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Theorem List for Intuitionistic Logic Explorer - 10801-10900   *Has distinct variable group(s)
TypeLabelDescription
Statement
 
Syntaxcabs 10801 Extend class notation to include a function for the absolute value (modulus) of a complex number.
class abs
 
Definitiondf-rsqrt 10802* 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 10803 Define the function for the absolute value (modulus) of a complex number. (Contributed by NM, 27-Jul-1999.)
abs = (𝑥 ∈ ℂ ↦ (√‘(𝑥 · (∗‘𝑥))))
 
Theoremsqrtrval 10804* Value of square root function. (Contributed by Jim Kingdon, 23-Aug-2020.)
(𝐴 ∈ ℝ → (√‘𝐴) = (𝑥 ∈ ℝ ((𝑥↑2) = 𝐴 ∧ 0 ≤ 𝑥)))
 
Theoremabsval 10805 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 10806 A real number does not lie on the negative imaginary axis. (Contributed by Mario Carneiro, 8-Jul-2013.)
(𝐴 ∈ ℝ → (i · 𝐴) ∉ ℝ+)
 
Theoremsqrt0rlem 10807 Lemma for sqrt0 10808. (Contributed by Jim Kingdon, 26-Aug-2020.)
((𝐴 ∈ ℝ ∧ ((𝐴↑2) = 0 ∧ 0 ≤ 𝐴)) ↔ 𝐴 = 0)
 
Theoremsqrt0 10808 Square root of zero. (Contributed by Mario Carneiro, 9-Jul-2013.)
(√‘0) = 0
 
Theoremresqrexlem1arp 10809 Lemma for resqrex 10830. 1 + 𝐴 is a positive real (expressed in a way that will help apply seqf 10265 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 10810* Lemma for resqrex 10830. Applying the recursion rule yields a positive real (expressed in a way that will help apply seqf 10265 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 10811* Lemma for resqrex 10830. 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 10812* Lemma for resqrex 10830. 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 10813* Lemma for resqrex 10830. 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 10814* Lemma for resqrex 10830. 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 10815* Lemma for resqrex 10830. 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 10816* Lemma for resqrex 10830. 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 10817* Lemma for resqrex 10830. 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 10818* Lemma for resqrex 10830. 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 10819* Lemma for resqrex 10830. 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 10820* Lemma for resqrex 10830. 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 10821* Lemma for resqrex 10830. 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 10822* Lemma for resqrex 10830. 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 10823* Lemma for resqrex 10830. 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 10824* Lemma for resqrex 10830. 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 10825* Lemma for resqrex 10830. 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 10826* Lemma for resqrex 10830. 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 10827* Lemma for resqrex 10830. 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 10828* Lemma for resqrex 10830. 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 10829* Lemma for resqrex 10830. 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 10830* 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 10831* 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 10832* 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 10833 Closure of the square root function. (Contributed by Mario Carneiro, 9-Jul-2013.)
((𝐴 ∈ ℝ ∧ 0 ≤ 𝐴) → (√‘𝐴) ∈ ℝ)
 
Theoremrersqrtthlem 10834 Lemma for resqrtth 10835. (Contributed by Jim Kingdon, 10-Aug-2021.)
((𝐴 ∈ ℝ ∧ 0 ≤ 𝐴) → (((√‘𝐴)↑2) = 𝐴 ∧ 0 ≤ (√‘𝐴)))
 
Theoremresqrtth 10835 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 10836 Square of square root. (Contributed by Mario Carneiro, 10-Jul-2013.)
((𝐴 ∈ ℝ ∧ 0 ≤ 𝐴) → ((√‘𝐴) · (√‘𝐴)) = 𝐴)
 
Theoremsqrtge0 10837 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 10838 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 10839 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 10840 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 10841 Square root is strictly monotonic. Closed form of sqrtlti 10941. (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 10842 Analogue to sqrt11 10843 but for apartness. (Contributed by Jim Kingdon, 11-Aug-2021.)
(((𝐴 ∈ ℝ ∧ 0 ≤ 𝐴) ∧ (𝐵 ∈ ℝ ∧ 0 ≤ 𝐵)) → ((√‘𝐴) # (√‘𝐵) ↔ 𝐴 # 𝐵))
 
Theoremsqrt11 10843 The square root function is one-to-one. Also see sqrt11ap 10842 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 10844 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 10845 The square root of a positive real is a positive real. (Contributed by NM, 22-Feb-2008.)
(𝐴 ∈ ℝ+ → (√‘𝐴) ∈ ℝ+)
 
Theoremsqrtdiv 10846 Square root distributes over division. (Contributed by Mario Carneiro, 5-May-2016.)
(((𝐴 ∈ ℝ ∧ 0 ≤ 𝐴) ∧ 𝐵 ∈ ℝ+) → (√‘(𝐴 / 𝐵)) = ((√‘𝐴) / (√‘𝐵)))
 
