2 12021 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 12022 One half of rpmulgcd2 12023, 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 12023 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 12024 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 12025* 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 12026 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 12027 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)
 
5.1.11  Cancellability of congruences
 
Theoremcongr 12028* 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 12029 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 12030* 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 12031 One direction of the bicondition in cncongr 12033. 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 12032 The other direction of the bicondition in cncongr 12033. (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 12033 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 12034 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)))
 
5.2  Elementary prime number theory
 
5.2.1  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 3702.

 
Syntaxcprime 12035 Extend the definition of a class to include the set of prime numbers.
class
 
Definitiondf-prm 12036* Define the set of prime numbers. (Contributed by Paul Chapman, 22-Jun-2011.)
ℙ = {𝑝 ∈ ℕ ∣ {𝑛 ∈ ℕ ∣ 𝑛𝑝} ≈ 2o}
 
Theoremisprm 12037* 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 ∣ 𝑛𝑃} ≈ 2o))
 
Theoremprmnn 12038 A prime number is a positive integer. (Contributed by Paul Chapman, 22-Jun-2011.)
(𝑃 ∈ ℙ → 𝑃 ∈ ℕ)
 
Theoremprmz 12039 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)
 
Theoremprmssnn 12040 The prime numbers are a subset of the positive integers. (Contributed by AV, 22-Jul-2020.)
ℙ ⊆ ℕ
 
Theoremprmex 12041 The set of prime numbers exists. (Contributed by AV, 22-Jul-2020.)
ℙ ∈ V
 
Theorem1nprm 12042 1 is not a prime number. (Contributed by Paul Chapman, 22-Jun-2011.) (Proof shortened by Fan Zheng, 3-Jul-2016.)
¬ 1 ∈ ℙ
 
Theorem1idssfct 12043* The positive divisors of a positive integer include 1 and itself. (Contributed by Paul Chapman, 22-Jun-2011.)
(𝑁 ∈ ℕ → {1, 𝑁} ⊆ {𝑛 ∈ ℕ ∣ 𝑛𝑁})
 
Theoremisprm2lem 12044* Lemma for isprm2 12045. (Contributed by Paul Chapman, 22-Jun-2011.)
((𝑃 ∈ ℕ ∧ 𝑃 ≠ 1) → ({𝑛 ∈ ℕ ∣ 𝑛𝑃} ≈ 2o ↔ {𝑛 ∈ ℕ ∣ 𝑛𝑃} = {1, 𝑃}))
 
Theoremisprm2 12045* The predicate "is a prime number". A prime number is an integer greater than or equal to 2 whose only positive divisors are 1 and itself. Definition in [ApostolNT] p. 16. (Contributed by Paul Chapman, 26-Oct-2012.)
(𝑃 ∈ ℙ ↔ (𝑃 ∈ (ℤ‘2) ∧ ∀𝑧 ∈ ℕ (𝑧𝑃 → (𝑧 = 1 ∨ 𝑧 = 𝑃))))
 
Theoremisprm3 12046* The predicate "is a prime number". A prime number is an integer greater than or equal to 2 with no divisors strictly between 1 and itself. (Contributed by Paul Chapman, 26-Oct-2012.)
(𝑃 ∈ ℙ ↔ (𝑃 ∈ (ℤ‘2) ∧ ∀𝑧 ∈ (2...(𝑃 − 1)) ¬ 𝑧𝑃))
 
Theoremisprm4 12047* The predicate "is a prime number". A prime number is an integer greater than or equal to 2 whose only divisor greater than or equal to 2 is itself. (Contributed by Paul Chapman, 26-Oct-2012.)
(𝑃 ∈ ℙ ↔ (𝑃 ∈ (ℤ‘2) ∧ ∀𝑧 ∈ (ℤ‘2)(𝑧𝑃𝑧 = 𝑃)))
 
Theoremprmind2 12048* A variation on prmind 12049 assuming complete induction for primes. (Contributed by Mario Carneiro, 20-Jun-2015.)
(𝑥 = 1 → (𝜑𝜓))    &   (𝑥 = 𝑦 → (𝜑𝜒))    &   (𝑥 = 𝑧 → (𝜑𝜃))    &   (𝑥 = (𝑦 · 𝑧) → (𝜑𝜏))    &   (𝑥 = 𝐴 → (𝜑𝜂))    &   𝜓    &   ((𝑥 ∈ ℙ ∧ ∀𝑦 ∈ (1...(𝑥 − 1))𝜒) → 𝜑)    &   ((𝑦 ∈ (ℤ‘2) ∧ 𝑧 ∈ (ℤ‘2)) → ((𝜒𝜃) → 𝜏))       (𝐴 ∈ ℕ → 𝜂)
 
