HomeHome Intuitionistic Logic Explorer
Theorem List (p. 104 of 114)
< Previous  Next >
Bad symbols? Try the
GIF version.

Mirrors  >  Metamath Home Page  >  ILE Home Page  >  Theorem List Contents  >  Recent Proofs       This page: Page List

Theorem List for Intuitionistic Logic Explorer - 10301-10400   *Has distinct variable group(s)
TypeLabelDescription
Statement
 
Theoremcjdivapd 10301 Complex conjugate distributes over division. (Contributed by Jim Kingdon, 15-Jun-2020.)
(𝜑𝐴 ∈ ℂ)    &   (𝜑𝐵 ∈ ℂ)    &   (𝜑𝐵 # 0)       (𝜑 → (∗‘(𝐴 / 𝐵)) = ((∗‘𝐴) / (∗‘𝐵)))
 
Theoremrered 10302 A real number equals its real part. One direction of Proposition 10-3.4(f) of [Gleason] p. 133. (Contributed by Mario Carneiro, 29-May-2016.)
(𝜑𝐴 ∈ ℝ)       (𝜑 → (ℜ‘𝐴) = 𝐴)
 
Theoremreim0d 10303 The imaginary part of a real number is 0. (Contributed by Mario Carneiro, 29-May-2016.)
(𝜑𝐴 ∈ ℝ)       (𝜑 → (ℑ‘𝐴) = 0)
 
Theoremcjred 10304 A real number equals its complex conjugate. Proposition 10-3.4(f) of [Gleason] p. 133. (Contributed by Mario Carneiro, 29-May-2016.)
(𝜑𝐴 ∈ ℝ)       (𝜑 → (∗‘𝐴) = 𝐴)
 
Theoremremul2d 10305 Real part of a product. (Contributed by Mario Carneiro, 29-May-2016.)
(𝜑𝐴 ∈ ℝ)    &   (𝜑𝐵 ∈ ℂ)       (𝜑 → (ℜ‘(𝐴 · 𝐵)) = (𝐴 · (ℜ‘𝐵)))
 
Theoremimmul2d 10306 Imaginary part of a product. (Contributed by Mario Carneiro, 29-May-2016.)
(𝜑𝐴 ∈ ℝ)    &   (𝜑𝐵 ∈ ℂ)       (𝜑 → (ℑ‘(𝐴 · 𝐵)) = (𝐴 · (ℑ‘𝐵)))
 
Theoremredivapd 10307 Real part of a division. Related to remul2 10206. (Contributed by Jim Kingdon, 15-Jun-2020.)
(𝜑𝐴 ∈ ℝ)    &   (𝜑𝐵 ∈ ℂ)    &   (𝜑𝐴 # 0)       (𝜑 → (ℜ‘(𝐵 / 𝐴)) = ((ℜ‘𝐵) / 𝐴))
 
Theoremimdivapd 10308 Imaginary part of a division. Related to remul2 10206. (Contributed by Jim Kingdon, 15-Jun-2020.)
(𝜑𝐴 ∈ ℝ)    &   (𝜑𝐵 ∈ ℂ)    &   (𝜑𝐴 # 0)       (𝜑 → (ℑ‘(𝐵 / 𝐴)) = ((ℑ‘𝐵) / 𝐴))
 
Theoremcrred 10309 The real part of a complex number representation. Definition 10-3.1 of [Gleason] p. 132. (Contributed by Mario Carneiro, 29-May-2016.)
(𝜑𝐴 ∈ ℝ)    &   (𝜑𝐵 ∈ ℝ)       (𝜑 → (ℜ‘(𝐴 + (i · 𝐵))) = 𝐴)
 
Theoremcrimd 10310 The imaginary part of a complex number representation. Definition 10-3.1 of [Gleason] p. 132. (Contributed by Mario Carneiro, 29-May-2016.)
(𝜑𝐴 ∈ ℝ)    &   (𝜑𝐵 ∈ ℝ)       (𝜑 → (ℑ‘(𝐴 + (i · 𝐵))) = 𝐵)
 
3.7.3  Sequence convergence
 
Theoremcaucvgrelemrec 10311* Two ways to express a reciprocal. (Contributed by Jim Kingdon, 20-Jul-2021.)
((𝐴 ∈ ℝ ∧ 𝐴 # 0) → (𝑟 ∈ ℝ (𝐴 · 𝑟) = 1) = (1 / 𝐴))
 
Theoremcaucvgrelemcau 10312* Lemma for caucvgre 10313. Converting the Cauchy condition. (Contributed by Jim Kingdon, 20-Jul-2021.)
(𝜑𝐹:ℕ⟶ℝ)    &   (𝜑 → ∀𝑛 ∈ ℕ ∀𝑘 ∈ (ℤ𝑛)((𝐹𝑛) < ((𝐹𝑘) + (1 / 𝑛)) ∧ (𝐹𝑘) < ((𝐹𝑛) + (1 / 𝑛))))       (𝜑 → ∀𝑛 ∈ ℕ ∀𝑘 ∈ ℕ (𝑛 < 𝑘 → ((𝐹𝑛) < ((𝐹𝑘) + (𝑟 ∈ ℝ (𝑛 · 𝑟) = 1)) ∧ (𝐹𝑘) < ((𝐹𝑛) + (𝑟 ∈ ℝ (𝑛 · 𝑟) = 1)))))
 
Theoremcaucvgre 10313* Convergence of real sequences.

A Cauchy sequence (as defined here, which has a rate of convergence built in) of real numbers converges to a real number. Specifically on rate of convergence, all terms after the nth term must be within 1 / 𝑛 of the nth term.

(Contributed by Jim Kingdon, 19-Jul-2021.)

