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Theorem List for Metamath Proof Explorer - 15901-16000   *Has distinct variable group(s)
TypeLabelDescription
Statement
 
Theoremcosbnd2 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))
 
Theoremef01bndlem 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)(𝐹𝑘)) < ((𝐴↑4) / 6))
 
Theoremsin01bnd 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))
 
Theoremcos01bnd 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))))
 
Theoremcos1bnd 15905 Bounds on the cosine of 1. (Contributed by Paul Chapman, 19-Jan-2008.)
((1 / 3) < (cos‘1) ∧ (cos‘1) < (2 / 3))
 
Theoremcos2bnd 15906 Bounds on the cosine of 2. (Contributed by Paul Chapman, 19-Jan-2008.)
(-(7 / 9) < (cos‘2) ∧ (cos‘2) < -(1 / 9))
 
Theoremsinltx 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)
 
Theoremsin01gt0 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))
 
Theoremcos01gt0 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))
 
Theoremsin02gt0 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))
 
Theoremsincos1sgn 15911 The signs of the sine and cosine of 1. (Contributed by Paul Chapman, 19-Jan-2008.)
(0 < (sin‘1) ∧ 0 < (cos‘1))
 
Theoremsincos2sgn 15912 The signs of the sine and cosine of 2. (Contributed by Paul Chapman, 19-Jan-2008.)
(0 < (sin‘2) ∧ (cos‘2) < 0)
 
Theoremsin4lt0 15913 The sine of 4 is negative. (Contributed by Paul Chapman, 19-Jan-2008.)
(sin‘4) < 0
 
Theoremabsefi 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)
 
Theoremabsef 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)))
 
Theoremabsefib 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))
 
Theoremefieq1re 15917 A number whose imaginary exponential is one is real. (Contributed by NM, 21-Aug-2008.)
((𝐴 ∈ ℂ ∧ (exp‘(i · 𝐴)) = 1) → 𝐴 ∈ ℝ)
 
Theoremdemoivre 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)))))
 
TheoremdemoivreALT 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)))))
 
5.11.1.1  The circle constant (tau = 2 pi)
 
Syntaxctau 15920 Extend class notation to include the constant tau, τ = 6.28318....
class τ
 
Definitiondf-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 . 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, < )
 
5.11.2  _e is irrational
 
Theoremeirrlem 15922* Lemma for eirr 15923. (Contributed by Paul Chapman, 9-Feb-2008.) (Revised by Mario Carneiro, 29-Apr-2014.)
𝐹 = (𝑛 ∈ ℕ0 ↦ (1 / (!‘𝑛)))    &   (𝜑𝑃 ∈ ℤ)    &   (𝜑𝑄 ∈ ℕ)    &   (𝜑 → e = (𝑃 / 𝑄))        ¬ 𝜑
 
Theoremeirr 15923 e is irrational. (Contributed by Paul Chapman, 9-Feb-2008.) (Proof shortened by Mario Carneiro, 29-Apr-2014.)
e ∉ ℚ
 
Theoremegt2lt3 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)
 
Theoremepos 15925 Euler's constant e is greater than 0. (Contributed by Jeff Hankins, 22-Nov-2008.)
0 < e
 
Theoremepr 15926 Euler's constant e is a positive real. (Contributed by Jeff Hankins, 22-Nov-2008.)
e ∈ ℝ+
 
Theoremene0 15927 e is not 0. (Contributed by David A. Wheeler, 17-Oct-2017.)
e ≠ 0
 
Theoremene1 15928 e is not 1. (Contributed by David A. Wheeler, 17-Oct-2017.)
e ≠ 1
 
5.12  Cardinality of real and complex number subsets
 
5.12.1  Countability of integers and rationals
 
Theoremxpnnen 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
 
Theoremznnen 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
 
Theoremqnnen 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
 
5.12.2  The reals are uncountable
 
Theoremrpnnen2lem1 15932* Lemma for rpnnen2 15944. (Contributed by Mario Carneiro, 13-May-2013.)
𝐹 = (𝑥 ∈ 𝒫 ℕ ↦ (𝑛 ∈ ℕ ↦ if(𝑛𝑥, ((1 / 3)↑𝑛), 0)))       ((𝐴 ⊆ ℕ ∧ 𝑁 ∈ ℕ) → ((𝐹𝐴)‘𝑁) = if(𝑁𝐴, ((1 / 3)↑𝑁), 0))
 