Theoremprmind 12049* Perform induction over the multiplicative structure of . If a property 𝜑(𝑥) holds for the primes and 1 and is preserved under multiplication, then it holds for every positive integer. (Contributed by Mario Carneiro, 20-Jun-2015.)
(𝑥 = 1 → (𝜑𝜓))    &   (𝑥 = 𝑦 → (𝜑𝜒))    &   (𝑥 = 𝑧 → (𝜑𝜃))    &   (𝑥 = (𝑦 · 𝑧) → (𝜑𝜏))    &   (𝑥 = 𝐴 → (𝜑𝜂))    &   𝜓    &   (𝑥 ∈ ℙ → 𝜑)    &   ((𝑦 ∈ (ℤ‘2) ∧ 𝑧 ∈ (ℤ‘2)) → ((𝜒𝜃) → 𝜏))       (𝐴 ∈ ℕ → 𝜂)
 
Theoremdvdsprime 12050 If 𝑀 divides a prime, then 𝑀 is either the prime or one. (Contributed by Scott Fenton, 8-Apr-2014.)
((𝑃 ∈ ℙ ∧ 𝑀 ∈ ℕ) → (𝑀𝑃 ↔ (𝑀 = 𝑃𝑀 = 1)))
 
Theoremnprm 12051 A product of two integers greater than one is composite. (Contributed by Mario Carneiro, 20-Jun-2015.)
((𝐴 ∈ (ℤ‘2) ∧ 𝐵 ∈ (ℤ‘2)) → ¬ (𝐴 · 𝐵) ∈ ℙ)
 
Theoremnprmi 12052 An inference for compositeness. (Contributed by Mario Carneiro, 18-Feb-2014.) (Revised by Mario Carneiro, 20-Jun-2015.)
𝐴 ∈ ℕ    &   𝐵 ∈ ℕ    &   1 < 𝐴    &   1 < 𝐵    &   (𝐴 · 𝐵) = 𝑁        ¬ 𝑁 ∈ ℙ
 
Theoremdvdsnprmd 12053 If a number is divisible by an integer greater than 1 and less then the number, the number is not prime. (Contributed by AV, 24-Jul-2021.)
(𝜑 → 1 < 𝐴)    &   (𝜑𝐴 < 𝑁)    &   (𝜑𝐴𝑁)       (𝜑 → ¬ 𝑁 ∈ ℙ)
 
Theoremprm2orodd 12054 A prime number is either 2 or odd. (Contributed by AV, 19-Jun-2021.)
(𝑃 ∈ ℙ → (𝑃 = 2 ∨ ¬ 2 ∥ 𝑃))
 
Theorem2prm 12055 2 is a prime number. (Contributed by Paul Chapman, 22-Jun-2011.) (Proof shortened by Fan Zheng, 16-Jun-2016.)
2 ∈ ℙ
 
Theorem3prm 12056 3 is a prime number. (Contributed by Paul Chapman, 22-Jun-2011.)
3 ∈ ℙ
 
Theorem4nprm 12057 4 is not a prime number. (Contributed by Paul Chapman, 22-Jun-2011.) (Proof shortened by Mario Carneiro, 18-Feb-2014.)
¬ 4 ∈ ℙ
 
Theoremprmdc 12058 Primality is decidable. (Contributed by Jim Kingdon, 30-Sep-2024.)
(𝑁 ∈ ℕ → DECID 𝑁 ∈ ℙ)
 
Theoremprmuz2 12059 A prime number is an integer greater than or equal to 2. (Contributed by Paul Chapman, 17-Nov-2012.)
(𝑃 ∈ ℙ → 𝑃 ∈ (ℤ‘2))
 
Theoremprmgt1 12060 A prime number is an integer greater than 1. (Contributed by Alexander van der Vekens, 17-May-2018.)
(𝑃 ∈ ℙ → 1 < 𝑃)
 
