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| Type | Label | Description |
|---|---|---|
| Statement | ||
| Theorem | ruclem7 16201* | Lemma for ruc 16208. Successor value for the interval sequence. (Contributed by Mario Carneiro, 28-May-2014.) |
| ⊢ (𝜑 → 𝐹:ℕ⟶ℝ) & ⊢ (𝜑 → 𝐷 = (𝑥 ∈ (ℝ × ℝ), 𝑦 ∈ ℝ ↦ ⦋(((1st ‘𝑥) + (2nd ‘𝑥)) / 2) / 𝑚⦌if(𝑚 < 𝑦, 〈(1st ‘𝑥), 𝑚〉, 〈((𝑚 + (2nd ‘𝑥)) / 2), (2nd ‘𝑥)〉))) & ⊢ 𝐶 = ({〈0, 〈0, 1〉〉} ∪ 𝐹) & ⊢ 𝐺 = seq0(𝐷, 𝐶) ⇒ ⊢ ((𝜑 ∧ 𝑁 ∈ ℕ0) → (𝐺‘(𝑁 + 1)) = ((𝐺‘𝑁)𝐷(𝐹‘(𝑁 + 1)))) | ||
| Theorem | ruclem8 16202* | Lemma for ruc 16208. The intervals of the 𝐺 sequence are all nonempty. (Contributed by Mario Carneiro, 28-May-2014.) |
| ⊢ (𝜑 → 𝐹:ℕ⟶ℝ) & ⊢ (𝜑 → 𝐷 = (𝑥 ∈ (ℝ × ℝ), 𝑦 ∈ ℝ ↦ ⦋(((1st ‘𝑥) + (2nd ‘𝑥)) / 2) / 𝑚⦌if(𝑚 < 𝑦, 〈(1st ‘𝑥), 𝑚〉, 〈((𝑚 + (2nd ‘𝑥)) / 2), (2nd ‘𝑥)〉))) & ⊢ 𝐶 = ({〈0, 〈0, 1〉〉} ∪ 𝐹) & ⊢ 𝐺 = seq0(𝐷, 𝐶) ⇒ ⊢ ((𝜑 ∧ 𝑁 ∈ ℕ0) → (1st ‘(𝐺‘𝑁)) < (2nd ‘(𝐺‘𝑁))) | ||
| Theorem | ruclem9 16203* | Lemma for ruc 16208. The first components of the 𝐺 sequence are increasing, and the second components are decreasing. (Contributed by Mario Carneiro, 28-May-2014.) |
| ⊢ (𝜑 → 𝐹:ℕ⟶ℝ) & ⊢ (𝜑 → 𝐷 = (𝑥 ∈ (ℝ × ℝ), 𝑦 ∈ ℝ ↦ ⦋(((1st ‘𝑥) + (2nd ‘𝑥)) / 2) / 𝑚⦌if(𝑚 < 𝑦, 〈(1st ‘𝑥), 𝑚〉, 〈((𝑚 + (2nd ‘𝑥)) / 2), (2nd ‘𝑥)〉))) & ⊢ 𝐶 = ({〈0, 〈0, 1〉〉} ∪ 𝐹) & ⊢ 𝐺 = seq0(𝐷, 𝐶) & ⊢ (𝜑 → 𝑀 ∈ ℕ0) & ⊢ (𝜑 → 𝑁 ∈ (ℤ≥‘𝑀)) ⇒ ⊢ (𝜑 → ((1st ‘(𝐺‘𝑀)) ≤ (1st ‘(𝐺‘𝑁)) ∧ (2nd ‘(𝐺‘𝑁)) ≤ (2nd ‘(𝐺‘𝑀)))) | ||
| Theorem | ruclem10 16204* | Lemma for ruc 16208. Every first component of the 𝐺 sequence is less than every second component. That is, the sequences form a chain a1 < a2 <... < b2 < b1, where ai are the first components and bi are the second components. (Contributed by Mario Carneiro, 28-May-2014.) |
| ⊢ (𝜑 → 𝐹:ℕ⟶ℝ) & ⊢ (𝜑 → 𝐷 = (𝑥 ∈ (ℝ × ℝ), 𝑦 ∈ ℝ ↦ ⦋(((1st ‘𝑥) + (2nd ‘𝑥)) / 2) / 𝑚⦌if(𝑚 < 𝑦, 〈(1st ‘𝑥), 𝑚〉, 〈((𝑚 + (2nd ‘𝑥)) / 2), (2nd ‘𝑥)〉))) & ⊢ 𝐶 = ({〈0, 〈0, 1〉〉} ∪ 𝐹) & ⊢ 𝐺 = seq0(𝐷, 𝐶) & ⊢ (𝜑 → 𝑀 ∈ ℕ0) & ⊢ (𝜑 → 𝑁 ∈ ℕ0) ⇒ ⊢ (𝜑 → (1st ‘(𝐺‘𝑀)) < (2nd ‘(𝐺‘𝑁))) | ||
| Theorem | ruclem11 16205* | Lemma for ruc 16208. Closure lemmas for supremum. (Contributed by Mario Carneiro, 28-May-2014.) |
| ⊢ (𝜑 → 𝐹:ℕ⟶ℝ) & ⊢ (𝜑 → 𝐷 = (𝑥 ∈ (ℝ × ℝ), 𝑦 ∈ ℝ ↦ ⦋(((1st ‘𝑥) + (2nd ‘𝑥)) / 2) / 𝑚⦌if(𝑚 < 𝑦, 〈(1st ‘𝑥), 𝑚〉, 〈((𝑚 + (2nd ‘𝑥)) / 2), (2nd ‘𝑥)〉))) & ⊢ 𝐶 = ({〈0, 〈0, 1〉〉} ∪ 𝐹) & ⊢ 𝐺 = seq0(𝐷, 𝐶) ⇒ ⊢ (𝜑 → (ran (1st ∘ 𝐺) ⊆ ℝ ∧ ran (1st ∘ 𝐺) ≠ ∅ ∧ ∀𝑧 ∈ ran (1st ∘ 𝐺)𝑧 ≤ 1)) | ||
| Theorem | ruclem12 16206* | Lemma for ruc 16208. The supremum of the increasing sequence 1st ∘ 𝐺 is a real number that is not in the range of 𝐹. (Contributed by Mario Carneiro, 28-May-2014.) |
| ⊢ (𝜑 → 𝐹:ℕ⟶ℝ) & ⊢ (𝜑 → 𝐷 = (𝑥 ∈ (ℝ × ℝ), 𝑦 ∈ ℝ ↦ ⦋(((1st ‘𝑥) + (2nd ‘𝑥)) / 2) / 𝑚⦌if(𝑚 < 𝑦, 〈(1st ‘𝑥), 𝑚〉, 〈((𝑚 + (2nd ‘𝑥)) / 2), (2nd ‘𝑥)〉))) & ⊢ 𝐶 = ({〈0, 〈0, 1〉〉} ∪ 𝐹) & ⊢ 𝐺 = seq0(𝐷, 𝐶) & ⊢ 𝑆 = sup(ran (1st ∘ 𝐺), ℝ, < ) ⇒ ⊢ (𝜑 → 𝑆 ∈ (ℝ ∖ ran 𝐹)) | ||
