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Theorem List for Metamath Proof Explorer - 14901-15000   *Has distinct variable group(s)
TypeLabelDescription
Statement
 
Theoremsin2t 14901 Double-angle formula for sine. (Contributed by Paul Chapman, 17-Jan-2008.)
(𝐴 ∈ ℂ → (sin‘(2 · 𝐴)) = (2 · ((sin‘𝐴) · (cos‘𝐴))))
 
Theoremcos2t 14902 Double-angle formula for cosine. (Contributed by Paul Chapman, 24-Jan-2008.)
(𝐴 ∈ ℂ → (cos‘(2 · 𝐴)) = ((2 · ((cos‘𝐴)↑2)) − 1))
 
Theoremcos2tsin 14903 Double-angle formula for cosine in terms of sine. (Contributed by NM, 12-Sep-2008.)
(𝐴 ∈ ℂ → (cos‘(2 · 𝐴)) = (1 − (2 · ((sin‘𝐴)↑2))))
 
Theoremsinbnd 14904 The sine of a real number lies between -1 and 1. Equation 18 of [Gleason] p. 311. (Contributed by NM, 16-Jan-2006.)
(𝐴 ∈ ℝ → (-1 ≤ (sin‘𝐴) ∧ (sin‘𝐴) ≤ 1))
 
Theoremcosbnd 14905 The cosine of a real number lies between -1 and 1. Equation 18 of [Gleason] p. 311. (Contributed by NM, 16-Jan-2006.)
(𝐴 ∈ ℝ → (-1 ≤ (cos‘𝐴) ∧ (cos‘𝐴) ≤ 1))
 
Theoremsinbnd2 14906 The sine of a real number is in the closed interval from -1 to 1. (Contributed by Mario Carneiro, 12-May-2014.)
(𝐴 ∈ ℝ → (sin‘𝐴) ∈ (-1[,]1))
 
Theoremcosbnd2 14907 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 14908* Lemma for sin01bnd 14909 and cos01bnd 14910. (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 14909 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 14910 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 14911 Bounds on the cosine of 1. (Contributed by Paul Chapman, 19-Jan-2008.)
((1 / 3) < (cos‘1) ∧ (cos‘1) < (2 / 3))
 
Theoremcos2bnd 14912 Bounds on the cosine of 2. (Contributed by Paul Chapman, 19-Jan-2008.)
(-(7 / 9) < (cos‘2) ∧ (cos‘2) < -(1 / 9))
 
Theoremsinltx 14913 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 14914 The sine of a positive real number less than or equal to 1 is positive. (Contributed by Paul Chapman, 19-Jan-2008.) Replace OLD theorem. (Revised by Wolf Lammen, 25-Sep-2020.)
(𝐴 ∈ (0(,]1) → 0 < (sin‘𝐴))
 
Theoremcos01gt0 14915 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 14916 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 14917 The signs of the sine and cosine of 1. (Contributed by Paul Chapman, 19-Jan-2008.)
(0 < (sin‘1) ∧ 0 < (cos‘1))
 
Theoremsincos2sgn 14918 The signs of the sine and cosine of 2. (Contributed by Paul Chapman, 19-Jan-2008.)
(0 < (sin‘2) ∧ (cos‘2) < 0)
 
Theoremsin4lt0 14919 The sine of 4 is negative. (Contributed by Paul Chapman, 19-Jan-2008.)
(sin‘4) < 0
 
Theoremabsefi 14920 The absolute value of the exponential function 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 14921 The absolute value of the exponential function is the exponential function of the real part. (Contributed by Paul Chapman, 13-Sep-2007.)
(𝐴 ∈ ℂ → (abs‘(exp‘𝐴)) = (exp‘(ℜ‘𝐴)))
 
Theoremabsefib 14922 A number is real iff its imaginary exponential has absolute value one. (Contributed by NM, 21-Aug-2008.)
(𝐴 ∈ ℂ → (𝐴 ∈ ℝ ↔ (abs‘(exp‘(i · 𝐴))) = 1))
 
Theoremefieq1re 14923 A number whose imaginary exponential is one is real. (Contributed by NM, 21-Aug-2008.)
((𝐴 ∈ ℂ ∧ (exp‘(i · 𝐴)) = 1) → 𝐴 ∈ ℝ)
 
Theoremdemoivre 14924 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 14925 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 14925 Alternate proof of demoivre 14924. 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)))))
 
