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Type | Label | Description |
---|---|---|
Statement | ||
Theorem | rpnnen2lem2 15401* | Lemma for rpnnen2 15412. (Contributed by Mario Carneiro, 13-May-2013.) (Revised by Mario Carneiro, 23-Aug-2014.) |
⊢ 𝐹 = (𝑥 ∈ 𝒫 ℕ ↦ (𝑛 ∈ ℕ ↦ if(𝑛 ∈ 𝑥, ((1 / 3)↑𝑛), 0))) ⇒ ⊢ (𝐴 ⊆ ℕ → (𝐹‘𝐴):ℕ⟶ℝ) | ||
Theorem | rpnnen2lem3 15402* | Lemma for rpnnen2 15412. (Contributed by Mario Carneiro, 13-May-2013.) |
⊢ 𝐹 = (𝑥 ∈ 𝒫 ℕ ↦ (𝑛 ∈ ℕ ↦ if(𝑛 ∈ 𝑥, ((1 / 3)↑𝑛), 0))) ⇒ ⊢ seq1( + , (𝐹‘ℕ)) ⇝ (1 / 2) | ||
Theorem | rpnnen2lem4 15403* | Lemma for rpnnen2 15412. (Contributed by Mario Carneiro, 13-May-2013.) (Revised by Mario Carneiro, 31-Aug-2014.) |
⊢ 𝐹 = (𝑥 ∈ 𝒫 ℕ ↦ (𝑛 ∈ ℕ ↦ if(𝑛 ∈ 𝑥, ((1 / 3)↑𝑛), 0))) ⇒ ⊢ ((𝐴 ⊆ 𝐵 ∧ 𝐵 ⊆ ℕ ∧ 𝑘 ∈ ℕ) → (0 ≤ ((𝐹‘𝐴)‘𝑘) ∧ ((𝐹‘𝐴)‘𝑘) ≤ ((𝐹‘𝐵)‘𝑘))) | ||
Theorem | rpnnen2lem5 15404* | Lemma for rpnnen2 15412. (Contributed by Mario Carneiro, 13-May-2013.) (Revised by Mario Carneiro, 30-Apr-2014.) |
⊢ 𝐹 = (𝑥 ∈ 𝒫 ℕ ↦ (𝑛 ∈ ℕ ↦ if(𝑛 ∈ 𝑥, ((1 / 3)↑𝑛), 0))) ⇒ ⊢ ((𝐴 ⊆ ℕ ∧ 𝑀 ∈ ℕ) → seq𝑀( + , (𝐹‘𝐴)) ∈ dom ⇝ ) | ||
Theorem | rpnnen2lem6 15405* | Lemma for rpnnen2 15412. (Contributed by Mario Carneiro, 13-May-2013.) (Revised by Mario Carneiro, 30-Apr-2014.) |
⊢ 𝐹 = (𝑥 ∈ 𝒫 ℕ ↦ (𝑛 ∈ ℕ ↦ if(𝑛 ∈ 𝑥, ((1 / 3)↑𝑛), 0))) ⇒ ⊢ ((𝐴 ⊆ ℕ ∧ 𝑀 ∈ ℕ) → Σ𝑘 ∈ (ℤ≥‘𝑀)((𝐹‘𝐴)‘𝑘) ∈ ℝ) | ||
Theorem | rpnnen2lem7 15406* | Lemma for rpnnen2 15412. (Contributed by Mario Carneiro, 13-May-2013.) (Revised by Mario Carneiro, 30-Apr-2014.) |
⊢ 𝐹 = (𝑥 ∈ 𝒫 ℕ ↦ (𝑛 ∈ ℕ ↦ if(𝑛 ∈ 𝑥, ((1 / 3)↑𝑛), 0))) ⇒ ⊢ ((𝐴 ⊆ 𝐵 ∧ 𝐵 ⊆ ℕ ∧ 𝑀 ∈ ℕ) → Σ𝑘 ∈ (ℤ≥‘𝑀)((𝐹‘𝐴)‘𝑘) ≤ Σ𝑘 ∈ (ℤ≥‘𝑀)((𝐹‘𝐵)‘𝑘)) | ||
Theorem | rpnnen2lem8 15407* | Lemma for rpnnen2 15412. (Contributed by Mario Carneiro, 13-May-2013.) (Revised by Mario Carneiro, 30-Apr-2014.) |
⊢ 𝐹 = (𝑥 ∈ 𝒫 ℕ ↦ (𝑛 ∈ ℕ ↦ if(𝑛 ∈ 𝑥, ((1 / 3)↑𝑛), 0))) ⇒ ⊢ ((𝐴 ⊆ ℕ ∧ 𝑀 ∈ ℕ) → Σ𝑘 ∈ ℕ ((𝐹‘𝐴)‘𝑘) = (Σ𝑘 ∈ (1...(𝑀 − 1))((𝐹‘𝐴)‘𝑘) + Σ𝑘 ∈ (ℤ≥‘𝑀)((𝐹‘𝐴)‘𝑘))) | ||
Theorem | rpnnen2lem9 15408* | Lemma for rpnnen2 15412. (Contributed by Mario Carneiro, 13-May-2013.) (Revised by Mario Carneiro, 30-Apr-2014.) |
⊢ 𝐹 = (𝑥 ∈ 𝒫 ℕ ↦ (𝑛 ∈ ℕ ↦ if(𝑛 ∈ 𝑥, ((1 / 3)↑𝑛), 0))) ⇒ ⊢ (𝑀 ∈ ℕ → Σ𝑘 ∈ (ℤ≥‘𝑀)((𝐹‘(ℕ ∖ {𝑀}))‘𝑘) = (0 + (((1 / 3)↑(𝑀 + 1)) / (1 − (1 / 3))))) | ||
Theorem | rpnnen2lem10 15409* | Lemma for rpnnen2 15412. (Contributed by Mario Carneiro, 13-May-2013.) (Revised by Mario Carneiro, 30-Apr-2014.) |
⊢ 𝐹 = (𝑥 ∈ 𝒫 ℕ ↦ (𝑛 ∈ ℕ ↦ if(𝑛 ∈ 𝑥, ((1 / 3)↑𝑛), 0))) & ⊢ (𝜑 → 𝐴 ⊆ ℕ) & ⊢ (𝜑 → 𝐵 ⊆ ℕ) & ⊢ (𝜑 → 𝑚 ∈ (𝐴 ∖ 𝐵)) & ⊢ (𝜑 → ∀𝑛 ∈ ℕ (𝑛 < 𝑚 → (𝑛 ∈ 𝐴 ↔ 𝑛 ∈ 𝐵))) & ⊢ (𝜓 ↔ Σ𝑘 ∈ ℕ ((𝐹‘𝐴)‘𝑘) = Σ𝑘 ∈ ℕ ((𝐹‘𝐵)‘𝑘)) ⇒ ⊢ ((𝜑 ∧ 𝜓) → Σ𝑘 ∈ (ℤ≥‘𝑚)((𝐹‘𝐴)‘𝑘) = Σ𝑘 ∈ (ℤ≥‘𝑚)((𝐹‘𝐵)‘𝑘)) | ||
Theorem | rpnnen2lem11 15410* | Lemma for rpnnen2 15412. (Contributed by Mario Carneiro, 13-May-2013.) |
