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Theorem List for Intuitionistic Logic Explorer - 11701-11800   *Has distinct variable group(s)
TypeLabelDescription
Statement
 
Theoremdvdssqim 11701 Unidirectional form of dvdssq 11708. (Contributed by Scott Fenton, 19-Apr-2014.)
((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → (𝑀𝑁 → (𝑀↑2) ∥ (𝑁↑2)))
 
Theoremdvdsmulgcd 11702 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))))
 
Theoremrpmulgcd 11703 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))
 
Theoremrplpwr 11704 If 𝐴 and 𝐵 are relatively prime, then so are 𝐴𝑁 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 → ((𝐴𝑁) gcd 𝐵) = 1))
 
Theoremrppwr 11705 If 𝐴 and 𝐵 are relatively prime, then so are 𝐴𝑁 and 𝐵𝑁. (Contributed by Scott Fenton, 12-Apr-2014.) (Revised by Mario Carneiro, 19-Apr-2014.)
((𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝑁 ∈ ℕ) → ((𝐴 gcd 𝐵) = 1 → ((𝐴𝑁) gcd (𝐵𝑁)) = 1))
 
Theoremsqgcd 11706 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)))
 
Theoremdvdssqlem 11707 Lemma for dvdssq 11708. (Contributed by Scott Fenton, 18-Apr-2014.) (Revised by Mario Carneiro, 19-Apr-2014.)
((𝑀 ∈ ℕ ∧ 𝑁 ∈ ℕ) → (𝑀𝑁 ↔ (𝑀↑2) ∥ (𝑁↑2)))
 
Theoremdvdssq 11708 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) → (𝑀𝑁 ↔ (𝑀↑2) ∥ (𝑁↑2)))
 
Theorembezoutr 11709 Partial converse to bezout 11688. 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)))
 
Theorembezoutr1 11710 Converse of bezout 11688 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))
 
5.1.6  Algorithms
 
Theoremnn0seqcvgd 11711* 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.)
(𝜑𝐹:ℕ0⟶ℕ0)    &   (𝜑𝑁 = (𝐹‘0))    &   ((𝜑𝑘 ∈ ℕ0) → ((𝐹‘(𝑘 + 1)) ≠ 0 → (𝐹‘(𝑘 + 1)) < (𝐹𝑘)))       (𝜑 → (𝐹𝑁) = 0)
 
Theoremialgrlem1st 11712 Lemma for ialgr0 11714. Expressing algrflemg 6120 in a form suitable for theorems such as seq3-1 10226 or seqf 10227. (Contributed by Jim Kingdon, 22-Jul-2021.)
(𝜑𝐹:𝑆𝑆)       ((𝜑 ∧ (𝑥𝑆𝑦𝑆)) → (𝑥(𝐹 ∘ 1st )𝑦) ∈ 𝑆)
 
Theoremialgrlemconst 11713 Lemma for ialgr0 11714. Closure of a constant function, in a form suitable for theorems such as seq3-1 10226 or seqf 10227. (Contributed by Jim Kingdon, 22-Jul-2021.)
𝑍 = (ℤ𝑀)    &   (𝜑𝐴𝑆)       ((𝜑𝑥 ∈ (ℤ𝑀)) → ((𝑍 × {𝐴})‘𝑥) ∈ 𝑆)
 
Theoremialgr0 11714 The value of the algorithm iterator 𝑅 at 0 is the initial state 𝐴. (Contributed by Paul Chapman, 31-Mar-2011.) (Revised by Jim Kingdon, 12-Mar-2023.)
𝑍 = (ℤ𝑀)    &   𝑅 = seq𝑀((𝐹 ∘ 1st ), (𝑍 × {𝐴}))    &   (𝜑𝑀 ∈ ℤ)    &   (𝜑𝐴𝑆)    &   (𝜑𝐹:𝑆𝑆)       (𝜑 → (𝑅𝑀) = 𝐴)
 