(𝜑𝐹:ℕ⟶ℝ)    &   (𝜑 → ∀𝑛 ∈ ℕ ∀𝑘 ∈ (ℤ𝑛)((𝐹𝑛) < ((𝐹𝑘) + (1 / 𝑛)) ∧ (𝐹𝑘) < ((𝐹𝑛) + (1 / 𝑛))))       (𝜑 → ∃𝑦 ∈ ℝ ∀𝑥 ∈ ℝ+𝑗 ∈ ℕ ∀𝑖 ∈ (ℤ𝑗)((𝐹𝑖) < (𝑦 + 𝑥) ∧ 𝑦 < ((𝐹𝑖) + 𝑥)))
 
Theoremcvg1nlemcxze 10314 Lemma for cvg1n 10318. Rearranging an expression related to the rate of convergence. (Contributed by Jim Kingdon, 6-Aug-2021.)
(𝜑𝐶 ∈ ℝ+)    &   (𝜑𝑋 ∈ ℝ+)    &   (𝜑𝑍 ∈ ℕ)    &   (𝜑𝐸 ∈ ℕ)    &   (𝜑𝐴 ∈ ℕ)    &   (𝜑 → ((((𝐶 · 2) / 𝑋) / 𝑍) + 𝐴) < 𝐸)       (𝜑 → (𝐶 / (𝐸 · 𝑍)) < (𝑋 / 2))
 
Theoremcvg1nlemf 10315* Lemma for cvg1n 10318. The modified sequence 𝐺 is a sequence. (Contributed by Jim Kingdon, 1-Aug-2021.)
(𝜑𝐹:ℕ⟶ℝ)    &   (𝜑𝐶 ∈ ℝ+)    &   (𝜑 → ∀𝑛 ∈ ℕ ∀𝑘 ∈ (ℤ𝑛)((𝐹𝑛) < ((𝐹𝑘) + (𝐶 / 𝑛)) ∧ (𝐹𝑘) < ((𝐹𝑛) + (𝐶 / 𝑛))))    &   𝐺 = (𝑗 ∈ ℕ ↦ (𝐹‘(𝑗 · 𝑍)))    &   (𝜑𝑍 ∈ ℕ)    &   (𝜑𝐶 < 𝑍)       (𝜑𝐺:ℕ⟶ℝ)
 
Theoremcvg1nlemcau 10316* Lemma for cvg1n 10318. By selecting spaced out terms for the modified sequence 𝐺, the terms are within 1 / 𝑛 (without the constant 𝐶). (Contributed by Jim Kingdon, 1-Aug-2021.)
(𝜑𝐹:ℕ⟶ℝ)    &   (𝜑𝐶 ∈ ℝ+)    &   (𝜑 → ∀𝑛 ∈ ℕ ∀𝑘 ∈ (ℤ𝑛)((𝐹𝑛) < ((𝐹𝑘) + (𝐶 / 𝑛)) ∧ (𝐹𝑘) < ((𝐹𝑛) + (𝐶 / 𝑛))))    &   𝐺 = (𝑗 ∈ ℕ ↦ (𝐹‘(𝑗 · 𝑍)))    &   (𝜑𝑍 ∈ ℕ)    &   (𝜑𝐶 < 𝑍)       (𝜑 → ∀𝑛 ∈ ℕ ∀𝑘 ∈ (ℤ𝑛)((𝐺𝑛) < ((𝐺𝑘) + (1 / 𝑛)) ∧ (𝐺𝑘) < ((𝐺𝑛) + (1 / 𝑛))))
 
Theoremcvg1nlemres 10317* Lemma for cvg1n 10318. The original sequence 𝐹 has a limit (turns out it is the same as the limit of the modified sequence 𝐺). (Contributed by Jim Kingdon, 1-Aug-2021.)
(𝜑𝐹:ℕ⟶ℝ)    &   (𝜑𝐶 ∈ ℝ+)    &   (𝜑 → ∀𝑛 ∈ ℕ ∀𝑘 ∈ (ℤ𝑛)((𝐹𝑛) < ((𝐹𝑘) + (𝐶 / 𝑛)) ∧ (𝐹𝑘) < ((𝐹𝑛) + (𝐶 / 𝑛))))    &   𝐺 = (𝑗 ∈ ℕ ↦ (𝐹‘(𝑗 · 𝑍)))    &   (𝜑𝑍 ∈ ℕ)    &   (𝜑𝐶 < 𝑍)       (𝜑 → ∃𝑦 ∈ ℝ ∀𝑥 ∈ ℝ+𝑗 ∈ ℕ ∀𝑖 ∈ (ℤ𝑗)((𝐹𝑖) < (𝑦 + 𝑥) ∧ 𝑦 < ((𝐹𝑖) + 𝑥)))
 
Theoremcvg1n 10318* Convergence of real sequences.

This is a version of caucvgre 10313 with a constant multiplier 𝐶 on the rate of convergence. That is, all terms after the nth term must be within 𝐶 / 𝑛 of the nth term.

(Contributed by Jim Kingdon, 1-Aug-2021.)

(𝜑𝐹:ℕ⟶ℝ)    &   (𝜑𝐶 ∈ ℝ+)    &   (𝜑 → ∀𝑛 ∈ ℕ ∀𝑘 ∈ (ℤ𝑛)((𝐹𝑛) < ((𝐹𝑘) + (𝐶 / 𝑛)) ∧ (𝐹𝑘) < ((𝐹𝑛) + (𝐶 / 𝑛))))       (𝜑 → ∃𝑦 ∈ ℝ ∀𝑥 ∈ ℝ+𝑗 ∈ ℕ ∀𝑖 ∈ (ℤ𝑗)((𝐹𝑖) < (𝑦 + 𝑥) ∧ 𝑦 < ((𝐹𝑖) + 𝑥)))
 
Theoremuzin2 10319 The upper integers are closed under intersection. (Contributed by Mario Carneiro, 24-Dec-2013.)
((𝐴 ∈ ran ℤ𝐵 ∈ ran ℤ) → (𝐴𝐵) ∈ ran ℤ)
 
Theoremrexanuz 10320* Combine two different upper integer properties into one. (Contributed by Mario Carneiro, 25-Dec-2013.)
(∃𝑗 ∈ ℤ ∀𝑘 ∈ (ℤ𝑗)(𝜑𝜓) ↔ (∃𝑗 ∈ ℤ ∀𝑘 ∈ (ℤ𝑗)𝜑 ∧ ∃𝑗 ∈ ℤ ∀𝑘 ∈ (ℤ𝑗)𝜓))
 
Theoremrexfiuz 10321* Combine finitely many different upper integer properties into one. (Contributed by Mario Carneiro, 6-Jun-2014.)
(𝐴 ∈ Fin → (∃𝑗 ∈ ℤ ∀𝑘 ∈ (ℤ𝑗)∀𝑛𝐴 𝜑 ↔ ∀𝑛𝐴𝑗 ∈ ℤ ∀𝑘 ∈ (ℤ𝑗)𝜑))
 
Theoremrexuz3 10322* Restrict the base of the upper integers set to another upper integers set. (Contributed by Mario Carneiro, 26-Dec-2013.)
𝑍 = (ℤ𝑀)       (𝑀 ∈ ℤ → (∃𝑗𝑍𝑘 ∈ (ℤ𝑗)𝜑 ↔ ∃𝑗 ∈ ℤ ∀𝑘 ∈ (ℤ𝑗)𝜑))
 
