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Type | Label | Description |
---|---|---|
Statement | ||
Theorem | cosbnd2 15901 | The cosine of a real number is in the closed interval from -1 to 1. (Contributed by Mario Carneiro, 12-May-2014.) |
⊢ (𝐴 ∈ ℝ → (cos‘𝐴) ∈ (-1[,]1)) | ||
Theorem | ef01bndlem 15902* | Lemma for sin01bnd 15903 and cos01bnd 15904. (Contributed by Paul Chapman, 19-Jan-2008.) |
⊢ 𝐹 = (𝑛 ∈ ℕ0 ↦ (((i · 𝐴)↑𝑛) / (!‘𝑛))) ⇒ ⊢ (𝐴 ∈ (0(,]1) → (abs‘Σ𝑘 ∈ (ℤ≥‘4)(𝐹‘𝑘)) < ((𝐴↑4) / 6)) | ||
Theorem | sin01bnd 15903 | Bounds on the sine of a positive real number less than or equal to 1. (Contributed by Paul Chapman, 19-Jan-2008.) (Revised by Mario Carneiro, 30-Apr-2014.) |
⊢ (𝐴 ∈ (0(,]1) → ((𝐴 − ((𝐴↑3) / 3)) < (sin‘𝐴) ∧ (sin‘𝐴) < 𝐴)) | ||
Theorem | cos01bnd 15904 | Bounds on the cosine of a positive real number less than or equal to 1. (Contributed by Paul Chapman, 19-Jan-2008.) (Revised by Mario Carneiro, 30-Apr-2014.) |
⊢ (𝐴 ∈ (0(,]1) → ((1 − (2 · ((𝐴↑2) / 3))) < (cos‘𝐴) ∧ (cos‘𝐴) < (1 − ((𝐴↑2) / 3)))) | ||
Theorem | cos1bnd 15905 | Bounds on the cosine of 1. (Contributed by Paul Chapman, 19-Jan-2008.) |
⊢ ((1 / 3) < (cos‘1) ∧ (cos‘1) < (2 / 3)) | ||
Theorem | cos2bnd 15906 | Bounds on the cosine of 2. (Contributed by Paul Chapman, 19-Jan-2008.) |
⊢ (-(7 / 9) < (cos‘2) ∧ (cos‘2) < -(1 / 9)) | ||
Theorem | sinltx 15907 | The sine of a positive real number is less than its argument. (Contributed by Mario Carneiro, 29-Jul-2014.) |
⊢ (𝐴 ∈ ℝ+ → (sin‘𝐴) < 𝐴) | ||
Theorem | sin01gt0 15908 | The sine of a positive real number less than or equal to 1 is positive. (Contributed by Paul Chapman, 19-Jan-2008.) (Revised by Wolf Lammen, 25-Sep-2020.) |
⊢ (𝐴 ∈ (0(,]1) → 0 < (sin‘𝐴)) | ||
Theorem | cos01gt0 15909 | The cosine of a positive real number less than or equal to 1 is positive. (Contributed by Paul Chapman, 19-Jan-2008.) |
⊢ (𝐴 ∈ (0(,]1) → 0 < (cos‘𝐴)) | ||
Theorem | sin02gt0 15910 | The sine of a positive real number less than or equal to 2 is positive. (Contributed by Paul Chapman, 19-Jan-2008.) |
⊢ (𝐴 ∈ (0(,]2) → 0 < (sin‘𝐴)) | ||
Theorem | sincos1sgn 15911 | The signs of the sine and cosine of 1. (Contributed by Paul Chapman, 19-Jan-2008.) |
⊢ (0 < (sin‘1) ∧ 0 < (cos‘1)) | ||
Theorem | sincos2sgn 15912 | The signs of the sine and cosine of 2. (Contributed by Paul Chapman, 19-Jan-2008.) |
⊢ (0 < (sin‘2) ∧ (cos‘2) < 0) | ||
Theorem | sin4lt0 15913 | The sine of 4 is negative. (Contributed by Paul Chapman, 19-Jan-2008.) |
⊢ (sin‘4) < 0 | ||
Theorem | absefi 15914 | 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 15915 | 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 15916 | 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 15917 | 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 15918 | 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 15919 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 15919 | Alternate proof of demoivre 15918. 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 15920 | Extend class notation to include the constant tau, τ = 6.28318.... |
class τ | ||
Definition | df-tau 15921 | 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 15922* | Lemma for eirr 15923. (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 15923 | e is irrational. (Contributed by Paul Chapman, 9-Feb-2008.) (Proof shortened by Mario Carneiro, 29-Apr-2014.) |
⊢ e ∉ ℚ | ||
Theorem | egt2lt3 15924 | 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 15925 | Euler's constant e is greater than 0. (Contributed by Jeff Hankins, 22-Nov-2008.) |
⊢ 0 < e | ||
Theorem | epr 15926 | Euler's constant e is a positive real. (Contributed by Jeff Hankins, 22-Nov-2008.) |
⊢ e ∈ ℝ+ | ||
Theorem | ene0 15927 | e is not 0. (Contributed by David A. Wheeler, 17-Oct-2017.) |
⊢ e ≠ 0 | ||
Theorem | ene1 15928 | e is not 1. (Contributed by David A. Wheeler, 17-Oct-2017.) |
⊢ e ≠ 1 | ||
