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Type | Label | Description |
---|---|---|
Statement | ||
Theorem | dvdssubr 11801 | An integer divides another iff it divides their difference. (Contributed by Paul Chapman, 31-Mar-2011.) |
⊢ ((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → (𝑀 ∥ 𝑁 ↔ 𝑀 ∥ (𝑁 − 𝑀))) | ||
Theorem | dvdsadd2b 11802 | Adding a multiple of the base does not affect divisibility. (Contributed by Stefan O'Rear, 23-Sep-2014.) |
⊢ ((𝐴 ∈ ℤ ∧ 𝐵 ∈ ℤ ∧ (𝐶 ∈ ℤ ∧ 𝐴 ∥ 𝐶)) → (𝐴 ∥ 𝐵 ↔ 𝐴 ∥ (𝐶 + 𝐵))) | ||
Theorem | dvdslelemd 11803 | Lemma for dvdsle 11804. (Contributed by Jim Kingdon, 8-Nov-2021.) |
⊢ (𝜑 → 𝑀 ∈ ℤ) & ⊢ (𝜑 → 𝑁 ∈ ℕ) & ⊢ (𝜑 → 𝐾 ∈ ℤ) & ⊢ (𝜑 → 𝑁 < 𝑀) ⇒ ⊢ (𝜑 → (𝐾 · 𝑀) ≠ 𝑁) | ||
Theorem | dvdsle 11804 | The divisors of a positive integer are bounded by it. The proof does not use /. (Contributed by Paul Chapman, 21-Mar-2011.) |
⊢ ((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℕ) → (𝑀 ∥ 𝑁 → 𝑀 ≤ 𝑁)) | ||
Theorem | dvdsleabs 11805 | The divisors of a nonzero integer are bounded by its absolute value. Theorem 1.1(i) in [ApostolNT] p. 14 (comparison property of the divides relation). (Contributed by Paul Chapman, 21-Mar-2011.) (Proof shortened by Fan Zheng, 3-Jul-2016.) |
⊢ ((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ ∧ 𝑁 ≠ 0) → (𝑀 ∥ 𝑁 → 𝑀 ≤ (abs‘𝑁))) | ||
Theorem | dvdsleabs2 11806 | Transfer divisibility to an order constraint on absolute values. (Contributed by Stefan O'Rear, 24-Sep-2014.) |
⊢ ((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ ∧ 𝑁 ≠ 0) → (𝑀 ∥ 𝑁 → (abs‘𝑀) ≤ (abs‘𝑁))) | ||
Theorem | dvdsabseq 11807 | If two integers divide each other, they must be equal, up to a difference in sign. Theorem 1.1(j) in [ApostolNT] p. 14. (Contributed by Mario Carneiro, 30-May-2014.) (Revised by AV, 7-Aug-2021.) |
⊢ ((𝑀 ∥ 𝑁 ∧ 𝑁 ∥ 𝑀) → (abs‘𝑀) = (abs‘𝑁)) | ||
Theorem | dvdseq 11808 | If two nonnegative integers divide each other, they must be equal. (Contributed by Mario Carneiro, 30-May-2014.) (Proof shortened by AV, 7-Aug-2021.) |
⊢ (((𝑀 ∈ ℕ0 ∧ 𝑁 ∈ ℕ0) ∧ (𝑀 ∥ 𝑁 ∧ 𝑁 ∥ 𝑀)) → 𝑀 = 𝑁) | ||
Theorem | divconjdvds 11809 | If a nonzero integer 𝑀 divides another integer 𝑁, the other integer 𝑁 divided by the nonzero integer 𝑀 (i.e. the divisor conjugate of 𝑁 to 𝑀) divides the other integer 𝑁. Theorem 1.1(k) in [ApostolNT] p. 14. (Contributed by AV, 7-Aug-2021.) |
⊢ ((𝑀 ∥ 𝑁 ∧ 𝑀 ≠ 0) → (𝑁 / 𝑀) ∥ 𝑁) | ||
Theorem | dvdsdivcl 11810* | The complement of a divisor of 𝑁 is also a divisor of 𝑁. (Contributed by Mario Carneiro, 2-Jul-2015.) (Proof shortened by AV, 9-Aug-2021.) |