| Theorem | ruclem13 16207 | Lemma for ruc 16208. There is no function that maps ℕ onto ℝ. (Use nex 1807 if you want this in the form ¬ ∃𝑓𝑓:ℕ–onto→ℝ.) (Contributed by NM, 14-Oct-2004.) (Proof shortened by Fan Zheng, 6-Jun-2016.) |
| ⊢ ¬ 𝐹:ℕ–onto→ℝ | ||
| Theorem | ruc 16208 | The set of positive integers is strictly dominated by the set of real numbers, i.e. the real numbers are uncountable. The proof consists of lemmas ruclem1 16196 through ruclem13 16207 and this final piece. Our proof is based on the proof of Theorem 5.18 of [Truss] p. 114. See ruclem13 16207 for the function existence version of this theorem. For an informal discussion of this proof, see mmcomplex.html#uncountable 16207. For an alternate proof see rucALT 16195. This is Metamath 100 proof #22. (Contributed by NM, 13-Oct-2004.) |
| ⊢ ℕ ≺ ℝ | ||
| Theorem | resdomq 16209 | The set of rationals is strictly less equinumerous than the set of reals (ℝ strictly dominates ℚ). (Contributed by NM, 18-Dec-2004.) |
| ⊢ ℚ ≺ ℝ | ||
| Theorem | aleph1re 16210 | There are at least aleph-one real numbers. (Contributed by NM, 2-Feb-2005.) |
| ⊢ (ℵ‘1o) ≼ ℝ | ||
| Theorem | aleph1irr 16211 | There are at least aleph-one irrationals. (Contributed by NM, 2-Feb-2005.) |
| ⊢ (ℵ‘1o) ≼ (ℝ ∖ ℚ) | ||
| Theorem | cnso 16212 | The complex numbers can be linearly ordered. (Contributed by Stefan O'Rear, 16-Nov-2014.) |
| ⊢ ∃𝑥 𝑥 Or ℂ | ||
Here we introduce elementary number theory, in particular the elementary properties of divisibility and elementary prime number theory. | ||
| Theorem | sqrt2irrlem 16213 | Lemma for sqrt2irr 16214. This is the core of the proof: if 𝐴 / 𝐵 = √(2), then 𝐴 and 𝐵 are even, so 𝐴 / 2 and 𝐵 / 2 are smaller representatives, which is absurd by the method of infinite descent (here implemented by strong induction). This is Metamath 100 proof #1. (Contributed by NM, 20-Aug-2001.) (Revised by Mario Carneiro, 12-Sep-2015.) (Proof shortened by JV, 4-Jan-2022.) |
| ⊢ (𝜑 → 𝐴 ∈ ℤ) & ⊢ (𝜑 → 𝐵 ∈ ℕ) & ⊢ (𝜑 → (√‘2) = (𝐴 / 𝐵)) ⇒ ⊢ (𝜑 → ((𝐴 / 2) ∈ ℤ ∧ (𝐵 / 2) ∈ ℕ)) | ||
| Theorem | sqrt2irr 16214 | The square root of 2 is irrational. See zsqrtelqelz 16726 for a generalization to all non-square integers. The proof's core is proven in sqrt2irrlem 16213, which shows that if 𝐴 / 𝐵 = √(2), then 𝐴 and 𝐵 are even, so 𝐴 / 2 and 𝐵 / 2 are smaller representatives, which is absurd. An older version of this proof was included in The Seventeen Provers of the World compiled by Freek Wiedijk. It is also the first of the "top 100" mathematical theorems whose formalization is tracked by Freek Wiedijk on his Formalizing 100 Theorems page at http://www.cs.ru.nl/~freek/100/ 16213. (Contributed by NM, 8-Jan-2002.) (Proof shortened by Mario Carneiro, 12-Sep-2015.) |
| ⊢ (√‘2) ∉ ℚ | ||
| Theorem | sqrt2re 16215 | The square root of 2 exists and is a real number. (Contributed by NM, 3-Dec-2004.) |
| ⊢ (√‘2) ∈ ℝ | ||
