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Theorem List for Intuitionistic Logic Explorer - 10501-10600   *Has distinct variable group(s)
TypeLabelDescription
Statement
 
Theoremhalfleoddlt 10501 An integer is greater than half of an odd number iff it is greater than or equal to the half of the odd number. (Contributed by AV, 1-Jul-2021.)
((𝑁 ∈ ℤ ∧ ¬ 2 ∥ 𝑁𝑀 ∈ ℤ) → ((𝑁 / 2) ≤ 𝑀 ↔ (𝑁 / 2) < 𝑀))
 
Theoremopoe 10502 The sum of two odds is even. (Contributed by Scott Fenton, 7-Apr-2014.) (Revised by Mario Carneiro, 19-Apr-2014.)
(((𝐴 ∈ ℤ ∧ ¬ 2 ∥ 𝐴) ∧ (𝐵 ∈ ℤ ∧ ¬ 2 ∥ 𝐵)) → 2 ∥ (𝐴 + 𝐵))
 
Theoremomoe 10503 The difference of two odds is even. (Contributed by Scott Fenton, 7-Apr-2014.) (Revised by Mario Carneiro, 19-Apr-2014.)
(((𝐴 ∈ ℤ ∧ ¬ 2 ∥ 𝐴) ∧ (𝐵 ∈ ℤ ∧ ¬ 2 ∥ 𝐵)) → 2 ∥ (𝐴𝐵))
 
Theoremopeo 10504 The sum of an odd and an even is odd. (Contributed by Scott Fenton, 7-Apr-2014.) (Revised by Mario Carneiro, 19-Apr-2014.)
(((𝐴 ∈ ℤ ∧ ¬ 2 ∥ 𝐴) ∧ (𝐵 ∈ ℤ ∧ 2 ∥ 𝐵)) → ¬ 2 ∥ (𝐴 + 𝐵))
 
Theoremomeo 10505 The difference of an odd and an even is odd. (Contributed by Scott Fenton, 7-Apr-2014.) (Revised by Mario Carneiro, 19-Apr-2014.)
(((𝐴 ∈ ℤ ∧ ¬ 2 ∥ 𝐴) ∧ (𝐵 ∈ ℤ ∧ 2 ∥ 𝐵)) → ¬ 2 ∥ (𝐴𝐵))
 
Theoremm1expe 10506 Exponentiation of -1 by an even power. Variant of m1expeven 9672. (Contributed by AV, 25-Jun-2021.)
(2 ∥ 𝑁 → (-1↑𝑁) = 1)
 
Theoremm1expo 10507 Exponentiation of -1 by an odd power. (Contributed by AV, 26-Jun-2021.)
((𝑁 ∈ ℤ ∧ ¬ 2 ∥ 𝑁) → (-1↑𝑁) = -1)
 
Theoremm1exp1 10508 Exponentiation of negative one is one iff the exponent is even. (Contributed by AV, 20-Jun-2021.)
(𝑁 ∈ ℤ → ((-1↑𝑁) = 1 ↔ 2 ∥ 𝑁))
 
Theoremnn0enne 10509 A positive integer is an even nonnegative integer iff it is an even positive integer. (Contributed by AV, 30-May-2020.)
(𝑁 ∈ ℕ → ((𝑁 / 2) ∈ ℕ0 ↔ (𝑁 / 2) ∈ ℕ))
 
Theoremnn0ehalf 10510 The half of an even nonnegative integer is a nonnegative integer. (Contributed by AV, 22-Jun-2020.) (Revised by AV, 28-Jun-2021.)
((𝑁 ∈ ℕ0 ∧ 2 ∥ 𝑁) → (𝑁 / 2) ∈ ℕ0)
 
Theoremnnehalf 10511 The half of an even positive integer is a positive integer. (Contributed by AV, 28-Jun-2021.)
((𝑁 ∈ ℕ ∧ 2 ∥ 𝑁) → (𝑁 / 2) ∈ ℕ)
 
Theoremnn0o1gt2 10512 An odd nonnegative integer is either 1 or greater than 2. (Contributed by AV, 2-Jun-2020.)
((𝑁 ∈ ℕ0 ∧ ((𝑁 + 1) / 2) ∈ ℕ0) → (𝑁 = 1 ∨ 2 < 𝑁))
 
Theoremnno 10513 An alternate characterization of an odd integer greater than 1. (Contributed by AV, 2-Jun-2020.)
((𝑁 ∈ (ℤ‘2) ∧ ((𝑁 + 1) / 2) ∈ ℕ0) → ((𝑁 − 1) / 2) ∈ ℕ)
 
Theoremnn0o 10514 An alternate characterization of an odd nonnegative integer. (Contributed by AV, 28-May-2020.) (Proof shortened by AV, 2-Jun-2020.)
((𝑁 ∈ ℕ0 ∧ ((𝑁 + 1) / 2) ∈ ℕ0) → ((𝑁 − 1) / 2) ∈ ℕ0)
 
Theoremnn0ob 10515 Alternate characterizations of an odd nonnegative integer. (Contributed by AV, 4-Jun-2020.)
(𝑁 ∈ ℕ0 → (((𝑁 + 1) / 2) ∈ ℕ0 ↔ ((𝑁 − 1) / 2) ∈ ℕ0))
 
Theoremnn0oddm1d2 10516 A positive integer is odd iff its predecessor divided by 2 is a positive integer. (Contributed by AV, 28-Jun-2021.)
(𝑁 ∈ ℕ0 → (¬ 2 ∥ 𝑁 ↔ ((𝑁 − 1) / 2) ∈ ℕ0))
 
