HomeHome Intuitionistic Logic Explorer
Theorem List (p. 111 of 148)
< 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 - 11001-11100   *Has distinct variable group(s)
TypeLabelDescription
Statement
 
Syntaxcabs 11001 Extend class notation to include a function for the absolute value (modulus) of a complex number.
class abs
 
Definitiondf-rsqrt 11002* 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 11003 Define the function for the absolute value (modulus) of a complex number. (Contributed by NM, 27-Jul-1999.)
abs = (𝑥 ∈ ℂ ↦ (√‘(𝑥 · (∗‘𝑥))))
 
Theoremsqrtrval 11004* Value of square root function. (Contributed by Jim Kingdon, 23-Aug-2020.)
(𝐴 ∈ ℝ → (√‘𝐴) = (𝑥 ∈ ℝ ((𝑥↑2) = 𝐴 ∧ 0 ≤ 𝑥)))
 
Theoremabsval 11005 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 11006 A real number does not lie on the negative imaginary axis. (Contributed by Mario Carneiro, 8-Jul-2013.)
(𝐴 ∈ ℝ → (i · 𝐴) ∉ ℝ+)
 
Theoremsqrt0rlem 11007 Lemma for sqrt0 11008. (Contributed by Jim Kingdon, 26-Aug-2020.)
((𝐴 ∈ ℝ ∧ ((𝐴↑2) = 0 ∧ 0 ≤ 𝐴)) ↔ 𝐴 = 0)
 
Theoremsqrt0 11008 Square root of zero. (Contributed by Mario Carneiro, 9-Jul-2013.)
(√‘0) = 0
 
Theoremresqrexlem1arp 11009 Lemma for resqrex 11030. 1 + 𝐴 is a positive real (expressed in a way that will help apply seqf 10458 and similar theorems). (Contributed by Jim Kingdon, 28-Jul-2021.) (Revised by Jim Kingdon, 16-Oct-2022.)
(𝜑𝐴 ∈ ℝ)    &   (𝜑 → 0 ≤ 𝐴)       ((𝜑𝑁 ∈ ℕ) → ((ℕ × {(1 + 𝐴)})‘𝑁) ∈ ℝ+)
 
Theoremresqrexlemp1rp 11010* Lemma for resqrex 11030. Applying the recursion rule yields a positive real (expressed in a way that will help apply seqf 10458 and similar theorems). (Contributed by Jim Kingdon, 28-Jul-2021.) (Revised by Jim Kingdon, 16-Oct-2022.)
(𝜑𝐴 ∈ ℝ)    &   (𝜑 → 0 ≤ 𝐴)       ((𝜑 ∧ (𝐵 ∈ ℝ+𝐶 ∈ ℝ+)) → (𝐵(𝑦 ∈ ℝ+, 𝑧 ∈ ℝ+ ↦ ((𝑦 + (𝐴 / 𝑦)) / 2))𝐶) ∈ ℝ+)
 
Theoremresqrexlemf 11011* Lemma for resqrex 11030. The sequence is a function. (Contributed by Mario Carneiro and Jim Kingdon, 27-Jul-2021.) (Revised by Jim Kingdon, 16-Oct-2022.)
𝐹 = seq1((𝑦 ∈ ℝ+, 𝑧 ∈ ℝ+ ↦ ((𝑦 + (𝐴 / 𝑦)) / 2)), (ℕ × {(1 + 𝐴)}))    &   (𝜑𝐴 ∈ ℝ)    &   (𝜑 → 0 ≤ 𝐴)       (𝜑𝐹:ℕ⟶ℝ+)
 
Theoremresqrexlemf1 11012* Lemma for resqrex 11030. 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.) (Revised by Jim Kingdon, 16-Oct-2022.)
𝐹 = seq1((𝑦 ∈ ℝ+, 𝑧 ∈ ℝ+ ↦ ((𝑦 + (𝐴 / 𝑦)) / 2)), (ℕ × {(1 + 𝐴)}))    &   (𝜑𝐴 ∈ ℝ)    &   (𝜑 → 0 ≤ 𝐴)       (𝜑 → (𝐹‘1) = (1 + 𝐴))
 