⊢ 𝐹 = (𝑥 ∈ 𝒫 ℕ ↦ (𝑛 ∈ ℕ ↦ if(𝑛 ∈ 𝑥, ((1 / 3)↑𝑛), 0))) & ⊢ (𝜑 → 𝐴 ⊆ ℕ) & ⊢ (𝜑 → 𝐵 ⊆ ℕ) & ⊢ (𝜑 → 𝑚 ∈ (𝐴 ∖ 𝐵)) & ⊢ (𝜑 → ∀𝑛 ∈ ℕ (𝑛 < 𝑚 → (𝑛 ∈ 𝐴 ↔ 𝑛 ∈ 𝐵))) & ⊢ (𝜓 ↔ Σ𝑘 ∈ ℕ ((𝐹‘𝐴)‘𝑘) = Σ𝑘 ∈ ℕ ((𝐹‘𝐵)‘𝑘)) ⇒ ⊢ (𝜑 → ¬ 𝜓) | ||
Theorem | rpnnen2lem12 15411* | Lemma for rpnnen2 15412. (Contributed by Mario Carneiro, 13-May-2013.) |
⊢ 𝐹 = (𝑥 ∈ 𝒫 ℕ ↦ (𝑛 ∈ ℕ ↦ if(𝑛 ∈ 𝑥, ((1 / 3)↑𝑛), 0))) ⇒ ⊢ 𝒫 ℕ ≼ (0[,]1) | ||
Theorem | rpnnen2 15412 |
The other half of rpnnen 15413, where we show an injection from sets of
positive integers to real numbers. The obvious choice for this is
binary expansion, but it has the unfortunate property that it does not
produce an injection on numbers which end with all 0's or all 1's (the
more well-known decimal version of this is 0.999... 15070). Instead, we
opt for a ternary expansion, which produces (a scaled version of) the
Cantor set. Since the Cantor set is riddled with gaps, we can show that
any two sequences that are not equal must differ somewhere, and when
they do, they are placed a finite distance apart, thus ensuring that the
map is injective.
Our map assigns to each subset 𝐴 of the positive integers the number Σ𝑘 ∈ 𝐴(3↑-𝑘) = Σ𝑘 ∈ ℕ((𝐹‘𝐴)‘𝑘), where ((𝐹‘𝐴)‘𝑘) = if(𝑘 ∈ 𝐴, (3↑-𝑘), 0)) (rpnnen2lem1 15400). This is an infinite sum of real numbers (rpnnen2lem2 15401), and since 𝐴 ⊆ 𝐵 implies (𝐹‘𝐴) ≤ (𝐹‘𝐵) (rpnnen2lem4 15403) and (𝐹‘ℕ) converges to 1 / 2 (rpnnen2lem3 15402) by geoisum1 15068, the sum is convergent to some real (rpnnen2lem5 15404 and rpnnen2lem6 15405) by the comparison test for convergence cvgcmp 15004. The comparison test also tells us that 𝐴 ⊆ 𝐵 implies Σ(𝐹‘𝐴) ≤ Σ(𝐹‘𝐵) (rpnnen2lem7 15406). Putting it all together, if we have two sets 𝑥 ≠ 𝑦, there must differ somewhere, and so there must be an 𝑚 such that ∀𝑛 < 𝑚(𝑛 ∈ 𝑥 ↔ 𝑛 ∈ 𝑦) but 𝑚 ∈ (𝑥 ∖ 𝑦) or vice versa. In this case, we split off the first 𝑚 − 1 terms (rpnnen2lem8 15407) and cancel them (rpnnen2lem10 15409), since these are the same for both sets. For the remaining terms, we use the subset property to establish that Σ(𝐹‘𝑦) ≤ Σ(𝐹‘(ℕ ∖ {𝑚})) and Σ(𝐹‘{𝑚}) ≤ Σ(𝐹‘𝑥) (where these sums are only over (ℤ≥‘𝑚)), and since Σ(𝐹‘(ℕ ∖ {𝑚})) = (3↑-𝑚) / 2 (rpnnen2lem9 15408) and Σ(𝐹‘{𝑚}) = (3↑-𝑚), we establish that Σ(𝐹‘𝑦) < Σ(𝐹‘𝑥) (rpnnen2lem11 15410) so that they must be different. By contraposition (rpnnen2lem12 15411), we find that this map is an injection. (Contributed by Mario Carneiro, 13-May-2013.) (Proof shortened by Mario Carneiro, 30-Apr-2014.) (Revised by NM, 17-Aug-2021.) |
⊢ 𝒫 ℕ ≼ (0[,]1) | ||