Theoremsqrtsq2 10847 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 10848 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 10849 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 10850 The square root of 1 is 1. (Contributed by NM, 31-Jul-1999.)
(√‘1) = 1
 
Theoremsqrt4 10851 The square root of 4 is 2. (Contributed by NM, 3-Aug-1999.)
(√‘4) = 2
 
Theoremsqrt9 10852 The square root of 9 is 3. (Contributed by NM, 11-May-2004.)
(√‘9) = 3
 
Theoremsqrt2gt1lt2 10853 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 10854 Absolute value of negative. (Contributed by NM, 27-Feb-2005.)
(𝐴 ∈ ℂ → (abs‘-𝐴) = (abs‘𝐴))
 
Theoremabscl 10855 Real closure of absolute value. (Contributed by NM, 3-Oct-1999.)
(𝐴 ∈ ℂ → (abs‘𝐴) ∈ ℝ)
 
Theoremabscj 10856 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 10857 Square of value of absolute value function. (Contributed by NM, 16-Jan-2006.)
(𝐴 ∈ ℂ → ((abs‘𝐴)↑2) = (𝐴 · (∗‘𝐴)))
 
Theoremabsvalsq2 10858 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 10859 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 10860 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 10861 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 10862 The absolute value of 0. (Contributed by NM, 26-Mar-2005.) (Revised by Mario Carneiro, 29-May-2016.)
(abs‘0) = 0
 
Theoremabsi 10863 The absolute value of the imaginary unit. (Contributed by NM, 26-Mar-2005.)
(abs‘i) = 1
 
Theoremabsge0 10864 Absolute value is nonnegative. (Contributed by NM, 20-Nov-2004.) (Revised by Mario Carneiro, 29-May-2016.)
(𝐴 ∈ ℂ → 0 ≤ (abs‘𝐴))
 
Theoremabsrpclap 10865 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 10866 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 10867 Strong extensionality for absolute value. (Contributed by Jim Kingdon, 12-Aug-2021.)
((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ) → ((abs‘𝐴) # (abs‘𝐵) → 𝐴 # 𝐵))
 
Theoremabs00 10868 The absolute value of a number is zero iff the number is zero. Also see abs00ap 10866 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 10869 A complex number is zero iff its absolute value is zero. Deduction form of abs00 10868. (Contributed by David Moews, 28-Feb-2017.)
(𝜑𝐴 ∈ ℂ)       (𝜑 → ((abs‘𝐴) = 0 ↔ 𝐴 = 0))
 
Theoremabs00bd 10870 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 10871 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 10872 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 10873 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 10874 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 10875 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 10876 The absolute value of 1. Common special case. (Contributed by David A. Wheeler, 16-Jul-2016.)
(abs‘1) = 1
 
Theoremabsnid 10877 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 10878 A real number is less than or equal to its absolute value. (Contributed by NM, 27-Feb-2005.)
(𝐴 ∈ ℝ → 𝐴 ≤ (abs‘𝐴))
 
Theoremqabsor 10879 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 10880 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 10881 Absolute value of a real number. (Contributed by NM, 17-Mar-2005.)
(𝐴 ∈ ℝ → (abs‘𝐴) = (√‘(𝐴↑2)))
 
Theoremabsresq 10882 Square of the absolute value of a real number. (Contributed by NM, 16-Jan-2006.)
(𝐴 ∈ ℝ → ((abs‘𝐴)↑2) = (𝐴↑2))
 
Theoremabsexp 10883 Absolute value of positive integer exponentiation. (Contributed by NM, 5-Jan-2006.)
((𝐴 ∈ ℂ ∧ 𝑁 ∈ ℕ0) → (abs‘(𝐴𝑁)) = ((abs‘𝐴)↑𝑁))
 
Theoremabsexpzap 10884 Absolute value of integer exponentiation. (Contributed by Jim Kingdon, 11-Aug-2021.)
((𝐴 ∈ ℂ ∧ 𝐴 # 0 ∧ 𝑁 ∈ ℤ) → (abs‘(𝐴𝑁)) = ((abs‘𝐴)↑𝑁))
 
Theoremabssq 10885 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 10886 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 10887 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 10888 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 10889 The absolute value of an integer is a nonnegative integer. (Contributed by NM, 27-Feb-2005.)
(𝐴 ∈ ℤ → (abs‘𝐴) ∈ ℕ0)
 
Theoremzabscl 10890 The absolute value of an integer is an integer. (Contributed by Stefan O'Rear, 24-Sep-2014.)
(𝐴 ∈ ℤ → (abs‘𝐴) ∈ ℤ)
 
Theoremltabs 10891 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 10892 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 10893 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 10894 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 10895 If the absolute value of a complex number is less than a real, its difference from the real is nonzero. See also abssubap0 10894 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 10896 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 10897 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 10898 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 10899 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 10900 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))
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