Theoremrexanuz2 10323* Combine two different upper integer properties into one. (Contributed by Mario Carneiro, 26-Dec-2013.)
𝑍 = (ℤ𝑀)       (∃𝑗𝑍𝑘 ∈ (ℤ𝑗)(𝜑𝜓) ↔ (∃𝑗𝑍𝑘 ∈ (ℤ𝑗)𝜑 ∧ ∃𝑗𝑍𝑘 ∈ (ℤ𝑗)𝜓))
 
Theoremr19.29uz 10324* A version of 19.29 1554 for upper integer quantifiers. (Contributed by Mario Carneiro, 10-Feb-2014.)
𝑍 = (ℤ𝑀)       ((∀𝑘𝑍 𝜑 ∧ ∃𝑗𝑍𝑘 ∈ (ℤ𝑗)𝜓) → ∃𝑗𝑍𝑘 ∈ (ℤ𝑗)(𝜑𝜓))
 
Theoremr19.2uz 10325* A version of r19.2m 3356 for upper integer quantifiers. (Contributed by Mario Carneiro, 15-Feb-2014.)
𝑍 = (ℤ𝑀)       (∃𝑗𝑍𝑘 ∈ (ℤ𝑗)𝜑 → ∃𝑘𝑍 𝜑)
 
Theoremrecvguniqlem 10326 Lemma for recvguniq 10327. Some of the rearrangements of the expressions. (Contributed by Jim Kingdon, 8-Aug-2021.)
(𝜑𝐹:ℕ⟶ℝ)    &   (𝜑𝐴 ∈ ℝ)    &   (𝜑𝐵 ∈ ℝ)    &   (𝜑𝐾 ∈ ℕ)    &   (𝜑𝐴 < ((𝐹𝐾) + ((𝐴𝐵) / 2)))    &   (𝜑 → (𝐹𝐾) < (𝐵 + ((𝐴𝐵) / 2)))       (𝜑 → ⊥)
 
Theoremrecvguniq 10327* Limits are unique. (Contributed by Jim Kingdon, 7-Aug-2021.)
(𝜑𝐹:ℕ⟶ℝ)    &   (𝜑𝐿 ∈ ℝ)    &   (𝜑 → ∀𝑥 ∈ ℝ+𝑗 ∈ ℕ ∀𝑘 ∈ (ℤ𝑗)((𝐹𝑘) < (𝐿 + 𝑥) ∧ 𝐿 < ((𝐹𝑘) + 𝑥)))    &   (𝜑𝑀 ∈ ℝ)    &   (𝜑 → ∀𝑥 ∈ ℝ+𝑗 ∈ ℕ ∀𝑘 ∈ (ℤ𝑗)((𝐹𝑘) < (𝑀 + 𝑥) ∧ 𝑀 < ((𝐹𝑘) + 𝑥)))       (𝜑𝐿 = 𝑀)
 
3.7.4  Square root; absolute value
 
Syntaxcsqrt 10328 Extend class notation to include square root of a complex number.
class
 
Syntaxcabs 10329 Extend class notation to include a function for the absolute value (modulus) of a complex number.
class abs
 
Definitiondf-rsqrt 10330* Define a function whose value is the square root of a nonnegative real number.

Defining the square root for complex numbers has one difficult part: choosing between the two roots. The usual way to define a principal square root for all complex numbers relies on excluded middle or something similar. But in the case of a nonnegative real number, we don't have the complications presented for general complex numbers, and we can choose the nonnegative root.

(Contributed by Jim Kingdon, 23-Aug-2020.)

√ = (𝑥 ∈ ℝ ↦ (𝑦 ∈ ℝ ((𝑦↑2) = 𝑥 ∧ 0 ≤ 𝑦)))
 
Definitiondf-abs 10331 Define the function for the absolute value (modulus) of a complex number. (Contributed by NM, 27-Jul-1999.)
abs = (𝑥 ∈ ℂ ↦ (√‘(𝑥 · (∗‘𝑥))))
 
Theoremsqrtrval 10332* Value of square root function. (Contributed by Jim Kingdon, 23-Aug-2020.)
(𝐴 ∈ ℝ → (√‘𝐴) = (𝑥 ∈ ℝ ((𝑥↑2) = 𝐴 ∧ 0 ≤ 𝑥)))
 
Theoremabsval 10333 The absolute value (modulus) of a complex number. Proposition 10-3.7(a) of [Gleason] p. 133. (Contributed by NM, 27-Jul-1999.) (Revised by Mario Carneiro, 7-Nov-2013.)
(𝐴 ∈ ℂ → (abs‘𝐴) = (√‘(𝐴 · (∗‘𝐴))))
 
Theoremrennim 10334 A real number does not lie on the negative imaginary axis. (Contributed by Mario Carneiro, 8-Jul-2013.)
(𝐴 ∈ ℝ → (i · 𝐴) ∉ ℝ+)
 
Theoremsqrt0rlem 10335 Lemma for sqrt0 10336. (Contributed by Jim Kingdon, 26-Aug-2020.)
((𝐴 ∈ ℝ ∧ ((𝐴↑2) = 0 ∧ 0 ≤ 𝐴)) ↔ 𝐴 = 0)
 
Theoremsqrt0 10336 Square root of zero. (Contributed by Mario Carneiro, 9-Jul-2013.)
(√‘0) = 0
 
Theoremresqrexlem1arp 10337* Lemma for resqrex 10358. 1 + 𝐴 is a positive real (expressed in a way that will help apply iseqfcl 9796 and similar theorems). (Contributed by Jim Kingdon, 28-Jul-2021.)
𝐹 = seq1((𝑦 ∈ ℝ+, 𝑧 ∈ ℝ+ ↦ ((𝑦 + (𝐴 / 𝑦)) / 2)), (ℕ × {(1 + 𝐴)}), ℝ+)    &   (𝜑𝐴 ∈ ℝ)    &   (𝜑 → 0 ≤ 𝐴)       ((𝜑𝑁 ∈ ℕ) → ((ℕ × {(1 + 𝐴)})‘𝑁) ∈ ℝ+)
 