| Theorem | sqrt2irr0 16216 | The square root of 2 is an irrational number. (Contributed by AV, 23-Dec-2022.) |
| ⊢ (√‘2) ∈ (ℝ ∖ ℚ) | ||
| Theorem | nthruc 16217 | The sequence ℕ, ℤ, ℚ, ℝ, and ℂ forms a chain of proper subsets. In each case the proper subset relationship is shown by demonstrating a number that belongs to one set but not the other. We show that zero belongs to ℤ but not ℕ, one-half belongs to ℚ but not ℤ, the square root of 2 belongs to ℝ but not ℚ, and finally that the imaginary number i belongs to ℂ but not ℝ. See nthruz 16218 for a further refinement. (Contributed by NM, 12-Jan-2002.) |
| ⊢ ((ℕ ⊊ ℤ ∧ ℤ ⊊ ℚ) ∧ (ℚ ⊊ ℝ ∧ ℝ ⊊ ℂ)) | ||
| Theorem | nthruz 16218 | The sequence ℕ, ℕ0, and ℤ forms a chain of proper subsets. In each case the proper subset relationship is shown by demonstrating a number that belongs to one set but not the other. We show that zero belongs to ℕ0 but not ℕ and minus one belongs to ℤ but not ℕ0. This theorem refines the chain of proper subsets nthruc 16217. (Contributed by NM, 9-May-2004.) |
| ⊢ (ℕ ⊊ ℕ0 ∧ ℕ0 ⊊ ℤ) | ||
| Syntax | cdvds 16219 | Extend the definition of a class to include the divides relation. See df-dvds 16220. |
| class ∥ | ||
| Definition | df-dvds 16220* | Define the divides relation, see definition in [ApostolNT] p. 14. (Contributed by Paul Chapman, 21-Mar-2011.) |
| ⊢ ∥ = {〈𝑥, 𝑦〉 ∣ ((𝑥 ∈ ℤ ∧ 𝑦 ∈ ℤ) ∧ ∃𝑛 ∈ ℤ (𝑛 · 𝑥) = 𝑦)} | ||
| Theorem | divides 16221* | Define the divides relation. 𝑀 ∥ 𝑁 means 𝑀 divides into 𝑁 with no remainder. For example, 3 ∥ 6 (ex-dvds 30551). As proven in dvdsval3 16223, 𝑀 ∥ 𝑁 ↔ (𝑁 mod 𝑀) = 0. See divides 16221 and dvdsval2 16222 for other equivalent expressions. (Contributed by Paul Chapman, 21-Mar-2011.) |
| ⊢ ((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → (𝑀 ∥ 𝑁 ↔ ∃𝑛 ∈ ℤ (𝑛 · 𝑀) = 𝑁)) | ||
| Theorem | dvdsval2 16222 | One nonzero integer divides another integer if and only if their quotient is an integer. (Contributed by Jeff Hankins, 29-Sep-2013.) |
| ⊢ ((𝑀 ∈ ℤ ∧ 𝑀 ≠ 0 ∧ 𝑁 ∈ ℤ) → (𝑀 ∥ 𝑁 ↔ (𝑁 / 𝑀) ∈ ℤ)) | ||
| Theorem | dvdsval3 16223 | One nonzero integer divides another integer if and only if the remainder upon division is zero, see remark in [ApostolNT] p. 106. (Contributed by Mario Carneiro, 22-Feb-2014.) (Revised by Mario Carneiro, 15-Jul-2014.) |
| ⊢ ((𝑀 ∈ ℕ ∧ 𝑁 ∈ ℤ) → (𝑀 ∥ 𝑁 ↔ (𝑁 mod 𝑀) = 0)) | ||
| Theorem | dvdszrcl 16224 | Reverse closure for the divisibility relation. (Contributed by Stefan O'Rear, 5-Sep-2015.) |
| ⊢ (𝑋 ∥ 𝑌 → (𝑋 ∈ ℤ ∧ 𝑌 ∈ ℤ)) | ||
| Theorem | dvdsmod0 16225 | If a positive integer divides another integer, then the remainder upon division is zero. (Contributed by AV, 3-Mar-2022.) |
| ⊢ ((𝑀 ∈ ℕ ∧ 𝑀 ∥ 𝑁) → (𝑁 mod 𝑀) = 0) | ||
| Theorem | p1modz1 16226 | If a number greater than 1 divides another number, the second number increased by 1 is 1 modulo the first number. (Contributed by AV, 19-Mar-2022.) |
| ⊢ ((𝑀 ∥ 𝐴 ∧ 1 < 𝑀) → ((𝐴 + 1) mod 𝑀) = 1) | ||