Theoremresqrexlemfp1 11013* Lemma for resqrex 11030. 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)}))    &   (𝜑𝐴 ∈ ℝ)    &   (𝜑 → 0 ≤ 𝐴)       ((𝜑𝑁 ∈ ℕ) → (𝐹‘(𝑁 + 1)) = (((𝐹𝑁) + (𝐴 / (𝐹𝑁))) / 2))
 
Theoremresqrexlemover 11014* Lemma for resqrex 11030. 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)}))    &   (𝜑𝐴 ∈ ℝ)    &   (𝜑 → 0 ≤ 𝐴)       ((𝜑𝑁 ∈ ℕ) → 𝐴 < ((𝐹𝑁)↑2))
 
Theoremresqrexlemdec 11015* Lemma for resqrex 11030. 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)}))    &   (𝜑𝐴 ∈ ℝ)    &   (𝜑 → 0 ≤ 𝐴)       ((𝜑𝑁 ∈ ℕ) → (𝐹‘(𝑁 + 1)) < (𝐹𝑁))
 
Theoremresqrexlemdecn 11016* Lemma for resqrex 11030. The sequence is decreasing. (Contributed by Jim Kingdon, 31-Jul-2021.)
𝐹 = seq1((𝑦 ∈ ℝ+, 𝑧 ∈ ℝ+ ↦ ((𝑦 + (𝐴 / 𝑦)) / 2)), (ℕ × {(1 + 𝐴)}))    &   (𝜑𝐴 ∈ ℝ)    &   (𝜑 → 0 ≤ 𝐴)    &   (𝜑𝑁 ∈ ℕ)    &   (𝜑𝑀 ∈ ℕ)    &   (𝜑𝑁 < 𝑀)       (𝜑 → (𝐹𝑀) < (𝐹𝑁))
 
Theoremresqrexlemlo 11017* Lemma for resqrex 11030. 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)}))    &   (𝜑𝐴 ∈ ℝ)    &   (𝜑 → 0 ≤ 𝐴)       ((𝜑𝑁 ∈ ℕ) → (1 / (2↑𝑁)) < (𝐹𝑁))
 
Theoremresqrexlemcalc1 11018* Lemma for resqrex 11030. 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)}))    &   (𝜑𝐴 ∈ ℝ)    &   (𝜑 → 0 ≤ 𝐴)       ((𝜑𝑁 ∈ ℕ) → (((𝐹‘(𝑁 + 1))↑2) − 𝐴) = (((((𝐹𝑁)↑2) − 𝐴)↑2) / (4 · ((𝐹𝑁)↑2))))
 
Theoremresqrexlemcalc2 11019* Lemma for resqrex 11030. 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)}))    &   (𝜑𝐴 ∈ ℝ)    &   (𝜑 → 0 ≤ 𝐴)       ((𝜑𝑁 ∈ ℕ) → (((𝐹‘(𝑁 + 1))↑2) − 𝐴) ≤ ((((𝐹𝑁)↑2) − 𝐴) / 4))
 
Theoremresqrexlemcalc3 11020* Lemma for resqrex 11030. 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)}))    &   (𝜑𝐴 ∈ ℝ)    &   (𝜑 → 0 ≤ 𝐴)       ((𝜑𝑁 ∈ ℕ) → (((𝐹𝑁)↑2) − 𝐴) ≤ (((𝐹‘1)↑2) / (4↑(𝑁 − 1))))
 