Theoremprmm2nn0 12061 Subtracting 2 from a prime number results in a nonnegative integer. (Contributed by Alexander van der Vekens, 30-Aug-2018.)
(𝑃 ∈ ℙ → (𝑃 − 2) ∈ ℕ0)
 
Theoremoddprmgt2 12062 An odd prime is greater than 2. (Contributed by AV, 20-Aug-2021.)
(𝑃 ∈ (ℙ ∖ {2}) → 2 < 𝑃)
 
Theoremoddprmge3 12063 An odd prime is greater than or equal to 3. (Contributed by Alexander van der Vekens, 7-Oct-2018.) (Revised by AV, 20-Aug-2021.)
(𝑃 ∈ (ℙ ∖ {2}) → 𝑃 ∈ (ℤ‘3))
 
Theoremsqnprm 12064 A square is never prime. (Contributed by Mario Carneiro, 20-Jun-2015.)
(𝐴 ∈ ℤ → ¬ (𝐴↑2) ∈ ℙ)
 
Theoremdvdsprm 12065 An integer greater than or equal to 2 divides a prime number iff it is equal to it. (Contributed by Paul Chapman, 26-Oct-2012.)
((𝑁 ∈ (ℤ‘2) ∧ 𝑃 ∈ ℙ) → (𝑁𝑃𝑁 = 𝑃))
 
Theoremexprmfct 12066* Every integer greater than or equal to 2 has a prime factor. (Contributed by Paul Chapman, 26-Oct-2012.) (Proof shortened by Mario Carneiro, 20-Jun-2015.)
(𝑁 ∈ (ℤ‘2) → ∃𝑝 ∈ ℙ 𝑝𝑁)
 
Theoremprmdvdsfz 12067* Each integer greater than 1 and less then or equal to a fixed number is divisible by a prime less then or equal to this fixed number. (Contributed by AV, 15-Aug-2020.)
((𝑁 ∈ ℕ ∧ 𝐼 ∈ (2...𝑁)) → ∃𝑝 ∈ ℙ (𝑝𝑁𝑝𝐼))
 
Theoremnprmdvds1 12068 No prime number divides 1. (Contributed by Paul Chapman, 17-Nov-2012.) (Proof shortened by Mario Carneiro, 2-Jul-2015.)
(𝑃 ∈ ℙ → ¬ 𝑃 ∥ 1)
 
Theoremisprm5lem 12069* Lemma for isprm5 12070. The interesting direction (showing that one only needs to check prime divisors up to the square root of 𝑃). (Contributed by Jim Kingdon, 20-Oct-2024.)
(𝜑𝑃 ∈ (ℤ‘2))    &   (𝜑 → ∀𝑧 ∈ ℙ ((𝑧↑2) ≤ 𝑃 → ¬ 𝑧𝑃))    &   (𝜑𝑋 ∈ (2...(𝑃 − 1)))       (𝜑 → ¬ 𝑋𝑃)
 
Theoremisprm5 12070* One need only check prime divisors of 𝑃 up to 𝑃 in order to ensure primality. (Contributed by Mario Carneiro, 18-Feb-2014.)
(𝑃 ∈ ℙ ↔ (𝑃 ∈ (ℤ‘2) ∧ ∀𝑧 ∈ ℙ ((𝑧↑2) ≤ 𝑃 → ¬ 𝑧𝑃)))
 
Theoremdivgcdodd 12071 Either 𝐴 / (𝐴 gcd 𝐵) is odd or 𝐵 / (𝐴 gcd 𝐵) is odd. (Contributed by Scott Fenton, 19-Apr-2014.)
((𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ) → (¬ 2 ∥ (𝐴 / (𝐴 gcd 𝐵)) ∨ ¬ 2 ∥ (𝐵 / (𝐴 gcd 𝐵))))
 
5.2.2  Coprimality and Euclid's lemma (cont.)

This section is about coprimality with respect to primes, and a special version of Euclid's lemma for primes is provided, see euclemma 12074.