Theoremresqrexlemp1rp 10338* Lemma for resqrex 10358. Applying the recursion rule yields a positive real (expressed in a way that will help apply iseqfcl 9796 and similar theorems). (Contributed by Jim Kingdon, 28-Jul-2021.)
𝐹 = seq1((𝑦 ∈ ℝ+, 𝑧 ∈ ℝ+ ↦ ((𝑦 + (𝐴 / 𝑦)) / 2)), (ℕ × {(1 + 𝐴)}), ℝ+)    &   (𝜑𝐴 ∈ ℝ)    &   (𝜑 → 0 ≤ 𝐴)       ((𝜑 ∧ (𝐵 ∈ ℝ+𝐶 ∈ ℝ+)) → (𝐵(𝑦 ∈ ℝ+, 𝑧 ∈ ℝ+ ↦ ((𝑦 + (𝐴 / 𝑦)) / 2))𝐶) ∈ ℝ+)
 
Theoremresqrexlemf 10339* Lemma for resqrex 10358. The sequence is a function. (Contributed by Mario Carneiro and Jim Kingdon, 27-Jul-2021.)
𝐹 = seq1((𝑦 ∈ ℝ+, 𝑧 ∈ ℝ+ ↦ ((𝑦 + (𝐴 / 𝑦)) / 2)), (ℕ × {(1 + 𝐴)}), ℝ+)    &   (𝜑𝐴 ∈ ℝ)    &   (𝜑 → 0 ≤ 𝐴)       (𝜑𝐹:ℕ⟶ℝ+)
 
Theoremresqrexlemf1 10340* Lemma for resqrex 10358. Initial value. Although this sequence converges to the square root with any positive initial value, this choice makes various steps in the proof of convergence easier. (Contributed by Mario Carneiro and Jim Kingdon, 27-Jul-2021.)
𝐹 = seq1((𝑦 ∈ ℝ+, 𝑧 ∈ ℝ+ ↦ ((𝑦 + (𝐴 / 𝑦)) / 2)), (ℕ × {(1 + 𝐴)}), ℝ+)    &   (𝜑𝐴 ∈ ℝ)    &   (𝜑 → 0 ≤ 𝐴)       (𝜑 → (𝐹‘1) = (1 + 𝐴))
 
Theoremresqrexlemfp1 10341* Lemma for resqrex 10358. Recursion rule. This sequence is the ancient method for computing square roots, often known as the babylonian method, although known to many ancient cultures. (Contributed by Mario Carneiro and Jim Kingdon, 27-Jul-2021.)
𝐹 = seq1((𝑦 ∈ ℝ+, 𝑧 ∈ ℝ+ ↦ ((𝑦 + (𝐴 / 𝑦)) / 2)), (ℕ × {(1 + 𝐴)}), ℝ+)    &   (𝜑𝐴 ∈ ℝ)    &   (𝜑 → 0 ≤ 𝐴)       ((𝜑𝑁 ∈ ℕ) → (𝐹‘(𝑁 + 1)) = (((𝐹𝑁) + (𝐴 / (𝐹𝑁))) / 2))
 
Theoremresqrexlemover 10342* Lemma for resqrex 10358. Each element of the sequence is an overestimate. (Contributed by Mario Carneiro and Jim Kingdon, 27-Jul-2021.)
𝐹 = seq1((𝑦 ∈ ℝ+, 𝑧 ∈ ℝ+ ↦ ((𝑦 + (𝐴 / 𝑦)) / 2)), (ℕ × {(1 + 𝐴)}), ℝ+)    &   (𝜑𝐴 ∈ ℝ)    &   (𝜑 → 0 ≤ 𝐴)       ((𝜑𝑁 ∈ ℕ) → 𝐴 < ((𝐹𝑁)↑2))
 
Theoremresqrexlemdec 10343* Lemma for resqrex 10358. The sequence is decreasing. (Contributed by Mario Carneiro and Jim Kingdon, 29-Jul-2021.)
𝐹 = seq1((𝑦 ∈ ℝ+, 𝑧 ∈ ℝ+ ↦ ((𝑦 + (𝐴 / 𝑦)) / 2)), (ℕ × {(1 + 𝐴)}), ℝ+)    &   (𝜑𝐴 ∈ ℝ)    &   (𝜑 → 0 ≤ 𝐴)       ((𝜑𝑁 ∈ ℕ) → (𝐹‘(𝑁 + 1)) < (𝐹𝑁))
 
Theoremresqrexlemdecn 10344* Lemma for resqrex 10358. The sequence is decreasing. (Contributed by Jim Kingdon, 31-Jul-2021.)
𝐹 = seq1((𝑦 ∈ ℝ+, 𝑧 ∈ ℝ+ ↦ ((𝑦 + (𝐴 / 𝑦)) / 2)), (ℕ × {(1 + 𝐴)}), ℝ+)    &   (𝜑𝐴 ∈ ℝ)    &   (𝜑 → 0 ≤ 𝐴)    &   (𝜑𝑁 ∈ ℕ)    &   (𝜑𝑀 ∈ ℕ)    &   (𝜑𝑁 < 𝑀)       (𝜑 → (𝐹𝑀) < (𝐹𝑁))
 
Theoremresqrexlemlo 10345* Lemma for resqrex 10358. A (variable) lower bound for each term of the sequence. (Contributed by Mario Carneiro and Jim Kingdon, 29-Jul-2021.)
𝐹 = seq1((𝑦 ∈ ℝ+, 𝑧 ∈ ℝ+ ↦ ((𝑦 + (𝐴 / 𝑦)) / 2)), (ℕ × {(1 + 𝐴)}), ℝ+)    &   (𝜑𝐴 ∈ ℝ)    &   (𝜑 → 0 ≤ 𝐴)       ((𝜑𝑁 ∈ ℕ) → (1 / (2↑𝑁)) < (𝐹𝑁))
 
Theoremresqrexlemcalc1 10346* Lemma for resqrex 10358. Some of the calculations involved in showing that the sequence converges. (Contributed by Mario Carneiro and Jim Kingdon, 29-Jul-2021.)
𝐹 = seq1((𝑦 ∈ ℝ+, 𝑧 ∈ ℝ+ ↦ ((𝑦 + (𝐴 / 𝑦)) / 2)), (ℕ × {(1 + 𝐴)}), ℝ+)    &   (𝜑𝐴 ∈ ℝ)    &   (𝜑 → 0 ≤ 𝐴)       ((𝜑𝑁 ∈ ℕ) → (((𝐹‘(𝑁 + 1))↑2) − 𝐴) = (((((𝐹𝑁)↑2) − 𝐴)↑2) / (4 · ((𝐹𝑁)↑2))))
 