| Theorem | dvdsmodexp 16227 | If a positive integer divides another integer, this other integer is equal to its positive powers modulo the positive integer. (Formerly part of the proof for fermltl 16752). (Contributed by Mario Carneiro, 28-Feb-2014.) (Revised by AV, 19-Mar-2022.) |
| ⊢ ((𝑁 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝑁 ∥ 𝐴) → ((𝐴↑𝐵) mod 𝑁) = (𝐴 mod 𝑁)) | ||
| Theorem | nndivdvds 16228 | Strong form of dvdsval2 16222 for positive integers. (Contributed by Stefan O'Rear, 13-Sep-2014.) |
| ⊢ ((𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ) → (𝐵 ∥ 𝐴 ↔ (𝐴 / 𝐵) ∈ ℕ)) | ||
| Theorem | nndivides 16229* | Definition of the divides relation for positive integers. (Contributed by AV, 26-Jul-2021.) |
| ⊢ ((𝑀 ∈ ℕ ∧ 𝑁 ∈ ℕ) → (𝑀 ∥ 𝑁 ↔ ∃𝑛 ∈ ℕ (𝑛 · 𝑀) = 𝑁)) | ||
| Theorem | moddvds 16230 | Two ways to say 𝐴≡𝐵 (mod 𝑁), see also definition in [ApostolNT] p. 106. (Contributed by Mario Carneiro, 18-Feb-2014.) |
| ⊢ ((𝑁 ∈ ℕ ∧ 𝐴 ∈ ℤ ∧ 𝐵 ∈ ℤ) → ((𝐴 mod 𝑁) = (𝐵 mod 𝑁) ↔ 𝑁 ∥ (𝐴 − 𝐵))) | ||
| Theorem | modm1div 16231 | An integer greater than one divides another integer minus one iff the second integer modulo the first integer is one. (Contributed by AV, 30-May-2023.) |
| ⊢ ((𝑁 ∈ (ℤ≥‘2) ∧ 𝐴 ∈ ℤ) → ((𝐴 mod 𝑁) = 1 ↔ 𝑁 ∥ (𝐴 − 1))) | ||
| Theorem | addmulmodb 16232 | An integer plus a product is itself modulo a positive integer iff the product is divisible by the positive integer. (Contributed by AV, 8-Sep-2025.) |
| ⊢ ((𝑁 ∈ ℕ ∧ (𝐴 ∈ ℤ ∧ 𝐵 ∈ ℤ ∧ 𝐶 ∈ ℤ)) → (𝑁 ∥ (𝐵 · 𝐶) ↔ ((𝐴 + (𝐵 · 𝐶)) mod 𝑁) = (𝐴 mod 𝑁))) | ||
| Theorem | dvds0lem 16233 | A lemma to assist theorems of ∥ with no antecedents. (Contributed by Paul Chapman, 21-Mar-2011.) |
| ⊢ (((𝐾 ∈ ℤ ∧ 𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) ∧ (𝐾 · 𝑀) = 𝑁) → 𝑀 ∥ 𝑁) | ||
| Theorem | dvds1lem 16234* | A lemma to assist theorems of ∥ with one antecedent. (Contributed by Paul Chapman, 21-Mar-2011.) |
| ⊢ (𝜑 → (𝐽 ∈ ℤ ∧ 𝐾 ∈ ℤ)) & ⊢ (𝜑 → (𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ)) & ⊢ ((𝜑 ∧ 𝑥 ∈ ℤ) → 𝑍 ∈ ℤ) & ⊢ ((𝜑 ∧ 𝑥 ∈ ℤ) → ((𝑥 · 𝐽) = 𝐾 → (𝑍 · 𝑀) = 𝑁)) ⇒ ⊢ (𝜑 → (𝐽 ∥ 𝐾 → 𝑀 ∥ 𝑁)) | ||
| Theorem | dvds2lem 16235* | A lemma to assist theorems of ∥ with two antecedents. (Contributed by Paul Chapman, 21-Mar-2011.) |
| ⊢ (𝜑 → (𝐼 ∈ ℤ ∧ 𝐽 ∈ ℤ)) & ⊢ (𝜑 → (𝐾 ∈ ℤ ∧ 𝐿 ∈ ℤ)) & ⊢ (𝜑 → (𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ)) & ⊢ ((𝜑 ∧ (𝑥 ∈ ℤ ∧ 𝑦 ∈ ℤ)) → 𝑍 ∈ ℤ) & ⊢ ((𝜑 ∧ (𝑥 ∈ ℤ ∧ 𝑦 ∈ ℤ)) → (((𝑥 · 𝐼) = 𝐽 ∧ (𝑦 · 𝐾) = 𝐿) → (𝑍 · 𝑀) = 𝑁)) ⇒ ⊢ (𝜑 → ((𝐼 ∥ 𝐽 ∧ 𝐾 ∥ 𝐿) → 𝑀 ∥ 𝑁)) | ||
| Theorem | iddvds 16236 | An integer divides itself. Theorem 1.1(a) in [ApostolNT] p. 14 (reflexive property of the divides relation). (Contributed by Paul Chapman, 21-Mar-2011.) |
| ⊢ (𝑁 ∈ ℤ → 𝑁 ∥ 𝑁) | ||
| Theorem | 1dvds 16237 | 1 divides any integer. Theorem 1.1(f) in [ApostolNT] p. 14. (Contributed by Paul Chapman, 21-Mar-2011.) |
| ⊢ (𝑁 ∈ ℤ → 1 ∥ 𝑁) | ||
