 Home Intuitionistic Logic ExplorerTheorem List (p. 109 of 114) < Previous  Next > Bad symbols? Try the GIF version. Mirrors  >  Metamath Home Page  >  ILE Home Page  >  Theorem List Contents  >  Recent Proofs       This page: Page List

Theorem List for Intuitionistic Logic Explorer - 10801-10900   *Has distinct variable group(s)
TypeLabelDescription
Statement

Theoremdivalglemqt 10801 Lemma for divalg 10806. 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 10802 Lemma for divalg 10806. 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 10803* Lemma for divalg 10806. 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 10804* Lemma for divalg 10806. 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 10805* Lemma for divalg 10806. 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 10806* 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 10807* Express the division algorithm as stated in divalg 10806 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 10808* 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 10809 The result of the mod operator satisfies the requirements for the remainder 𝑅 in the division algorithm for a positive divisor (compare divalg2 10808 and divalgb 10807). 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 10810 The result of the mod operator satisfies the requirements for the remainder 𝑅 in the division algorithm for a positive divisor. Variant of divalgmod 10809. (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 10811* 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 10812 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 10813 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 10814 Special case of ndvdsadd 10813. 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 10815 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 10816 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 10817 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 10818 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 10819 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 10820 Extend the definition of a class to include the greatest common divisor operator.
class gcd

Definitiondf-gcd 10821* Define the gcd operator. For example, (-6 gcd 9) = 3 (ex-gcd 11103). (Contributed by Paul Chapman, 21-Mar-2011.)
gcd = (𝑥 ∈ ℤ, 𝑦 ∈ ℤ ↦ if((𝑥 = 0 ∧ 𝑦 = 0), 0, sup({𝑛 ∈ ℤ ∣ (𝑛𝑥𝑛𝑦)}, ℝ, < )))

Theoremgcdmndc 10822 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 10823* Lemma for zsupcl 10825. 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 10824* Lemma for zsupcl 10825. 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 10825* 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 nonempty 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 10826* 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 10827* 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 10828* 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 10829* 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 10830* 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 10831* 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 10832* Closure of the supremum used in defining gcd. A lemma for gcdval 10833 and gcdn0cl 10836. (Contributed by Jim Kingdon, 11-Dec-2021.)
(((𝑋 ∈ ℤ ∧ 𝑌 ∈ ℤ) ∧ ¬ (𝑋 = 0 ∧ 𝑌 = 0)) → sup({𝑛 ∈ ℤ ∣ (𝑛𝑋𝑛𝑌)}, ℝ, < ) ∈ ℕ)

Theoremgcdval 10833* 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 10834 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 10835* 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 10836 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 10837 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 10838 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 10839 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 10840 Closure of the gcd operator. (Contributed by Paul Chapman, 21-Mar-2011.)
((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → (𝑀 gcd 𝑁) ∈ ℕ0)

Theoremgcdnncl 10841 Closure of the gcd operator. (Contributed by Thierry Arnoux, 2-Feb-2020.)
((𝑀 ∈ ℕ ∧ 𝑁 ∈ ℕ) → (𝑀 gcd 𝑁) ∈ ℕ)

Theoremgcdcld 10842 Closure of the gcd operator. (Contributed by Mario Carneiro, 29-May-2016.)
(𝜑𝑀 ∈ ℤ)    &   (𝜑𝑁 ∈ ℤ)       (𝜑 → (𝑀 gcd 𝑁) ∈ ℕ0)

Theoremgcd2n0cl 10843 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 10844* 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 10845 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 10846 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 10847 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 10848 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 10849 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 10850 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 10851 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 10852 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 10853 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 10854 The gcd of a nonnegative integer with 0 is itself. (Contributed by Paul Chapman, 31-Mar-2011.)
(𝑁 ∈ ℕ0 → (𝑁 gcd 0) = 𝑁)

Theoremgcdneg 10855 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 10856 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 10857 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 10858 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 10859 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 10860 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 10861 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 10862 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 10863 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 10864 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 10865 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 10866 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 10867* 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 10868* 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 10869* 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 10870* 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 10871* 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 10872* 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 10873* Lemma for Bézout's identity. Like bezoutlemex 10872 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 10874* Lemma for Bézout's identity. Like bezoutlemzz 10873 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 10875* Lemma for Bézout's identity. Like bezoutlemaz 10874 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))))