Theoremrpnnen2lem2 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(𝑛𝑥, ((1 / 3)↑𝑛), 0)))       (𝐴 ⊆ ℕ → (𝐹𝐴):ℕ⟶ℝ)
 
Theoremrpnnen2lem3 15934* Lemma for rpnnen2 15944. (Contributed by Mario Carneiro, 13-May-2013.)
𝐹 = (𝑥 ∈ 𝒫 ℕ ↦ (𝑛 ∈ ℕ ↦ if(𝑛𝑥, ((1 / 3)↑𝑛), 0)))       seq1( + , (𝐹‘ℕ)) ⇝ (1 / 2)
 
Theoremrpnnen2lem4 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(𝑛𝑥, ((1 / 3)↑𝑛), 0)))       ((𝐴𝐵𝐵 ⊆ ℕ ∧ 𝑘 ∈ ℕ) → (0 ≤ ((𝐹𝐴)‘𝑘) ∧ ((𝐹𝐴)‘𝑘) ≤ ((𝐹𝐵)‘𝑘)))
 
Theoremrpnnen2lem5 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(𝑛𝑥, ((1 / 3)↑𝑛), 0)))       ((𝐴 ⊆ ℕ ∧ 𝑀 ∈ ℕ) → seq𝑀( + , (𝐹𝐴)) ∈ dom ⇝ )
 
Theoremrpnnen2lem6 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(𝑛𝑥, ((1 / 3)↑𝑛), 0)))       ((𝐴 ⊆ ℕ ∧ 𝑀 ∈ ℕ) → Σ𝑘 ∈ (ℤ𝑀)((𝐹𝐴)‘𝑘) ∈ ℝ)
 
Theoremrpnnen2lem7 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(𝑛𝑥, ((1 / 3)↑𝑛), 0)))       ((𝐴𝐵𝐵 ⊆ ℕ ∧ 𝑀 ∈ ℕ) → Σ𝑘 ∈ (ℤ𝑀)((𝐹𝐴)‘𝑘) ≤ Σ𝑘 ∈ (ℤ𝑀)((𝐹𝐵)‘𝑘))
 
Theoremrpnnen2lem8 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(𝑛𝑥, ((1 / 3)↑𝑛), 0)))       ((𝐴 ⊆ ℕ ∧ 𝑀 ∈ ℕ) → Σ𝑘 ∈ ℕ ((𝐹𝐴)‘𝑘) = (Σ𝑘 ∈ (1...(𝑀 − 1))((𝐹𝐴)‘𝑘) + Σ𝑘 ∈ (ℤ𝑀)((𝐹𝐴)‘𝑘)))
 
Theoremrpnnen2lem9 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(𝑛𝑥, ((1 / 3)↑𝑛), 0)))       (𝑀 ∈ ℕ → Σ𝑘 ∈ (ℤ𝑀)((𝐹‘(ℕ ∖ {𝑀}))‘𝑘) = (0 + (((1 / 3)↑(𝑀 + 1)) / (1 − (1 / 3)))))
 
Theoremrpnnen2lem10 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(𝑛𝑥, ((1 / 3)↑𝑛), 0)))    &   (𝜑𝐴 ⊆ ℕ)    &   (𝜑𝐵 ⊆ ℕ)    &   (𝜑𝑚 ∈ (𝐴𝐵))    &   (𝜑 → ∀𝑛 ∈ ℕ (𝑛 < 𝑚 → (𝑛𝐴𝑛𝐵)))    &   (𝜓 ↔ Σ𝑘 ∈ ℕ ((𝐹𝐴)‘𝑘) = Σ𝑘 ∈ ℕ ((𝐹𝐵)‘𝑘))       ((𝜑𝜓) → Σ𝑘 ∈ (ℤ𝑚)((𝐹𝐴)‘𝑘) = Σ𝑘 ∈ (ℤ𝑚)((𝐹𝐵)‘𝑘))
 