Theoremalgrf 11715 An algorithm is a step function 𝐹:𝑆𝑆 on a state space 𝑆. An algorithm acts on an initial state 𝐴𝑆 by iteratively applying 𝐹 to give 𝐴, (𝐹𝐴), (𝐹‘(𝐹𝐴)) 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:ℕ0𝑆 "runs" the algorithm 𝐹 so that (𝑅𝑘) 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 = (ℤ𝑀)    &   𝑅 = seq𝑀((𝐹 ∘ 1st ), (𝑍 × {𝐴}))    &   (𝜑𝑀 ∈ ℤ)    &   (𝜑𝐴𝑆)    &   (𝜑𝐹:𝑆𝑆)       (𝜑𝑅:𝑍𝑆)
 
Theoremalgrp1 11716 The value of the algorithm iterator 𝑅 at (𝐾 + 1). (Contributed by Paul Chapman, 31-Mar-2011.) (Revised by Jim Kingdon, 12-Mar-2023.)
𝑍 = (ℤ𝑀)    &   𝑅 = seq𝑀((𝐹 ∘ 1st ), (𝑍 × {𝐴}))    &   (𝜑𝑀 ∈ ℤ)    &   (𝜑𝐴𝑆)    &   (𝜑𝐹:𝑆𝑆)       ((𝜑𝐾𝑍) → (𝑅‘(𝐾 + 1)) = (𝐹‘(𝑅𝐾)))
 
Theoremalginv 11717* If 𝐼 is an invariant of 𝐹, then 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    &   (𝑥𝑆 → (𝐼‘(𝐹𝑥)) = (𝐼𝑥))       ((𝐴𝑆𝐾 ∈ ℕ0) → (𝐼‘(𝑅𝐾)) = (𝐼‘(𝑅‘0)))
 
Theoremalgcvg 11718* 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 11719 Lemma for algcvgb 11720. (Contributed by Paul Chapman, 31-Mar-2011.)
((𝑀 ∈ ℕ0𝑁 ∈ ℕ0) → ((𝑁 ≠ 0 → 𝑁 < 𝑀) ↔ ((𝑀 ≠ 0 → 𝑁 < 𝑀) ∧ (𝑀 = 0 → 𝑁 = 0))))
 
Theoremalgcvgb 11720 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 11721* 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 11722* 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)))
 
5.1.7  Euclid's Algorithm
 
Theoremeucalgval2 11723* 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 11724* Euclid's Algorithm eucalg 11729 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 11725* 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 11726* 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 11727* 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 11728* 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 11729* 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))
 
5.1.8  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."

In this section, an operation calculating the least common multiple of two integers (df-lcm 11731). The definition is valid for all integers, including negative integers and 0, obeying the above mentioned convention.

 
Syntaxclcm 11730 Extend the definition of a class to include the least common multiple operator.
class lcm
 
Definitiondf-lcm 11731* 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 ∣ (𝑥𝑛𝑦𝑛)}, ℝ, < )))
 
Theoremlcmmndc 11732 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))
 
Theoremlcmval 11733* 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 11625 and gcdval 11637. (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 11734 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 11735 The value, by convention, of the lcm operator when either operand is 0. (Use lcmcom 11734 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 11736* 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 11737* Lemma for lcmn0cl 11738 and dvdslcm 11739. (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 11738 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 11739 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 11740 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 11741 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 11742 Closure of the lcm operator. (Contributed by Steve Rodriguez, 20-Jan-2020.)
((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → (𝑀 lcm 𝑁) ∈ ℕ0)
 
Theoremgcddvdslcm 11743 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 11744 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 11745 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 11746 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 11747 Lemma for lcmgcd 11748 and lcmdvds 11749. 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 11748 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 11688; see e.g. https://proofwiki.org/wiki/Product_of_GCD_and_LCM 11688 and https://math.stackexchange.com/a/470827 11688. 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 11735 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 11749 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 11750 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 11751 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 11752 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 11753 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 11754 Biconditional form of lcmdvds 11749. (Contributed by Steve Rodriguez, 20-Jan-2020.)
((𝐾 ∈ ℤ ∧ 𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → ((𝑀𝐾𝑁𝐾) ↔ (𝑀 lcm 𝑁) ∥ 𝐾))
 
Theoremlcmass 11755 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 11756 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 11757 The least common multiple of six and four is twelve. (Contributed by AV, 27-Aug-2020.)
(6 lcm 4) = 12
 
5.1.9  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 11758. The negation, i.e. two integers are not coprime, can be expressed either by (𝐴 gcd 𝐵) ≠ 1, see ncoprmgcdne1b 11759, or equivalently by 1 < (𝐴 gcd 𝐵), see ncoprmgcdgt1b 11760.