Theorembezoutlembi 10876* Lemma for Bézout's identity. Like bezoutlembz 10875 but the greatest common divisor condition is a biconditional, not just an implication. (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))))

Theorembezoutlemmo 10877* Lemma for Bézout's identity. There is at most one nonnegative integer meeting the greatest common divisor condition. (Contributed by Mario Carneiro and Jim Kingdon, 9-Jan-2022.)
(𝜑𝐴 ∈ ℤ)    &   (𝜑𝐵 ∈ ℤ)    &   (𝜑𝐷 ∈ ℕ0)    &   (𝜑 → ∀𝑧 ∈ ℤ (𝑧𝐷 ↔ (𝑧𝐴𝑧𝐵)))    &   (𝜑𝐸 ∈ ℕ0)    &   (𝜑 → ∀𝑧 ∈ ℤ (𝑧𝐸 ↔ (𝑧𝐴𝑧𝐵)))       (𝜑𝐷 = 𝐸)

Theorembezoutlemeu 10878* Lemma for Bézout's identity. There is exactly one nonnegative integer meeting the greatest common divisor condition. (Contributed by Mario Carneiro and Jim Kingdon, 9-Jan-2022.)
(𝜑𝐴 ∈ ℤ)    &   (𝜑𝐵 ∈ ℤ)    &   (𝜑𝐷 ∈ ℕ0)    &   (𝜑 → ∀𝑧 ∈ ℤ (𝑧𝐷 ↔ (𝑧𝐴𝑧𝐵)))       (𝜑 → ∃!𝑑 ∈ ℕ0𝑧 ∈ ℤ (𝑧𝑑 ↔ (𝑧𝐴𝑧𝐵)))

Theorembezoutlemle 10879* Lemma for Bézout's identity. The number satisfying the greatest common divisor condition is the largest number which divides both 𝐴 and 𝐵. (Contributed by Mario Carneiro and Jim Kingdon, 9-Jan-2022.)
(𝜑𝐴 ∈ ℤ)    &   (𝜑𝐵 ∈ ℤ)    &   (𝜑𝐷 ∈ ℕ0)    &   (𝜑 → ∀𝑧 ∈ ℤ (𝑧𝐷 ↔ (𝑧𝐴𝑧𝐵)))    &   (𝜑 → ¬ (𝐴 = 0 ∧ 𝐵 = 0))       (𝜑 → ∀𝑧 ∈ ℤ ((𝑧𝐴𝑧𝐵) → 𝑧𝐷))

Theorembezoutlemsup 10880* Lemma for Bézout's identity. The number satisfying the greatest common divisor condition is the supremum of divisors of both 𝐴 and 𝐵. (Contributed by Mario Carneiro and Jim Kingdon, 9-Jan-2022.)
(𝜑𝐴 ∈ ℤ)    &   (𝜑𝐵 ∈ ℤ)    &   (𝜑𝐷 ∈ ℕ0)    &   (𝜑 → ∀𝑧 ∈ ℤ (𝑧𝐷 ↔ (𝑧𝐴𝑧𝐵)))    &   (𝜑 → ¬ (𝐴 = 0 ∧ 𝐵 = 0))       (𝜑𝐷 = sup({𝑧 ∈ ℤ ∣ (𝑧𝐴𝑧𝐵)}, ℝ, < ))

Theoremdfgcd3 10881* Alternate definition of the gcd operator. (Contributed by Jim Kingdon, 31-Dec-2021.)
((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → (𝑀 gcd 𝑁) = (𝑑 ∈ ℕ0𝑧 ∈ ℤ (𝑧𝑑 ↔ (𝑧𝑀𝑧𝑁))))

Theorembezout 10882* Bézout's identity: For any integers 𝐴 and 𝐵, there are integers 𝑥, 𝑦 such that (𝐴 gcd 𝐵) = 𝐴 · 𝑥 + 𝐵 · 𝑦. This is Metamath 100 proof #60.