Theorem | rpnnen 15413 | The cardinality of the continuum is the same as the powerset of ω. This is a stronger statement than ruc 15429, which only asserts that ℝ is uncountable, i.e. has a cardinality larger than ω. The main proof is in two parts, rpnnen1 12232 and rpnnen2 15412, each showing an injection in one direction, and this last part uses sbth 8484 to prove that the sets are equinumerous. By constructing explicit injections, we avoid the use of AC. (Contributed by Mario Carneiro, 13-May-2013.) (Revised by Mario Carneiro, 23-Aug-2014.) |
⊢ ℝ ≈ 𝒫 ℕ | ||
Theorem | rexpen 15414 | The real numbers are equinumerous to their own Cartesian product, even though it is not necessarily true that ℝ is well-orderable (so we cannot use infxpidm2 9289 directly). (Contributed by NM, 30-Jul-2004.) (Revised by Mario Carneiro, 16-Jun-2013.) |
⊢ (ℝ × ℝ) ≈ ℝ | ||
Theorem | cpnnen 15415 | The complex numbers are equinumerous to the powerset of the positive integers. (Contributed by Mario Carneiro, 16-Jun-2013.) |
⊢ ℂ ≈ 𝒫 ℕ | ||
Theorem | rucALT 15416 | Alternate proof of ruc 15429. This proof is a simple corollary of rpnnen 15413, which determines the exact cardinality of the reals. For an alternate proof discussed at mmcomplex.html#uncountable 15413, see ruc 15429. (Contributed by NM, 13-Oct-2004.) (Revised by Mario Carneiro, 13-May-2013.) (Proof modification is discouraged.) (New usage is discouraged.) |
⊢ ℕ ≺ ℝ | ||
Theorem | ruclem1 15417* | Lemma for ruc 15429 (the reals are uncountable). Substitutions for the function 𝐷. (Contributed by Mario Carneiro, 28-May-2014.) (Revised by Fan Zheng, 6-Jun-2016.) |
⊢ (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)〉))) & ⊢ (𝜑 → 𝐴 ∈ ℝ) & ⊢ (𝜑 → 𝐵 ∈ ℝ) & ⊢ (𝜑 → 𝑀 ∈ ℝ) & ⊢ 𝑋 = (1st ‘(〈𝐴, 𝐵〉𝐷𝑀)) & ⊢ 𝑌 = (2nd ‘(〈𝐴, 𝐵〉𝐷𝑀)) ⇒ ⊢ (𝜑 → ((〈𝐴, 𝐵〉𝐷𝑀) ∈ (ℝ × ℝ) ∧ 𝑋 = if(((𝐴 + 𝐵) / 2) < 𝑀, 𝐴, ((((𝐴 + 𝐵) / 2) + 𝐵) / 2)) ∧ 𝑌 = if(((𝐴 + 𝐵) / 2) < 𝑀, ((𝐴 + 𝐵) / 2), 𝐵))) | ||
Theorem | ruclem2 15418* | Lemma for ruc 15429. Ordering property for the input to 𝐷. (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)〉))) & ⊢ (𝜑 → 𝐴 ∈ ℝ) & ⊢ (𝜑 → 𝐵 ∈ ℝ) & ⊢ (𝜑 → 𝑀 ∈ ℝ) & ⊢ 𝑋 = (1st ‘(〈𝐴, 𝐵〉𝐷𝑀)) & ⊢ 𝑌 = (2nd ‘(〈𝐴, 𝐵〉𝐷𝑀)) & ⊢ (𝜑 → 𝐴 < 𝐵) ⇒ ⊢ (𝜑 → (𝐴 ≤ 𝑋 ∧ 𝑋 < 𝑌 ∧ 𝑌 ≤ 𝐵)) | ||
Theorem | ruclem3 15419* | Lemma for ruc 15429. The constructed interval [𝑋, 𝑌] always excludes 𝑀. (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)〉))) & ⊢ (𝜑 → 𝐴 ∈ ℝ) & ⊢ (𝜑 → 𝐵 ∈ ℝ) & ⊢ (𝜑 → 𝑀 ∈ ℝ) & ⊢ 𝑋 = (1st ‘(〈𝐴, 𝐵〉𝐷𝑀)) & ⊢ 𝑌 = (2nd ‘(〈𝐴, 𝐵〉𝐷𝑀)) & ⊢ (𝜑 → 𝐴 < 𝐵) ⇒ ⊢ (𝜑 → (𝑀 < 𝑋 ∨ 𝑌 < 𝑀)) | ||
Theorem | ruclem4 15420* | Lemma for ruc 15429. Initial value of 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 → (𝐺‘0) = 〈0, 1〉) | ||
Theorem | ruclem6 15421* | Lemma for ruc 15429. Domain and range of 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 → 𝐺:ℕ0⟶(ℝ × ℝ)) | ||