| Theorem | dvds0 16238 | Any integer divides 0. Theorem 1.1(g) in [ApostolNT] p. 14. (Contributed by Paul Chapman, 21-Mar-2011.) |
| ⊢ (𝑁 ∈ ℤ → 𝑁 ∥ 0) | ||
| Theorem | negdvdsb 16239 | An integer divides another iff its negation does. (Contributed by Paul Chapman, 21-Mar-2011.) |
| ⊢ ((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → (𝑀 ∥ 𝑁 ↔ -𝑀 ∥ 𝑁)) | ||
| Theorem | dvdsnegb 16240 | An integer divides another iff it divides its negation. (Contributed by Paul Chapman, 21-Mar-2011.) |
| ⊢ ((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → (𝑀 ∥ 𝑁 ↔ 𝑀 ∥ -𝑁)) | ||
| Theorem | absdvdsb 16241 | An integer divides another iff its absolute value does. (Contributed by Paul Chapman, 21-Mar-2011.) |
| ⊢ ((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → (𝑀 ∥ 𝑁 ↔ (abs‘𝑀) ∥ 𝑁)) | ||
| Theorem | dvdsabsb 16242 | An integer divides another iff it divides its absolute value. (Contributed by Paul Chapman, 21-Mar-2011.) |
| ⊢ ((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → (𝑀 ∥ 𝑁 ↔ 𝑀 ∥ (abs‘𝑁))) | ||
| Theorem | 0dvds 16243 | Only 0 is divisible by 0. Theorem 1.1(h) in [ApostolNT] p. 14. (Contributed by Paul Chapman, 21-Mar-2011.) |
| ⊢ (𝑁 ∈ ℤ → (0 ∥ 𝑁 ↔ 𝑁 = 0)) | ||
| Theorem | dvdsmul1 16244 | An integer divides a multiple of itself. (Contributed by Paul Chapman, 21-Mar-2011.) |
| ⊢ ((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → 𝑀 ∥ (𝑀 · 𝑁)) | ||
| Theorem | dvdsmul2 16245 | An integer divides a multiple of itself. (Contributed by Paul Chapman, 21-Mar-2011.) |
| ⊢ ((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → 𝑁 ∥ (𝑀 · 𝑁)) | ||
| Theorem | iddvdsexp 16246 | An integer divides a positive integer power of itself. (Contributed by Paul Chapman, 26-Oct-2012.) |
| ⊢ ((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℕ) → 𝑀 ∥ (𝑀↑𝑁)) | ||
| Theorem | muldvds1 16247 | If a product divides an integer, so does one of its factors. (Contributed by Paul Chapman, 21-Mar-2011.) |
| ⊢ ((𝐾 ∈ ℤ ∧ 𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → ((𝐾 · 𝑀) ∥ 𝑁 → 𝐾 ∥ 𝑁)) | ||
| Theorem | muldvds2 16248 | If a product divides an integer, so does one of its factors. (Contributed by Paul Chapman, 21-Mar-2011.) |
| ⊢ ((𝐾 ∈ ℤ ∧ 𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → ((𝐾 · 𝑀) ∥ 𝑁 → 𝑀 ∥ 𝑁)) | ||
| Theorem | dvdscmul 16249 | Multiplication by a constant maintains the divides relation. Theorem 1.1(d) in [ApostolNT] p. 14 (multiplication property of the divides relation). (Contributed by Paul Chapman, 21-Mar-2011.) |
| ⊢ ((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ ∧ 𝐾 ∈ ℤ) → (𝑀 ∥ 𝑁 → (𝐾 · 𝑀) ∥ (𝐾 · 𝑁))) | ||
| Theorem | dvdsmulc 16250 | Multiplication by a constant maintains the divides relation. (Contributed by Paul Chapman, 21-Mar-2011.) |
| ⊢ ((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ ∧ 𝐾 ∈ ℤ) → (𝑀 ∥ 𝑁 → (𝑀 · 𝐾) ∥ (𝑁 · 𝐾))) | ||
| Theorem | dvdscmulr 16251 | Cancellation law for the divides relation. Theorem 1.1(e) in [ApostolNT] p. 14. (Contributed by Paul Chapman, 21-Mar-2011.) |