Theoremresqrexlemnmsq 11021* Lemma for resqrex 11030. 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)}))    &   (𝜑𝐴 ∈ ℝ)    &   (𝜑 → 0 ≤ 𝐴)    &   (𝜑𝑁 ∈ ℕ)    &   (𝜑𝑀 ∈ ℕ)    &   (𝜑𝑁𝑀)       (𝜑 → (((𝐹𝑁)↑2) − ((𝐹𝑀)↑2)) < (((𝐹‘1)↑2) / (4↑(𝑁 − 1))))
 
Theoremresqrexlemnm 11022* Lemma for resqrex 11030. 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)}))    &   (𝜑𝐴 ∈ ℝ)    &   (𝜑 → 0 ≤ 𝐴)    &   (𝜑𝑁 ∈ ℕ)    &   (𝜑𝑀 ∈ ℕ)    &   (𝜑𝑁𝑀)       (𝜑 → ((𝐹𝑁) − (𝐹𝑀)) < ((((𝐹‘1)↑2) · 2) / (2↑(𝑁 − 1))))
 
Theoremresqrexlemcvg 11023* Lemma for resqrex 11030. 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)}))    &   (𝜑𝐴 ∈ ℝ)    &   (𝜑 → 0 ≤ 𝐴)       (𝜑 → ∃𝑟 ∈ ℝ ∀𝑥 ∈ ℝ+𝑗 ∈ ℕ ∀𝑖 ∈ (ℤ𝑗)((𝐹𝑖) < (𝑟 + 𝑥) ∧ 𝑟 < ((𝐹𝑖) + 𝑥)))
 
Theoremresqrexlemgt0 11024* Lemma for resqrex 11030. A limit is nonnegative. (Contributed by Jim Kingdon, 7-Aug-2021.)
𝐹 = seq1((𝑦 ∈ ℝ+, 𝑧 ∈ ℝ+ ↦ ((𝑦 + (𝐴 / 𝑦)) / 2)), (ℕ × {(1 + 𝐴)}))    &   (𝜑𝐴 ∈ ℝ)    &   (𝜑 → 0 ≤ 𝐴)    &   (𝜑𝐿 ∈ ℝ)    &   (𝜑 → ∀𝑒 ∈ ℝ+𝑗 ∈ ℕ ∀𝑖 ∈ (ℤ𝑗)((𝐹𝑖) < (𝐿 + 𝑒) ∧ 𝐿 < ((𝐹𝑖) + 𝑒)))       (𝜑 → 0 ≤ 𝐿)
 
Theoremresqrexlemoverl 11025* Lemma for resqrex 11030. 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)}))    &   (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 11026* Lemma for resqrex 11030. 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)}))    &   (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 11027* Lemma for resqrex 11030. 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)}))    &   (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 11028* Lemma for resqrex 11030. 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)}))    &   (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 11029* Lemma for resqrex 11030. 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)}))    &   (𝜑𝐴 ∈ ℝ)    &   (𝜑 → 0 ≤ 𝐴)       (𝜑 → ∃𝑥 ∈ ℝ (0 ≤ 𝑥 ∧ (𝑥↑2) = 𝐴))
 
Theoremresqrex 11030* 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 11031* 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 11032* 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 11033 Closure of the square root function. (Contributed by Mario Carneiro, 9-Jul-2013.)
((𝐴 ∈ ℝ ∧ 0 ≤ 𝐴) → (√‘𝐴) ∈ ℝ)
 
Theoremrersqrtthlem 11034 Lemma for resqrtth 11035. (Contributed by Jim Kingdon, 10-Aug-2021.)
((𝐴 ∈ ℝ ∧ 0 ≤ 𝐴) → (((√‘𝐴)↑2) = 𝐴 ∧ 0 ≤ (√‘𝐴)))
 
Theoremresqrtth 11035 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 11036 Square of square root. (Contributed by Mario Carneiro, 10-Jul-2013.)
((𝐴 ∈ ℝ ∧ 0 ≤ 𝐴) → ((√‘𝐴) · (√‘𝐴)) = 𝐴)
 