Theorem | xpnnen 15929 | 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 15930 | 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 15931 | 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 15932* | Lemma for rpnnen2 15944. (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 15933* | Lemma for rpnnen2 15944. (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 15934* | Lemma for rpnnen2 15944. (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 15935* | Lemma for rpnnen2 15944. (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 15936* | Lemma for rpnnen2 15944. (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 15937* | Lemma for rpnnen2 15944. (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 15938* | Lemma for rpnnen2 15944. (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 15939* | Lemma for rpnnen2 15944. (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 15940* | Lemma for rpnnen2 15944. (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 15941* | Lemma for rpnnen2 15944. (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 15942* | Lemma for rpnnen2 15944. (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 15943* | Lemma for rpnnen2 15944. (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 15944 |
The other half of rpnnen 15945, 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... 15602). 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 15932). This is an infinite sum of real numbers (rpnnen2lem2 15933), and since 𝐴 ⊆ 𝐵 implies (𝐹‘𝐴) ≤ (𝐹‘𝐵) (rpnnen2lem4 15935) and (𝐹‘ℕ) converges to 1 / 2 (rpnnen2lem3 15934) by geoisum1 15600, the sum is convergent to some real (rpnnen2lem5 15936 and rpnnen2lem6 15937) by the comparison test for convergence cvgcmp 15537. The comparison test also tells us that 𝐴 ⊆ 𝐵 implies Σ(𝐹‘𝐴) ≤ Σ(𝐹‘𝐵) (rpnnen2lem7 15938). 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 15939) and cancel them (rpnnen2lem10 15941), 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 15940) and Σ(𝐹‘{𝑚}) = (3↑-𝑚), we establish that Σ(𝐹‘𝑦) < Σ(𝐹‘𝑥) (rpnnen2lem11 15942) so that they must be different. By contraposition (rpnnen2lem12 15943), 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 15945 | The cardinality of the continuum is the same as the powerset of ω. This is a stronger statement than ruc 15961, which only asserts that ℝ is uncountable, i.e. has a cardinality larger than ω. The main proof is in two parts, rpnnen1 12732 and rpnnen2 15944, each showing an injection in one direction, and this last part uses sbth 8889 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 15946 | 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 9782 directly). (Contributed by NM, 30-Jul-2004.) (Revised by Mario Carneiro, 16-Jun-2013.) |
⊢ (ℝ × ℝ) ≈ ℝ | ||
Theorem | cpnnen 15947 | The complex numbers are equinumerous to the powerset of the positive integers. (Contributed by Mario Carneiro, 16-Jun-2013.) |
⊢ ℂ ≈ 𝒫 ℕ | ||
Theorem | rucALT 15948 | Alternate proof of ruc 15961. This proof is a simple corollary of rpnnen 15945, which determines the exact cardinality of the reals. For an alternate proof discussed at mmcomplex.html#uncountable 15945, see ruc 15961. (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 15949* | Lemma for ruc 15961 (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 15950* | Lemma for ruc 15961. 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 15951* | Lemma for ruc 15961. 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 15952* | Lemma for ruc 15961. 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 15953* | Lemma for ruc 15961. 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 15954* | Lemma for ruc 15961. 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 15955* | Lemma for ruc 15961. 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 15956* | Lemma for ruc 15961. 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 15957* | Lemma for ruc 15961. 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 15958* | Lemma for ruc 15961. 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 15959* | Lemma for ruc 15961. 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 15960 | Lemma for ruc 15961. There is no function that maps ℕ onto ℝ. (Use nex 1803 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 15961 | 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 15949 through ruclem13 15960 and this final piece. Our proof is based on the proof of Theorem 5.18 of [Truss] p. 114. See ruclem13 15960 for the function existence version of this theorem. For an informal discussion of this proof, see mmcomplex.html#uncountable 15960. For an alternate proof see rucALT 15948. This is Metamath 100 proof #22. (Contributed by NM, 13-Oct-2004.) |