Theorem | ruclem7 15422* | Lemma for ruc 15429. 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 15423* | Lemma for ruc 15429. 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 15424* | Lemma for ruc 15429. 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 15425* | Lemma for ruc 15429. 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 15426* | Lemma for ruc 15429. 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 15427* | Lemma for ruc 15429. 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 15428 | Lemma for ruc 15429. There is no function that maps ℕ onto ℝ. (Use nex 1782 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 15429 | 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 15417 through ruclem13 15428 and this final piece. Our proof is based on the proof of Theorem 5.18 of [Truss] p. 114. See ruclem13 15428 for the function existence version of this theorem. For an informal discussion of this proof, see mmcomplex.html#uncountable 15428. For an alternate proof see rucALT 15416. This is Metamath 100 proof #22. (Contributed by NM, 13-Oct-2004.) |
⊢ ℕ ≺ ℝ | ||
Theorem | resdomq 15430 | 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 15431 | There are at least aleph-one real numbers. (Contributed by NM, 2-Feb-2005.) |
⊢ (ℵ‘1o) ≼ ℝ | ||
Theorem | aleph1irr 15432 | There are at least aleph-one irrationals. (Contributed by NM, 2-Feb-2005.) |
⊢ (ℵ‘1o) ≼ (ℝ ∖ ℚ) | ||
Theorem | cnso 15433 | 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 15434 | Lemma for sqrt2irr 15435. 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 15435 | The square root of 2 is irrational. See zsqrtelqelz 15927 for a generalization to all non-square integers. The proof's core is proven in sqrt2irrlem 15434, 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/ 15434. (Contributed by NM, 8-Jan-2002.) (Proof shortened by Mario Carneiro, 12-Sep-2015.) |
⊢ (√‘2) ∉ ℚ | ||
Theorem | sqrt2re 15436 | The square root of 2 exists and is a real number. (Contributed by NM, 3-Dec-2004.) |
⊢ (√‘2) ∈ ℝ | ||
Theorem | sqrt2irr0 15437 | The square root of 2 is an irrational number. (Contributed by AV, 23-Dec-2022.) |
⊢ (√‘2) ∈ (ℝ ∖ ℚ) | ||
Theorem | nthruc 15438 | 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 15439 for a further refinement. (Contributed by NM, 12-Jan-2002.) |
⊢ ((ℕ ⊊ ℤ ∧ ℤ ⊊ ℚ) ∧ (ℚ ⊊ ℝ ∧ ℝ ⊊ ℂ)) | ||
Theorem | nthruz 15439 | 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 15438. (Contributed by NM, 9-May-2004.) |
⊢ (ℕ ⊊ ℕ0 ∧ ℕ0 ⊊ ℤ) | ||
Syntax | cdvds 15440 | Extend the definition of a class to include the divides relation. See df-dvds 15441. |