⊢ ((𝑁 ∈ ℕ ∧ 𝐴 ∈ {𝑥 ∈ ℕ ∣ 𝑥 ∥ 𝑁}) → (𝑁 / 𝐴) ∈ {𝑥 ∈ ℕ ∣ 𝑥 ∥ 𝑁}) | ||
Theorem | dvdsflip 11811* | An involution of the divisors of a number. (Contributed by Stefan O'Rear, 12-Sep-2015.) (Proof shortened by Mario Carneiro, 13-May-2016.) |
⊢ 𝐴 = {𝑥 ∈ ℕ ∣ 𝑥 ∥ 𝑁} & ⊢ 𝐹 = (𝑦 ∈ 𝐴 ↦ (𝑁 / 𝑦)) ⇒ ⊢ (𝑁 ∈ ℕ → 𝐹:𝐴–1-1-onto→𝐴) | ||
Theorem | dvdsssfz1 11812* | The set of divisors of a number is a subset of a finite set. (Contributed by Mario Carneiro, 22-Sep-2014.) |
⊢ (𝐴 ∈ ℕ → {𝑝 ∈ ℕ ∣ 𝑝 ∥ 𝐴} ⊆ (1...𝐴)) | ||
Theorem | dvds1 11813 | The only nonnegative integer that divides 1 is 1. (Contributed by Mario Carneiro, 2-Jul-2015.) |
⊢ (𝑀 ∈ ℕ0 → (𝑀 ∥ 1 ↔ 𝑀 = 1)) | ||
Theorem | alzdvds 11814* | Only 0 is divisible by all integers. (Contributed by Paul Chapman, 21-Mar-2011.) |
⊢ (𝑁 ∈ ℤ → (∀𝑥 ∈ ℤ 𝑥 ∥ 𝑁 ↔ 𝑁 = 0)) | ||
Theorem | dvdsext 11815* | Poset extensionality for division. (Contributed by Stefan O'Rear, 6-Sep-2015.) |
⊢ ((𝐴 ∈ ℕ0 ∧ 𝐵 ∈ ℕ0) → (𝐴 = 𝐵 ↔ ∀𝑥 ∈ ℕ0 (𝐴 ∥ 𝑥 ↔ 𝐵 ∥ 𝑥))) | ||
Theorem | fzm1ndvds 11816 | No number between 1 and 𝑀 − 1 divides 𝑀. (Contributed by Mario Carneiro, 24-Jan-2015.) |
⊢ ((𝑀 ∈ ℕ ∧ 𝑁 ∈ (1...(𝑀 − 1))) → ¬ 𝑀 ∥ 𝑁) | ||
Theorem | fzo0dvdseq 11817 | Zero is the only one of the first 𝐴 nonnegative integers that is divisible by 𝐴. (Contributed by Stefan O'Rear, 6-Sep-2015.) |
⊢ (𝐵 ∈ (0..^𝐴) → (𝐴 ∥ 𝐵 ↔ 𝐵 = 0)) | ||
Theorem | fzocongeq 11818 | Two different elements of a half-open range are not congruent mod its length. (Contributed by Stefan O'Rear, 6-Sep-2015.) |
⊢ ((𝐴 ∈ (𝐶..^𝐷) ∧ 𝐵 ∈ (𝐶..^𝐷)) → ((𝐷 − 𝐶) ∥ (𝐴 − 𝐵) ↔ 𝐴 = 𝐵)) | ||
Theorem | addmodlteqALT 11819 | Two nonnegative integers less than the modulus are equal iff the sums of these integer with another integer are equal modulo the modulus. Shorter proof of addmodlteq 10354 based on the "divides" relation. (Contributed by AV, 14-Mar-2021.) (New usage is discouraged.) (Proof modification is discouraged.) |
⊢ ((𝐼 ∈ (0..^𝑁) ∧ 𝐽 ∈ (0..^𝑁) ∧ 𝑆 ∈ ℤ) → (((𝐼 + 𝑆) mod 𝑁) = ((𝐽 + 𝑆) mod 𝑁) ↔ 𝐼 = 𝐽)) | ||
Theorem | dvdsfac 11820 | A positive integer divides any greater factorial. (Contributed by Paul Chapman, 28-Nov-2012.) |
⊢ ((𝐾 ∈ ℕ ∧ 𝑁 ∈ (ℤ≥‘𝐾)) → 𝐾 ∥ (!‘𝑁)) | ||
Theorem | dvdsexp 11821 | A power divides a power with a greater exponent. (Contributed by Mario Carneiro, 23-Feb-2014.) |
⊢ ((𝐴 ∈ ℤ ∧ 𝑀 ∈ ℕ0 ∧ 𝑁 ∈ (ℤ≥‘𝑀)) → (𝐴↑𝑀) ∥ (𝐴↑𝑁)) | ||