Theoremnnoddm1d2 10517 A positive integer is odd iff its successor divided by 2 is a positive integer. (Contributed by AV, 28-Jun-2021.)
(𝑁 ∈ ℕ → (¬ 2 ∥ 𝑁 ↔ ((𝑁 + 1) / 2) ∈ ℕ))
 
Theoremz0even 10518 0 is even. (Contributed by AV, 11-Feb-2020.) (Revised by AV, 23-Jun-2021.)
2 ∥ 0
 
Theoremn2dvds1 10519 2 does not divide 1 (common case). That means 1 is odd. (Contributed by David A. Wheeler, 8-Dec-2018.)
¬ 2 ∥ 1
 
Theoremn2dvdsm1 10520 2 does not divide -1. That means -1 is odd. (Contributed by AV, 15-Aug-2021.)
¬ 2 ∥ -1
 
Theoremz2even 10521 2 is even. (Contributed by AV, 12-Feb-2020.) (Revised by AV, 23-Jun-2021.)
2 ∥ 2
 
Theoremn2dvds3 10522 2 does not divide 3, i.e. 3 is an odd number. (Contributed by AV, 28-Feb-2021.)
¬ 2 ∥ 3
 
Theoremz4even 10523 4 is an even number. (Contributed by AV, 23-Jul-2020.) (Revised by AV, 4-Jul-2021.)
2 ∥ 4
 
Theorem4dvdseven 10524 An integer which is divisible by 4 is an even integer. (Contributed by AV, 4-Jul-2021.)
(4 ∥ 𝑁 → 2 ∥ 𝑁)
 
4.1.3  The division algorithm
 
Theoremdivalglemnn 10525* Lemma for divalg 10531. Existence for a positive denominator. (Contributed by Jim Kingdon, 30-Nov-2021.)
((𝑁 ∈ ℤ ∧ 𝐷 ∈ ℕ) → ∃𝑟 ∈ ℤ ∃𝑞 ∈ ℤ (0 ≤ 𝑟𝑟 < (abs‘𝐷) ∧ 𝑁 = ((𝑞 · 𝐷) + 𝑟)))
 
Theoremdivalglemqt 10526 Lemma for divalg 10531. The 𝑄 = 𝑇 case involved in showing uniqueness. (Contributed by Jim Kingdon, 5-Dec-2021.)
(𝜑𝐷 ∈ ℤ)    &   (𝜑𝑅 ∈ ℤ)    &   (𝜑𝑆 ∈ ℤ)    &   (𝜑𝑄 ∈ ℤ)    &   (𝜑𝑇 ∈ ℤ)    &   (𝜑𝑄 = 𝑇)    &   (𝜑 → ((𝑄 · 𝐷) + 𝑅) = ((𝑇 · 𝐷) + 𝑆))       (𝜑𝑅 = 𝑆)
 
Theoremdivalglemnqt 10527 Lemma for divalg 10531. The 𝑄 < 𝑇 case involved in showing uniqueness. (Contributed by Jim Kingdon, 4-Dec-2021.)
(𝜑𝐷 ∈ ℕ)    &   (𝜑𝑅 ∈ ℤ)    &   (𝜑𝑆 ∈ ℤ)    &   (𝜑𝑄 ∈ ℤ)    &   (𝜑𝑇 ∈ ℤ)    &   (𝜑 → 0 ≤ 𝑆)    &   (𝜑𝑅 < 𝐷)    &   (𝜑 → ((𝑄 · 𝐷) + 𝑅) = ((𝑇 · 𝐷) + 𝑆))       (𝜑 → ¬ 𝑄 < 𝑇)
 
Theoremdivalglemeunn 10528* Lemma for divalg 10531. Uniqueness for a positive denominator. (Contributed by Jim Kingdon, 4-Dec-2021.)
((𝑁 ∈ ℤ ∧ 𝐷 ∈ ℕ) → ∃!𝑟 ∈ ℤ ∃𝑞 ∈ ℤ (0 ≤ 𝑟𝑟 < (abs‘𝐷) ∧ 𝑁 = ((𝑞 · 𝐷) + 𝑟)))
 
Theoremdivalglemex 10529* Lemma for divalg 10531. The quotient and remainder exist. (Contributed by Jim Kingdon, 30-Nov-2021.)
((𝑁 ∈ ℤ ∧ 𝐷 ∈ ℤ ∧ 𝐷 ≠ 0) → ∃𝑟 ∈ ℤ ∃𝑞 ∈ ℤ (0 ≤ 𝑟𝑟 < (abs‘𝐷) ∧ 𝑁 = ((𝑞 · 𝐷) + 𝑟)))
 
Theoremdivalglemeuneg 10530* Lemma for divalg 10531. Uniqueness for a negative denominator. (Contributed by Jim Kingdon, 4-Dec-2021.)
((𝑁 ∈ ℤ ∧ 𝐷 ∈ ℤ ∧ 𝐷 < 0) → ∃!𝑟 ∈ ℤ ∃𝑞 ∈ ℤ (0 ≤ 𝑟𝑟 < (abs‘𝐷) ∧ 𝑁 = ((𝑞 · 𝐷) + 𝑟)))
 