 
Theoremcoprm 12072 A prime number either divides an integer or is coprime to it, but not both. Theorem 1.8 in [ApostolNT] p. 17. (Contributed by Paul Chapman, 22-Jun-2011.)
((𝑃 ∈ ℙ ∧ 𝑁 ∈ ℤ) → (¬ 𝑃𝑁 ↔ (𝑃 gcd 𝑁) = 1))
 
Theoremprmrp 12073 Unequal prime numbers are relatively prime. (Contributed by Mario Carneiro, 23-Feb-2014.)
((𝑃 ∈ ℙ ∧ 𝑄 ∈ ℙ) → ((𝑃 gcd 𝑄) = 1 ↔ 𝑃𝑄))
 
Theoremeuclemma 12074 Euclid's lemma. A prime number divides the product of two integers iff it divides at least one of them. Theorem 1.9 in [ApostolNT] p. 17. (Contributed by Paul Chapman, 17-Nov-2012.)
((𝑃 ∈ ℙ ∧ 𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → (𝑃 ∥ (𝑀 · 𝑁) ↔ (𝑃𝑀𝑃𝑁)))
 
Theoremisprm6 12075* A number is prime iff it satisfies Euclid's lemma euclemma 12074. (Contributed by Mario Carneiro, 6-Sep-2015.)
(𝑃 ∈ ℙ ↔ (𝑃 ∈ (ℤ‘2) ∧ ∀𝑥 ∈ ℤ ∀𝑦 ∈ ℤ (𝑃 ∥ (𝑥 · 𝑦) → (𝑃𝑥𝑃𝑦))))
 
Theoremprmdvdsexp 12076 A prime divides a positive power of an integer iff it divides the integer. (Contributed by Mario Carneiro, 24-Feb-2014.) (Revised by Mario Carneiro, 17-Jul-2014.)
((𝑃 ∈ ℙ ∧ 𝐴 ∈ ℤ ∧ 𝑁 ∈ ℕ) → (𝑃 ∥ (𝐴𝑁) ↔ 𝑃𝐴))
 
Theoremprmdvdsexpb 12077 A prime divides a positive power of another iff they are equal. (Contributed by Paul Chapman, 30-Nov-2012.) (Revised by Mario Carneiro, 24-Feb-2014.)
((𝑃 ∈ ℙ ∧ 𝑄 ∈ ℙ ∧ 𝑁 ∈ ℕ) → (𝑃 ∥ (𝑄𝑁) ↔ 𝑃 = 𝑄))
 
Theoremprmdvdsexpr 12078 If a prime divides a nonnegative power of another, then they are equal. (Contributed by Mario Carneiro, 16-Jan-2015.)
((𝑃 ∈ ℙ ∧ 𝑄 ∈ ℙ ∧ 𝑁 ∈ ℕ0) → (𝑃 ∥ (𝑄𝑁) → 𝑃 = 𝑄))
 
Theoremprmexpb 12079 Two positive prime powers are equal iff the primes and the powers are equal. (Contributed by Paul Chapman, 30-Nov-2012.)
(((𝑃 ∈ ℙ ∧ 𝑄 ∈ ℙ) ∧ (𝑀 ∈ ℕ ∧ 𝑁 ∈ ℕ)) → ((𝑃𝑀) = (𝑄𝑁) ↔ (𝑃 = 𝑄𝑀 = 𝑁)))
 
Theoremprmfac1 12080 The factorial of a number only contains primes less than the base. (Contributed by Mario Carneiro, 6-Mar-2014.)
((𝑁 ∈ ℕ0𝑃 ∈ ℙ ∧ 𝑃 ∥ (!‘𝑁)) → 𝑃𝑁)
 
Theoremrpexp 12081 If two numbers 𝐴 and 𝐵 are relatively prime, then they are still relatively prime if raised to a power. (Contributed by Mario Carneiro, 24-Feb-2014.)
((𝐴 ∈ ℤ ∧ 𝐵 ∈ ℤ ∧ 𝑁 ∈ ℕ) → (((𝐴𝑁) gcd 𝐵) = 1 ↔ (𝐴 gcd 𝐵) = 1))
 
Theoremrpexp1i 12082 Relative primality passes to asymmetric powers. (Contributed by Stefan O'Rear, 27-Sep-2014.)
((𝐴 ∈ ℤ ∧ 𝐵 ∈ ℤ ∧ 𝑀 ∈ ℕ0) → ((𝐴 gcd 𝐵) = 1 → ((𝐴𝑀) gcd 𝐵) = 1))
 