Theoremrpnnen2lem11 15942* Lemma for rpnnen2 15944. (Contributed by Mario Carneiro, 13-May-2013.)
𝐹 = (𝑥 ∈ 𝒫 ℕ ↦ (𝑛 ∈ ℕ ↦ if(𝑛𝑥, ((1 / 3)↑𝑛), 0)))    &   (𝜑𝐴 ⊆ ℕ)    &   (𝜑𝐵 ⊆ ℕ)    &   (𝜑𝑚 ∈ (𝐴𝐵))    &   (𝜑 → ∀𝑛 ∈ ℕ (𝑛 < 𝑚 → (𝑛𝐴𝑛𝐵)))    &   (𝜓 ↔ Σ𝑘 ∈ ℕ ((𝐹𝐴)‘𝑘) = Σ𝑘 ∈ ℕ ((𝐹𝐵)‘𝑘))       (𝜑 → ¬ 𝜓)
 
Theoremrpnnen2lem12 15943* Lemma for rpnnen2 15944. (Contributed by Mario Carneiro, 13-May-2013.)
𝐹 = (𝑥 ∈ 𝒫 ℕ ↦ (𝑛 ∈ ℕ ↦ if(𝑛𝑥, ((1 / 3)↑𝑛), 0)))       𝒫 ℕ ≼ (0[,]1)
 
Theoremrpnnen2 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 Σ𝑘𝐴(3↑-𝑘) = Σ𝑘 ∈ ℕ((𝐹𝐴)‘𝑘), where ((𝐹𝐴)‘𝑘) = if(𝑘𝐴, (3↑-𝑘), 0)) (rpnnen2lem1 15932). This is an infinite sum of real numbers (rpnnen2lem2 15933), and since 𝐴𝐵 implies (𝐹𝐴) ≤ (𝐹𝐵) (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 𝐴𝐵 implies Σ(𝐹𝐴) ≤ Σ(𝐹𝐵) (rpnnen2lem7 15938).

Putting it all together, if we have two sets 𝑥𝑦, there must differ somewhere, and so there must be an 𝑚 such that 𝑛 < 𝑚(𝑛𝑥𝑛𝑦) but 𝑚 ∈ (𝑥𝑦) 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(𝐹𝑦) ≤ Σ(𝐹‘(ℕ ∖ {𝑚})) and Σ(𝐹‘{𝑚}) ≤ Σ(𝐹𝑥) (where these sums are only over (ℤ𝑚)), and since Σ(𝐹‘(ℕ ∖ {𝑚})) = (3↑-𝑚) / 2 (rpnnen2lem9 15940) and Σ(𝐹‘{𝑚}) = (3↑-𝑚), we establish that Σ(𝐹𝑦) < Σ(𝐹𝑥) (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)
 
Theoremrpnnen 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 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
 
Theoremrexpen 15946 The real numbers are equinumerous to their own Cartesian product, even though it is not necessarily true that 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
 
Theoremcpnnen 15947 The complex numbers are equinumerous to the powerset of the positive integers. (Contributed by Mario Carneiro, 16-Jun-2013.)
ℂ ≈ 𝒫 ℕ
 
TheoremrucALT 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
 
Theoremruclem1 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.)
(𝜑𝐹:ℕ⟶ℝ)    &   (𝜑𝐷 = (𝑥 ∈ (ℝ × ℝ), 𝑦 ∈ ℝ ↦ (((1st𝑥) + (2nd𝑥)) / 2) / 𝑚if(𝑚 < 𝑦, ⟨(1st𝑥), 𝑚⟩, ⟨((𝑚 + (2nd𝑥)) / 2), (2nd𝑥)⟩)))    &   (𝜑𝐴 ∈ ℝ)    &   (𝜑𝐵 ∈ ℝ)    &   (𝜑𝑀 ∈ ℝ)    &   𝑋 = (1st ‘(⟨𝐴, 𝐵𝐷𝑀))    &   𝑌 = (2nd ‘(⟨𝐴, 𝐵𝐷𝑀))       (𝜑 → ((⟨𝐴, 𝐵𝐷𝑀) ∈ (ℝ × ℝ) ∧ 𝑋 = if(((𝐴 + 𝐵) / 2) < 𝑀, 𝐴, ((((𝐴 + 𝐵) / 2) + 𝐵) / 2)) ∧ 𝑌 = if(((𝐴 + 𝐵) / 2) < 𝑀, ((𝐴 + 𝐵) / 2), 𝐵)))
 