| ⊢ ((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ ∧ (𝐾 ∈ ℤ ∧ 𝐾 ≠ 0)) → ((𝐾 · 𝑀) ∥ (𝐾 · 𝑁) ↔ 𝑀 ∥ 𝑁)) | ||
| Theorem | dvdsmulcr 16252 | Cancellation law for the divides relation. (Contributed by Paul Chapman, 21-Mar-2011.) |
| ⊢ ((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ ∧ (𝐾 ∈ ℤ ∧ 𝐾 ≠ 0)) → ((𝑀 · 𝐾) ∥ (𝑁 · 𝐾) ↔ 𝑀 ∥ 𝑁)) | ||
| Theorem | summodnegmod 16253 | The sum of two integers modulo a positive integer equals zero iff the first of the two integers equals the negative of the other integer modulo the positive integer. (Contributed by AV, 25-Jul-2021.) |
| ⊢ ((𝐴 ∈ ℤ ∧ 𝐵 ∈ ℤ ∧ 𝑁 ∈ ℕ) → (((𝐴 + 𝐵) mod 𝑁) = 0 ↔ (𝐴 mod 𝑁) = (-𝐵 mod 𝑁))) | ||
| Theorem | difmod0 16254 | The difference of two integers modulo a positive integer equals zero iff the two integers are equal modulo the positive integer. (Contributed by AV, 15-Nov-2025.) |
| ⊢ ((𝐴 ∈ ℤ ∧ 𝐵 ∈ ℤ ∧ 𝑁 ∈ ℕ) → (((𝐴 − 𝐵) mod 𝑁) = 0 ↔ (𝐴 mod 𝑁) = (𝐵 mod 𝑁))) | ||
| Theorem | modmulconst 16255 | Constant multiplication in a modulo operation, see theorem 5.3 in [ApostolNT] p. 108. (Contributed by AV, 21-Jul-2021.) |
| ⊢ (((𝐴 ∈ ℤ ∧ 𝐵 ∈ ℤ ∧ 𝐶 ∈ ℕ) ∧ 𝑀 ∈ ℕ) → ((𝐴 mod 𝑀) = (𝐵 mod 𝑀) ↔ ((𝐶 · 𝐴) mod (𝐶 · 𝑀)) = ((𝐶 · 𝐵) mod (𝐶 · 𝑀)))) | ||
| Theorem | dvds2ln 16256 | If an integer divides each of two other integers, it divides any linear combination of them. Theorem 1.1(c) in [ApostolNT] p. 14 (linearity property of the divides relation). (Contributed by Paul Chapman, 21-Mar-2011.) |
| ⊢ (((𝐼 ∈ ℤ ∧ 𝐽 ∈ ℤ) ∧ (𝐾 ∈ ℤ ∧ 𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ)) → ((𝐾 ∥ 𝑀 ∧ 𝐾 ∥ 𝑁) → 𝐾 ∥ ((𝐼 · 𝑀) + (𝐽 · 𝑁)))) | ||
| Theorem | dvds2add 16257 | If an integer divides each of two other integers, it divides their sum. (Contributed by Paul Chapman, 21-Mar-2011.) |
| ⊢ ((𝐾 ∈ ℤ ∧ 𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → ((𝐾 ∥ 𝑀 ∧ 𝐾 ∥ 𝑁) → 𝐾 ∥ (𝑀 + 𝑁))) | ||
| Theorem | dvds2sub 16258 | If an integer divides each of two other integers, it divides their difference. (Contributed by Paul Chapman, 21-Mar-2011.) |
| ⊢ ((𝐾 ∈ ℤ ∧ 𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → ((𝐾 ∥ 𝑀 ∧ 𝐾 ∥ 𝑁) → 𝐾 ∥ (𝑀 − 𝑁))) | ||
| Theorem | dvds2addd 16259 | Deduction form of dvds2add 16257. (Contributed by SN, 21-Aug-2024.) |
| ⊢ (𝜑 → 𝐾 ∈ ℤ) & ⊢ (𝜑 → 𝑀 ∈ ℤ) & ⊢ (𝜑 → 𝑁 ∈ ℤ) & ⊢ (𝜑 → 𝐾 ∥ 𝑀) & ⊢ (𝜑 → 𝐾 ∥ 𝑁) ⇒ ⊢ (𝜑 → 𝐾 ∥ (𝑀 + 𝑁)) | ||
| Theorem | dvds2subd 16260 | Deduction form of dvds2sub 16258. (Contributed by Stanislas Polu, 9-Mar-2020.) |
| ⊢ (𝜑 → 𝐾 ∈ ℤ) & ⊢ (𝜑 → 𝑀 ∈ ℤ) & ⊢ (𝜑 → 𝑁 ∈ ℤ) & ⊢ (𝜑 → 𝐾 ∥ 𝑀) & ⊢ (𝜑 → 𝐾 ∥ 𝑁) ⇒ ⊢ (𝜑 → 𝐾 ∥ (𝑀 − 𝑁)) | ||
| Theorem | dvdstr 16261 | The divides relation is transitive. Theorem 1.1(b) in [ApostolNT] p. 14 (transitive property of the divides relation). (Contributed by Paul Chapman, 21-Mar-2011.) |
| ⊢ ((𝐾 ∈ ℤ ∧ 𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → ((𝐾 ∥ 𝑀 ∧ 𝑀 ∥ 𝑁) → 𝐾 ∥ 𝑁)) | ||
