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Theorem List for Intuitionistic Logic Explorer - 10001-10100   *Has distinct variable group(s)
TypeLabelDescription
Statement
 
Theoremcrred 10001 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 10002 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 10003* 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 10004* Lemma for caucvgre 10005. 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 10005* 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 10006 Lemma for cvg1n 10010. 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 10007* Lemma for cvg1n 10010. 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 10008* Lemma for cvg1n 10010. 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 10009* Lemma for cvg1n 10010. 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 10010* Convergence of real sequences.

This is a version of caucvgre 10005 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 10011 The upper integers are closed under intersection. (Contributed by Mario Carneiro, 24-Dec-2013.)
((𝐴 ∈ ran ℤ𝐵 ∈ ran ℤ) → (𝐴𝐵) ∈ ran ℤ)
 
Theoremrexanuz 10012* 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 10013* 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 10014* 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 10015* 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 10016* A version of 19.29 1552 for upper integer quantifiers. (Contributed by Mario Carneiro, 10-Feb-2014.)
𝑍 = (ℤ𝑀)       ((∀𝑘𝑍 𝜑 ∧ ∃𝑗𝑍𝑘 ∈ (ℤ𝑗)𝜓) → ∃𝑗𝑍𝑘 ∈ (ℤ𝑗)(𝜑𝜓))
 
Theoremr19.2uz 10017* A version of r19.2m 3336 for upper integer quantifiers. (Contributed by Mario Carneiro, 15-Feb-2014.)
𝑍 = (ℤ𝑀)       (∃𝑗𝑍𝑘 ∈ (ℤ𝑗)𝜑 → ∃𝑘𝑍 𝜑)
 
Theoremrecvguniqlem 10018 Lemma for recvguniq 10019. 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 10019* 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 10020 Extend class notation to include square root of a complex number.
class
 
Syntaxcabs 10021 Extend class notation to include a function for the absolute value (modulus) of a complex number.
class abs
 
Definitiondf-rsqrt 10022* 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 10023 Define the function for the absolute value (modulus) of a complex number. (Contributed by NM, 27-Jul-1999.)
abs = (𝑥 ∈ ℂ ↦ (√‘(𝑥 · (∗‘𝑥))))
 
Theoremsqrtrval 10024* Value of square root function. (Contributed by Jim Kingdon, 23-Aug-2020.)
(𝐴 ∈ ℝ → (√‘𝐴) = (𝑥 ∈ ℝ ((𝑥↑2) = 𝐴 ∧ 0 ≤ 𝑥)))
 
Theoremabsval 10025 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 10026 A real number does not lie on the negative imaginary axis. (Contributed by Mario Carneiro, 8-Jul-2013.)
(𝐴 ∈ ℝ → (i · 𝐴) ∉ ℝ+)
 
Theoremsqrt0rlem 10027 Lemma for sqrt0 10028. (Contributed by Jim Kingdon, 26-Aug-2020.)
((𝐴 ∈ ℝ ∧ ((𝐴↑2) = 0 ∧ 0 ≤ 𝐴)) ↔ 𝐴 = 0)
 
Theoremsqrt0 10028 Square root of zero. (Contributed by Mario Carneiro, 9-Jul-2013.)
(√‘0) = 0
 
Theoremresqrexlem1arp 10029* Lemma for resqrex 10050. 1 + 𝐴 is a positive real (expressed in a way that will help apply iseqfcl 9535 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 10030* Lemma for resqrex 10050. Applying the recursion rule yields a positive real (expressed in a way that will help apply iseqfcl 9535 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 10031* Lemma for resqrex 10050. 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 10032* Lemma for resqrex 10050. 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 10033* Lemma for resqrex 10050. 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 10034* Lemma for resqrex 10050. 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 10035* Lemma for resqrex 10050. 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 10036* Lemma for resqrex 10050. 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 10037* Lemma for resqrex 10050. 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 10038* Lemma for resqrex 10050. 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 10039* Lemma for resqrex 10050. 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 10040* Lemma for resqrex 10050. 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 10041* Lemma for resqrex 10050. 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 10042* Lemma for resqrex 10050. 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 10043* Lemma for resqrex 10050. 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 10044* Lemma for resqrex 10050. 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 10045* Lemma for resqrex 10050. 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 10046* Lemma for resqrex 10050. 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 10047* Lemma for resqrex 10050. 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 10048* Lemma for resqrex 10050. 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 10049* Lemma for resqrex 10050. 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 10050* 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 10051* 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 10052* 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 10053 Closure of the square root function. (Contributed by Mario Carneiro, 9-Jul-2013.)
((𝐴 ∈ ℝ ∧ 0 ≤ 𝐴) → (√‘𝐴) ∈ ℝ)
 
Theoremrersqrtthlem 10054 Lemma for resqrtth 10055. (Contributed by Jim Kingdon, 10-Aug-2021.)
((𝐴 ∈ ℝ ∧ 0 ≤ 𝐴) → (((√‘𝐴)↑2) = 𝐴 ∧ 0 ≤ (√‘𝐴)))
 
Theoremresqrtth 10055 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 10056 Square of square root. (Contributed by Mario Carneiro, 10-Jul-2013.)
((𝐴 ∈ ℝ ∧ 0 ≤ 𝐴) → ((√‘𝐴) · (√‘𝐴)) = 𝐴)
 
Theoremsqrtge0 10057 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 10058 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 10059 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 10060 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 10061 Square root is strictly monotonic. Closed form of sqrtlti 10161. (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 10062 Analogue to sqrt11 10063 but for apartness. (Contributed by Jim Kingdon, 11-Aug-2021.)
(((𝐴 ∈ ℝ ∧ 0 ≤ 𝐴) ∧ (𝐵 ∈ ℝ ∧ 0 ≤ 𝐵)) → ((√‘𝐴) # (√‘𝐵) ↔ 𝐴 # 𝐵))
 