Theoremdivalg 10531* The division algorithm (theorem). Dividing an integer 𝑁 by a nonzero integer 𝐷 produces a (unique) quotient 𝑞 and a unique remainder 0 ≤ 𝑟 < (abs‘𝐷). Theorem 1.14 in [ApostolNT] p. 19. (Contributed by Paul Chapman, 21-Mar-2011.)
((𝑁 ∈ ℤ ∧ 𝐷 ∈ ℤ ∧ 𝐷 ≠ 0) → ∃!𝑟 ∈ ℤ ∃𝑞 ∈ ℤ (0 ≤ 𝑟𝑟 < (abs‘𝐷) ∧ 𝑁 = ((𝑞 · 𝐷) + 𝑟)))
 
Theoremdivalgb 10532* Express the division algorithm as stated in divalg 10531 in terms of . (Contributed by Paul Chapman, 31-Mar-2011.)
((𝑁 ∈ ℤ ∧ 𝐷 ∈ ℤ ∧ 𝐷 ≠ 0) → (∃!𝑟 ∈ ℤ ∃𝑞 ∈ ℤ (0 ≤ 𝑟𝑟 < (abs‘𝐷) ∧ 𝑁 = ((𝑞 · 𝐷) + 𝑟)) ↔ ∃!𝑟 ∈ ℕ0 (𝑟 < (abs‘𝐷) ∧ 𝐷 ∥ (𝑁𝑟))))
 
Theoremdivalg2 10533* The division algorithm (theorem) for a positive divisor. (Contributed by Paul Chapman, 21-Mar-2011.)
((𝑁 ∈ ℤ ∧ 𝐷 ∈ ℕ) → ∃!𝑟 ∈ ℕ0 (𝑟 < 𝐷𝐷 ∥ (𝑁𝑟)))
 
Theoremdivalgmod 10534 The result of the mod operator satisfies the requirements for the remainder 𝑅 in the division algorithm for a positive divisor (compare divalg2 10533 and divalgb 10532). This demonstration theorem justifies the use of mod to yield an explicit remainder from this point forward. (Contributed by Paul Chapman, 31-Mar-2011.) (Revised by AV, 21-Aug-2021.)
((𝑁 ∈ ℤ ∧ 𝐷 ∈ ℕ) → (𝑅 = (𝑁 mod 𝐷) ↔ (𝑅 ∈ ℕ0 ∧ (𝑅 < 𝐷𝐷 ∥ (𝑁𝑅)))))
 
Theoremdivalgmodcl 10535 The result of the mod operator satisfies the requirements for the remainder 𝑅 in the division algorithm for a positive divisor. Variant of divalgmod 10534. (Contributed by Stefan O'Rear, 17-Oct-2014.) (Proof shortened by AV, 21-Aug-2021.)
((𝑁 ∈ ℤ ∧ 𝐷 ∈ ℕ ∧ 𝑅 ∈ ℕ0) → (𝑅 = (𝑁 mod 𝐷) ↔ (𝑅 < 𝐷𝐷 ∥ (𝑁𝑅))))
 
Theoremmodremain 10536* The result of the modulo operation is the remainder of the division algorithm. (Contributed by AV, 19-Aug-2021.)
((𝑁 ∈ ℤ ∧ 𝐷 ∈ ℕ ∧ (𝑅 ∈ ℕ0𝑅 < 𝐷)) → ((𝑁 mod 𝐷) = 𝑅 ↔ ∃𝑧 ∈ ℤ ((𝑧 · 𝐷) + 𝑅) = 𝑁))
 
Theoremndvdssub 10537 Corollary of the division algorithm. If an integer 𝐷 greater than 1 divides 𝑁, then it does not divide any of 𝑁 − 1, 𝑁 − 2... 𝑁 − (𝐷 − 1). (Contributed by Paul Chapman, 31-Mar-2011.)
((𝑁 ∈ ℤ ∧ 𝐷 ∈ ℕ ∧ (𝐾 ∈ ℕ ∧ 𝐾 < 𝐷)) → (𝐷𝑁 → ¬ 𝐷 ∥ (𝑁𝐾)))
 
Theoremndvdsadd 10538 Corollary of the division algorithm. If an integer 𝐷 greater than 1 divides 𝑁, then it does not divide any of 𝑁 + 1, 𝑁 + 2... 𝑁 + (𝐷 − 1). (Contributed by Paul Chapman, 31-Mar-2011.)
((𝑁 ∈ ℤ ∧ 𝐷 ∈ ℕ ∧ (𝐾 ∈ ℕ ∧ 𝐾 < 𝐷)) → (𝐷𝑁 → ¬ 𝐷 ∥ (𝑁 + 𝐾)))
 
Theoremndvdsp1 10539 Special case of ndvdsadd 10538. If an integer 𝐷 greater than 1 divides 𝑁, it does not divide 𝑁 + 1. (Contributed by Paul Chapman, 31-Mar-2011.)
((𝑁 ∈ ℤ ∧ 𝐷 ∈ ℕ ∧ 1 < 𝐷) → (𝐷𝑁 → ¬ 𝐷 ∥ (𝑁 + 1)))
 
Theoremndvdsi 10540 A quick test for non-divisibility. (Contributed by Mario Carneiro, 18-Feb-2014.)
𝐴 ∈ ℕ    &   𝑄 ∈ ℕ0    &   𝑅 ∈ ℕ    &   ((𝐴 · 𝑄) + 𝑅) = 𝐵    &   𝑅 < 𝐴        ¬ 𝐴𝐵
 