A proof of Euclid's lemma based on coprimality is provided in coprmdvds 11762 (as opposed to Euclid's lemma for primes).

 
Theoremcoprmgcdb 11758* 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 11759* 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 11760* 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 11761 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 11762 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 11763 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 11764 One half of rpmulgcd2 11765, 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 11765 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 11766 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 11767* 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 11768 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 11769 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)
 
5.1.10  Cancellability of congruences
 
Theoremcongr 11770* 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 11771 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 11772* 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 11773 One direction of the bicondition in cncongr 11775. 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 11774 The other direction of the bicondition in cncongr 11775. (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 11775 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 11776 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 3645.

 
Syntaxcprime 11777 Extend the definition of a class to include the set of prime numbers.
class
 
Definitiondf-prm 11778* Define the set of prime numbers. (Contributed by Paul Chapman, 22-Jun-2011.)
ℙ = {𝑝 ∈ ℕ ∣ {𝑛 ∈ ℕ ∣ 𝑛𝑝} ≈ 2o}
 
Theoremisprm 11779* 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 11780 A prime number is a positive integer. (Contributed by Paul Chapman, 22-Jun-2011.)
(𝑃 ∈ ℙ → 𝑃 ∈ ℕ)
 
Theoremprmz 11781 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 11782 The prime numbers are a subset of the positive integers. (Contributed by AV, 22-Jul-2020.)
ℙ ⊆ ℕ
 
Theoremprmex 11783 The set of prime numbers exists. (Contributed by AV, 22-Jul-2020.)
ℙ ∈ V
 
Theorem1nprm 11784 1 is not a prime number. (Contributed by Paul Chapman, 22-Jun-2011.) (Proof shortened by Fan Zheng, 3-Jul-2016.)
¬ 1 ∈ ℙ
 
Theorem1idssfct 11785* 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 11786* Lemma for isprm2 11787. (Contributed by Paul Chapman, 22-Jun-2011.)
((𝑃 ∈ ℕ ∧ 𝑃 ≠ 1) → ({𝑛 ∈ ℕ ∣ 𝑛𝑃} ≈ 2o ↔ {𝑛 ∈ ℕ ∣ 𝑛𝑃} = {1, 𝑃}))
 
Theoremisprm2 11787* 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 11788* 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 11789* 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 11790* A variation on prmind 11791 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 11791* 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 11792 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 11793 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 11794 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 11795 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 11796 A prime number is either 2 or odd. (Contributed by AV, 19-Jun-2021.)
(𝑃 ∈ ℙ → (𝑃 = 2 ∨ ¬ 2 ∥ 𝑃))
 
Theorem2prm 11797 2 is a prime number. (Contributed by Paul Chapman, 22-Jun-2011.) (Proof shortened by Fan Zheng, 16-Jun-2016.)
2 ∈ ℙ
 
Theorem3prm 11798 3 is a prime number. (Contributed by Paul Chapman, 22-Jun-2011.)
3 ∈ ℙ
 
Theorem4nprm 11799 4 is not a prime number. (Contributed by Paul Chapman, 22-Jun-2011.) (Proof shortened by Mario Carneiro, 18-Feb-2014.)
¬ 4 ∈ ℙ
 
Theoremprmuz2 11800 A prime number is an integer greater than or equal to 2. (Contributed by Paul Chapman, 17-Nov-2012.)
(𝑃 ∈ ℙ → 𝑃 ∈ (ℤ‘2))
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