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Type | Label | Description |
---|---|---|
Statement | ||
Theorem | sincos1sgn 16001 | The signs of the sine and cosine of 1. (Contributed by Paul Chapman, 19-Jan-2008.) |
⊢ (0 < (sin‘1) ∧ 0 < (cos‘1)) | ||
Theorem | sincos2sgn 16002 | The signs of the sine and cosine of 2. (Contributed by Paul Chapman, 19-Jan-2008.) |
⊢ (0 < (sin‘2) ∧ (cos‘2) < 0) | ||
Theorem | sin4lt0 16003 | The sine of 4 is negative. (Contributed by Paul Chapman, 19-Jan-2008.) |
⊢ (sin‘4) < 0 | ||
Theorem | absefi 16004 | The absolute value of the exponential of an imaginary number is one. Equation 48 of [Rudin] p. 167. (Contributed by Jason Orendorff, 9-Feb-2007.) |
⊢ (𝐴 ∈ ℝ → (abs‘(exp‘(i · 𝐴))) = 1) | ||
Theorem | absef 16005 | The absolute value of the exponential is the exponential of the real part. (Contributed by Paul Chapman, 13-Sep-2007.) |
⊢ (𝐴 ∈ ℂ → (abs‘(exp‘𝐴)) = (exp‘(ℜ‘𝐴))) | ||
Theorem | absefib 16006 | A complex number is real iff the exponential of its product with i has absolute value one. (Contributed by NM, 21-Aug-2008.) |
⊢ (𝐴 ∈ ℂ → (𝐴 ∈ ℝ ↔ (abs‘(exp‘(i · 𝐴))) = 1)) | ||
Theorem | efieq1re 16007 | A number whose imaginary exponential is one is real. (Contributed by NM, 21-Aug-2008.) |
⊢ ((𝐴 ∈ ℂ ∧ (exp‘(i · 𝐴)) = 1) → 𝐴 ∈ ℝ) | ||
Theorem | demoivre 16008 | De Moivre's Formula. Proof by induction given at http://en.wikipedia.org/wiki/De_Moivre's_formula, but restricted to nonnegative integer powers. See also demoivreALT 16009 for an alternate longer proof not using the exponential function. (Contributed by NM, 24-Jul-2007.) |
⊢ ((𝐴 ∈ ℂ ∧ 𝑁 ∈ ℤ) → (((cos‘𝐴) + (i · (sin‘𝐴)))↑𝑁) = ((cos‘(𝑁 · 𝐴)) + (i · (sin‘(𝑁 · 𝐴))))) | ||
Theorem | demoivreALT 16009 | Alternate proof of demoivre 16008. It is longer but does not use the exponential function. This is Metamath 100 proof #17. (Contributed by Steve Rodriguez, 10-Nov-2006.) (Proof modification is discouraged.) (New usage is discouraged.) |
⊢ ((𝐴 ∈ ℂ ∧ 𝑁 ∈ ℕ0) → (((cos‘𝐴) + (i · (sin‘𝐴)))↑𝑁) = ((cos‘(𝑁 · 𝐴)) + (i · (sin‘(𝑁 · 𝐴))))) | ||
Syntax | ctau 16010 | Extend class notation to include the constant tau, τ = 6.28318.... |
class τ | ||
Definition | df-tau 16011 | Define the circle constant tau, τ = 6.28318..., which is the smallest positive real number whose cosine is one. Various notations have been used or proposed for this number including τ, a three-legged variant of π, or 2π. Note the difference between this constant τ and the formula variable 𝜏. Following our convention, the constant is displayed in upright font while the variable is in italic font; furthermore, the colors are different. (Contributed by Jim Kingdon, 9-Apr-2018.) (Revised by AV, 1-Oct-2020.) |
⊢ τ = inf((ℝ+ ∩ (◡cos “ {1})), ℝ, < ) | ||
Theorem | eirrlem 16012* | Lemma for eirr 16013. (Contributed by Paul Chapman, 9-Feb-2008.) (Revised by Mario Carneiro, 29-Apr-2014.) |
⊢ 𝐹 = (𝑛 ∈ ℕ0 ↦ (1 / (!‘𝑛))) & ⊢ (𝜑 → 𝑃 ∈ ℤ) & ⊢ (𝜑 → 𝑄 ∈ ℕ) & ⊢ (𝜑 → e = (𝑃 / 𝑄)) ⇒ ⊢ ¬ 𝜑 | ||
Theorem | eirr 16013 | e is irrational. (Contributed by Paul Chapman, 9-Feb-2008.) (Proof shortened by Mario Carneiro, 29-Apr-2014.) |
⊢ e ∉ ℚ | ||
Theorem | egt2lt3 16014 | Euler's constant e = 2.71828... is strictly bounded below by 2 and above by 3. (Contributed by NM, 28-Nov-2008.) (Revised by Mario Carneiro, 29-Apr-2014.) |
⊢ (2 < e ∧ e < 3) | ||
Theorem | epos 16015 | Euler's constant e is greater than 0. (Contributed by Jeff Hankins, 22-Nov-2008.) |
⊢ 0 < e | ||
Theorem | epr 16016 | Euler's constant e is a positive real. (Contributed by Jeff Hankins, 22-Nov-2008.) |
⊢ e ∈ ℝ+ | ||