Theoremflodddiv4 10541 The floor of an odd integer divided by 4. (Contributed by AV, 17-Jun-2021.)
((𝑀 ∈ ℤ ∧ 𝑁 = ((2 · 𝑀) + 1)) → (⌊‘(𝑁 / 4)) = if(2 ∥ 𝑀, (𝑀 / 2), ((𝑀 − 1) / 2)))
 
Theoremfldivndvdslt 10542 The floor of an integer divided by a nonzero integer not dividing the first integer is less than the integer divided by the positive integer. (Contributed by AV, 4-Jul-2021.)
((𝐾 ∈ ℤ ∧ (𝐿 ∈ ℤ ∧ 𝐿 ≠ 0) ∧ ¬ 𝐿𝐾) → (⌊‘(𝐾 / 𝐿)) < (𝐾 / 𝐿))
 
Theoremflodddiv4lt 10543 The floor of an odd number divided by 4 is less than the odd number divided by 4. (Contributed by AV, 4-Jul-2021.)
((𝑁 ∈ ℤ ∧ ¬ 2 ∥ 𝑁) → (⌊‘(𝑁 / 4)) < (𝑁 / 4))
 
Theoremflodddiv4t2lthalf 10544 The floor of an odd number divided by 4, multiplied by 2 is less than the half of the odd number. (Contributed by AV, 4-Jul-2021.)
((𝑁 ∈ ℤ ∧ ¬ 2 ∥ 𝑁) → ((⌊‘(𝑁 / 4)) · 2) < (𝑁 / 2))
 
4.1.4  The greatest common divisor operator
 
Syntaxcgcd 10545 Extend the definition of a class to include the greatest common divisor operator.
class gcd
 
Definitiondf-gcd 10546* Define the gcd operator. For example, (-6 gcd 9) = 3 (ex-gcd 10828). (Contributed by Paul Chapman, 21-Mar-2011.)
gcd = (𝑥 ∈ ℤ, 𝑦 ∈ ℤ ↦ if((𝑥 = 0 ∧ 𝑦 = 0), 0, sup({𝑛 ∈ ℤ ∣ (𝑛𝑥𝑛𝑦)}, ℝ, < )))
 
Theoremgcdmndc 10547 Decidablity lemma used in various proofs related to gcd. (Contributed by Jim Kingdon, 12-Dec-2021.)
((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → DECID (𝑀 = 0 ∧ 𝑁 = 0))
 
Theoremzsupcllemstep 10548* Lemma for zsupcl 10550. Induction step. (Contributed by Jim Kingdon, 7-Dec-2021.)
((𝜑𝑛 ∈ (ℤ𝑀)) → DECID 𝜓)       (𝐾 ∈ (ℤ𝑀) → (((𝜑 ∧ ∀𝑛 ∈ (ℤ𝐾) ¬ 𝜓) → ∃𝑥 ∈ ℤ (∀𝑦 ∈ {𝑛 ∈ ℤ ∣ 𝜓} ¬ 𝑥 < 𝑦 ∧ ∀𝑦 ∈ ℝ (𝑦 < 𝑥 → ∃𝑧 ∈ {𝑛 ∈ ℤ ∣ 𝜓}𝑦 < 𝑧))) → ((𝜑 ∧ ∀𝑛 ∈ (ℤ‘(𝐾 + 1)) ¬ 𝜓) → ∃𝑥 ∈ ℤ (∀𝑦 ∈ {𝑛 ∈ ℤ ∣ 𝜓} ¬ 𝑥 < 𝑦 ∧ ∀𝑦 ∈ ℝ (𝑦 < 𝑥 → ∃𝑧 ∈ {𝑛 ∈ ℤ ∣ 𝜓}𝑦 < 𝑧)))))
 
Theoremzsupcllemex 10549* Lemma for zsupcl 10550. Existence of the supremum. (Contributed by Jim Kingdon, 7-Dec-2021.)
(𝜑𝑀 ∈ ℤ)    &   (𝑛 = 𝑀 → (𝜓𝜒))    &   (𝜑𝜒)    &   ((𝜑𝑛 ∈ (ℤ𝑀)) → DECID 𝜓)    &   (𝜑 → ∃𝑗 ∈ (ℤ𝑀)∀𝑛 ∈ (ℤ𝑗) ¬ 𝜓)       (𝜑 → ∃𝑥 ∈ ℤ (∀𝑦 ∈ {𝑛 ∈ ℤ ∣ 𝜓} ¬ 𝑥 < 𝑦 ∧ ∀𝑦 ∈ ℝ (𝑦 < 𝑥 → ∃𝑧 ∈ {𝑛 ∈ ℤ ∣ 𝜓}𝑦 < 𝑧)))
 
Theoremzsupcl 10550* Closure of supremum for decidable integer properties. The property which defines the set we are taking the supremum of must (a) be true at 𝑀 (which corresponds to the non-empty condition of classical supremum theorems), (b) decidable at each value after 𝑀, and (c) be false after 𝑗 (which corresponds to the upper bound condition found in classical supremum theorems). (Contributed by Jim Kingdon, 7-Dec-2021.)
(𝜑𝑀 ∈ ℤ)    &   (𝑛 = 𝑀 → (𝜓𝜒))    &   (𝜑𝜒)    &   ((𝜑𝑛 ∈ (ℤ𝑀)) → DECID 𝜓)    &   (𝜑 → ∃𝑗 ∈ (ℤ𝑀)∀𝑛 ∈ (ℤ𝑗) ¬ 𝜓)       (𝜑 → sup({𝑛 ∈ ℤ ∣ 𝜓}, ℝ, < ) ∈ (ℤ𝑀))
 