Theorem | dvdsmod 11822 | Any number 𝐾 whose mod base 𝑁 is divisible by a divisor 𝑃 of the base is also divisible by 𝑃. This means that primes will also be relatively prime to the base when reduced mod 𝑁 for any base. (Contributed by Mario Carneiro, 13-Mar-2014.) |
⊢ (((𝑃 ∈ ℕ ∧ 𝑁 ∈ ℕ ∧ 𝐾 ∈ ℤ) ∧ 𝑃 ∥ 𝑁) → (𝑃 ∥ (𝐾 mod 𝑁) ↔ 𝑃 ∥ 𝐾)) | ||
Theorem | mulmoddvds 11823 | If an integer is divisible by a positive integer, the product of this integer with another integer modulo the positive integer is 0. (Contributed by Alexander van der Vekens, 30-Aug-2018.) |
⊢ ((𝑁 ∈ ℕ ∧ 𝐴 ∈ ℤ ∧ 𝐵 ∈ ℤ) → (𝑁 ∥ 𝐴 → ((𝐴 · 𝐵) mod 𝑁) = 0)) | ||
Theorem | 3dvdsdec 11824 | A decimal number is divisible by three iff the sum of its two "digits" is divisible by three. The term "digits" in its narrow sense is only correct if 𝐴 and 𝐵 actually are digits (i.e. nonnegative integers less than 10). However, this theorem holds for arbitrary nonnegative integers 𝐴 and 𝐵, especially if 𝐴 is itself a decimal number, e.g., 𝐴 = ;𝐶𝐷. (Contributed by AV, 14-Jun-2021.) (Revised by AV, 8-Sep-2021.) |
⊢ 𝐴 ∈ ℕ0 & ⊢ 𝐵 ∈ ℕ0 ⇒ ⊢ (3 ∥ ;𝐴𝐵 ↔ 3 ∥ (𝐴 + 𝐵)) | ||
Theorem | 3dvds2dec 11825 | A decimal number is divisible by three iff the sum of its three "digits" is divisible by three. The term "digits" in its narrow sense is only correct if 𝐴, 𝐵 and 𝐶 actually are digits (i.e. nonnegative integers less than 10). However, this theorem holds for arbitrary nonnegative integers 𝐴, 𝐵 and 𝐶. (Contributed by AV, 14-Jun-2021.) (Revised by AV, 1-Aug-2021.) |
⊢ 𝐴 ∈ ℕ0 & ⊢ 𝐵 ∈ ℕ0 & ⊢ 𝐶 ∈ ℕ0 ⇒ ⊢ (3 ∥ ;;𝐴𝐵𝐶 ↔ 3 ∥ ((𝐴 + 𝐵) + 𝐶)) | ||
The set ℤ of integers can be partitioned into the set of even numbers and the set of odd numbers, see zeo4 11829. Instead of defining new class variables Even and Odd to represent these sets, we use the idiom 2 ∥ 𝑁 to say that "𝑁 is even" (which implies 𝑁 ∈ ℤ, see evenelz 11826) and ¬ 2 ∥ 𝑁 to say that "𝑁 is odd" (under the assumption that 𝑁 ∈ ℤ). The previously proven theorems about even and odd numbers, like zneo 9313, zeo 9317, zeo2 9318, etc. use different representations, which are equivalent with the representations using the divides relation, see evend2 11848 and oddp1d2 11849. The corresponding theorems are zeneo 11830, zeo3 11827 and zeo4 11829. | ||
Theorem | evenelz 11826 | An even number is an integer. This follows immediately from the reverse closure of the divides relation, see dvdszrcl 11754. (Contributed by AV, 22-Jun-2021.) |
⊢ (2 ∥ 𝑁 → 𝑁 ∈ ℤ) | ||
Theorem | zeo3 11827 | An integer is even or odd. (Contributed by AV, 17-Jun-2021.) |
⊢ (𝑁 ∈ ℤ → (2 ∥ 𝑁 ∨ ¬ 2 ∥ 𝑁)) | ||