class ∥ | ||
Definition | df-dvds 15441* | 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 15442* | Define the divides relation. 𝑀 ∥ 𝑁 means 𝑀 divides into 𝑁 with no remainder. For example, 3 ∥ 6 (ex-dvds 27927). As proven in dvdsval3 15444, 𝑀 ∥ 𝑁 ↔ (𝑁 mod 𝑀) = 0. See divides 15442 and dvdsval2 15443 for other equivalent expressions. (Contributed by Paul Chapman, 21-Mar-2011.) |
⊢ ((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → (𝑀 ∥ 𝑁 ↔ ∃𝑛 ∈ ℤ (𝑛 · 𝑀) = 𝑁)) | ||
Theorem | dvdsval2 15443 | 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 15444 | 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 15445 | Reverse closure for the divisibility relation. (Contributed by Stefan O'Rear, 5-Sep-2015.) |
⊢ (𝑋 ∥ 𝑌 → (𝑋 ∈ ℤ ∧ 𝑌 ∈ ℤ)) | ||
Theorem | dvdsmod0 15446 | 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 15447 | 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 15448 | 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 15950). (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 15449 | Strong form of dvdsval2 15443 for positive integers. (Contributed by Stefan O'Rear, 13-Sep-2014.) |
⊢ ((𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ) → (𝐵 ∥ 𝐴 ↔ (𝐴 / 𝐵) ∈ ℕ)) | ||
Theorem | nndivides 15450* | 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 15451 | 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 15452 | A number greater than 1 divides an integer minus 1 iff the integer modulo the number is 1. (Contributed by AV, 30-May-2023.) |
⊢ ((𝑁 ∈ (ℤ≥‘2) ∧ 𝐴 ∈ ℤ) → ((𝐴 mod 𝑁) = 1 ↔ 𝑁 ∥ (𝐴 − 1))) | ||
Theorem | dvds0lem 15453 | 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 15454* | 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 15455* | 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 15456 | 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 15457 | 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 15458 | 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 15459 | 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 15460 | 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 15461 | 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 15462 | 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 15463 | 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 15464 | An integer divides a multiple of itself. (Contributed by Paul Chapman, 21-Mar-2011.) |
⊢ ((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → 𝑀 ∥ (𝑀 · 𝑁)) | ||
Theorem | dvdsmul2 15465 | An integer divides a multiple of itself. (Contributed by Paul Chapman, 21-Mar-2011.) |
⊢ ((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → 𝑁 ∥ (𝑀 · 𝑁)) | ||