Theoremresqrexlemcalc2 10347* Lemma for resqrex 10358. Some of the calculations involved in showing that the sequence converges. (Contributed by Mario Carneiro and Jim Kingdon, 29-Jul-2021.)
𝐹 = seq1((𝑦 ∈ ℝ+, 𝑧 ∈ ℝ+ ↦ ((𝑦 + (𝐴 / 𝑦)) / 2)), (ℕ × {(1 + 𝐴)}), ℝ+)    &   (𝜑𝐴 ∈ ℝ)    &   (𝜑 → 0 ≤ 𝐴)       ((𝜑𝑁 ∈ ℕ) → (((𝐹‘(𝑁 + 1))↑2) − 𝐴) ≤ ((((𝐹𝑁)↑2) − 𝐴) / 4))
 
Theoremresqrexlemcalc3 10348* Lemma for resqrex 10358. Some of the calculations involved in showing that the sequence converges. (Contributed by Mario Carneiro and Jim Kingdon, 29-Jul-2021.)
𝐹 = seq1((𝑦 ∈ ℝ+, 𝑧 ∈ ℝ+ ↦ ((𝑦 + (𝐴 / 𝑦)) / 2)), (ℕ × {(1 + 𝐴)}), ℝ+)    &   (𝜑𝐴 ∈ ℝ)    &   (𝜑 → 0 ≤ 𝐴)       ((𝜑𝑁 ∈ ℕ) → (((𝐹𝑁)↑2) − 𝐴) ≤ (((𝐹‘1)↑2) / (4↑(𝑁 − 1))))
 
Theoremresqrexlemnmsq 10349* Lemma for resqrex 10358. The difference between the squares of two terms of the sequence. (Contributed by Mario Carneiro and Jim Kingdon, 30-Jul-2021.)
𝐹 = seq1((𝑦 ∈ ℝ+, 𝑧 ∈ ℝ+ ↦ ((𝑦 + (𝐴 / 𝑦)) / 2)), (ℕ × {(1 + 𝐴)}), ℝ+)    &   (𝜑𝐴 ∈ ℝ)    &   (𝜑 → 0 ≤ 𝐴)    &   (𝜑𝑁 ∈ ℕ)    &   (𝜑𝑀 ∈ ℕ)    &   (𝜑𝑁𝑀)       (𝜑 → (((𝐹𝑁)↑2) − ((𝐹𝑀)↑2)) < (((𝐹‘1)↑2) / (4↑(𝑁 − 1))))
 
Theoremresqrexlemnm 10350* Lemma for resqrex 10358. The difference between two terms of the sequence. (Contributed by Mario Carneiro and Jim Kingdon, 31-Jul-2021.)
𝐹 = seq1((𝑦 ∈ ℝ+, 𝑧 ∈ ℝ+ ↦ ((𝑦 + (𝐴 / 𝑦)) / 2)), (ℕ × {(1 + 𝐴)}), ℝ+)    &   (𝜑𝐴 ∈ ℝ)    &   (𝜑 → 0 ≤ 𝐴)    &   (𝜑𝑁 ∈ ℕ)    &   (𝜑𝑀 ∈ ℕ)    &   (𝜑𝑁𝑀)       (𝜑 → ((𝐹𝑁) − (𝐹𝑀)) < ((((𝐹‘1)↑2) · 2) / (2↑(𝑁 − 1))))
 
Theoremresqrexlemcvg 10351* Lemma for resqrex 10358. The sequence has a limit. (Contributed by Jim Kingdon, 6-Aug-2021.)
𝐹 = seq1((𝑦 ∈ ℝ+, 𝑧 ∈ ℝ+ ↦ ((𝑦 + (𝐴 / 𝑦)) / 2)), (ℕ × {(1 + 𝐴)}), ℝ+)    &   (𝜑𝐴 ∈ ℝ)    &   (𝜑 → 0 ≤ 𝐴)       (𝜑 → ∃𝑟 ∈ ℝ ∀𝑥 ∈ ℝ+𝑗 ∈ ℕ ∀𝑖 ∈ (ℤ𝑗)((𝐹𝑖) < (𝑟 + 𝑥) ∧ 𝑟 < ((𝐹𝑖) + 𝑥)))
 
Theoremresqrexlemgt0 10352* Lemma for resqrex 10358. A limit is nonnegative. (Contributed by Jim Kingdon, 7-Aug-2021.)
𝐹 = seq1((𝑦 ∈ ℝ+, 𝑧 ∈ ℝ+ ↦ ((𝑦 + (𝐴 / 𝑦)) / 2)), (ℕ × {(1 + 𝐴)}), ℝ+)    &   (𝜑𝐴 ∈ ℝ)    &   (𝜑 → 0 ≤ 𝐴)    &   (𝜑𝐿 ∈ ℝ)    &   (𝜑 → ∀𝑒 ∈ ℝ+𝑗 ∈ ℕ ∀𝑖 ∈ (ℤ𝑗)((𝐹𝑖) < (𝐿 + 𝑒) ∧ 𝐿 < ((𝐹𝑖) + 𝑒)))       (𝜑 → 0 ≤ 𝐿)
 
Theoremresqrexlemoverl 10353* Lemma for resqrex 10358. Every term in the sequence is an overestimate compared with the limit 𝐿. Although this theorem is stated in terms of a particular sequence the proof could be adapted for any decreasing convergent sequence. (Contributed by Jim Kingdon, 9-Aug-2021.)
𝐹 = seq1((𝑦 ∈ ℝ+, 𝑧 ∈ ℝ+ ↦ ((𝑦 + (𝐴 / 𝑦)) / 2)), (ℕ × {(1 + 𝐴)}), ℝ+)    &   (𝜑𝐴 ∈ ℝ)    &   (𝜑 → 0 ≤ 𝐴)    &   (𝜑𝐿 ∈ ℝ)    &   (𝜑 → ∀𝑒 ∈ ℝ+𝑗 ∈ ℕ ∀𝑖 ∈ (ℤ𝑗)((𝐹𝑖) < (𝐿 + 𝑒) ∧ 𝐿 < ((𝐹𝑖) + 𝑒)))    &   (𝜑𝐾 ∈ ℕ)       (𝜑𝐿 ≤ (𝐹𝐾))
 