5.11.2  _e is irrational
 
Theoremeirrlem 14926* Lemma for eirr 14927. (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 14927 e is irrational. (Contributed by Paul Chapman, 9-Feb-2008.) (Proof shortened by Mario Carneiro, 29-Apr-2014.)
e ∉ ℚ
 
Theoremegt2lt3 14928 Euler's constant e = 2.71828... is bounded by 2 and 3. (Contributed by NM, 28-Nov-2008.) (Revised by Mario Carneiro, 29-Apr-2014.)
(2 < e ∧ e < 3)
 
Theoremepos 14929 Euler's constant e is greater than 0. (Contributed by Jeff Hankins, 22-Nov-2008.)
0 < e
 
Theoremepr 14930 Euler's constant e is a positive real. (Contributed by Jeff Hankins, 22-Nov-2008.)
e ∈ ℝ+
 
Theoremene0 14931 e is not 0. (Contributed by David A. Wheeler, 17-Oct-2017.)
e ≠ 0
 
Theoremene1 14932 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 14933 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
 
Theoremznnenlem 14934 Lemma for znnen 14935. (Contributed by NM, 31-Jul-2004.)
(((0 ≤ 𝑥 ∧ ¬ 0 ≤ 𝑦) ∧ (𝑥 ∈ ℤ ∧ 𝑦 ∈ ℤ)) → (𝑥 = 𝑦 ↔ (2 · 𝑥) = ((-2 · 𝑦) + 1)))
 
Theoremznnen 14935 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 14936 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 14937* Lemma for rpnnen2 14949. (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 14938* Lemma for rpnnen2 14949. (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 14939* Lemma for rpnnen2 14949. (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 14940* Lemma for rpnnen2 14949. (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 14941* Lemma for rpnnen2 14949. (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 14942* Lemma for rpnnen2 14949. (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 14943* Lemma for rpnnen2 14949. (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 14944* Lemma for rpnnen2 14949. (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 14945* Lemma for rpnnen2 14949. (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 14946* Lemma for rpnnen2 14949. (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 14947* Lemma for rpnnen2 14949. (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 14948* Lemma for rpnnen2 14949. (Contributed by Mario Carneiro, 13-May-2013.)
𝐹 = (𝑥 ∈ 𝒫 ℕ ↦ (𝑛 ∈ ℕ ↦ if(𝑛𝑥, ((1 / 3)↑𝑛), 0)))       𝒫 ℕ ≼ (0[,]1)
 
Theoremrpnnen2 14949 The other half of rpnnen 14950, 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... 14606). 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 14937). This is an infinite sum of real numbers (rpnnen2lem2 14938), and since 𝐴𝐵 implies (𝐹𝐴) ≤ (𝐹𝐵) (rpnnen2lem4 14940) and (𝐹‘ℕ) converges to 1 / 2 (rpnnen2lem3 14939) by geoisum1 14604, the sum is convergent to some real (rpnnen2lem5 14941 and rpnnen2lem6 14942) by the comparison test for convergence cvgcmp 14542. The comparison test also tells us that 𝐴𝐵 implies Σ(𝐹𝐴) ≤ Σ(𝐹𝐵) (rpnnen2lem7 14943).

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 14944) and cancel them (rpnnen2lem10 14946), 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 14945) and Σ(𝐹‘{𝑚}) = (3↑-𝑚), we establish that Σ(𝐹𝑦) < Σ(𝐹𝑥) (rpnnen2lem11 14947) so that they must be different. By contraposition (rpnnen2lem12 14948), 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 14950 The cardinality of the continuum is the same as the powerset of ω. This is a stronger statement than ruc 14966, which only asserts that is uncountable, i.e. has a cardinality larger than ω. The main proof is in two parts, rpnnen1 11817 and rpnnen2 14949, each showing an injection in one direction, and this last part uses sbth 8077 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 14951 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 8837 directly). (Contributed by NM, 30-Jul-2004.) (Revised by Mario Carneiro, 16-Jun-2013.)
(ℝ × ℝ) ≈ ℝ
 
Theoremcpnnen 14952 The complex numbers are equinumerous to the powerset of the positive integers. (Contributed by Mario Carneiro, 16-Jun-2013.)
ℂ ≈ 𝒫 ℕ
 
TheoremrucALT 14953 Alternate proof of ruc 14966. This proof is a simple corollary of rpnnen 14950, which determines the exact cardinality of the reals. For an alternate proof discussed at mmcomplex.html#uncountable, see ruc 14966. (Contributed by NM, 13-Oct-2004.) (Revised by Mario Carneiro, 13-May-2013.) (Proof modification is discouraged.) (New usage is discouraged.)
ℕ ≺ ℝ
 
