P. 118 of Theorem List - Intuitionistic Logic Explorer

Theorem List for Intuitionistic Logic Explorer - 11701-11800   *Has distinct variable group(s)

Type	Label	Description
Statement
Theorem	resqrexlemf1 11701*	Lemma for resqrex 11719. 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)}))    &   ⊢ (𝜑 → 𝐴 ∈ ℝ)    &   ⊢ (𝜑 → 0 ≤ 𝐴)    ⇒   ⊢ (𝜑 → (𝐹‘1) = (1 + 𝐴))
Theorem	resqrexlemfp1 11702*	Lemma for resqrex 11719. 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)}))    &   ⊢ (𝜑 → 𝐴 ∈ ℝ)    &   ⊢ (𝜑 → 0 ≤ 𝐴)    ⇒   ⊢ ((𝜑 ∧ 𝑁 ∈ ℕ) → (𝐹‘(𝑁 + 1)) = (((𝐹‘𝑁) + (𝐴 / (𝐹‘𝑁))) / 2))
Theorem	resqrexlemover 11703*	Lemma for resqrex 11719. 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)}))    &   ⊢ (𝜑 → 𝐴 ∈ ℝ)    &   ⊢ (𝜑 → 0 ≤ 𝐴)    ⇒   ⊢ ((𝜑 ∧ 𝑁 ∈ ℕ) → 𝐴 < ((𝐹‘𝑁)↑2))
Theorem	resqrexlemdec 11704*	Lemma for resqrex 11719. 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)}))    &   ⊢ (𝜑 → 𝐴 ∈ ℝ)    &   ⊢ (𝜑 → 0 ≤ 𝐴)    ⇒   ⊢ ((𝜑 ∧ 𝑁 ∈ ℕ) → (𝐹‘(𝑁 + 1)) < (𝐹‘𝑁))
Theorem	resqrexlemdecn 11705*	Lemma for resqrex 11719. The sequence is decreasing. (Contributed by Jim Kingdon, 31-Jul-2021.)
⊢ 𝐹 = seq1((𝑦 ∈ ℝ+, 𝑧 ∈ ℝ+ ↦ ((𝑦 + (𝐴 / 𝑦)) / 2)), (ℕ × {(1 + 𝐴)}))    &   ⊢ (𝜑 → 𝐴 ∈ ℝ)    &   ⊢ (𝜑 → 0 ≤ 𝐴)    &   ⊢ (𝜑 → 𝑁 ∈ ℕ)    &   ⊢ (𝜑 → 𝑀 ∈ ℕ)    &   ⊢ (𝜑 → 𝑁 < 𝑀)    ⇒   ⊢ (𝜑 → (𝐹‘𝑀) < (𝐹‘𝑁))
Theorem	resqrexlemlo 11706*	Lemma for resqrex 11719. 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)}))    &   ⊢ (𝜑 → 𝐴 ∈ ℝ)    &   ⊢ (𝜑 → 0 ≤ 𝐴)    ⇒   ⊢ ((𝜑 ∧ 𝑁 ∈ ℕ) → (1 / (2↑𝑁)) < (𝐹‘𝑁))
Theorem	resqrexlemcalc1 11707*	Lemma for resqrex 11719. 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)}))    &   ⊢ (𝜑 → 𝐴 ∈ ℝ)    &   ⊢ (𝜑 → 0 ≤ 𝐴)    ⇒   ⊢ ((𝜑 ∧ 𝑁 ∈ ℕ) → (((𝐹‘(𝑁 + 1))↑2) − 𝐴) = (((((𝐹‘𝑁)↑2) − 𝐴)↑2) / (4 · ((𝐹‘𝑁)↑2))))
Theorem	resqrexlemcalc2 11708*	Lemma for resqrex 11719. 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)}))    &   ⊢ (𝜑 → 𝐴 ∈ ℝ)    &   ⊢ (𝜑 → 0 ≤ 𝐴)    ⇒   ⊢ ((𝜑 ∧ 𝑁 ∈ ℕ) → (((𝐹‘(𝑁 + 1))↑2) − 𝐴) ≤ ((((𝐹‘𝑁)↑2) − 𝐴) / 4))
Theorem	resqrexlemcalc3 11709*	Lemma for resqrex 11719. 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)}))    &   ⊢ (𝜑 → 𝐴 ∈ ℝ)    &   ⊢ (𝜑 → 0 ≤ 𝐴)    ⇒   ⊢ ((𝜑 ∧ 𝑁 ∈ ℕ) → (((𝐹‘𝑁)↑2) − 𝐴) ≤ (((𝐹‘1)↑2) / (4↑(𝑁 − 1))))