Theoremresqrexlemglsq 10354* Lemma for resqrex 10358. The sequence formed by squaring each term of 𝐹 converges to (𝐿↑2). (Contributed by Mario Carneiro and Jim Kingdon, 8-Aug-2021.)
𝐹 = seq1((𝑦 ∈ ℝ+, 𝑧 ∈ ℝ+ ↦ ((𝑦 + (𝐴 / 𝑦)) / 2)), (ℕ × {(1 + 𝐴)}), ℝ+)    &   (𝜑𝐴 ∈ ℝ)    &   (𝜑 → 0 ≤ 𝐴)    &   (𝜑𝐿 ∈ ℝ)    &   (𝜑 → ∀𝑒 ∈ ℝ+𝑗 ∈ ℕ ∀𝑖 ∈ (ℤ𝑗)((𝐹𝑖) < (𝐿 + 𝑒) ∧ 𝐿 < ((𝐹𝑖) + 𝑒)))    &   𝐺 = (𝑥 ∈ ℕ ↦ ((𝐹𝑥)↑2))       (𝜑 → ∀𝑒 ∈ ℝ+𝑗 ∈ ℕ ∀𝑘 ∈ (ℤ𝑗)((𝐺𝑘) < ((𝐿↑2) + 𝑒) ∧ (𝐿↑2) < ((𝐺𝑘) + 𝑒)))
 
Theoremresqrexlemga 10355* Lemma for resqrex 10358. The sequence formed by squaring each term of 𝐹 converges to 𝐴. (Contributed by Mario Carneiro and Jim Kingdon, 8-Aug-2021.)
𝐹 = seq1((𝑦 ∈ ℝ+, 𝑧 ∈ ℝ+ ↦ ((𝑦 + (𝐴 / 𝑦)) / 2)), (ℕ × {(1 + 𝐴)}), ℝ+)    &   (𝜑𝐴 ∈ ℝ)    &   (𝜑 → 0 ≤ 𝐴)    &   (𝜑𝐿 ∈ ℝ)    &   (𝜑 → ∀𝑒 ∈ ℝ+𝑗 ∈ ℕ ∀𝑖 ∈ (ℤ𝑗)((𝐹𝑖) < (𝐿 + 𝑒) ∧ 𝐿 < ((𝐹𝑖) + 𝑒)))    &   𝐺 = (𝑥 ∈ ℕ ↦ ((𝐹𝑥)↑2))       (𝜑 → ∀𝑒 ∈ ℝ+𝑗 ∈ ℕ ∀𝑘 ∈ (ℤ𝑗)((𝐺𝑘) < (𝐴 + 𝑒) ∧ 𝐴 < ((𝐺𝑘) + 𝑒)))
 
Theoremresqrexlemsqa 10356* Lemma for resqrex 10358. The square of a limit is 𝐴. (Contributed by Jim Kingdon, 7-Aug-2021.)
𝐹 = seq1((𝑦 ∈ ℝ+, 𝑧 ∈ ℝ+ ↦ ((𝑦 + (𝐴 / 𝑦)) / 2)), (ℕ × {(1 + 𝐴)}), ℝ+)    &   (𝜑𝐴 ∈ ℝ)    &   (𝜑 → 0 ≤ 𝐴)    &   (𝜑𝐿 ∈ ℝ)    &   (𝜑 → ∀𝑒 ∈ ℝ+𝑗 ∈ ℕ ∀𝑖 ∈ (ℤ𝑗)((𝐹𝑖) < (𝐿 + 𝑒) ∧ 𝐿 < ((𝐹𝑖) + 𝑒)))       (𝜑 → (𝐿↑2) = 𝐴)
 
Theoremresqrexlemex 10357* Lemma for resqrex 10358. Existence of square root given a sequence which converges to the square root. (Contributed by Mario Carneiro and Jim Kingdon, 27-Jul-2021.)
𝐹 = seq1((𝑦 ∈ ℝ+, 𝑧 ∈ ℝ+ ↦ ((𝑦 + (𝐴 / 𝑦)) / 2)), (ℕ × {(1 + 𝐴)}), ℝ+)    &   (𝜑𝐴 ∈ ℝ)    &   (𝜑 → 0 ≤ 𝐴)       (𝜑 → ∃𝑥 ∈ ℝ (0 ≤ 𝑥 ∧ (𝑥↑2) = 𝐴))
 
Theoremresqrex 10358* Existence of a square root for positive reals. (Contributed by Mario Carneiro, 9-Jul-2013.)
((𝐴 ∈ ℝ ∧ 0 ≤ 𝐴) → ∃𝑥 ∈ ℝ (0 ≤ 𝑥 ∧ (𝑥↑2) = 𝐴))
 
Theoremrsqrmo 10359* Uniqueness for the square root function. (Contributed by Jim Kingdon, 10-Aug-2021.)
((𝐴 ∈ ℝ ∧ 0 ≤ 𝐴) → ∃*𝑥 ∈ ℝ ((𝑥↑2) = 𝐴 ∧ 0 ≤ 𝑥))
 
Theoremrersqreu 10360* Existence and uniqueness for the real square root function. (Contributed by Jim Kingdon, 10-Aug-2021.)
((𝐴 ∈ ℝ ∧ 0 ≤ 𝐴) → ∃!𝑥 ∈ ℝ ((𝑥↑2) = 𝐴 ∧ 0 ≤ 𝑥))
 
Theoremresqrtcl 10361 Closure of the square root function. (Contributed by Mario Carneiro, 9-Jul-2013.)
((𝐴 ∈ ℝ ∧ 0 ≤ 𝐴) → (√‘𝐴) ∈ ℝ)
 
Theoremrersqrtthlem 10362 Lemma for resqrtth 10363. (Contributed by Jim Kingdon, 10-Aug-2021.)
((𝐴 ∈ ℝ ∧ 0 ≤ 𝐴) → (((√‘𝐴)↑2) = 𝐴 ∧ 0 ≤ (√‘𝐴)))
 
Theoremresqrtth 10363 Square root theorem over the reals. Theorem I.35 of [Apostol] p. 29. (Contributed by Mario Carneiro, 9-Jul-2013.)
((𝐴 ∈ ℝ ∧ 0 ≤ 𝐴) → ((√‘𝐴)↑2) = 𝐴)
 
Theoremremsqsqrt 10364 Square of square root. (Contributed by Mario Carneiro, 10-Jul-2013.)
((𝐴 ∈ ℝ ∧ 0 ≤ 𝐴) → ((√‘𝐴) · (√‘𝐴)) = 𝐴)
 
Theoremsqrtge0 10365 The square root function is nonnegative for nonnegative input. (Contributed by NM, 26-May-1999.) (Revised by Mario Carneiro, 9-Jul-2013.)
((𝐴 ∈ ℝ ∧ 0 ≤ 𝐴) → 0 ≤ (√‘𝐴))
 