Theoremsqrtge0 11037 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 11038 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 11039 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 11040 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 11041 Square root is strictly monotonic. Closed form of sqrtlti 11141. (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 11042 Analogue to sqrt11 11043 but for apartness. (Contributed by Jim Kingdon, 11-Aug-2021.)
(((𝐴 ∈ ℝ ∧ 0 ≤ 𝐴) ∧ (𝐵 ∈ ℝ ∧ 0 ≤ 𝐵)) → ((√‘𝐴) # (√‘𝐵) ↔ 𝐴 # 𝐵))
 
Theoremsqrt11 11043 The square root function is one-to-one. Also see sqrt11ap 11042 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 11044 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 11045 The square root of a positive real is a positive real. (Contributed by NM, 22-Feb-2008.)
(𝐴 ∈ ℝ+ → (√‘𝐴) ∈ ℝ+)
 
Theoremsqrtdiv 11046 Square root distributes over division. (Contributed by Mario Carneiro, 5-May-2016.)
(((𝐴 ∈ ℝ ∧ 0 ≤ 𝐴) ∧ 𝐵 ∈ ℝ+) → (√‘(𝐴 / 𝐵)) = ((√‘𝐴) / (√‘𝐵)))
 
Theoremsqrtsq2 11047 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 11048 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 11049 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 11050 The square root of 1 is 1. (Contributed by NM, 31-Jul-1999.)
(√‘1) = 1
 
Theoremsqrt4 11051 The square root of 4 is 2. (Contributed by NM, 3-Aug-1999.)
(√‘4) = 2
 
Theoremsqrt9 11052 The square root of 9 is 3. (Contributed by NM, 11-May-2004.)
(√‘9) = 3
 
Theoremsqrt2gt1lt2 11053 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 11054 Absolute value of negative. (Contributed by NM, 27-Feb-2005.)
(𝐴 ∈ ℂ → (abs‘-𝐴) = (abs‘𝐴))
 
Theoremabscl 11055 Real closure of absolute value. (Contributed by NM, 3-Oct-1999.)
(𝐴 ∈ ℂ → (abs‘𝐴) ∈ ℝ)
 
Theoremabscj 11056 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 11057 Square of value of absolute value function. (Contributed by NM, 16-Jan-2006.)
(𝐴 ∈ ℂ → ((abs‘𝐴)↑2) = (𝐴 · (∗‘𝐴)))
 
Theoremabsvalsq2 11058 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 11059 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 11060 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 11061 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 11062 The absolute value of 0. (Contributed by NM, 26-Mar-2005.) (Revised by Mario Carneiro, 29-May-2016.)
(abs‘0) = 0
 
Theoremabsi 11063 The absolute value of the imaginary unit. (Contributed by NM, 26-Mar-2005.)
(abs‘i) = 1
 
Theoremabsge0 11064 Absolute value is nonnegative. (Contributed by NM, 20-Nov-2004.) (Revised by Mario Carneiro, 29-May-2016.)
(𝐴 ∈ ℂ → 0 ≤ (abs‘𝐴))
 
Theoremabsrpclap 11065 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 11066 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 11067 Strong extensionality for absolute value. (Contributed by Jim Kingdon, 12-Aug-2021.)
((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ) → ((abs‘𝐴) # (abs‘𝐵) → 𝐴 # 𝐵))
 
Theoremabs00 11068 The absolute value of a number is zero iff the number is zero. Also see abs00ap 11066 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 11069 A complex number is zero iff its absolute value is zero. Deduction form of abs00 11068. (Contributed by David Moews, 28-Feb-2017.)
(𝜑𝐴 ∈ ℂ)       (𝜑 → ((abs‘𝐴) = 0 ↔ 𝐴 = 0))
 
Theoremabs00bd 11070 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 11071 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 11072 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 11073 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 11074 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 11075 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 11076 The absolute value of 1. Common special case. (Contributed by David A. Wheeler, 16-Jul-2016.)
(abs‘1) = 1
 