Theoremruclem1 14954* Lemma for ruc 14966 (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 14955* Lemma for ruc 14966. 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 14956* Lemma for ruc 14966. 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 14957* Lemma for ruc 14966. 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 14958* Lemma for ruc 14966. 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 14959* Lemma for ruc 14966. 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 14960* Lemma for ruc 14966. 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 14961* Lemma for ruc 14966. 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 14962* Lemma for ruc 14966. 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 14963* Lemma for ruc 14966. 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 14964* Lemma for ruc 14966. 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 14965 Lemma for ruc 14966. There is no function that maps onto . (Use nex 1730 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 14966 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 14954 through ruclem13 14965 and this final piece. Our proof is based on the proof of Theorem 5.18 of [Truss] p. 114. See ruclem13 14965 for the function existence version of this theorem. For an informal discussion of this proof, see mmcomplex.html#uncountable. For an alternate proof see rucALT 14953. This is Metamath 100 proof #22. (Contributed by NM, 13-Oct-2004.)
ℕ ≺ ℝ
 
Theoremresdomq 14967 The set of rationals is strictly less equinumerous than the set of reals ( strictly dominates ). (Contributed by NM, 18-Dec-2004.)
ℚ ≺ ℝ
 
Theoremaleph1re 14968 There are at least aleph-one real numbers. (Contributed by NM, 2-Feb-2005.)
(ℵ‘1𝑜) ≼ ℝ
 
Theoremaleph1irr 14969 There are at least aleph-one irrationals. (Contributed by NM, 2-Feb-2005.)
(ℵ‘1𝑜) ≼ (ℝ ∖ ℚ)
 
Theoremcnso 14970 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 14971 Lemma for sqrt2irr 14973. 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))
 
Theoremsqrt2irrlemOLD 14972 Obsolete proof of sqrt2irrlem 14971 as of 4-Jan-2022. (Contributed by NM, 20-Aug-2001.) (Revised by Mario Carneiro, 12-Sep-2015.) (Proof modification is discouraged.) (New usage is discouraged.)
(𝜑𝐴 ∈ ℤ)    &   (𝜑𝐵 ∈ ℕ)    &   (𝜑 → (√‘2) = (𝐴 / 𝐵))       (𝜑 → ((𝐴 / 2) ∈ ℤ ∧ (𝐵 / 2) ∈ ℕ))
 
Theoremsqrt2irr 14973 The square root of 2 is irrational. See zsqrtelqelz 15460 for a generalization to all non-square integers. The proof's core is proven in sqrt2irrlem 14971, 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/. (Contributed by NM, 8-Jan-2002.) (Proof shortened by Mario Carneiro, 12-Sep-2015.)
(√‘2) ∉ ℚ
 
Theoremsqrt2re 14974 The square root of 2 exists and is a real number. (Contributed by NM, 3-Dec-2004.)
(√‘2) ∈ ℝ
 
6.1.2  Some Number sets are chains of proper subsets
 
Theoremnthruc 14975 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 14976 for a further refinement. (Contributed by NM, 12-Jan-2002.)
((ℕ ⊊ ℤ ∧ ℤ ⊊ ℚ) ∧ (ℚ ⊊ ℝ ∧ ℝ ⊊ ℂ))
 
Theoremnthruz 14976 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 14975. (Contributed by NM, 9-May-2004.)
(ℕ ⊊ ℕ0 ∧ ℕ0 ⊊ ℤ)
 
6.1.3  The divides relation
 
Syntaxcdvds 14977 Extend the definition of a class to include the divides relation. See df-dvds 14978.
class
 
Definitiondf-dvds 14978* 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 14979* Define the divides relation. 𝑀𝑁 means 𝑀 divides into 𝑁 with no remainder. For example, 3 ∥ 6 (ex-dvds 27297). As proven in dvdsval3 14981, 𝑀𝑁 ↔ (𝑁 mod 𝑀) = 0. See divides 14979 and dvdsval2 14980 for other equivalent expressions. (Contributed by Paul Chapman, 21-Mar-2011.)
((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → (𝑀𝑁 ↔ ∃𝑛 ∈ ℤ (𝑛 · 𝑀) = 𝑁))
 
Theoremdvdsval2 14980 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 14981 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 14982 Reverse closure for the divisibility relation. (Contributed by Stefan O'Rear, 5-Sep-2015.)
(𝑋𝑌 → (𝑋 ∈ ℤ ∧ 𝑌 ∈ ℤ))
 