Theoremsqrtgt0 10366 The square root function is positive for positive input. (Contributed by Mario Carneiro, 10-Jul-2013.) (Revised by Mario Carneiro, 6-Sep-2013.)
((𝐴 ∈ ℝ ∧ 0 < 𝐴) → 0 < (√‘𝐴))
 
Theoremsqrtmul 10367 Square root distributes over multiplication. (Contributed by NM, 30-Jul-1999.) (Revised by Mario Carneiro, 29-May-2016.)
(((𝐴 ∈ ℝ ∧ 0 ≤ 𝐴) ∧ (𝐵 ∈ ℝ ∧ 0 ≤ 𝐵)) → (√‘(𝐴 · 𝐵)) = ((√‘𝐴) · (√‘𝐵)))
 
Theoremsqrtle 10368 Square root is monotonic. (Contributed by NM, 17-Mar-2005.) (Proof shortened by Mario Carneiro, 29-May-2016.)
(((𝐴 ∈ ℝ ∧ 0 ≤ 𝐴) ∧ (𝐵 ∈ ℝ ∧ 0 ≤ 𝐵)) → (𝐴𝐵 ↔ (√‘𝐴) ≤ (√‘𝐵)))
 
Theoremsqrtlt 10369 Square root is strictly monotonic. Closed form of sqrtlti 10469. (Contributed by Scott Fenton, 17-Apr-2014.) (Proof shortened by Mario Carneiro, 29-May-2016.)
(((𝐴 ∈ ℝ ∧ 0 ≤ 𝐴) ∧ (𝐵 ∈ ℝ ∧ 0 ≤ 𝐵)) → (𝐴 < 𝐵 ↔ (√‘𝐴) < (√‘𝐵)))
 
Theoremsqrt11ap 10370 Analogue to sqrt11 10371 but for apartness. (Contributed by Jim Kingdon, 11-Aug-2021.)
(((𝐴 ∈ ℝ ∧ 0 ≤ 𝐴) ∧ (𝐵 ∈ ℝ ∧ 0 ≤ 𝐵)) → ((√‘𝐴) # (√‘𝐵) ↔ 𝐴 # 𝐵))
 
Theoremsqrt11 10371 The square root function is one-to-one. Also see sqrt11ap 10370 which would follow easily from this given excluded middle, but which is proved another way without it. (Contributed by Scott Fenton, 11-Jun-2013.)
(((𝐴 ∈ ℝ ∧ 0 ≤ 𝐴) ∧ (𝐵 ∈ ℝ ∧ 0 ≤ 𝐵)) → ((√‘𝐴) = (√‘𝐵) ↔ 𝐴 = 𝐵))
 
Theoremsqrt00 10372 A square root is zero iff its argument is 0. (Contributed by NM, 27-Jul-1999.) (Proof shortened by Mario Carneiro, 29-May-2016.)
((𝐴 ∈ ℝ ∧ 0 ≤ 𝐴) → ((√‘𝐴) = 0 ↔ 𝐴 = 0))
 
Theoremrpsqrtcl 10373 The square root of a positive real is a positive real. (Contributed by NM, 22-Feb-2008.)
(𝐴 ∈ ℝ+ → (√‘𝐴) ∈ ℝ+)
 
Theoremsqrtdiv 10374 Square root distributes over division. (Contributed by Mario Carneiro, 5-May-2016.)
(((𝐴 ∈ ℝ ∧ 0 ≤ 𝐴) ∧ 𝐵 ∈ ℝ+) → (√‘(𝐴 / 𝐵)) = ((√‘𝐴) / (√‘𝐵)))
 
Theoremsqrtsq2 10375 Relationship between square root and squares. (Contributed by NM, 31-Jul-1999.) (Revised by Mario Carneiro, 29-May-2016.)
(((𝐴 ∈ ℝ ∧ 0 ≤ 𝐴) ∧ (𝐵 ∈ ℝ ∧ 0 ≤ 𝐵)) → ((√‘𝐴) = 𝐵𝐴 = (𝐵↑2)))
 
Theoremsqrtsq 10376 Square root of square. (Contributed by NM, 14-Jan-2006.) (Revised by Mario Carneiro, 29-May-2016.)
((𝐴 ∈ ℝ ∧ 0 ≤ 𝐴) → (√‘(𝐴↑2)) = 𝐴)
 
Theoremsqrtmsq 10377 Square root of square. (Contributed by NM, 2-Aug-1999.) (Revised by Mario Carneiro, 29-May-2016.)
((𝐴 ∈ ℝ ∧ 0 ≤ 𝐴) → (√‘(𝐴 · 𝐴)) = 𝐴)
 
Theoremsqrt1 10378 The square root of 1 is 1. (Contributed by NM, 31-Jul-1999.)
(√‘1) = 1
 
Theoremsqrt4 10379 The square root of 4 is 2. (Contributed by NM, 3-Aug-1999.)
(√‘4) = 2
 
Theoremsqrt9 10380 The square root of 9 is 3. (Contributed by NM, 11-May-2004.)
(√‘9) = 3
 
Theoremsqrt2gt1lt2 10381 The square root of 2 is bounded by 1 and 2. (Contributed by Roy F. Longton, 8-Aug-2005.) (Revised by Mario Carneiro, 6-Sep-2013.)
(1 < (√‘2) ∧ (√‘2) < 2)
 
Theoremabsneg 10382 Absolute value of negative. (Contributed by NM, 27-Feb-2005.)
(𝐴 ∈ ℂ → (abs‘-𝐴) = (abs‘𝐴))
 
Theoremabscl 10383 Real closure of absolute value. (Contributed by NM, 3-Oct-1999.)
(𝐴 ∈ ℂ → (abs‘𝐴) ∈ ℝ)
 
Theoremabscj 10384 The absolute value of a number and its conjugate are the same. Proposition 10-3.7(b) of [Gleason] p. 133. (Contributed by NM, 28-Apr-2005.)
(𝐴 ∈ ℂ → (abs‘(∗‘𝐴)) = (abs‘𝐴))
 
Theoremabsvalsq 10385 Square of value of absolute value function. (Contributed by NM, 16-Jan-2006.)
(𝐴 ∈ ℂ → ((abs‘𝐴)↑2) = (𝐴 · (∗‘𝐴)))
 
Theoremabsvalsq2 10386 Square of value of absolute value function. (Contributed by NM, 1-Feb-2007.)
(𝐴 ∈ ℂ → ((abs‘𝐴)↑2) = (((ℜ‘𝐴)↑2) + ((ℑ‘𝐴)↑2)))
 