Theoremruclem2 15950* Lemma for ruc 15961. Ordering property for the input to 𝐷. (Contributed by Mario Carneiro, 28-May-2014.)
(𝜑𝐹:ℕ⟶ℝ)    &   (𝜑𝐷 = (𝑥 ∈ (ℝ × ℝ), 𝑦 ∈ ℝ ↦ (((1st𝑥) + (2nd𝑥)) / 2) / 𝑚if(𝑚 < 𝑦, ⟨(1st𝑥), 𝑚⟩, ⟨((𝑚 + (2nd𝑥)) / 2), (2nd𝑥)⟩)))    &   (𝜑𝐴 ∈ ℝ)    &   (𝜑𝐵 ∈ ℝ)    &   (𝜑𝑀 ∈ ℝ)    &   𝑋 = (1st ‘(⟨𝐴, 𝐵𝐷𝑀))    &   𝑌 = (2nd ‘(⟨𝐴, 𝐵𝐷𝑀))    &   (𝜑𝐴 < 𝐵)       (𝜑 → (𝐴𝑋𝑋 < 𝑌𝑌𝐵))
 
Theoremruclem3 15951* Lemma for ruc 15961. The constructed interval [𝑋, 𝑌] always excludes 𝑀. (Contributed by Mario Carneiro, 28-May-2014.)
(𝜑𝐹:ℕ⟶ℝ)    &   (𝜑𝐷 = (𝑥 ∈ (ℝ × ℝ), 𝑦 ∈ ℝ ↦ (((1st𝑥) + (2nd𝑥)) / 2) / 𝑚if(𝑚 < 𝑦, ⟨(1st𝑥), 𝑚⟩, ⟨((𝑚 + (2nd𝑥)) / 2), (2nd𝑥)⟩)))    &   (𝜑𝐴 ∈ ℝ)    &   (𝜑𝐵 ∈ ℝ)    &   (𝜑𝑀 ∈ ℝ)    &   𝑋 = (1st ‘(⟨𝐴, 𝐵𝐷𝑀))    &   𝑌 = (2nd ‘(⟨𝐴, 𝐵𝐷𝑀))    &   (𝜑𝐴 < 𝐵)       (𝜑 → (𝑀 < 𝑋𝑌 < 𝑀))
 
Theoremruclem4 15952* Lemma for ruc 15961. Initial value of the interval sequence. (Contributed by Mario Carneiro, 28-May-2014.)
(𝜑𝐹:ℕ⟶ℝ)    &   (𝜑𝐷 = (𝑥 ∈ (ℝ × ℝ), 𝑦 ∈ ℝ ↦ (((1st𝑥) + (2nd𝑥)) / 2) / 𝑚if(𝑚 < 𝑦, ⟨(1st𝑥), 𝑚⟩, ⟨((𝑚 + (2nd𝑥)) / 2), (2nd𝑥)⟩)))    &   𝐶 = ({⟨0, ⟨0, 1⟩⟩} ∪ 𝐹)    &   𝐺 = seq0(𝐷, 𝐶)       (𝜑 → (𝐺‘0) = ⟨0, 1⟩)
 