Theoremzssinfcl 10551* The infimum of a set of integers is an element of the set. (Contributed by Jim Kingdon, 16-Jan-2022.)
(𝜑 → ∃𝑥 ∈ ℝ (∀𝑦𝐵 ¬ 𝑦 < 𝑥 ∧ ∀𝑦 ∈ ℝ (𝑥 < 𝑦 → ∃𝑧𝐵 𝑧 < 𝑦)))    &   (𝜑𝐵 ⊆ ℤ)    &   (𝜑 → inf(𝐵, ℝ, < ) ∈ ℤ)       (𝜑 → inf(𝐵, ℝ, < ) ∈ 𝐵)
 
Theoreminfssuzex 10552* Existence of the infimum of a subset of an upper set of integers. (Contributed by Jim Kingdon, 13-Jan-2022.)
(𝜑𝑀 ∈ ℤ)    &   𝑆 = {𝑛 ∈ (ℤ𝑀) ∣ 𝜓}    &   (𝜑𝐴𝑆)    &   ((𝜑𝑛 ∈ (𝑀...𝐴)) → DECID 𝜓)       (𝜑 → ∃𝑥 ∈ ℝ (∀𝑦𝑆 ¬ 𝑦 < 𝑥 ∧ ∀𝑦 ∈ ℝ (𝑥 < 𝑦 → ∃𝑧𝑆 𝑧 < 𝑦)))
 
Theoreminfssuzledc 10553* The infimum of a subset of an upper set of integers is less than or equal to all members of the subset. (Contributed by Jim Kingdon, 13-Jan-2022.)
(𝜑𝑀 ∈ ℤ)    &   𝑆 = {𝑛 ∈ (ℤ𝑀) ∣ 𝜓}    &   (𝜑𝐴𝑆)    &   ((𝜑𝑛 ∈ (𝑀...𝐴)) → DECID 𝜓)       (𝜑 → inf(𝑆, ℝ, < ) ≤ 𝐴)
 
Theoreminfssuzcldc 10554* The infimum of a subset of an upper set of integers belongs to the subset. (Contributed by Jim Kingdon, 20-Jan-2022.)
(𝜑𝑀 ∈ ℤ)    &   𝑆 = {𝑛 ∈ (ℤ𝑀) ∣ 𝜓}    &   (𝜑𝐴𝑆)    &   ((𝜑𝑛 ∈ (𝑀...𝐴)) → DECID 𝜓)       (𝜑 → inf(𝑆, ℝ, < ) ∈ 𝑆)
 
Theoremdvdsbnd 10555* There is an upper bound to the divisors of a nonzero integer. (Contributed by Jim Kingdon, 11-Dec-2021.)
((𝐴 ∈ ℤ ∧ 𝐴 ≠ 0) → ∃𝑛 ∈ ℕ ∀𝑚 ∈ (ℤ𝑛) ¬ 𝑚𝐴)
 
Theoremgcdsupex 10556* Existence of the supremum used in defining gcd. (Contributed by Jim Kingdon, 12-Dec-2021.)
(((𝑋 ∈ ℤ ∧ 𝑌 ∈ ℤ) ∧ ¬ (𝑋 = 0 ∧ 𝑌 = 0)) → ∃𝑥 ∈ ℤ (∀𝑦 ∈ {𝑛 ∈ ℤ ∣ (𝑛𝑋𝑛𝑌)} ¬ 𝑥 < 𝑦 ∧ ∀𝑦 ∈ ℝ (𝑦 < 𝑥 → ∃𝑧 ∈ {𝑛 ∈ ℤ ∣ (𝑛𝑋𝑛𝑌)}𝑦 < 𝑧)))
 
Theoremgcdsupcl 10557* Closure of the supremum used in defining gcd. A lemma for gcdval 10558 and gcdn0cl 10561. (Contributed by Jim Kingdon, 11-Dec-2021.)
(((𝑋 ∈ ℤ ∧ 𝑌 ∈ ℤ) ∧ ¬ (𝑋 = 0 ∧ 𝑌 = 0)) → sup({𝑛 ∈ ℤ ∣ (𝑛𝑋𝑛𝑌)}, ℝ, < ) ∈ ℕ)
 
Theoremgcdval 10558* The value of the gcd operator. (𝑀 gcd 𝑁) is the greatest common divisor of 𝑀 and 𝑁. If 𝑀 and 𝑁 are both 0, the result is defined conventionally as 0. (Contributed by Paul Chapman, 21-Mar-2011.) (Revised by Mario Carneiro, 10-Nov-2013.)
((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → (𝑀 gcd 𝑁) = if((𝑀 = 0 ∧ 𝑁 = 0), 0, sup({𝑛 ∈ ℤ ∣ (𝑛𝑀𝑛𝑁)}, ℝ, < )))
 
Theoremgcd0val 10559 The value, by convention, of the gcd operator when both operands are 0. (Contributed by Paul Chapman, 21-Mar-2011.)
(0 gcd 0) = 0
 
Theoremgcdn0val 10560* The value of the gcd operator when at least one operand is nonzero. (Contributed by Paul Chapman, 21-Mar-2011.)
(((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) ∧ ¬ (𝑀 = 0 ∧ 𝑁 = 0)) → (𝑀 gcd 𝑁) = sup({𝑛 ∈ ℤ ∣ (𝑛𝑀𝑛𝑁)}, ℝ, < ))
 