| Theorem | dvdstrd 16262 | The divides relation is transitive, a deduction version of dvdstr 16261. (Contributed by metakunt, 12-May-2024.) |
| ⊢ (𝜑 → 𝐾 ∈ ℤ) & ⊢ (𝜑 → 𝑀 ∈ ℤ) & ⊢ (𝜑 → 𝑁 ∈ ℤ) & ⊢ (𝜑 → 𝐾 ∥ 𝑀) & ⊢ (𝜑 → 𝑀 ∥ 𝑁) ⇒ ⊢ (𝜑 → 𝐾 ∥ 𝑁) | ||
| Theorem | dvdsmultr1 16263 | If an integer divides another, it divides a multiple of it. (Contributed by Paul Chapman, 17-Nov-2012.) |
| ⊢ ((𝐾 ∈ ℤ ∧ 𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → (𝐾 ∥ 𝑀 → 𝐾 ∥ (𝑀 · 𝑁))) | ||
| Theorem | dvdsmultr1d 16264 | Deduction form of dvdsmultr1 16263. (Contributed by Stanislas Polu, 9-Mar-2020.) |
| ⊢ (𝜑 → 𝐾 ∈ ℤ) & ⊢ (𝜑 → 𝑀 ∈ ℤ) & ⊢ (𝜑 → 𝑁 ∈ ℤ) & ⊢ (𝜑 → 𝐾 ∥ 𝑀) ⇒ ⊢ (𝜑 → 𝐾 ∥ (𝑀 · 𝑁)) | ||
| Theorem | dvdsmultr2 16265 | If an integer divides another, it divides a multiple of it. (Contributed by Paul Chapman, 17-Nov-2012.) |
| ⊢ ((𝐾 ∈ ℤ ∧ 𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → (𝐾 ∥ 𝑁 → 𝐾 ∥ (𝑀 · 𝑁))) | ||
| Theorem | dvdsmultr2d 16266 | Deduction form of dvdsmultr2 16265. (Contributed by SN, 23-Aug-2024.) |
| ⊢ (𝜑 → 𝐾 ∈ ℤ) & ⊢ (𝜑 → 𝑀 ∈ ℤ) & ⊢ (𝜑 → 𝑁 ∈ ℤ) & ⊢ (𝜑 → 𝐾 ∥ 𝑁) ⇒ ⊢ (𝜑 → 𝐾 ∥ (𝑀 · 𝑁)) | ||
| Theorem | ordvdsmul 16267 | If an integer divides either of two others, it divides their product. (Contributed by Paul Chapman, 17-Nov-2012.) (Proof shortened by Mario Carneiro, 17-Jul-2014.) |
| ⊢ ((𝐾 ∈ ℤ ∧ 𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → ((𝐾 ∥ 𝑀 ∨ 𝐾 ∥ 𝑁) → 𝐾 ∥ (𝑀 · 𝑁))) | ||
| Theorem | dvdssub2 16268 | If an integer divides a difference, then it divides one term iff it divides the other. (Contributed by Mario Carneiro, 13-Jul-2014.) |
| ⊢ (((𝐾 ∈ ℤ ∧ 𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) ∧ 𝐾 ∥ (𝑀 − 𝑁)) → (𝐾 ∥ 𝑀 ↔ 𝐾 ∥ 𝑁)) | ||
| Theorem | dvdsadd 16269 | An integer divides another iff it divides their sum. (Contributed by Paul Chapman, 31-Mar-2011.) (Revised by Mario Carneiro, 13-Jul-2014.) |
| ⊢ ((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → (𝑀 ∥ 𝑁 ↔ 𝑀 ∥ (𝑀 + 𝑁))) | ||
| Theorem | dvdsaddr 16270 | An integer divides another iff it divides their sum. (Contributed by Paul Chapman, 31-Mar-2011.) |
| ⊢ ((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → (𝑀 ∥ 𝑁 ↔ 𝑀 ∥ (𝑁 + 𝑀))) | ||
| Theorem | dvdssub 16271 | An integer divides another iff it divides their difference. (Contributed by Paul Chapman, 31-Mar-2011.) |
| ⊢ ((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → (𝑀 ∥ 𝑁 ↔ 𝑀 ∥ (𝑀 − 𝑁))) | ||
| Theorem | dvdssubr 16272 | 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 16273 | 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 | dvdsaddre2b 16274 | Adding a multiple of the base does not affect divisibility. Variant of dvdsadd2b 16273 only requiring 𝐵 to be a real number (not necessarily an integer). (Contributed by AV, 19-Jul-2021.) |
| ⊢ ((𝐴 ∈ ℤ ∧ 𝐵 ∈ ℝ ∧ (𝐶 ∈ ℤ ∧ 𝐴 ∥ 𝐶)) → (𝐴 ∥ 𝐵 ↔ 𝐴 ∥ (𝐶 + 𝐵))) | ||