Theorem | zeoxor 11828 | An integer is even or odd but not both. (Contributed by Jim Kingdon, 10-Nov-2021.) |
⊢ (𝑁 ∈ ℤ → (2 ∥ 𝑁 ⊻ ¬ 2 ∥ 𝑁)) | ||
Theorem | zeo4 11829 | An integer is even or odd but not both. (Contributed by AV, 17-Jun-2021.) |
⊢ (𝑁 ∈ ℤ → (2 ∥ 𝑁 ↔ ¬ ¬ 2 ∥ 𝑁)) | ||
Theorem | zeneo 11830 | No even integer equals an odd integer (i.e. no integer can be both even and odd). Exercise 10(a) of [Apostol] p. 28. This variant of zneo 9313 follows immediately from the fact that a contradiction implies anything, see pm2.21i 641. (Contributed by AV, 22-Jun-2021.) |
⊢ ((𝐴 ∈ ℤ ∧ 𝐵 ∈ ℤ) → ((2 ∥ 𝐴 ∧ ¬ 2 ∥ 𝐵) → 𝐴 ≠ 𝐵)) | ||
Theorem | odd2np1lem 11831* | Lemma for odd2np1 11832. (Contributed by Scott Fenton, 3-Apr-2014.) (Revised by Mario Carneiro, 19-Apr-2014.) |
⊢ (𝑁 ∈ ℕ0 → (∃𝑛 ∈ ℤ ((2 · 𝑛) + 1) = 𝑁 ∨ ∃𝑘 ∈ ℤ (𝑘 · 2) = 𝑁)) | ||
Theorem | odd2np1 11832* | An integer is odd iff it is one plus twice another integer. (Contributed by Scott Fenton, 3-Apr-2014.) (Revised by Mario Carneiro, 19-Apr-2014.) |
⊢ (𝑁 ∈ ℤ → (¬ 2 ∥ 𝑁 ↔ ∃𝑛 ∈ ℤ ((2 · 𝑛) + 1) = 𝑁)) | ||
Theorem | even2n 11833* | An integer is even iff it is twice another integer. (Contributed by AV, 25-Jun-2020.) |
⊢ (2 ∥ 𝑁 ↔ ∃𝑛 ∈ ℤ (2 · 𝑛) = 𝑁) | ||
Theorem | oddm1even 11834 | An integer is odd iff its predecessor is even. (Contributed by Mario Carneiro, 5-Sep-2016.) |
⊢ (𝑁 ∈ ℤ → (¬ 2 ∥ 𝑁 ↔ 2 ∥ (𝑁 − 1))) | ||
Theorem | oddp1even 11835 | An integer is odd iff its successor is even. (Contributed by Mario Carneiro, 5-Sep-2016.) |
⊢ (𝑁 ∈ ℤ → (¬ 2 ∥ 𝑁 ↔ 2 ∥ (𝑁 + 1))) | ||
Theorem | oexpneg 11836 | The exponential of the negative of a number, when the exponent is odd. (Contributed by Mario Carneiro, 25-Apr-2015.) |
⊢ ((𝐴 ∈ ℂ ∧ 𝑁 ∈ ℕ ∧ ¬ 2 ∥ 𝑁) → (-𝐴↑𝑁) = -(𝐴↑𝑁)) | ||
Theorem | mod2eq0even 11837 | An integer is 0 modulo 2 iff it is even (i.e. divisible by 2), see example 2 in [ApostolNT] p. 107. (Contributed by AV, 21-Jul-2021.) |
⊢ (𝑁 ∈ ℤ → ((𝑁 mod 2) = 0 ↔ 2 ∥ 𝑁)) | ||
Theorem | mod2eq1n2dvds 11838 | An integer is 1 modulo 2 iff it is odd (i.e. not divisible by 2), see example 3 in [ApostolNT] p. 107. (Contributed by AV, 24-May-2020.) |
⊢ (𝑁 ∈ ℤ → ((𝑁 mod 2) = 1 ↔ ¬ 2 ∥ 𝑁)) | ||
Theorem | oddnn02np1 11839* | A nonnegative integer is odd iff it is one plus twice another nonnegative integer. (Contributed by AV, 19-Jun-2021.) |
⊢ (𝑁 ∈ ℕ0 → (¬ 2 ∥ 𝑁 ↔ ∃𝑛 ∈ ℕ0 ((2 · 𝑛) + 1) = 𝑁)) | ||
Theorem | oddge22np1 11840* | An integer greater than one is odd iff it is one plus twice a positive integer. (Contributed by AV, 16-Aug-2021.) |