Theoremnndivdvds 14983 Strong form of dvdsval2 14980 for positive integers. (Contributed by Stefan O'Rear, 13-Sep-2014.)
((𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ) → (𝐵𝐴 ↔ (𝐴 / 𝐵) ∈ ℕ))
 
Theoremnndivides 14984* 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 14985 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)))
 
Theoremdvds0lem 14986 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 14987* 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 14988* 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 14989 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 14990 1 divides any integer. Theorem 1.1(f) in [ApostolNT] p. 14. (Contributed by Paul Chapman, 21-Mar-2011.)
(𝑁 ∈ ℤ → 1 ∥ 𝑁)
 
Theoremdvds0 14991 Any integer divides 0. Theorem 1.1(g) in [ApostolNT] p. 14. (Contributed by Paul Chapman, 21-Mar-2011.)
(𝑁 ∈ ℤ → 𝑁 ∥ 0)
 
Theoremnegdvdsb 14992 An integer divides another iff its negation does. (Contributed by Paul Chapman, 21-Mar-2011.)
((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → (𝑀𝑁 ↔ -𝑀𝑁))
 
Theoremdvdsnegb 14993 An integer divides another iff it divides its negation. (Contributed by Paul Chapman, 21-Mar-2011.)
((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → (𝑀𝑁𝑀 ∥ -𝑁))
 
Theoremabsdvdsb 14994 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 14995 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 14996 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 14997 An integer divides a multiple of itself. (Contributed by Paul Chapman, 21-Mar-2011.)
((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → 𝑀 ∥ (𝑀 · 𝑁))
 
Theoremdvdsmul2 14998 An integer divides a multiple of itself. (Contributed by Paul Chapman, 21-Mar-2011.)
((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ) → 𝑁 ∥ (𝑀 · 𝑁))
 
Theoremiddvdsexp 14999 An integer divides a positive integer power of itself. (Contributed by Paul Chapman, 26-Oct-2012.)
((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℕ) → 𝑀 ∥ (𝑀𝑁))
 
Theoremmuldvds1 15000 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))
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144 14301-14400 145 14401-14500 146 14501-14600 147 14601-14700 148 14701-14800 149 14801-14900 150 14901-15000 151 15001-15100 152 15101-15200 153 15201-15300 154 15301-15400 155 15401-15500 156 15501-15600 157 15601-15700 158 15701-15800 159 15801-15900 160 15901-16000 161 16001-16100 162 16101-16200 163 16201-16300 164 16301-16400 165 16401-16500 166 16501-16600 167 16601-16700 168 16701-16800 169 16801-16900 170 16901-17000 171 17001-17100 172 17101-17200 173 17201-17300 174 17301-17400 175 17401-17500 176 17501-17600 177 17601-17700 178 17701-17800 179 17801-17900 180 17901-18000 181 18001-18100 182 18101-18200 183 18201-18300 184 18301-18400 185 18401-18500 186 18501-18600 187 18601-18700 188 18701-18800 189 18801-18900 190 18901-19000 191 19001-19100 192 19101-19200 193 19201-19300 194 19301-19400 195 19401-19500 196 19501-19600 197 19601-19700 198 19701-19800 199 19801-19900 200 19901-20000 201 20001-20100 202 20101-20200 203 20201-20300 204 20301-20400 205 20401-20500 206 20501-20600 207 20601-20700 208 20701-20800 209 20801-20900 210 20901-21000 211 21001-21100 212 21101-21200 213 21201-21300 214 21301-21400 215 21401-21500 216 21501-21600 217 21601-21700 218 21701-21800 219 21801-21900 220 21901-22000 221 22001-22100 222 22101-22200 223 22201-22300 224 22301-22400 225 22401-22500 226 22501-22600 227 22601-22700 228 22701-22800 229 22801-22900 230 22901-23000 231 23001-23100 232 23101-23200 233 23201-23300 234 23301-23400 235 23401-23500 236 23501-23600 237 23601-23700 238 23701-23800 239 23801-23900 240 23901-24000 241 24001-24100 242 24101-24200 243 24201-24300 244 24301-24400 245 24401-24500 246 24501-24600 247 24601-24700 248 24701-24800 249 24801-24900 250 24901-25000 251 25001-25100 252 25101-25200 253 25201-25300 254 25301-25400 255 25401-25500 256 25501-25600 257 25601-25700 258 25701-25800 259 25801-25900 260 25901-26000 261 26001-26100 262 26101-26200 263 26201-26300 264 26301-26400 265 26401-26500 266 26501-26600 267 26601-26700 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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