Theoremsqabsadd 10387 Square of absolute value of sum. Proposition 10-3.7(g) of [Gleason] p. 133. (Contributed by NM, 21-Jan-2007.)
((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ) → ((abs‘(𝐴 + 𝐵))↑2) = ((((abs‘𝐴)↑2) + ((abs‘𝐵)↑2)) + (2 · (ℜ‘(𝐴 · (∗‘𝐵))))))
 
Theoremsqabssub 10388 Square of absolute value of difference. (Contributed by NM, 21-Jan-2007.)
((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ) → ((abs‘(𝐴𝐵))↑2) = ((((abs‘𝐴)↑2) + ((abs‘𝐵)↑2)) − (2 · (ℜ‘(𝐴 · (∗‘𝐵))))))
 
Theoremabsval2 10389 Value of absolute value function. Definition 10.36 of [Gleason] p. 133. (Contributed by NM, 17-Mar-2005.)
(𝐴 ∈ ℂ → (abs‘𝐴) = (√‘(((ℜ‘𝐴)↑2) + ((ℑ‘𝐴)↑2))))
 
Theoremabs0 10390 The absolute value of 0. (Contributed by NM, 26-Mar-2005.) (Revised by Mario Carneiro, 29-May-2016.)
(abs‘0) = 0
 
Theoremabsi 10391 The absolute value of the imaginary unit. (Contributed by NM, 26-Mar-2005.)
(abs‘i) = 1
 
Theoremabsge0 10392 Absolute value is nonnegative. (Contributed by NM, 20-Nov-2004.) (Revised by Mario Carneiro, 29-May-2016.)
(𝐴 ∈ ℂ → 0 ≤ (abs‘𝐴))
 
Theoremabsrpclap 10393 The absolute value of a number apart from zero is a positive real. (Contributed by Jim Kingdon, 11-Aug-2021.)
((𝐴 ∈ ℂ ∧ 𝐴 # 0) → (abs‘𝐴) ∈ ℝ+)
 
Theoremabs00ap 10394 The absolute value of a number is apart from zero iff the number is apart from zero. (Contributed by Jim Kingdon, 11-Aug-2021.)
(𝐴 ∈ ℂ → ((abs‘𝐴) # 0 ↔ 𝐴 # 0))
 
Theoremabsext 10395 Strong extensionality for absolute value. (Contributed by Jim Kingdon, 12-Aug-2021.)
((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ) → ((abs‘𝐴) # (abs‘𝐵) → 𝐴 # 𝐵))
 
Theoremabs00 10396 The absolute value of a number is zero iff the number is zero. Also see abs00ap 10394 which is similar but for apartness. Proposition 10-3.7(c) of [Gleason] p. 133. (Contributed by NM, 26-Sep-2005.) (Proof shortened by Mario Carneiro, 29-May-2016.)
(𝐴 ∈ ℂ → ((abs‘𝐴) = 0 ↔ 𝐴 = 0))
 
Theoremabs00ad 10397 A complex number is zero iff its absolute value is zero. Deduction form of abs00 10396. (Contributed by David Moews, 28-Feb-2017.)
(𝜑𝐴 ∈ ℂ)       (𝜑 → ((abs‘𝐴) = 0 ↔ 𝐴 = 0))
 
Theoremabs00bd 10398 If a complex number is zero, its absolute value is zero. (Contributed by David Moews, 28-Feb-2017.)
(𝜑𝐴 = 0)       (𝜑 → (abs‘𝐴) = 0)
 
Theoremabsreimsq 10399 Square of the absolute value of a number that has been decomposed into real and imaginary parts. (Contributed by NM, 1-Feb-2007.)
((𝐴 ∈ ℝ ∧ 𝐵 ∈ ℝ) → ((abs‘(𝐴 + (i · 𝐵)))↑2) = ((𝐴↑2) + (𝐵↑2)))
 
Theoremabsreim 10400 Absolute value of a number that has been decomposed into real and imaginary parts. (Contributed by NM, 14-Jan-2006.)
((𝐴 ∈ ℝ ∧ 𝐵 ∈ ℝ) → (abs‘(𝐴 + (i · 𝐵))) = (√‘((𝐴↑2) + (𝐵↑2))))
    < Previous  Next >

Page List
Jump to page: Contents  1 1-100 2 101-200 3 201-300 4 301-400 5 401-500 6 501-600 7 601-700 8 701-800 9 801-900 10 901-1000 11 1001-1100 12 1101-1200 13 1201-1300 14 1301-1400 15 1401-1500 16 1501-1600 17 1601-1700 18 1701-1800 19 1801-1900 20 1901-2000 21 2001-2100 22 2101-2200 23 2201-2300 24 2301-2400 25 2401-2500 26 2501-2600 27 2601-2700 28 2701-2800 29 2801-2900 30 2901-3000 31 3001-3100 32 3101-3200 33 3201-3300 34 3301-3400 35 3401-3500 36 3501-3600 37 3601-3700 38 3701-3800 39 3801-3900 40 3901-4000 41 4001-4100 42 4101-4200 43 4201-4300 44 4301-4400 45 4401-4500 46 4501-4600 47 4601-4700 48 4701-4800 49 4801-4900 50 4901-5000 51 5001-5100 52 5101-5200 53 5201-5300 54 5301-5400 55 5401-5500 56 5501-5600 57 5601-5700 58 5701-5800 59 5801-5900 60 5901-6000 61 6001-6100 62 6101-6200 63 6201-6300 64 6301-6400 65 6401-6500 66 6501-6600 67 6601-6700 68 6701-6800 69 6801-6900 70 6901-7000 71 7001-7100 72 7101-7200 73 7201-7300 74 7301-7400 75 7401-7500 76 7501-7600 77 7601-7700 78 7701-7800 79 7801-7900 80 7901-8000 81 8001-8100 82 8101-8200 83 8201-8300 84 8301-8400 85 8401-8500 86 8501-8600 87 8601-8700 88 8701-8800 89 8801-8900 90 8901-9000 91 9001-9100 92 9101-9200 93 9201-9300 94 9301-9400 95 9401-9500 96 9501-9600 97 9601-9700 98 9701-9800 99 9801-9900 100 9901-10000 101 10001-10100 102 10101-10200 103 10201-10300 104 10301-10400 105 10401-10500 106 10501-10600 107 10601-10700 108 10701-10800 109 10801-10900 110 10901-11000 111 11001-11100 112 11101-11200 113 11201-11300 114 11301-11394
  Copyright terms: Public domain < Previous  Next >