⊢ (𝑁 ∈ (ℤ≥‘2) → (¬ 2 ∥ 𝑁 ↔ ∃𝑛 ∈ ℕ ((2 · 𝑛) + 1) = 𝑁)) | ||
Theorem | evennn02n 11841* | A nonnegative integer is even iff it is twice another nonnegative integer. (Contributed by AV, 12-Aug-2021.) |
⊢ (𝑁 ∈ ℕ0 → (2 ∥ 𝑁 ↔ ∃𝑛 ∈ ℕ0 (2 · 𝑛) = 𝑁)) | ||
Theorem | evennn2n 11842* | A positive integer is even iff it is twice another positive integer. (Contributed by AV, 12-Aug-2021.) |
⊢ (𝑁 ∈ ℕ → (2 ∥ 𝑁 ↔ ∃𝑛 ∈ ℕ (2 · 𝑛) = 𝑁)) | ||
Theorem | 2tp1odd 11843 | A number which is twice an integer increased by 1 is odd. (Contributed by AV, 16-Jul-2021.) |
⊢ ((𝐴 ∈ ℤ ∧ 𝐵 = ((2 · 𝐴) + 1)) → ¬ 2 ∥ 𝐵) | ||
Theorem | mulsucdiv2z 11844 | An integer multiplied with its successor divided by 2 yields an integer, i.e. an integer multiplied with its successor is even. (Contributed by AV, 19-Jul-2021.) |
⊢ (𝑁 ∈ ℤ → ((𝑁 · (𝑁 + 1)) / 2) ∈ ℤ) | ||
Theorem | sqoddm1div8z 11845 | A squared odd number minus 1 divided by 8 is an integer. (Contributed by AV, 19-Jul-2021.) |
⊢ ((𝑁 ∈ ℤ ∧ ¬ 2 ∥ 𝑁) → (((𝑁↑2) − 1) / 8) ∈ ℤ) | ||
Theorem | 2teven 11846 | A number which is twice an integer is even. (Contributed by AV, 16-Jul-2021.) |
⊢ ((𝐴 ∈ ℤ ∧ 𝐵 = (2 · 𝐴)) → 2 ∥ 𝐵) | ||
Theorem | zeo5 11847 | An integer is either even or odd, version of zeo3 11827 avoiding the negation of the representation of an odd number. (Proposed by BJ, 21-Jun-2021.) (Contributed by AV, 26-Jun-2020.) |
⊢ (𝑁 ∈ ℤ → (2 ∥ 𝑁 ∨ 2 ∥ (𝑁 + 1))) | ||
Theorem | evend2 11848 | An integer is even iff its quotient with 2 is an integer. This is a representation of even numbers without using the divides relation, see zeo 9317 and zeo2 9318. (Contributed by AV, 22-Jun-2021.) |
⊢ (𝑁 ∈ ℤ → (2 ∥ 𝑁 ↔ (𝑁 / 2) ∈ ℤ)) | ||
Theorem | oddp1d2 11849 | An integer is odd iff its successor divided by 2 is an integer. This is a representation of odd numbers without using the divides relation, see zeo 9317 and zeo2 9318. (Contributed by AV, 22-Jun-2021.) |
⊢ (𝑁 ∈ ℤ → (¬ 2 ∥ 𝑁 ↔ ((𝑁 + 1) / 2) ∈ ℤ)) | ||
Theorem | zob 11850 | Alternate characterizations of an odd number. (Contributed by AV, 7-Jun-2020.) |
⊢ (𝑁 ∈ ℤ → (((𝑁 + 1) / 2) ∈ ℤ ↔ ((𝑁 − 1) / 2) ∈ ℤ)) | ||
Theorem | oddm1d2 11851 | An integer is odd iff its predecessor divided by 2 is an integer. This is another representation of odd numbers without using the divides relation. (Contributed by AV, 18-Jun-2021.) (Proof shortened by AV, 22-Jun-2021.) |
⊢ (𝑁 ∈ ℤ → (¬ 2 ∥ 𝑁 ↔ ((𝑁 − 1) / 2) ∈ ℤ)) | ||