Theoremabsnid 11077 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 11078 A real number is less than or equal to its absolute value. (Contributed by NM, 27-Feb-2005.)
(𝐴 ∈ ℝ → 𝐴 ≤ (abs‘𝐴))
 
Theoremqabsor 11079 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 11080 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))
 
Theoremabsre 11081 Absolute value of a real number. (Contributed by NM, 17-Mar-2005.)
(𝐴 ∈ ℝ → (abs‘𝐴) = (√‘(𝐴↑2)))
 
Theoremabsresq 11082 Square of the absolute value of a real number. (Contributed by NM, 16-Jan-2006.)
(𝐴 ∈ ℝ → ((abs‘𝐴)↑2) = (𝐴↑2))
 
Theoremabsexp 11083 Absolute value of positive integer exponentiation. (Contributed by NM, 5-Jan-2006.)
((𝐴 ∈ ℂ ∧ 𝑁 ∈ ℕ0) → (abs‘(𝐴𝑁)) = ((abs‘𝐴)↑𝑁))
 
Theoremabsexpzap 11084 Absolute value of integer exponentiation. (Contributed by Jim Kingdon, 11-Aug-2021.)
((𝐴 ∈ ℂ ∧ 𝐴 # 0 ∧ 𝑁 ∈ ℤ) → (abs‘(𝐴𝑁)) = ((abs‘𝐴)↑𝑁))
 
Theoremabssq 11085 Square can be moved in and out of absolute value. (Contributed by Scott Fenton, 18-Apr-2014.) (Proof shortened by Mario Carneiro, 29-May-2016.)
(𝐴 ∈ ℂ → ((abs‘𝐴)↑2) = (abs‘(𝐴↑2)))
 
Theoremsqabs 11086 The squares of two reals are equal iff their absolute values are equal. (Contributed by NM, 6-Mar-2009.)
((𝐴 ∈ ℝ ∧ 𝐵 ∈ ℝ) → ((𝐴↑2) = (𝐵↑2) ↔ (abs‘𝐴) = (abs‘𝐵)))
 
Theoremabsrele 11087 The absolute value of a complex number is greater than or equal to the absolute value of its real part. (Contributed by NM, 1-Apr-2005.)
(𝐴 ∈ ℂ → (abs‘(ℜ‘𝐴)) ≤ (abs‘𝐴))
 
Theoremabsimle 11088 The absolute value of a complex number is greater than or equal to the absolute value of its imaginary part. (Contributed by NM, 17-Mar-2005.) (Proof shortened by Mario Carneiro, 29-May-2016.)
(𝐴 ∈ ℂ → (abs‘(ℑ‘𝐴)) ≤ (abs‘𝐴))
 
Theoremnn0abscl 11089 The absolute value of an integer is a nonnegative integer. (Contributed by NM, 27-Feb-2005.)
(𝐴 ∈ ℤ → (abs‘𝐴) ∈ ℕ0)
 
Theoremzabscl 11090 The absolute value of an integer is an integer. (Contributed by Stefan O'Rear, 24-Sep-2014.)
(𝐴 ∈ ℤ → (abs‘𝐴) ∈ ℤ)
 
Theoremltabs 11091 A number which is less than its absolute value is negative. (Contributed by Jim Kingdon, 12-Aug-2021.)
((𝐴 ∈ ℝ ∧ 𝐴 < (abs‘𝐴)) → 𝐴 < 0)
 
Theoremabslt 11092 Absolute value and 'less than' relation. (Contributed by NM, 6-Apr-2005.) (Revised by Mario Carneiro, 29-May-2016.)
((𝐴 ∈ ℝ ∧ 𝐵 ∈ ℝ) → ((abs‘𝐴) < 𝐵 ↔ (-𝐵 < 𝐴𝐴 < 𝐵)))
 
