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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))))

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