⊢ ℕ ≺ ℝ | ||
Theorem | resdomq 15962 | 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 15963 | There are at least aleph-one real numbers. (Contributed by NM, 2-Feb-2005.) |
⊢ (ℵ‘1o) ≼ ℝ | ||
Theorem | aleph1irr 15964 | There are at least aleph-one irrationals. (Contributed by NM, 2-Feb-2005.) |
⊢ (ℵ‘1o) ≼ (ℝ ∖ ℚ) | ||
Theorem | cnso 15965 | 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 15966 | Lemma for sqrt2irr 15967. 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 15967 | The square root of 2 is irrational. See zsqrtelqelz 16471 for a generalization to all non-square integers. The proof's core is proven in sqrt2irrlem 15966, 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/ 15966. (Contributed by NM, 8-Jan-2002.) (Proof shortened by Mario Carneiro, 12-Sep-2015.) |
⊢ (√‘2) ∉ ℚ | ||
Theorem | sqrt2re 15968 | The square root of 2 exists and is a real number. (Contributed by NM, 3-Dec-2004.) |
⊢ (√‘2) ∈ ℝ | ||
Theorem | sqrt2irr0 15969 | The square root of 2 is an irrational number. (Contributed by AV, 23-Dec-2022.) |
⊢ (√‘2) ∈ (ℝ ∖ ℚ) | ||
Theorem | nthruc 15970 | 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 15971 for a further refinement. (Contributed by NM, 12-Jan-2002.) |
⊢ ((ℕ ⊊ ℤ ∧ ℤ ⊊ ℚ) ∧ (ℚ ⊊ ℝ ∧ ℝ ⊊ ℂ)) | ||
Theorem | nthruz 15971 | 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 15970. (Contributed by NM, 9-May-2004.) |
⊢ (ℕ ⊊ ℕ0 ∧ ℕ0 ⊊ ℤ) | ||
Syntax | cdvds 15972 | Extend the definition of a class to include the divides relation. See df-dvds 15973. |
class ∥ | ||
Definition | df-dvds 15973* | 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 15974* | Define the divides relation. 𝑀 ∥ 𝑁 means 𝑀 divides into 𝑁 with no remainder. For example, 3 ∥ 6 (ex-dvds 28829). As proven in dvdsval3 15976, 𝑀 ∥ 𝑁 ↔ (𝑁 mod 𝑀) = 0. See divides 15974 and dvdsval2 15975 for other equivalent expressions. (Contributed by Paul Chapman, 21-Mar-2011.) |
⊢ ((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → (𝑀 ∥ 𝑁 ↔ ∃𝑛 ∈ ℤ (𝑛 · 𝑀) = 𝑁)) | ||
Theorem | dvdsval2 15975 | 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 15976 | 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 15977 | Reverse closure for the divisibility relation. (Contributed by Stefan O'Rear, 5-Sep-2015.) |
⊢ (𝑋 ∥ 𝑌 → (𝑋 ∈ ℤ ∧ 𝑌 ∈ ℤ)) | ||
Theorem | dvdsmod0 15978 | 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 15979 | 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 15980 | 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 16494). (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 15981 | Strong form of dvdsval2 15975 for positive integers. (Contributed by Stefan O'Rear, 13-Sep-2014.) |
⊢ ((𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ) → (𝐵 ∥ 𝐴 ↔ (𝐴 / 𝐵) ∈ ℕ)) | ||
Theorem | nndivides 15982* | 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 15983 | 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 15984 | 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 15985 | 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 15986* | 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 15987* | 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 15988 | 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 15989 | 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 15990 | 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 15991 | 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 15992 | 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 15993 | 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 15994 | 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 15995 | 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 15996 | An integer divides a multiple of itself. (Contributed by Paul Chapman, 21-Mar-2011.) |
⊢ ((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → 𝑀 ∥ (𝑀 · 𝑁)) | ||
Theorem | dvdsmul2 15997 | An integer divides a multiple of itself. (Contributed by Paul Chapman, 21-Mar-2011.) |
⊢ ((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → 𝑁 ∥ (𝑀 · 𝑁)) | ||
Theorem | iddvdsexp 15998 | 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 15999 | 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 16000 | 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)) |
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