Theoremruclem6 15953* Lemma for ruc 15961. Domain and range of the interval sequence. (Contributed by Mario Carneiro, 28-May-2014.)
(𝜑𝐹:ℕ⟶ℝ)    &   (𝜑𝐷 = (𝑥 ∈ (ℝ × ℝ), 𝑦 ∈ ℝ ↦ (((1st𝑥) + (2nd𝑥)) / 2) / 𝑚if(𝑚 < 𝑦, ⟨(1st𝑥), 𝑚⟩, ⟨((𝑚 + (2nd𝑥)) / 2), (2nd𝑥)⟩)))    &   𝐶 = ({⟨0, ⟨0, 1⟩⟩} ∪ 𝐹)    &   𝐺 = seq0(𝐷, 𝐶)       (𝜑𝐺:ℕ0⟶(ℝ × ℝ))
 
Theoremruclem7 15954* Lemma for ruc 15961. Successor value for the interval sequence. (Contributed by Mario Carneiro, 28-May-2014.)
(𝜑𝐹:ℕ⟶ℝ)    &   (𝜑𝐷 = (𝑥 ∈ (ℝ × ℝ), 𝑦 ∈ ℝ ↦ (((1st𝑥) + (2nd𝑥)) / 2) / 𝑚if(𝑚 < 𝑦, ⟨(1st𝑥), 𝑚⟩, ⟨((𝑚 + (2nd𝑥)) / 2), (2nd𝑥)⟩)))    &   𝐶 = ({⟨0, ⟨0, 1⟩⟩} ∪ 𝐹)    &   𝐺 = seq0(𝐷, 𝐶)       ((𝜑𝑁 ∈ ℕ0) → (𝐺‘(𝑁 + 1)) = ((𝐺𝑁)𝐷(𝐹‘(𝑁 + 1))))
 
Theoremruclem8 15955* Lemma for ruc 15961. The intervals of the 𝐺 sequence are all nonempty. (Contributed by Mario Carneiro, 28-May-2014.)
(𝜑𝐹:ℕ⟶ℝ)    &   (𝜑𝐷 = (𝑥 ∈ (ℝ × ℝ), 𝑦 ∈ ℝ ↦ (((1st𝑥) + (2nd𝑥)) / 2) / 𝑚if(𝑚 < 𝑦, ⟨(1st𝑥), 𝑚⟩, ⟨((𝑚 + (2nd𝑥)) / 2), (2nd𝑥)⟩)))    &   𝐶 = ({⟨0, ⟨0, 1⟩⟩} ∪ 𝐹)    &   𝐺 = seq0(𝐷, 𝐶)       ((𝜑𝑁 ∈ ℕ0) → (1st ‘(𝐺𝑁)) < (2nd ‘(𝐺𝑁)))
 
Theoremruclem9 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.)
(𝜑𝐹:ℕ⟶ℝ)    &   (𝜑𝐷 = (𝑥 ∈ (ℝ × ℝ), 𝑦 ∈ ℝ ↦ (((1st𝑥) + (2nd𝑥)) / 2) / 𝑚if(𝑚 < 𝑦, ⟨(1st𝑥), 𝑚⟩, ⟨((𝑚 + (2nd𝑥)) / 2), (2nd𝑥)⟩)))    &   𝐶 = ({⟨0, ⟨0, 1⟩⟩} ∪ 𝐹)    &   𝐺 = seq0(𝐷, 𝐶)    &   (𝜑𝑀 ∈ ℕ0)    &   (𝜑𝑁 ∈ (ℤ𝑀))       (𝜑 → ((1st ‘(𝐺𝑀)) ≤ (1st ‘(𝐺𝑁)) ∧ (2nd ‘(𝐺𝑁)) ≤ (2nd ‘(𝐺𝑀))))
 
Theoremruclem10 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.)
(𝜑𝐹:ℕ⟶ℝ)    &   (𝜑𝐷 = (𝑥 ∈ (ℝ × ℝ), 𝑦 ∈ ℝ ↦ (((1st𝑥) + (2nd𝑥)) / 2) / 𝑚if(𝑚 < 𝑦, ⟨(1st𝑥), 𝑚⟩, ⟨((𝑚 + (2nd𝑥)) / 2), (2nd𝑥)⟩)))    &   𝐶 = ({⟨0, ⟨0, 1⟩⟩} ∪ 𝐹)    &   𝐺 = seq0(𝐷, 𝐶)    &   (𝜑𝑀 ∈ ℕ0)    &   (𝜑𝑁 ∈ ℕ0)       (𝜑 → (1st ‘(𝐺𝑀)) < (2nd ‘(𝐺𝑁)))
 