Theoremabsle 11093 Absolute value and 'less than or equal to' relation. (Contributed by NM, 6-Apr-2005.) (Revised by Mario Carneiro, 29-May-2016.)
((𝐴 ∈ ℝ ∧ 𝐵 ∈ ℝ) → ((abs‘𝐴) ≤ 𝐵 ↔ (-𝐵𝐴𝐴𝐵)))
 
Theoremabssubap0 11094 If the absolute value of a complex number is less than a real, its difference from the real is apart from zero. (Contributed by Jim Kingdon, 12-Aug-2021.)
((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℝ ∧ (abs‘𝐴) < 𝐵) → (𝐵𝐴) # 0)
 
Theoremabssubne0 11095 If the absolute value of a complex number is less than a real, its difference from the real is nonzero. See also abssubap0 11094 which is the same with not equal changed to apart. (Contributed by NM, 2-Nov-2007.)
((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℝ ∧ (abs‘𝐴) < 𝐵) → (𝐵𝐴) ≠ 0)
 
Theoremabsdiflt 11096 The absolute value of a difference and 'less than' relation. (Contributed by Paul Chapman, 18-Sep-2007.)
((𝐴 ∈ ℝ ∧ 𝐵 ∈ ℝ ∧ 𝐶 ∈ ℝ) → ((abs‘(𝐴𝐵)) < 𝐶 ↔ ((𝐵𝐶) < 𝐴𝐴 < (𝐵 + 𝐶))))
 
Theoremabsdifle 11097 The absolute value of a difference and 'less than or equal to' relation. (Contributed by Paul Chapman, 18-Sep-2007.)
((𝐴 ∈ ℝ ∧ 𝐵 ∈ ℝ ∧ 𝐶 ∈ ℝ) → ((abs‘(𝐴𝐵)) ≤ 𝐶 ↔ ((𝐵𝐶) ≤ 𝐴𝐴 ≤ (𝐵 + 𝐶))))
 
Theoremelicc4abs 11098 Membership in a symmetric closed real interval. (Contributed by Stefan O'Rear, 16-Nov-2014.)
((𝐴 ∈ ℝ ∧ 𝐵 ∈ ℝ ∧ 𝐶 ∈ ℝ) → (𝐶 ∈ ((𝐴𝐵)[,](𝐴 + 𝐵)) ↔ (abs‘(𝐶𝐴)) ≤ 𝐵))
 
Theoremlenegsq 11099 Comparison to a nonnegative number based on comparison to squares. (Contributed by NM, 16-Jan-2006.)
((𝐴 ∈ ℝ ∧ 𝐵 ∈ ℝ ∧ 0 ≤ 𝐵) → ((𝐴𝐵 ∧ -𝐴𝐵) ↔ (𝐴↑2) ≤ (𝐵↑2)))
 
Theoremreleabs 11100 The real part of a number is less than or equal to its absolute value. Proposition 10-3.7(d) of [Gleason] p. 133. (Contributed by NM, 1-Apr-2005.)
(𝐴 ∈ ℂ → (ℜ‘𝐴) ≤ (abs‘𝐴))
    < 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 11301-11400 115 11401-11500 116 11501-11600 117 11601-11700 118 11701-11800 119 11801-11900 120 11901-12000 121 12001-12100 122 12101-12200 123 12201-12300 124 12301-12400 125 12401-12500 126 12501-12600 127 12601-12700 128 12701-12800 129 12801-12900 130 12901-13000 131 13001-13100 132 13101-13200 133 13201-13300 134 13301-13400 135 13401-13500 136 13501-13600 137 13601-13700 138 13701-13800 139 13801-13900 140 13901-14000 141 14001-14100 142 14101-14200 143 14201-14300 144 14301-14400 145 14401-14500 146 14501-14600 147 14601-14700 148 14701-14751
  Copyright terms: Public domain < Previous  Next >