Theorem | ene0 16017 | e is not 0. (Contributed by David A. Wheeler, 17-Oct-2017.) |
⊢ e ≠ 0 | ||
Theorem | ene1 16018 | e is not 1. (Contributed by David A. Wheeler, 17-Oct-2017.) |
⊢ e ≠ 1 | ||
Theorem | xpnnen 16019 | The Cartesian product of the set of positive integers with itself is equinumerous to the set of positive integers. (Contributed by NM, 1-Aug-2004.) (Revised by Mario Carneiro, 9-Mar-2013.) |
⊢ (ℕ × ℕ) ≈ ℕ | ||
Theorem | znnen 16020 | The set of integers and the set of positive integers are equinumerous. Exercise 1 of [Gleason] p. 140. (Contributed by NM, 31-Jul-2004.) (Proof shortened by Mario Carneiro, 13-Jun-2014.) |
⊢ ℤ ≈ ℕ | ||
Theorem | qnnen 16021 | The rational numbers are countable. This proof does not use the Axiom of Choice, even though it uses an onto function, because the base set (ℤ × ℕ) is numerable. Exercise 2 of [Enderton] p. 133. For purposes of the Metamath 100 list, we are considering Mario Carneiro's revision as the date this proof was completed. This is Metamath 100 proof #3. (Contributed by NM, 31-Jul-2004.) (Revised by Mario Carneiro, 3-Mar-2013.) |
⊢ ℚ ≈ ℕ | ||
Theorem | rpnnen2lem1 16022* | Lemma for rpnnen2 16034. (Contributed by Mario Carneiro, 13-May-2013.) |
⊢ 𝐹 = (𝑥 ∈ 𝒫 ℕ ↦ (𝑛 ∈ ℕ ↦ if(𝑛 ∈ 𝑥, ((1 / 3)↑𝑛), 0))) ⇒ ⊢ ((𝐴 ⊆ ℕ ∧ 𝑁 ∈ ℕ) → ((𝐹‘𝐴)‘𝑁) = if(𝑁 ∈ 𝐴, ((1 / 3)↑𝑁), 0)) | ||
Theorem | rpnnen2lem2 16023* | Lemma for rpnnen2 16034. (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 16024* | Lemma for rpnnen2 16034. (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 16025* | Lemma for rpnnen2 16034. (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 16026* | Lemma for rpnnen2 16034. (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 16027* | Lemma for rpnnen2 16034. (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 16028* | Lemma for rpnnen2 16034. (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 16029* | Lemma for rpnnen2 16034. (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 16030* | Lemma for rpnnen2 16034. (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 16031* | Lemma for rpnnen2 16034. (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 16032* | Lemma for rpnnen2 16034. (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 16033* | Lemma for rpnnen2 16034. (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 16034 |
The other half of rpnnen 16035, 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... 15692). 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 16022). This is an infinite sum of real numbers (rpnnen2lem2 16023), and since 𝐴 ⊆ 𝐵 implies (𝐹‘𝐴) ≤ (𝐹‘𝐵) (rpnnen2lem4 16025) and (𝐹‘ℕ) converges to 1 / 2 (rpnnen2lem3 16024) by geoisum1 15690, the sum is convergent to some real (rpnnen2lem5 16026 and rpnnen2lem6 16027) by the comparison test for convergence cvgcmp 15627. The comparison test also tells us that 𝐴 ⊆ 𝐵 implies Σ(𝐹‘𝐴) ≤ Σ(𝐹‘𝐵) (rpnnen2lem7 16028). 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 16029) and cancel them (rpnnen2lem10 16031), 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 16030) and Σ(𝐹‘{𝑚}) = (3↑-𝑚), we establish that Σ(𝐹‘𝑦) < Σ(𝐹‘𝑥) (rpnnen2lem11 16032) so that they must be different. By contraposition (rpnnen2lem12 16033), 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 16035 | The cardinality of the continuum is the same as the powerset of ω. This is a stronger statement than ruc 16051, which only asserts that ℝ is uncountable, i.e. has a cardinality larger than ω. The main proof is in two parts, rpnnen1 12824 and rpnnen2 16034, each showing an injection in one direction, and this last part uses sbth 8958 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 16036 | 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 9874 directly). (Contributed by NM, 30-Jul-2004.) (Revised by Mario Carneiro, 16-Jun-2013.) |