Theorem	resqrexlemnmsq 11710*	Lemma for resqrex 11719. 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)}))    &   ⊢ (𝜑 → 𝐴 ∈ ℝ)    &   ⊢ (𝜑 → 0 ≤ 𝐴)    &   ⊢ (𝜑 → 𝑁 ∈ ℕ)    &   ⊢ (𝜑 → 𝑀 ∈ ℕ)    &   ⊢ (𝜑 → 𝑁 ≤ 𝑀)    ⇒   ⊢ (𝜑 → (((𝐹‘𝑁)↑2) − ((𝐹‘𝑀)↑2)) < (((𝐹‘1)↑2) / (4↑(𝑁 − 1))))
Theorem	resqrexlemnm 11711*	Lemma for resqrex 11719. 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)}))    &   ⊢ (𝜑 → 𝐴 ∈ ℝ)    &   ⊢ (𝜑 → 0 ≤ 𝐴)    &   ⊢ (𝜑 → 𝑁 ∈ ℕ)    &   ⊢ (𝜑 → 𝑀 ∈ ℕ)    &   ⊢ (𝜑 → 𝑁 ≤ 𝑀)    ⇒   ⊢ (𝜑 → ((𝐹‘𝑁) − (𝐹‘𝑀)) < ((((𝐹‘1)↑2) · 2) / (2↑(𝑁 − 1))))
Theorem	resqrexlemcvg 11712*	Lemma for resqrex 11719. 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)}))    &   ⊢ (𝜑 → 𝐴 ∈ ℝ)    &   ⊢ (𝜑 → 0 ≤ 𝐴)    ⇒   ⊢ (𝜑 → ∃𝑟 ∈ ℝ ∀𝑥 ∈ ℝ+ ∃𝑗 ∈ ℕ ∀𝑖 ∈ (ℤ≥‘𝑗)((𝐹‘𝑖) < (𝑟 + 𝑥) ∧ 𝑟 < ((𝐹‘𝑖) + 𝑥)))
Theorem	resqrexlemgt0 11713*	Lemma for resqrex 11719. A limit is nonnegative. (Contributed by Jim Kingdon, 7-Aug-2021.)
⊢ 𝐹 = seq1((𝑦 ∈ ℝ+, 𝑧 ∈ ℝ+ ↦ ((𝑦 + (𝐴 / 𝑦)) / 2)), (ℕ × {(1 + 𝐴)}))    &   ⊢ (𝜑 → 𝐴 ∈ ℝ)    &   ⊢ (𝜑 → 0 ≤ 𝐴)    &   ⊢ (𝜑 → 𝐿 ∈ ℝ)    &   ⊢ (𝜑 → ∀𝑒 ∈ ℝ+ ∃𝑗 ∈ ℕ ∀𝑖 ∈ (ℤ≥‘𝑗)((𝐹‘𝑖) < (𝐿 + 𝑒) ∧ 𝐿 < ((𝐹‘𝑖) + 𝑒)))    ⇒   ⊢ (𝜑 → 0 ≤ 𝐿)
Theorem	resqrexlemoverl 11714*	Lemma for resqrex 11719. 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)}))    &   ⊢ (𝜑 → 𝐴 ∈ ℝ)    &   ⊢ (𝜑 → 0 ≤ 𝐴)    &   ⊢ (𝜑 → 𝐿 ∈ ℝ)    &   ⊢ (𝜑 → ∀𝑒 ∈ ℝ+ ∃𝑗 ∈ ℕ ∀𝑖 ∈ (ℤ≥‘𝑗)((𝐹‘𝑖) < (𝐿 + 𝑒) ∧ 𝐿 < ((𝐹‘𝑖) + 𝑒)))    &   ⊢ (𝜑 → 𝐾 ∈ ℕ)    ⇒   ⊢ (𝜑 → 𝐿 ≤ (𝐹‘𝐾))
Theorem	resqrexlemglsq 11715*	Lemma for resqrex 11719. 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)}))    &   ⊢ (𝜑 → 𝐴 ∈ ℝ)    &   ⊢ (𝜑 → 0 ≤ 𝐴)    &   ⊢ (𝜑 → 𝐿 ∈ ℝ)    &   ⊢ (𝜑 → ∀𝑒 ∈ ℝ+ ∃𝑗 ∈ ℕ ∀𝑖 ∈ (ℤ≥‘𝑗)((𝐹‘𝑖) < (𝐿 + 𝑒) ∧ 𝐿 < ((𝐹‘𝑖) + 𝑒)))    &   ⊢ 𝐺 = (𝑥 ∈ ℕ ↦ ((𝐹‘𝑥)↑2))    ⇒   ⊢ (𝜑 → ∀𝑒 ∈ ℝ+ ∃𝑗 ∈ ℕ ∀𝑘 ∈ (ℤ≥‘𝑗)((𝐺‘𝑘) < ((𝐿↑2) + 𝑒) ∧ (𝐿↑2) < ((𝐺‘𝑘) + 𝑒)))