Theoremruclem11 15958* Lemma for ruc 15961. Closure lemmas for supremum. (Contributed by Mario Carneiro, 28-May-2014.)
(𝜑𝐹:ℕ⟶ℝ)    &   (𝜑𝐷 = (𝑥 ∈ (ℝ × ℝ), 𝑦 ∈ ℝ ↦ (((1st𝑥) + (2nd𝑥)) / 2) / 𝑚if(𝑚 < 𝑦, ⟨(1st𝑥), 𝑚⟩, ⟨((𝑚 + (2nd𝑥)) / 2), (2nd𝑥)⟩)))    &   𝐶 = ({⟨0, ⟨0, 1⟩⟩} ∪ 𝐹)    &   𝐺 = seq0(𝐷, 𝐶)       (𝜑 → (ran (1st𝐺) ⊆ ℝ ∧ ran (1st𝐺) ≠ ∅ ∧ ∀𝑧 ∈ ran (1st𝐺)𝑧 ≤ 1))
 
Theoremruclem12 15959* Lemma for ruc 15961. The supremum of the increasing sequence 1st𝐺 is a real number that is not in the range of 𝐹. (Contributed by Mario Carneiro, 28-May-2014.)
(𝜑𝐹:ℕ⟶ℝ)    &   (𝜑𝐷 = (𝑥 ∈ (ℝ × ℝ), 𝑦 ∈ ℝ ↦ (((1st𝑥) + (2nd𝑥)) / 2) / 𝑚if(𝑚 < 𝑦, ⟨(1st𝑥), 𝑚⟩, ⟨((𝑚 + (2nd𝑥)) / 2), (2nd𝑥)⟩)))    &   𝐶 = ({⟨0, ⟨0, 1⟩⟩} ∪ 𝐹)    &   𝐺 = seq0(𝐷, 𝐶)    &   𝑆 = sup(ran (1st𝐺), ℝ, < )       (𝜑𝑆 ∈ (ℝ ∖ ran 𝐹))
 
Theoremruclem13 15960 Lemma for ruc 15961. There is no function that maps onto . (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
 
Theoremruc 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
 
Theoremresdomq 15962 The set of rationals is strictly less equinumerous than the set of reals ( strictly dominates ). (Contributed by NM, 18-Dec-2004.)
ℚ ≺ ℝ
 
Theoremaleph1re 15963 There are at least aleph-one real numbers. (Contributed by NM, 2-Feb-2005.)
(ℵ‘1o) ≼ ℝ
 
Theoremaleph1irr 15964 There are at least aleph-one irrationals. (Contributed by NM, 2-Feb-2005.)
(ℵ‘1o) ≼ (ℝ ∖ ℚ)
 
Theoremcnso 15965 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
 
Theoremsqrt2irrlem 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.)
(𝜑𝐴 ∈ ℤ)    &   (𝜑𝐵 ∈ ℕ)    &   (𝜑 → (√‘2) = (𝐴 / 𝐵))       (𝜑 → ((𝐴 / 2) ∈ ℤ ∧ (𝐵 / 2) ∈ ℕ))
 
Theoremsqrt2irr 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
 
Theoremsqrt2re 15968 The square root of 2 exists and is a real number. (Contributed by NM, 3-Dec-2004.)
(√‘2) ∈ ℝ
 
Theoremsqrt2irr0 15969 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
 
Theoremnthruc 15970 The sequence , , , , and 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 but not , one-half belongs to but not , the square root of 2 belongs to but not , and finally that the imaginary number i belongs to but not . See nthruz 15971 for a further refinement. (Contributed by NM, 12-Jan-2002.)
((ℕ ⊊ ℤ ∧ ℤ ⊊ ℚ) ∧ (ℚ ⊊ ℝ ∧ ℝ ⊊ ℂ))
 