⊢ (ℝ × ℝ) ≈ ℝ | ||
Theorem | cpnnen 16037 | The complex numbers are equinumerous to the powerset of the positive integers. (Contributed by Mario Carneiro, 16-Jun-2013.) |
⊢ ℂ ≈ 𝒫 ℕ | ||
Theorem | rucALT 16038 | Alternate proof of ruc 16051. This proof is a simple corollary of rpnnen 16035, which determines the exact cardinality of the reals. For an alternate proof discussed at mmcomplex.html#uncountable 16035, see ruc 16051. (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 16039* | Lemma for ruc 16051 (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 16040* | Lemma for ruc 16051. 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 16041* | Lemma for ruc 16051. 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 16042* | Lemma for ruc 16051. 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 16043* | Lemma for ruc 16051. Domain and codomain 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 16044* | Lemma for ruc 16051. 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 16045* | Lemma for ruc 16051. 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 16046* | Lemma for ruc 16051. 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 16047* | Lemma for ruc 16051. 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 16048* | Lemma for ruc 16051. 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 16049* | Lemma for ruc 16051. 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 16050 | Lemma for ruc 16051. There is no function that maps ℕ onto ℝ. (Use nex 1801 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 16051 | 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 16039 through ruclem13 16050 and this final piece. Our proof is based on the proof of Theorem 5.18 of [Truss] p. 114. See ruclem13 16050 for the function existence version of this theorem. For an informal discussion of this proof, see mmcomplex.html#uncountable 16050. For an alternate proof see rucALT 16038. This is Metamath 100 proof #22. (Contributed by NM, 13-Oct-2004.) |
⊢ ℕ ≺ ℝ | ||
Theorem | resdomq 16052 | 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 16053 | There are at least aleph-one real numbers. (Contributed by NM, 2-Feb-2005.) |
⊢ (ℵ‘1o) ≼ ℝ | ||
Theorem | aleph1irr 16054 | There are at least aleph-one irrationals. (Contributed by NM, 2-Feb-2005.) |
⊢ (ℵ‘1o) ≼ (ℝ ∖ ℚ) | ||
Theorem | cnso 16055 | 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 16056 | Lemma for sqrt2irr 16057. 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 16057 | The square root of 2 is irrational. See zsqrtelqelz 16559 for a generalization to all non-square integers. The proof's core is proven in sqrt2irrlem 16056, 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/ 16056. (Contributed by NM, 8-Jan-2002.) (Proof shortened by Mario Carneiro, 12-Sep-2015.) |
⊢ (√‘2) ∉ ℚ | ||
Theorem | sqrt2re 16058 | The square root of 2 exists and is a real number. (Contributed by NM, 3-Dec-2004.) |
⊢ (√‘2) ∈ ℝ | ||
Theorem | sqrt2irr0 16059 | The square root of 2 is an irrational number. (Contributed by AV, 23-Dec-2022.) |
⊢ (√‘2) ∈ (ℝ ∖ ℚ) | ||
Theorem | nthruc 16060 | 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 16061 for a further refinement. (Contributed by NM, 12-Jan-2002.) |
⊢ ((ℕ ⊊ ℤ ∧ ℤ ⊊ ℚ) ∧ (ℚ ⊊ ℝ ∧ ℝ ⊊ ℂ)) | ||
Theorem | nthruz 16061 | 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 16060. (Contributed by NM, 9-May-2004.) |