The proof is constructive, in the sense that it applies the Extended Euclidian Algorithm to constuct a number which can be shown to be (𝐴 gcd 𝐵) and which satisfies the rest of the theorem. In the presence of excluded middle, it is common to prove Bézout's identity by taking the smallest number which satisfies the Bézout condition, and showing it is the greatest common divisor. But we do not have the ability to show that number exists other than by providing a way to determine it. (Contributed by Mario Carneiro, 22-Feb-2014.)

((𝐴 ∈ ℤ ∧ 𝐵 ∈ ℤ) → ∃𝑥 ∈ ℤ ∃𝑦 ∈ ℤ (𝐴 gcd 𝐵) = ((𝐴 · 𝑥) + (𝐵 · 𝑦)))

Theoremdvdsgcd 10883 An integer which divides each of two others also divides their gcd. (Contributed by Paul Chapman, 22-Jun-2011.) (Revised by Mario Carneiro, 30-May-2014.)
((𝐾 ∈ ℤ ∧ 𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → ((𝐾𝑀𝐾𝑁) → 𝐾 ∥ (𝑀 gcd 𝑁)))

Theoremdvdsgcdb 10884 Biconditional form of dvdsgcd 10883. (Contributed by Scott Fenton, 2-Apr-2014.) (Revised by Mario Carneiro, 19-Apr-2014.)
((𝐾 ∈ ℤ ∧ 𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → ((𝐾𝑀𝐾𝑁) ↔ 𝐾 ∥ (𝑀 gcd 𝑁)))

Theoremdfgcd2 10885* Alternate definition of the gcd operator, see definition in [ApostolNT] p. 15. (Contributed by AV, 8-Aug-2021.)
((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → (𝐷 = (𝑀 gcd 𝑁) ↔ (0 ≤ 𝐷 ∧ (𝐷𝑀𝐷𝑁) ∧ ∀𝑒 ∈ ℤ ((𝑒𝑀𝑒𝑁) → 𝑒𝐷))))

Theoremgcdass 10886 Associative law for gcd operator. Theorem 1.4(b) in [ApostolNT] p. 16. (Contributed by Scott Fenton, 2-Apr-2014.) (Revised by Mario Carneiro, 19-Apr-2014.)
((𝑁 ∈ ℤ ∧ 𝑀 ∈ ℤ ∧ 𝑃 ∈ ℤ) → ((𝑁 gcd 𝑀) gcd 𝑃) = (𝑁 gcd (𝑀 gcd 𝑃)))

Theoremmulgcd 10887 Distribute multiplication by a nonnegative integer over gcd. (Contributed by Paul Chapman, 22-Jun-2011.) (Proof shortened by Mario Carneiro, 30-May-2014.)
((𝐾 ∈ ℕ0𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → ((𝐾 · 𝑀) gcd (𝐾 · 𝑁)) = (𝐾 · (𝑀 gcd 𝑁)))

Theoremabsmulgcd 10888 Distribute absolute value of multiplication over gcd. Theorem 1.4(c) in [ApostolNT] p. 16. (Contributed by Paul Chapman, 22-Jun-2011.)
((𝐾 ∈ ℤ ∧ 𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → ((𝐾 · 𝑀) gcd (𝐾 · 𝑁)) = (abs‘(𝐾 · (𝑀 gcd 𝑁))))

Theoremmulgcdr 10889 Reverse distribution law for the gcd operator. (Contributed by Scott Fenton, 2-Apr-2014.) (Revised by Mario Carneiro, 19-Apr-2014.)
((𝐴 ∈ ℤ ∧ 𝐵 ∈ ℤ ∧ 𝐶 ∈ ℕ0) → ((𝐴 · 𝐶) gcd (𝐵 · 𝐶)) = ((𝐴 gcd 𝐵) · 𝐶))

Theoremgcddiv 10890 Division law for GCD. (Contributed by Scott Fenton, 18-Apr-2014.) (Revised by Mario Carneiro, 19-Apr-2014.)
(((𝐴 ∈ ℤ ∧ 𝐵 ∈ ℤ ∧ 𝐶 ∈ ℕ) ∧ (𝐶𝐴𝐶𝐵)) → ((𝐴 gcd 𝐵) / 𝐶) = ((𝐴 / 𝐶) gcd (𝐵 / 𝐶)))