Theoremgcdn0cl 10561 Closure of the gcd operator. (Contributed by Paul Chapman, 21-Mar-2011.)
(((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) ∧ ¬ (𝑀 = 0 ∧ 𝑁 = 0)) → (𝑀 gcd 𝑁) ∈ ℕ)
 
Theoremgcddvds 10562 The gcd of two integers divides each of them. (Contributed by Paul Chapman, 21-Mar-2011.)
((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → ((𝑀 gcd 𝑁) ∥ 𝑀 ∧ (𝑀 gcd 𝑁) ∥ 𝑁))
 
Theoremdvdslegcd 10563 An integer which divides both operands of the gcd operator is bounded by it. (Contributed by Paul Chapman, 21-Mar-2011.)
(((𝐾 ∈ ℤ ∧ 𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) ∧ ¬ (𝑀 = 0 ∧ 𝑁 = 0)) → ((𝐾𝑀𝐾𝑁) → 𝐾 ≤ (𝑀 gcd 𝑁)))
 
Theoremnndvdslegcd 10564 A positive integer which divides both positive operands of the gcd operator is bounded by it. (Contributed by AV, 9-Aug-2020.)
((𝐾 ∈ ℕ ∧ 𝑀 ∈ ℕ ∧ 𝑁 ∈ ℕ) → ((𝐾𝑀𝐾𝑁) → 𝐾 ≤ (𝑀 gcd 𝑁)))
 
Theoremgcdcl 10565 Closure of the gcd operator. (Contributed by Paul Chapman, 21-Mar-2011.)
((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → (𝑀 gcd 𝑁) ∈ ℕ0)
 
Theoremgcdnncl 10566 Closure of the gcd operator. (Contributed by Thierry Arnoux, 2-Feb-2020.)
((𝑀 ∈ ℕ ∧ 𝑁 ∈ ℕ) → (𝑀 gcd 𝑁) ∈ ℕ)
 
Theoremgcdcld 10567 Closure of the gcd operator. (Contributed by Mario Carneiro, 29-May-2016.)
(𝜑𝑀 ∈ ℤ)    &   (𝜑𝑁 ∈ ℤ)       (𝜑 → (𝑀 gcd 𝑁) ∈ ℕ0)
 
Theoremgcd2n0cl 10568 Closure of the gcd operator if the second operand is not 0. (Contributed by AV, 10-Jul-2021.)
((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ ∧ 𝑁 ≠ 0) → (𝑀 gcd 𝑁) ∈ ℕ)
 
Theoremzeqzmulgcd 10569* An integer is the product of an integer and the gcd of it and another integer. (Contributed by AV, 11-Jul-2021.)
((𝐴 ∈ ℤ ∧ 𝐵 ∈ ℤ) → ∃𝑛 ∈ ℤ 𝐴 = (𝑛 · (𝐴 gcd 𝐵)))
 
Theoremdivgcdz 10570 An integer divided by the gcd of it and a nonzero integer is an integer. (Contributed by AV, 11-Jul-2021.)
((𝐴 ∈ ℤ ∧ 𝐵 ∈ ℤ ∧ 𝐵 ≠ 0) → (𝐴 / (𝐴 gcd 𝐵)) ∈ ℤ)
 
Theoremgcdf 10571 Domain and codomain of the gcd operator. (Contributed by Paul Chapman, 31-Mar-2011.) (Revised by Mario Carneiro, 16-Nov-2013.)
gcd :(ℤ × ℤ)⟶ℕ0
 
Theoremgcdcom 10572 The gcd operator is commutative. Theorem 1.4(a) in [ApostolNT] p. 16. (Contributed by Paul Chapman, 21-Mar-2011.)
((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → (𝑀 gcd 𝑁) = (𝑁 gcd 𝑀))
 
Theoremdivgcdnn 10573 A positive integer divided by the gcd of it and another integer is a positive integer. (Contributed by AV, 10-Jul-2021.)
((𝐴 ∈ ℕ ∧ 𝐵 ∈ ℤ) → (𝐴 / (𝐴 gcd 𝐵)) ∈ ℕ)
 
Theoremdivgcdnnr 10574 A positive integer divided by the gcd of it and another integer is a positive integer. (Contributed by AV, 10-Jul-2021.)
((𝐴 ∈ ℕ ∧ 𝐵 ∈ ℤ) → (𝐴 / (𝐵 gcd 𝐴)) ∈ ℕ)
 
Theoremgcdeq0 10575 The gcd of two integers is zero iff they are both zero. (Contributed by Paul Chapman, 21-Mar-2011.)
((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → ((𝑀 gcd 𝑁) = 0 ↔ (𝑀 = 0 ∧ 𝑁 = 0)))
 
Theoremgcdn0gt0 10576 The gcd of two integers is positive (nonzero) iff they are not both zero. (Contributed by Paul Chapman, 22-Jun-2011.)
((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → (¬ (𝑀 = 0 ∧ 𝑁 = 0) ↔ 0 < (𝑀 gcd 𝑁)))
 
Theoremgcd0id 10577 The gcd of 0 and an integer is the integer's absolute value. (Contributed by Paul Chapman, 21-Mar-2011.)
(𝑁 ∈ ℤ → (0 gcd 𝑁) = (abs‘𝑁))
 
Theoremgcdid0 10578 The gcd of an integer and 0 is the integer's absolute value. Theorem 1.4(d)2 in [ApostolNT] p. 16. (Contributed by Paul Chapman, 31-Mar-2011.)
(𝑁 ∈ ℤ → (𝑁 gcd 0) = (abs‘𝑁))
 