| Theorem | fsumdvds 16275* | If every term in a sum is divisible by 𝑁, then so is the sum. (Contributed by Mario Carneiro, 17-Jan-2015.) |
| ⊢ (𝜑 → 𝐴 ∈ Fin) & ⊢ (𝜑 → 𝑁 ∈ ℤ) & ⊢ ((𝜑 ∧ 𝑘 ∈ 𝐴) → 𝐵 ∈ ℤ) & ⊢ ((𝜑 ∧ 𝑘 ∈ 𝐴) → 𝑁 ∥ 𝐵) ⇒ ⊢ (𝜑 → 𝑁 ∥ Σ𝑘 ∈ 𝐴 𝐵) | ||
| Theorem | dvdslelem 16276 | Lemma for dvdsle 16277. (Contributed by Paul Chapman, 21-Mar-2011.) |
| ⊢ 𝑀 ∈ ℤ & ⊢ 𝑁 ∈ ℕ & ⊢ 𝐾 ∈ ℤ ⇒ ⊢ (𝑁 < 𝑀 → (𝐾 · 𝑀) ≠ 𝑁) | ||
| Theorem | dvdsle 16277 | 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 16278 | 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 16279 | 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 16280 | 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 16281 | 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 16282 | 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 16283* | 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 16284* | 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 16285* | 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 16286 | The only nonnegative integer that divides 1 is 1. (Contributed by Mario Carneiro, 2-Jul-2015.) |
| ⊢ (𝑀 ∈ ℕ0 → (𝑀 ∥ 1 ↔ 𝑀 = 1)) | ||
| Theorem | alzdvds 16287* | Only 0 is divisible by all integers. (Contributed by Paul Chapman, 21-Mar-2011.) |
| ⊢ (𝑁 ∈ ℤ → (∀𝑥 ∈ ℤ 𝑥 ∥ 𝑁 ↔ 𝑁 = 0)) | ||
| Theorem | dvdsext 16288* | 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 16289 | 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 16290 | 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 16291 | 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 16292 | 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 13906 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 16293 | A positive integer divides any greater factorial. (Contributed by Paul Chapman, 28-Nov-2012.) |
| ⊢ ((𝐾 ∈ ℕ ∧ 𝑁 ∈ (ℤ≥‘𝐾)) → 𝐾 ∥ (!‘𝑁)) | ||
| Theorem | dvdsexp2im 16294 | If an integer divides another integer, then it also divides any of its powers. (Contributed by Scott Fenton, 7-Apr-2014.) (Revised by Mario Carneiro, 19-Apr-2014.) |
| ⊢ ((𝐾 ∈ ℤ ∧ 𝑀 ∈ ℤ ∧ 𝑁 ∈ ℕ) → (𝐾 ∥ 𝑀 → 𝐾 ∥ (𝑀↑𝑁))) | ||
| Theorem | dvdsexp 16295 | 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 16296 | 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 16297 | 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.) (Proof shortened by AV, 18-Mar-2022.) |
| ⊢ ((𝑁 ∈ ℕ ∧ 𝐴 ∈ ℤ ∧ 𝐵 ∈ ℤ) → (𝑁 ∥ 𝐴 → ((𝐴 · 𝐵) mod 𝑁) = 0)) | ||
| Theorem | 3dvds 16298* | A rule for divisibility by 3 of a number written in base 10. This is Metamath 100 proof #85. (Contributed by Mario Carneiro, 14-Jul-2014.) (Revised by Mario Carneiro, 17-Jan-2015.) (Revised by AV, 8-Sep-2021.) |
| ⊢ ((𝑁 ∈ ℕ0 ∧ 𝐹:(0...𝑁)⟶ℤ) → (3 ∥ Σ𝑘 ∈ (0...𝑁)((𝐹‘𝑘) · (;10↑𝑘)) ↔ 3 ∥ Σ𝑘 ∈ (0...𝑁)(𝐹‘𝑘))) | ||
| Theorem | 3dvdsdec 16299 | 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 16300 | 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)) | ||
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