Theoremgcdmultiple 10891 The GCD of a multiple of a number is the number itself. (Contributed by Scott Fenton, 12-Apr-2014.) (Revised by Mario Carneiro, 19-Apr-2014.)
((𝑀 ∈ ℕ ∧ 𝑁 ∈ ℕ) → (𝑀 gcd (𝑀 · 𝑁)) = 𝑀)

Theoremgcdmultiplez 10892 Extend gcdmultiple 10891 so 𝑁 can be an integer. (Contributed by Scott Fenton, 18-Apr-2014.) (Revised by Mario Carneiro, 19-Apr-2014.)
((𝑀 ∈ ℕ ∧ 𝑁 ∈ ℤ) → (𝑀 gcd (𝑀 · 𝑁)) = 𝑀)

Theoremgcdzeq 10893 A positive integer 𝐴 is equal to its gcd with an integer 𝐵 if and only if 𝐴 divides 𝐵. Generalization of gcdeq 10894. (Contributed by AV, 1-Jul-2020.)
((𝐴 ∈ ℕ ∧ 𝐵 ∈ ℤ) → ((𝐴 gcd 𝐵) = 𝐴𝐴𝐵))

Theoremgcdeq 10894 𝐴 is equal to its gcd with 𝐵 if and only if 𝐴 divides 𝐵. (Contributed by Mario Carneiro, 23-Feb-2014.) (Proof shortened by AV, 8-Aug-2021.)
((𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ) → ((𝐴 gcd 𝐵) = 𝐴𝐴𝐵))

Theoremdvdssqim 10895 Unidirectional form of dvdssq 10902. (Contributed by Scott Fenton, 19-Apr-2014.)
((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → (𝑀𝑁 → (𝑀↑2) ∥ (𝑁↑2)))

Theoremdvdsmulgcd 10896 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 10897 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 10898 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 10899 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 10900 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)))

Page List
Jump to page: Contents  1 1-100 2 101-200 3 201-300 4 301-400 5 401-500 6 501-600 7 601-700 8 701-800 9 801-900 10 901-1000 11 1001-1100 12 1101-1200 13 1201-1300 14 1301-1400 15 1401-1500 16 1501-1600 17 1601-1700 18 1701-1800 19 1801-1900 20 1901-2000 21 2001-2100 22 2101-2200 23 2201-2300 24 2301-2400 25 2401-2500 26 2501-2600 27 2601-2700 28 2701-2800 29 2801-2900 30 2901-3000 31 3001-3100 32 3101-3200 33 3201-3300 34 3301-3400 35 3401-3500 36 3501-3600 37 3601-3700 38 3701-3800 39 3801-3900 40 3901-4000 41 4001-4100 42 4101-4200 43 4201-4300 44 4301-4400 45 4401-4500 46 4501-4600 47 4601-4700 48 4701-4800 49 4801-4900 50 4901-5000 51 5001-5100 52 5101-5200 53 5201-5300 54 5301-5400 55 5401-5500 56 5501-5600 57 5601-5700 58 5701-5800 59 5801-5900 60 5901-6000 61 6001-6100 62 6101-6200 63 6201-6300 64 6301-6400 65 6401-6500 66 6501-6600 67 6601-6700 68 6701-6800 69 6801-6900 70 6901-7000 71 7001-7100 72 7101-7200 73 7201-7300 74 7301-7400 75 7401-7500 76 7501-7600 77 7601-7700 78 7701-7800 79 7801-7900 80 7901-8000 81 8001-8100 82 8101-8200 83 8201-8300 84 8301-8400 85 8401-8500 86 8501-8600 87 8601-8700 88 8701-8800 89 8801-8900 90 8901-9000 91 9001-9100 92 9101-9200 93 9201-9300 94 9301-9400 95 9401-9500 96 9501-9600 97 9601-9700 98 9701-9800 99 9801-9900 100 9901-10000 101 10001-10100 102 10101-10200 103 10201-10300 104 10301-10400 105 10401-10500 106 10501-10600 107 10601-10700 108 10701-10800 109 10801-10900 110 10901-11000 111 11001-11100 112 11101-11200 113 11201-11300 114 11301-11370
 Copyright terms: Public domain < Previous  Next >