Theoremsqrt11 10063 The square root function is one-to-one. Also see sqrt11ap 10062 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 10064 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 10065 The square root of a positive real is a positive real. (Contributed by NM, 22-Feb-2008.)
(𝐴 ∈ ℝ+ → (√‘𝐴) ∈ ℝ+)
 
Theoremsqrtdiv 10066 Square root distributes over division. (Contributed by Mario Carneiro, 5-May-2016.)
(((𝐴 ∈ ℝ ∧ 0 ≤ 𝐴) ∧ 𝐵 ∈ ℝ+) → (√‘(𝐴 / 𝐵)) = ((√‘𝐴) / (√‘𝐵)))
 
Theoremsqrtsq2 10067 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 10068 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 10069 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 10070 The square root of 1 is 1. (Contributed by NM, 31-Jul-1999.)
(√‘1) = 1
 
Theoremsqrt4 10071 The square root of 4 is 2. (Contributed by NM, 3-Aug-1999.)
(√‘4) = 2
 
Theoremsqrt9 10072 The square root of 9 is 3. (Contributed by NM, 11-May-2004.)
(√‘9) = 3
 
Theoremsqrt2gt1lt2 10073 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 10074 Absolute value of negative. (Contributed by NM, 27-Feb-2005.)
(𝐴 ∈ ℂ → (abs‘-𝐴) = (abs‘𝐴))
 
Theoremabscl 10075 Real closure of absolute value. (Contributed by NM, 3-Oct-1999.)
(𝐴 ∈ ℂ → (abs‘𝐴) ∈ ℝ)
 
Theoremabscj 10076 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 10077 Square of value of absolute value function. (Contributed by NM, 16-Jan-2006.)
(𝐴 ∈ ℂ → ((abs‘𝐴)↑2) = (𝐴 · (∗‘𝐴)))
 
Theoremabsvalsq2 10078 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 10079 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 10080 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 10081 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 10082 The absolute value of 0. (Contributed by NM, 26-Mar-2005.) (Revised by Mario Carneiro, 29-May-2016.)
(abs‘0) = 0
 
Theoremabsi 10083 The absolute value of the imaginary unit. (Contributed by NM, 26-Mar-2005.)
(abs‘i) = 1
 
Theoremabsge0 10084 Absolute value is nonnegative. (Contributed by NM, 20-Nov-2004.) (Revised by Mario Carneiro, 29-May-2016.)
(𝐴 ∈ ℂ → 0 ≤ (abs‘𝐴))
 
Theoremabsrpclap 10085 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 10086 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 10087 Strong extensionality for absolute value. (Contributed by Jim Kingdon, 12-Aug-2021.)
((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ) → ((abs‘𝐴) # (abs‘𝐵) → 𝐴 # 𝐵))
 
Theoremabs00 10088 The absolute value of a number is zero iff the number is zero. Also see abs00ap 10086 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 10089 A complex number is zero iff its absolute value is zero. Deduction form of abs00 10088. (Contributed by David Moews, 28-Feb-2017.)
(𝜑𝐴 ∈ ℂ)       (𝜑 → ((abs‘𝐴) = 0 ↔ 𝐴 = 0))
 
Theoremabs00bd 10090 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 10091 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 10092 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))))
 
Theoremabsmul 10093 Absolute value distributes over multiplication. Proposition 10-3.7(f) of [Gleason] p. 133. (Contributed by NM, 11-Oct-1999.) (Revised by Mario Carneiro, 29-May-2016.)
((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ) → (abs‘(𝐴 · 𝐵)) = ((abs‘𝐴) · (abs‘𝐵)))
 
Theoremabsdivap 10094 Absolute value distributes over division. (Contributed by Jim Kingdon, 11-Aug-2021.)
((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐵 # 0) → (abs‘(𝐴 / 𝐵)) = ((abs‘𝐴) / (abs‘𝐵)))
 
Theoremabsid 10095 A nonnegative number is its own absolute value. (Contributed by NM, 11-Oct-1999.) (Revised by Mario Carneiro, 29-May-2016.)
((𝐴 ∈ ℝ ∧ 0 ≤ 𝐴) → (abs‘𝐴) = 𝐴)
 
Theoremabs1 10096 The absolute value of 1. Common special case. (Contributed by David A. Wheeler, 16-Jul-2016.)
(abs‘1) = 1
 
Theoremabsnid 10097 A negative number is the negative of its own absolute value. (Contributed by NM, 27-Feb-2005.)
((𝐴 ∈ ℝ ∧ 𝐴 ≤ 0) → (abs‘𝐴) = -𝐴)
 
Theoremleabs 10098 A real number is less than or equal to its absolute value. (Contributed by NM, 27-Feb-2005.)
(𝐴 ∈ ℝ → 𝐴 ≤ (abs‘𝐴))
 
Theoremqabsor 10099 The absolute value of a rational number is either that number or its negative. (Contributed by Jim Kingdon, 8-Nov-2021.)
(𝐴 ∈ ℚ → ((abs‘𝐴) = 𝐴 ∨ (abs‘𝐴) = -𝐴))
 
Theoremqabsord 10100 The absolute value of a rational number is either that number or its negative. (Contributed by Jim Kingdon, 8-Nov-2021.)
(𝜑𝐴 ∈ ℚ)       (𝜑 → ((abs‘𝐴) = 𝐴 ∨ (abs‘𝐴) = -𝐴))
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