Theoremnn0gcdid0 10579 The gcd of a nonnegative integer with 0 is itself. (Contributed by Paul Chapman, 31-Mar-2011.)
(𝑁 ∈ ℕ0 → (𝑁 gcd 0) = 𝑁)
 
Theoremgcdneg 10580 Negating one operand of the gcd operator does not alter the result. (Contributed by Paul Chapman, 21-Mar-2011.)
((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → (𝑀 gcd -𝑁) = (𝑀 gcd 𝑁))
 
Theoremneggcd 10581 Negating one operand of the gcd operator does not alter the result. (Contributed by Paul Chapman, 22-Jun-2011.)
((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → (-𝑀 gcd 𝑁) = (𝑀 gcd 𝑁))
 
Theoremgcdaddm 10582 Adding a multiple of one operand of the gcd operator to the other does not alter the result. (Contributed by Paul Chapman, 31-Mar-2011.)
((𝐾 ∈ ℤ ∧ 𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → (𝑀 gcd 𝑁) = (𝑀 gcd (𝑁 + (𝐾 · 𝑀))))
 
Theoremgcdadd 10583 The GCD of two numbers is the same as the GCD of the left and their sum. (Contributed by Scott Fenton, 20-Apr-2014.)
((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → (𝑀 gcd 𝑁) = (𝑀 gcd (𝑁 + 𝑀)))
 
Theoremgcdid 10584 The gcd of a number and itself is its absolute value. (Contributed by Paul Chapman, 31-Mar-2011.)
(𝑁 ∈ ℤ → (𝑁 gcd 𝑁) = (abs‘𝑁))
 
Theoremgcd1 10585 The gcd of a number with 1 is 1. Theorem 1.4(d)1 in [ApostolNT] p. 16. (Contributed by Mario Carneiro, 19-Feb-2014.)
(𝑀 ∈ ℤ → (𝑀 gcd 1) = 1)
 
Theoremgcdabs 10586 The gcd of two integers is the same as that of their absolute values. (Contributed by Paul Chapman, 31-Mar-2011.)
((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → ((abs‘𝑀) gcd (abs‘𝑁)) = (𝑀 gcd 𝑁))
 
Theoremgcdabs1 10587 gcd of the absolute value of the first operator. (Contributed by Scott Fenton, 2-Apr-2014.) (Revised by Mario Carneiro, 19-Apr-2014.)
((𝑁 ∈ ℤ ∧ 𝑀 ∈ ℤ) → ((abs‘𝑁) gcd 𝑀) = (𝑁 gcd 𝑀))
 
Theoremgcdabs2 10588 gcd of the absolute value of the second operator. (Contributed by Scott Fenton, 2-Apr-2014.) (Revised by Mario Carneiro, 19-Apr-2014.)
((𝑁 ∈ ℤ ∧ 𝑀 ∈ ℤ) → (𝑁 gcd (abs‘𝑀)) = (𝑁 gcd 𝑀))
 
Theoremmodgcd 10589 The gcd remains unchanged if one operand is replaced with its remainder modulo the other. (Contributed by Paul Chapman, 31-Mar-2011.)
((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℕ) → ((𝑀 mod 𝑁) gcd 𝑁) = (𝑀 gcd 𝑁))
 
Theorem1gcd 10590 The GCD of one and an integer is one. (Contributed by Scott Fenton, 17-Apr-2014.) (Revised by Mario Carneiro, 19-Apr-2014.)
(𝑀 ∈ ℤ → (1 gcd 𝑀) = 1)
 
Theorem6gcd4e2 10591 The greatest common divisor of six and four is two. To calculate this gcd, a simple form of Euclid's algorithm is used: (6 gcd 4) = ((4 + 2) gcd 4) = (2 gcd 4) and (2 gcd 4) = (2 gcd (2 + 2)) = (2 gcd 2) = 2. (Contributed by AV, 27-Aug-2020.)
(6 gcd 4) = 2
 
4.1.5  Bézout's identity
 
Theorembezoutlemnewy 10592* Lemma for Bézout's identity. The is-bezout predicate holds for (𝑦 mod 𝑊). (Contributed by Jim Kingdon, 6-Jan-2022.)
(𝜑 ↔ ∃𝑠 ∈ ℤ ∃𝑡 ∈ ℤ 𝑟 = ((𝐴 · 𝑠) + (𝐵 · 𝑡)))    &   (𝜃𝐴 ∈ ℕ0)    &   (𝜃𝐵 ∈ ℕ0)    &   (𝜃𝑊 ∈ ℕ)    &   (𝜃 → [𝑦 / 𝑟]𝜑)    &   (𝜃𝑦 ∈ ℕ0)    &   (𝜃[𝑊 / 𝑟]𝜑)       (𝜃[(𝑦 mod 𝑊) / 𝑟]𝜑)
 