Theorem | ltoddhalfle 11852 | An integer is less than half of an odd number iff it is less than or equal to the half of the predecessor of the odd number (which is an even number). (Contributed by AV, 29-Jun-2021.) |
⊢ ((𝑁 ∈ ℤ ∧ ¬ 2 ∥ 𝑁 ∧ 𝑀 ∈ ℤ) → (𝑀 < (𝑁 / 2) ↔ 𝑀 ≤ ((𝑁 − 1) / 2))) | ||
Theorem | halfleoddlt 11853 | 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 ∥ 𝑁 ∧ 𝑀 ∈ ℤ) → ((𝑁 / 2) ≤ 𝑀 ↔ (𝑁 / 2) < 𝑀)) | ||
Theorem | opoe 11854 | 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)) | ||
Theorem | omoe 11855 | 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 ∥ (𝐴 − 𝐵)) | ||
Theorem | opeo 11856 | 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)) | ||
Theorem | omeo 11857 | 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 ∥ (𝐴 − 𝐵)) | ||
Theorem | m1expe 11858 | Exponentiation of -1 by an even power. Variant of m1expeven 10523. (Contributed by AV, 25-Jun-2021.) |
⊢ (2 ∥ 𝑁 → (-1↑𝑁) = 1) | ||
Theorem | m1expo 11859 | Exponentiation of -1 by an odd power. (Contributed by AV, 26-Jun-2021.) |
⊢ ((𝑁 ∈ ℤ ∧ ¬ 2 ∥ 𝑁) → (-1↑𝑁) = -1) | ||
Theorem | m1exp1 11860 | Exponentiation of negative one is one iff the exponent is even. (Contributed by AV, 20-Jun-2021.) |
⊢ (𝑁 ∈ ℤ → ((-1↑𝑁) = 1 ↔ 2 ∥ 𝑁)) | ||
Theorem | nn0enne 11861 | 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)) | ||
Theorem | nn0ehalf 11862 | 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) | ||
Theorem | nnehalf 11863 | The half of an even positive integer is a positive integer. (Contributed by AV, 28-Jun-2021.) |
⊢ ((𝑁 ∈ ℕ ∧ 2 ∥ 𝑁) → (𝑁 / 2) ∈ ℕ) | ||
Theorem | nn0o1gt2 11864 | 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)) | ||
Theorem | nno 11865 | 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) | ||
Theorem | nn0o 11866 | 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) | ||
Theorem | nn0ob 11867 | Alternate characterizations of an odd nonnegative integer. (Contributed by AV, 4-Jun-2020.) |
⊢ (𝑁 ∈ ℕ0 → (((𝑁 + 1) / 2) ∈ ℕ0 ↔ ((𝑁 − 1) / 2) ∈ ℕ0)) | ||
Theorem | nn0oddm1d2 11868 | 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)) | ||
Theorem | nnoddm1d2 11869 | 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)) | ||
Theorem | z0even 11870 | 0 is even. (Contributed by AV, 11-Feb-2020.) (Revised by AV, 23-Jun-2021.) |
⊢ 2 ∥ 0 | ||
Theorem | n2dvds1 11871 | 2 does not divide 1 (common case). That means 1 is odd. (Contributed by David A. Wheeler, 8-Dec-2018.) |
⊢ ¬ 2 ∥ 1 | ||
Theorem | n2dvdsm1 11872 | 2 does not divide -1. That means -1 is odd. (Contributed by AV, 15-Aug-2021.) |
⊢ ¬ 2 ∥ -1 | ||