Theorem | iddvdsexp 15466 | 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 15467 | 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 15468 | 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 15469 | 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 15470 | 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 15471 | 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 15472 | 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 15473 | 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 | modmulconst 15474 | 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 15475 | 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 15476 | 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 15477 | 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 | dvds2subd 15478 | Natural deduction form of dvds2sub 15477. (Contributed by Stanislas Polu, 9-Mar-2020.) |
⊢ (𝜑 → 𝐾 ∈ ℤ) & ⊢ (𝜑 → 𝐾 ∥ 𝑀) & ⊢ (𝜑 → 𝐾 ∥ 𝑁) & ⊢ (𝜑 → 𝑀 ∈ ℤ) & ⊢ (𝜑 → 𝑁 ∈ ℤ) ⇒ ⊢ (𝜑 → 𝐾 ∥ (𝑀 − 𝑁)) | ||
Theorem | dvdstr 15479 | 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 | dvdsmultr1 15480 | 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 15481 | Natural deduction form of dvdsmultr1 15480. (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 15482 | 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 | ordvdsmul 15483 | 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 15484 | 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 15485 | 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 15486 | 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 15487 | 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 15488 | 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 15489 | 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 15490 | Adding a multiple of the base does not affect divisibility. Variant of dvdsadd2b 15489 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 15491* | 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 15492 | Lemma for dvdsle 15493. (Contributed by Paul Chapman, 21-Mar-2011.) |
⊢ 𝑀 ∈ ℤ & ⊢ 𝑁 ∈ ℕ & ⊢ 𝐾 ∈ ℤ ⇒ ⊢ (𝑁 < 𝑀 → (𝐾 · 𝑀) ≠ 𝑁) | ||
Theorem | dvdsle 15493 | 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 15494 | 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 15495 | 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 15496 | 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 15497 | 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 15498 | 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 15499* | 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 15500* | 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) |
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