⊢ (ℕ ⊊ ℕ0 ∧ ℕ0 ⊊ ℤ) | ||
Syntax | cdvds 16062 | Extend the definition of a class to include the divides relation. See df-dvds 16063. |
class ∥ | ||
Definition | df-dvds 16063* | 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 16064* | Define the divides relation. 𝑀 ∥ 𝑁 means 𝑀 divides into 𝑁 with no remainder. For example, 3 ∥ 6 (ex-dvds 29108). As proven in dvdsval3 16066, 𝑀 ∥ 𝑁 ↔ (𝑁 mod 𝑀) = 0. See divides 16064 and dvdsval2 16065 for other equivalent expressions. (Contributed by Paul Chapman, 21-Mar-2011.) |
⊢ ((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → (𝑀 ∥ 𝑁 ↔ ∃𝑛 ∈ ℤ (𝑛 · 𝑀) = 𝑁)) | ||
Theorem | dvdsval2 16065 | 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 16066 | 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 16067 | Reverse closure for the divisibility relation. (Contributed by Stefan O'Rear, 5-Sep-2015.) |
⊢ (𝑋 ∥ 𝑌 → (𝑋 ∈ ℤ ∧ 𝑌 ∈ ℤ)) | ||
Theorem | dvdsmod0 16068 | 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 16069 | 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 16070 | 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 16582). (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 16071 | Strong form of dvdsval2 16065 for positive integers. (Contributed by Stefan O'Rear, 13-Sep-2014.) |
⊢ ((𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ) → (𝐵 ∥ 𝐴 ↔ (𝐴 / 𝐵) ∈ ℕ)) | ||
Theorem | nndivides 16072* | 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 16073 | 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 16074 | 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 | dvds0lem 16075 | 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 16076* | 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 16077* | 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 16078 | 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 16079 | 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 16080 | 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 16081 | 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 16082 | 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 16083 | 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 16084 | 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 16085 | 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 16086 | An integer divides a multiple of itself. (Contributed by Paul Chapman, 21-Mar-2011.) |
⊢ ((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → 𝑀 ∥ (𝑀 · 𝑁)) | ||
Theorem | dvdsmul2 16087 | An integer divides a multiple of itself. (Contributed by Paul Chapman, 21-Mar-2011.) |
⊢ ((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → 𝑁 ∥ (𝑀 · 𝑁)) | ||
Theorem | iddvdsexp 16088 | 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 16089 | 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 16090 | 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 16091 | 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 16092 | 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 16093 | 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 16094 | 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 16095 | 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 16096 | 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 16097 | 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 16098 | 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 16099 | 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 16100 | Deduction form of dvds2add 16098. (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)) |
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