Theoremrpexp12i 12083 Relative primality passes to symmetric powers. (Contributed by Stefan O'Rear, 27-Sep-2014.)
((𝐴 ∈ ℤ ∧ 𝐵 ∈ ℤ ∧ (𝑀 ∈ ℕ0𝑁 ∈ ℕ0)) → ((𝐴 gcd 𝐵) = 1 → ((𝐴𝑀) gcd (𝐵𝑁)) = 1))
 
Theoremprmndvdsfaclt 12084 A prime number does not divide the factorial of a nonnegative integer less than the prime number. (Contributed by AV, 13-Jul-2021.)
((𝑃 ∈ ℙ ∧ 𝑁 ∈ ℕ0) → (𝑁 < 𝑃 → ¬ 𝑃 ∥ (!‘𝑁)))
 
Theoremcncongrprm 12085 Corollary 2 of Cancellability of Congruences: Two products with a common factor are congruent modulo a prime number not dividing the common factor iff the other factors are congruent modulo the prime number. (Contributed by AV, 13-Jul-2021.)
(((𝐴 ∈ ℤ ∧ 𝐵 ∈ ℤ ∧ 𝐶 ∈ ℤ) ∧ (𝑃 ∈ ℙ ∧ ¬ 𝑃𝐶)) → (((𝐴 · 𝐶) mod 𝑃) = ((𝐵 · 𝐶) mod 𝑃) ↔ (𝐴 mod 𝑃) = (𝐵 mod 𝑃)))
 
Theoremisevengcd2 12086 The predicate "is an even number". An even number and 2 have 2 as greatest common divisor. (Contributed by AV, 1-Jul-2020.) (Revised by AV, 8-Aug-2021.)
(𝑍 ∈ ℤ → (2 ∥ 𝑍 ↔ (2 gcd 𝑍) = 2))
 
Theoremisoddgcd1 12087 The predicate "is an odd number". An odd number and 2 have 1 as greatest common divisor. (Contributed by AV, 1-Jul-2020.) (Revised by AV, 8-Aug-2021.)
(𝑍 ∈ ℤ → (¬ 2 ∥ 𝑍 ↔ (2 gcd 𝑍) = 1))
 
Theorem3lcm2e6 12088 The least common multiple of three and two is six. The operands are unequal primes and thus coprime, so the result is (the absolute value of) their product. (Contributed by Steve Rodriguez, 20-Jan-2020.) (Proof shortened by AV, 27-Aug-2020.)
(3 lcm 2) = 6
 
5.2.3  Non-rationality of square root of 2
 
Theoremsqrt2irrlem 12089 Lemma for sqrt2irr 12090. This is the core of the proof: - if 𝐴 / 𝐵 = √(2), then 𝐴 and 𝐵 are even, so 𝐴 / 2 and 𝐵 / 2 are smaller representatives, which is absurd by the method of infinite descent (here implemented by strong induction). (Contributed by NM, 20-Aug-2001.) (Revised by Mario Carneiro, 12-Sep-2015.)
(𝜑𝐴 ∈ ℤ)    &   (𝜑𝐵 ∈ ℕ)    &   (𝜑 → (√‘2) = (𝐴 / 𝐵))       (𝜑 → ((𝐴 / 2) ∈ ℤ ∧ (𝐵 / 2) ∈ ℕ))
 
Theoremsqrt2irr 12090 The square root of 2 is not rational. That is, for any rational number, (√‘2) does not equal it. However, if we were to say "the square root of 2 is irrational" that would mean something stronger: "for any rational number, (√‘2) is apart from it" (the two statements are equivalent given excluded middle). See sqrt2irrap 12108 for the proof that the square root of two is irrational.

The proof's core is proven in sqrt2irrlem 12089, which shows that if 𝐴 / 𝐵 = √(2), then 𝐴 and 𝐵 are even, so 𝐴 / 2 and 𝐵 / 2 are smaller representatives, which is absurd. (Contributed by NM, 8-Jan-2002.) (Proof shortened by Mario Carneiro, 12-Sep-2015.)