Theorembezoutlemstep 10593* Lemma for Bézout's identity. This is the induction step for the proof by induction. (Contributed by Jim Kingdon, 3-Jan-2022.)
(𝜑 ↔ ∃𝑠 ∈ ℤ ∃𝑡 ∈ ℤ 𝑟 = ((𝐴 · 𝑠) + (𝐵 · 𝑡)))    &   (𝜃𝐴 ∈ ℕ0)    &   (𝜃𝐵 ∈ ℕ0)    &   (𝜃𝑊 ∈ ℕ)    &   (𝜃 → [𝑦 / 𝑟]𝜑)    &   (𝜃𝑦 ∈ ℕ0)    &   (𝜃[𝑊 / 𝑟]𝜑)    &   (𝜓 ↔ ∀𝑧 ∈ ℕ0 (𝑧𝑟 → (𝑧𝑥𝑧𝑦)))    &   ((𝜃[(𝑦 mod 𝑊) / 𝑟]𝜑) → ∃𝑟 ∈ ℕ0 ([(𝑦 mod 𝑊) / 𝑥][𝑊 / 𝑦]𝜓𝜑))    &   𝑥𝜃    &   𝑟𝜃       (𝜃 → ∃𝑟 ∈ ℕ0 ([𝑊 / 𝑥]𝜓𝜑))
 
Theorembezoutlemmain 10594* Lemma for Bézout's identity. This is the main result which we prove by induction and which represents the application of the Extended Euclidean algorithm. (Contributed by Jim Kingdon, 30-Dec-2021.)
(𝜑 ↔ ∃𝑠 ∈ ℤ ∃𝑡 ∈ ℤ 𝑟 = ((𝐴 · 𝑠) + (𝐵 · 𝑡)))    &   (𝜓 ↔ ∀𝑧 ∈ ℕ0 (𝑧𝑟 → (𝑧𝑥𝑧𝑦)))    &   (𝜃𝐴 ∈ ℕ0)    &   (𝜃𝐵 ∈ ℕ0)       (𝜃 → ∀𝑥 ∈ ℕ0 ([𝑥 / 𝑟]𝜑 → ∀𝑦 ∈ ℕ0 ([𝑦 / 𝑟]𝜑 → ∃𝑟 ∈ ℕ0 (𝜓𝜑))))
 
Theorembezoutlema 10595* Lemma for Bézout's identity. The is-bezout condition is satisfied by 𝐴. (Contributed by Jim Kingdon, 30-Dec-2021.)
(𝜑 ↔ ∃𝑠 ∈ ℤ ∃𝑡 ∈ ℤ 𝑟 = ((𝐴 · 𝑠) + (𝐵 · 𝑡)))    &   (𝜃𝐴 ∈ ℕ0)    &   (𝜃𝐵 ∈ ℕ0)       (𝜃[𝐴 / 𝑟]𝜑)
 
Theorembezoutlemb 10596* Lemma for Bézout's identity. The is-bezout condition is satisfied by 𝐵. (Contributed by Jim Kingdon, 30-Dec-2021.)
(𝜑 ↔ ∃𝑠 ∈ ℤ ∃𝑡 ∈ ℤ 𝑟 = ((𝐴 · 𝑠) + (𝐵 · 𝑡)))    &   (𝜃𝐴 ∈ ℕ0)    &   (𝜃𝐵 ∈ ℕ0)       (𝜃[𝐵 / 𝑟]𝜑)
 
Theorembezoutlemex 10597* Lemma for Bézout's identity. Existence of a number which we will later show to be the greater common divisor and its decomposition into cofactors. (Contributed by Mario Carneiro and Jim Kingdon, 3-Jan-2022.)
((𝐴 ∈ ℕ0𝐵 ∈ ℕ0) → ∃𝑑 ∈ ℕ0 (∀𝑧 ∈ ℕ0 (𝑧𝑑 → (𝑧𝐴𝑧𝐵)) ∧ ∃𝑥 ∈ ℤ ∃𝑦 ∈ ℤ 𝑑 = ((𝐴 · 𝑥) + (𝐵 · 𝑦))))
 
Theorembezoutlemzz 10598* Lemma for Bézout's identity. Like bezoutlemex 10597 but where ' z ' is any integer, not just a nonnegative one. (Contributed by Mario Carneiro and Jim Kingdon, 8-Jan-2022.)
((𝐴 ∈ ℕ0𝐵 ∈ ℕ0) → ∃𝑑 ∈ ℕ0 (∀𝑧 ∈ ℤ (𝑧𝑑 → (𝑧𝐴𝑧𝐵)) ∧ ∃𝑥 ∈ ℤ ∃𝑦 ∈ ℤ 𝑑 = ((𝐴 · 𝑥) + (𝐵 · 𝑦))))
 
Theorembezoutlemaz 10599* Lemma for Bézout's identity. Like bezoutlemzz 10598 but where ' A ' can be any integer, not just a nonnegative one. (Contributed by Mario Carneiro and Jim Kingdon, 8-Jan-2022.)
((𝐴 ∈ ℤ ∧ 𝐵 ∈ ℕ0) → ∃𝑑 ∈ ℕ0 (∀𝑧 ∈ ℤ (𝑧𝑑 → (𝑧𝐴𝑧𝐵)) ∧ ∃𝑥 ∈ ℤ ∃𝑦 ∈ ℤ 𝑑 = ((𝐴 · 𝑥) + (𝐵 · 𝑦))))
 
Theorembezoutlembz 10600* Lemma for Bézout's identity. Like bezoutlemaz 10599 but where ' B ' can be any integer, not just a nonnegative one. (Contributed by Mario Carneiro and Jim Kingdon, 8-Jan-2022.)
((𝐴 ∈ ℤ ∧ 𝐵 ∈ ℤ) → ∃𝑑 ∈ ℕ0 (∀𝑧 ∈ ℤ (𝑧𝑑 → (𝑧𝐴𝑧𝐵)) ∧ ∃𝑥 ∈ ℤ ∃𝑦 ∈ ℤ 𝑑 = ((𝐴 · 𝑥) + (𝐵 · 𝑦))))
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