Theoremnthruz 15971 The sequence , 0, and 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 0 but not and minus one belongs to but not 0. This theorem refines the chain of proper subsets nthruc 15970. (Contributed by NM, 9-May-2004.)
(ℕ ⊊ ℕ0 ∧ ℕ0 ⊊ ℤ)
 
6.1.3  The divides relation
 
Syntaxcdvds 15972 Extend the definition of a class to include the divides relation. See df-dvds 15973.
class
 
Definitiondf-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)}
 
Theoremdivides 15974* Define the divides relation. 𝑀𝑁 means 𝑀 divides into 𝑁 with no remainder. For example, 3 ∥ 6 (ex-dvds 28829). As proven in dvdsval3 15976, 𝑀𝑁 ↔ (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) → (𝑀𝑁 ↔ ∃𝑛 ∈ ℤ (𝑛 · 𝑀) = 𝑁))
 
Theoremdvdsval2 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) → (𝑀𝑁 ↔ (𝑁 / 𝑀) ∈ ℤ))
 
Theoremdvdsval3 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) → (𝑀𝑁 ↔ (𝑁 mod 𝑀) = 0))
 
Theoremdvdszrcl 15977 Reverse closure for the divisibility relation. (Contributed by Stefan O'Rear, 5-Sep-2015.)
(𝑋𝑌 → (𝑋 ∈ ℤ ∧ 𝑌 ∈ ℤ))
 
Theoremdvdsmod0 15978 If a positive integer divides another integer, then the remainder upon division is zero. (Contributed by AV, 3-Mar-2022.)
((𝑀 ∈ ℕ ∧ 𝑀𝑁) → (𝑁 mod 𝑀) = 0)
 
Theoremp1modz1 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.)
((𝑀𝐴 ∧ 1 < 𝑀) → ((𝐴 + 1) mod 𝑀) = 1)
 
Theoremdvdsmodexp 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 ∧ 𝑁𝐴) → ((𝐴𝐵) mod 𝑁) = (𝐴 mod 𝑁))
 
Theoremnndivdvds 15981 Strong form of dvdsval2 15975 for positive integers. (Contributed by Stefan O'Rear, 13-Sep-2014.)
((𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ) → (𝐵𝐴 ↔ (𝐴 / 𝐵) ∈ ℕ))
 
Theoremnndivides 15982* Definition of the divides relation for positive integers. (Contributed by AV, 26-Jul-2021.)
((𝑀 ∈ ℕ ∧ 𝑁 ∈ ℕ) → (𝑀𝑁 ↔ ∃𝑛 ∈ ℕ (𝑛 · 𝑀) = 𝑁))
 
Theoremmoddvds 15983 Two ways to say 𝐴𝐵 (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 ∥ (𝐴𝐵)))
 
Theoremmodm1div 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)))
 
Theoremdvds0lem 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) → 𝑀𝑁)
 
Theoremdvds1lem 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))    &   ((𝜑𝑥 ∈ ℤ) → 𝑍 ∈ ℤ)    &   ((𝜑𝑥 ∈ ℤ) → ((𝑥 · 𝐽) = 𝐾 → (𝑍 · 𝑀) = 𝑁))       (𝜑 → (𝐽𝐾𝑀𝑁))
 
Theoremdvds2lem 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 → ((𝐼𝐽𝐾𝐿) → 𝑀𝑁))
 
Theoremiddvds 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 → 𝑁𝑁)
 
Theorem1dvds 15989 1 divides any integer. Theorem 1.1(f) in [ApostolNT] p. 14. (Contributed by Paul Chapman, 21-Mar-2011.)
(𝑁 ∈ ℤ → 1 ∥ 𝑁)
 
Theoremdvds0 15990 Any integer divides 0. Theorem 1.1(g) in [ApostolNT] p. 14. (Contributed by Paul Chapman, 21-Mar-2011.)
(𝑁 ∈ ℤ → 𝑁 ∥ 0)
 
Theoremnegdvdsb 15991 An integer divides another iff its negation does. (Contributed by Paul Chapman, 21-Mar-2011.)
((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → (𝑀𝑁 ↔ -𝑀𝑁))
 
