P. 155 of Theorem List - Metamath Proof Explorer

Theorem List for Metamath Proof Explorer - 15401-15500   *Has distinct variable group(s)

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
PART 6  ELEMENTARY NUMBER THEORY Here we introduce elementary number theory, in particular the elementary properties of divisibility and elementary prime number theory.
6.1  Elementary properties of divisibility
6.1.1  Irrationality of square root of 2
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)
6.1.2  Some Number sets are chains of proper subsets
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)
6.1.3  The divides relation
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)