(√‘2) ∉ ℚ
 
Theoremsqrt2re 12091 The square root of 2 exists and is a real number. (Contributed by NM, 3-Dec-2004.)
(√‘2) ∈ ℝ
 
Theoremsqrt2irr0 12092 The square root of 2 is not rational. (Contributed by AV, 23-Dec-2022.)
(√‘2) ∈ (ℝ ∖ ℚ)
 
Theorempw2dvdslemn 12093* Lemma for pw2dvds 12094. If a natural number has some power of two which does not divide it, there is a highest power of two which does divide it. (Contributed by Jim Kingdon, 14-Nov-2021.)
((𝑁 ∈ ℕ ∧ 𝐴 ∈ ℕ ∧ ¬ (2↑𝐴) ∥ 𝑁) → ∃𝑚 ∈ ℕ0 ((2↑𝑚) ∥ 𝑁 ∧ ¬ (2↑(𝑚 + 1)) ∥ 𝑁))
 
Theorempw2dvds 12094* A natural number has a highest power of two which divides it. (Contributed by Jim Kingdon, 14-Nov-2021.)
(𝑁 ∈ ℕ → ∃𝑚 ∈ ℕ0 ((2↑𝑚) ∥ 𝑁 ∧ ¬ (2↑(𝑚 + 1)) ∥ 𝑁))
 
Theorempw2dvdseulemle 12095 Lemma for pw2dvdseu 12096. Powers of two which do and do not divide a natural number. (Contributed by Jim Kingdon, 17-Nov-2021.)
(𝜑𝑁 ∈ ℕ)    &   (𝜑𝐴 ∈ ℕ0)    &   (𝜑𝐵 ∈ ℕ0)    &   (𝜑 → (2↑𝐴) ∥ 𝑁)    &   (𝜑 → ¬ (2↑(𝐵 + 1)) ∥ 𝑁)       (𝜑𝐴𝐵)
 
Theorempw2dvdseu 12096* A natural number has a unique highest power of two which divides it. (Contributed by Jim Kingdon, 16-Nov-2021.)
(𝑁 ∈ ℕ → ∃!𝑚 ∈ ℕ0 ((2↑𝑚) ∥ 𝑁 ∧ ¬ (2↑(𝑚 + 1)) ∥ 𝑁))
 
Theoremoddpwdclemxy 12097* Lemma for oddpwdc 12102. Another way of stating that decomposing a natural number into a power of two and an odd number is unique. (Contributed by Jim Kingdon, 16-Nov-2021.)
((((𝑋 ∈ ℕ ∧ ¬ 2 ∥ 𝑋) ∧ 𝑌 ∈ ℕ0) ∧ 𝐴 = ((2↑𝑌) · 𝑋)) → (𝑋 = (𝐴 / (2↑(𝑧 ∈ ℕ0 ((2↑𝑧) ∥ 𝐴 ∧ ¬ (2↑(𝑧 + 1)) ∥ 𝐴)))) ∧ 𝑌 = (𝑧 ∈ ℕ0 ((2↑𝑧) ∥ 𝐴 ∧ ¬ (2↑(𝑧 + 1)) ∥ 𝐴))))
 
Theoremoddpwdclemdvds 12098* Lemma for oddpwdc 12102. A natural number is divisible by the highest power of two which divides it. (Contributed by Jim Kingdon, 17-Nov-2021.)
(𝐴 ∈ ℕ → (2↑(𝑧 ∈ ℕ0 ((2↑𝑧) ∥ 𝐴 ∧ ¬ (2↑(𝑧 + 1)) ∥ 𝐴))) ∥ 𝐴)
 
Theoremoddpwdclemndvds 12099* Lemma for oddpwdc 12102. A natural number is not divisible by one more than the highest power of two which divides it. (Contributed by Jim Kingdon, 17-Nov-2021.)
(𝐴 ∈ ℕ → ¬ (2↑((𝑧 ∈ ℕ0 ((2↑𝑧) ∥ 𝐴 ∧ ¬ (2↑(𝑧 + 1)) ∥ 𝐴)) + 1)) ∥ 𝐴)
 
Theoremoddpwdclemodd 12100* Lemma for oddpwdc 12102. Removing the powers of two from a natural number produces an odd number. (Contributed by Jim Kingdon, 16-Nov-2021.)
(𝐴 ∈ ℕ → ¬ 2 ∥ (𝐴 / (2↑(𝑧 ∈ ℕ0 ((2↑𝑧) ∥ 𝐴 ∧ ¬ (2↑(𝑧 + 1)) ∥ 𝐴)))))
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