Theoremdvdsnegb 15992 An integer divides another iff it divides its negation. (Contributed by Paul Chapman, 21-Mar-2011.)
((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → (𝑀𝑁𝑀 ∥ -𝑁))
 
Theoremabsdvdsb 15993 An integer divides another iff its absolute value does. (Contributed by Paul Chapman, 21-Mar-2011.)
((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → (𝑀𝑁 ↔ (abs‘𝑀) ∥ 𝑁))
 
Theoremdvdsabsb 15994 An integer divides another iff it divides its absolute value. (Contributed by Paul Chapman, 21-Mar-2011.)
((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → (𝑀𝑁𝑀 ∥ (abs‘𝑁)))
 
Theorem0dvds 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 ∥ 𝑁𝑁 = 0))
 
Theoremdvdsmul1 15996 An integer divides a multiple of itself. (Contributed by Paul Chapman, 21-Mar-2011.)
((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → 𝑀 ∥ (𝑀 · 𝑁))
 
Theoremdvdsmul2 15997 An integer divides a multiple of itself. (Contributed by Paul Chapman, 21-Mar-2011.)
((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → 𝑁 ∥ (𝑀 · 𝑁))
 
Theoremiddvdsexp 15998 An integer divides a positive integer power of itself. (Contributed by Paul Chapman, 26-Oct-2012.)
((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℕ) → 𝑀 ∥ (𝑀𝑁))
 
Theoremmuldvds1 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) ∥ 𝑁𝐾𝑁))
 
Theoremmuldvds2 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) ∥ 𝑁𝑀𝑁))
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268 26701-26800 269 26801-26900 270 26901-27000 271 27001-27100 272 27101-27200 273 27201-27300 274 27301-27400 275 27401-27500 276 27501-27600 277 27601-27700 278 27701-27800 279 27801-27900 280 27901-28000 281 28001-28100 282 28101-28200 283 28201-28300 284 28301-28400 285 28401-28500 286 28501-28600 287 28601-28700 288 28701-28800 289 28801-28900 290 28901-29000 291 29001-29100 292 29101-29200 293 29201-29300 294 29301-29400 295 29401-29500 296 29501-29600 297 29601-29700 298 29701-29800 299 29801-29900 300 29901-30000 301 30001-30100 302 30101-30200 303 30201-30300 304 30301-30400 305 30401-30500 306 30501-30600 307 30601-30700 308 30701-30800 309 30801-30900 310 30901-31000 311 31001-31100 312 31101-31200 313 31201-31300 314 31301-31400 315 31401-31500 316 31501-31600 317 31601-31700 318 31701-31800 319 31801-31900 320 31901-32000 321 32001-32100 322 32101-32200 323 32201-32300 324 32301-32400 325 32401-32500 326 32501-32600 327 32601-32700 328 32701-32800 329 32801-32900 330 32901-33000 331 33001-33100 332 33101-33200 333 33201-33300 334 33301-33400 335 33401-33500 336 33501-33600 337 33601-33700 338 33701-33800 339 33801-33900 340 33901-34000 341 34001-34100 342 34101-34200 343 34201-34300 344 34301-34400 345 34401-34500 346 34501-34600 347 34601-34700 348 34701-34800 349 34801-34900 350 34901-35000 351 35001-35100 352 35101-35200 353 35201-35300 354 35301-35400 355 35401-35500 356 35501-35600 357 35601-35700 358 35701-35800 359 35801-35900 360 35901-36000 361 36001-36100 362 36101-36200 363 36201-36300 364 36301-36400 365 36401-36500 366 36501-36600 367 36601-36700 368 36701-36800 369 36801-36900 370 36901-37000 371 37001-37100 372 37101-37200 373 37201-37300 374 37301-37400 375 37401-37500 376 37501-37600 377 37601-37700 378 37701-37800 379 37801-37900 380 37901-38000 381 38001-38100 382 38101-38200 383 38201-38300 384 38301-38400 385 38401-38500 386 38501-38600 387 38601-38700 388 38701-38800 389 38801-38900 390 38901-39000 391 39001-39100 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