Theorem | z2even 11873 | 2 is even. (Contributed by AV, 12-Feb-2020.) (Revised by AV, 23-Jun-2021.) |
⊢ 2 ∥ 2 | ||
Theorem | n2dvds3 11874 | 2 does not divide 3, i.e. 3 is an odd number. (Contributed by AV, 28-Feb-2021.) |
⊢ ¬ 2 ∥ 3 | ||
Theorem | z4even 11875 | 4 is an even number. (Contributed by AV, 23-Jul-2020.) (Revised by AV, 4-Jul-2021.) |
⊢ 2 ∥ 4 | ||
Theorem | 4dvdseven 11876 | An integer which is divisible by 4 is an even integer. (Contributed by AV, 4-Jul-2021.) |
⊢ (4 ∥ 𝑁 → 2 ∥ 𝑁) | ||
Theorem | divalglemnn 11877* | Lemma for divalg 11883. Existence for a positive denominator. (Contributed by Jim Kingdon, 30-Nov-2021.) |
⊢ ((𝑁 ∈ ℤ ∧ 𝐷 ∈ ℕ) → ∃𝑟 ∈ ℤ ∃𝑞 ∈ ℤ (0 ≤ 𝑟 ∧ 𝑟 < (abs‘𝐷) ∧ 𝑁 = ((𝑞 · 𝐷) + 𝑟))) | ||
Theorem | divalglemqt 11878 | Lemma for divalg 11883. The 𝑄 = 𝑇 case involved in showing uniqueness. (Contributed by Jim Kingdon, 5-Dec-2021.) |
⊢ (𝜑 → 𝐷 ∈ ℤ) & ⊢ (𝜑 → 𝑅 ∈ ℤ) & ⊢ (𝜑 → 𝑆 ∈ ℤ) & ⊢ (𝜑 → 𝑄 ∈ ℤ) & ⊢ (𝜑 → 𝑇 ∈ ℤ) & ⊢ (𝜑 → 𝑄 = 𝑇) & ⊢ (𝜑 → ((𝑄 · 𝐷) + 𝑅) = ((𝑇 · 𝐷) + 𝑆)) ⇒ ⊢ (𝜑 → 𝑅 = 𝑆) | ||
Theorem | divalglemnqt 11879 | Lemma for divalg 11883. The 𝑄 < 𝑇 case involved in showing uniqueness. (Contributed by Jim Kingdon, 4-Dec-2021.) |
⊢ (𝜑 → 𝐷 ∈ ℕ) & ⊢ (𝜑 → 𝑅 ∈ ℤ) & ⊢ (𝜑 → 𝑆 ∈ ℤ) & ⊢ (𝜑 → 𝑄 ∈ ℤ) & ⊢ (𝜑 → 𝑇 ∈ ℤ) & ⊢ (𝜑 → 0 ≤ 𝑆) & ⊢ (𝜑 → 𝑅 < 𝐷) & ⊢ (𝜑 → ((𝑄 · 𝐷) + 𝑅) = ((𝑇 · 𝐷) + 𝑆)) ⇒ ⊢ (𝜑 → ¬ 𝑄 < 𝑇) | ||
Theorem | divalglemeunn 11880* | Lemma for divalg 11883. Uniqueness for a positive denominator. (Contributed by Jim Kingdon, 4-Dec-2021.) |
⊢ ((𝑁 ∈ ℤ ∧ 𝐷 ∈ ℕ) → ∃!𝑟 ∈ ℤ ∃𝑞 ∈ ℤ (0 ≤ 𝑟 ∧ 𝑟 < (abs‘𝐷) ∧ 𝑁 = ((𝑞 · 𝐷) + 𝑟))) | ||
Theorem | divalglemex 11881* | Lemma for divalg 11883. The quotient and remainder exist. (Contributed by Jim Kingdon, 30-Nov-2021.) |
⊢ ((𝑁 ∈ ℤ ∧ 𝐷 ∈ ℤ ∧ 𝐷 ≠ 0) → ∃𝑟 ∈ ℤ ∃𝑞 ∈ ℤ (0 ≤ 𝑟 ∧ 𝑟 < (abs‘𝐷) ∧ 𝑁 = ((𝑞 · 𝐷) + 𝑟))) | ||
Theorem | divalglemeuneg 11882* | Lemma for divalg 11883. Uniqueness for a negative denominator. (Contributed by Jim Kingdon, 4-Dec-2021.) |
⊢ ((𝑁 ∈ ℤ ∧ 𝐷 ∈ ℤ ∧ 𝐷 < 0) → ∃!𝑟 ∈ ℤ ∃𝑞 ∈ ℤ (0 ≤ 𝑟 ∧ 𝑟 < (abs‘𝐷) ∧ 𝑁 = ((𝑞 · 𝐷) + 𝑟))) | ||
Theorem | divalg 11883* | 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 ≤ 𝑟 ∧ 𝑟 < (abs‘𝐷) ∧ 𝑁 = ((𝑞 · 𝐷) + 𝑟))) | ||
Theorem | divalgb 11884* | Express the division algorithm as stated in divalg 11883 in terms of ∥. (Contributed by Paul Chapman, 31-Mar-2011.) |
⊢ ((𝑁 ∈ ℤ ∧ 𝐷 ∈ ℤ ∧ 𝐷 ≠ 0) → (∃!𝑟 ∈ ℤ ∃𝑞 ∈ ℤ (0 ≤ 𝑟 ∧ 𝑟 < (abs‘𝐷) ∧ 𝑁 = ((𝑞 · 𝐷) + 𝑟)) ↔ ∃!𝑟 ∈ ℕ0 (𝑟 < (abs‘𝐷) ∧ 𝐷 ∥ (𝑁 − 𝑟)))) | ||
Theorem | divalg2 11885* | The division algorithm (theorem) for a positive divisor. (Contributed by Paul Chapman, 21-Mar-2011.) |
⊢ ((𝑁 ∈ ℤ ∧ 𝐷 ∈ ℕ) → ∃!𝑟 ∈ ℕ0 (𝑟 < 𝐷 ∧ 𝐷 ∥ (𝑁 − 𝑟))) | ||
Theorem | divalgmod 11886 | The result of the mod operator satisfies the requirements for the remainder 𝑅 in the division algorithm for a positive divisor (compare divalg2 11885 and divalgb 11884). 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 < 𝐷 ∧ 𝐷 ∥ (𝑁 − 𝑅))))) | ||
Theorem | divalgmodcl 11887 | The result of the mod operator satisfies the requirements for the remainder 𝑅 in the division algorithm for a positive divisor. Variant of divalgmod 11886. (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 < 𝐷 ∧ 𝐷 ∥ (𝑁 − 𝑅)))) | ||
Theorem | modremain 11888* | The result of the modulo operation is the remainder of the division algorithm. (Contributed by AV, 19-Aug-2021.) |
⊢ ((𝑁 ∈ ℤ ∧ 𝐷 ∈ ℕ ∧ (𝑅 ∈ ℕ0 ∧ 𝑅 < 𝐷)) → ((𝑁 mod 𝐷) = 𝑅 ↔ ∃𝑧 ∈ ℤ ((𝑧 · 𝐷) + 𝑅) = 𝑁)) | ||
Theorem | ndvdssub 11889 | 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)) → (𝐷 ∥ 𝑁 → ¬ 𝐷 ∥ (𝑁 − 𝐾))) | ||
Theorem | ndvdsadd 11890 | 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)) → (𝐷 ∥ 𝑁 → ¬ 𝐷 ∥ (𝑁 + 𝐾))) | ||
Theorem | ndvdsp1 11891 | Special case of ndvdsadd 11890. 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) → (𝐷 ∥ 𝑁 → ¬ 𝐷 ∥ (𝑁 + 1))) | ||
Theorem | ndvdsi 11892 | 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 ⇒ ⊢ ¬ 𝐴 ∥ 𝐵 | ||
Theorem | flodddiv4 11893 | 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))) | ||
Theorem | fldivndvdslt 11894 | 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) ∧ ¬ 𝐿 ∥ 𝐾) → (⌊‘(𝐾 / 𝐿)) < (𝐾 / 𝐿)) | ||
Theorem | flodddiv4lt 11895 | 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)) | ||
Theorem | flodddiv4t2lthalf 11896 | 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)) | ||
Syntax | cgcd 11897 | Extend the definition of a class to include the greatest common divisor operator. |
class gcd | ||
Definition | df-gcd 11898* | Define the gcd operator. For example, (-6 gcd 9) = 3 (ex-gcd 13766). (Contributed by Paul Chapman, 21-Mar-2011.) |
⊢ gcd = (𝑥 ∈ ℤ, 𝑦 ∈ ℤ ↦ if((𝑥 = 0 ∧ 𝑦 = 0), 0, sup({𝑛 ∈ ℤ ∣ (𝑛 ∥ 𝑥 ∧ 𝑛 ∥ 𝑦)}, ℝ, < ))) | ||
Theorem | gcdmndc 11899 | 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)) | ||
Theorem | zsupcllemstep 11900* | Lemma for zsupcl 11902. Induction step. (Contributed by Jim Kingdon, 7-Dec-2021.) |
⊢ ((𝜑 ∧ 𝑛 ∈ (ℤ≥‘𝑀)) → DECID 𝜓) ⇒ ⊢ (𝐾 ∈ (ℤ≥‘𝑀) → (((𝜑 ∧ ∀𝑛 ∈ (ℤ≥‘𝐾) ¬ 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))))) |
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