Theorem	resqrexlemga 11716*	Lemma for resqrex 11719. 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)}))    &   ⊢ (𝜑 → 𝐴 ∈ ℝ)    &   ⊢ (𝜑 → 0 ≤ 𝐴)    &   ⊢ (𝜑 → 𝐿 ∈ ℝ)    &   ⊢ (𝜑 → ∀𝑒 ∈ ℝ+ ∃𝑗 ∈ ℕ ∀𝑖 ∈ (ℤ≥‘𝑗)((𝐹‘𝑖) < (𝐿 + 𝑒) ∧ 𝐿 < ((𝐹‘𝑖) + 𝑒)))    &   ⊢ 𝐺 = (𝑥 ∈ ℕ ↦ ((𝐹‘𝑥)↑2))    ⇒   ⊢ (𝜑 → ∀𝑒 ∈ ℝ+ ∃𝑗 ∈ ℕ ∀𝑘 ∈ (ℤ≥‘𝑗)((𝐺‘𝑘) < (𝐴 + 𝑒) ∧ 𝐴 < ((𝐺‘𝑘) + 𝑒)))
Theorem	resqrexlemsqa 11717*	Lemma for resqrex 11719. 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)}))    &   ⊢ (𝜑 → 𝐴 ∈ ℝ)    &   ⊢ (𝜑 → 0 ≤ 𝐴)    &   ⊢ (𝜑 → 𝐿 ∈ ℝ)    &   ⊢ (𝜑 → ∀𝑒 ∈ ℝ+ ∃𝑗 ∈ ℕ ∀𝑖 ∈ (ℤ≥‘𝑗)((𝐹‘𝑖) < (𝐿 + 𝑒) ∧ 𝐿 < ((𝐹‘𝑖) + 𝑒)))    ⇒   ⊢ (𝜑 → (𝐿↑2) = 𝐴)
Theorem	resqrexlemex 11718*	Lemma for resqrex 11719. 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)}))    &   ⊢ (𝜑 → 𝐴 ∈ ℝ)    &   ⊢ (𝜑 → 0 ≤ 𝐴)    ⇒   ⊢ (𝜑 → ∃𝑥 ∈ ℝ (0 ≤ 𝑥 ∧ (𝑥↑2) = 𝐴))
Theorem	resqrex 11719*	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))
Theorem	rsqrmo 11720*	Uniqueness for the square root function. (Contributed by Jim Kingdon, 10-Aug-2021.)
⊢ ((𝐴 ∈ ℝ ∧ 0 ≤ 𝐴) → ∃*𝑥 ∈ ℝ ((𝑥↑2) = 𝐴 ∧ 0 ≤ 𝑥))
Theorem	rersqreu 11721*	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))
Theorem	resqrtcl 11722	Closure of the square root function. (Contributed by Mario Carneiro, 9-Jul-2013.)
⊢ ((𝐴 ∈ ℝ ∧ 0 ≤ 𝐴) → (√‘𝐴) ∈ ℝ)
Theorem	rersqrtthlem 11723	Lemma for resqrtth 11724. (Contributed by Jim Kingdon, 10-Aug-2021.)
⊢ ((𝐴 ∈ ℝ ∧ 0 ≤ 𝐴) → (((√‘𝐴)↑2) = 𝐴 ∧ 0 ≤ (√‘𝐴)))
Theorem	resqrtth 11724	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)
Theorem	remsqsqrt 11725	Square of square root. (Contributed by Mario Carneiro, 10-Jul-2013.)
⊢ ((𝐴 ∈ ℝ ∧ 0 ≤ 𝐴) → ((√‘𝐴) · (√‘𝐴)) = 𝐴)
Theorem	sqrtge0 11726	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))
Theorem	sqrtgt0 11727	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))
Theorem	sqrtmul 11728	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)))
Theorem	sqrtle 11729	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)) → (𝐴 ≤ 𝐵 ↔ (√‘𝐴) ≤ (√‘𝐵)))
Theorem	sqrtlt 11730	Square root is strictly monotonic. Closed form of sqrtlti 11830. (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)))
Theorem	sqrt11ap 11731	Analogue to sqrt11 11732 but for apartness. (Contributed by Jim Kingdon, 11-Aug-2021.)
⊢ (((𝐴 ∈ ℝ ∧ 0 ≤ 𝐴) ∧ (𝐵 ∈ ℝ ∧ 0 ≤ 𝐵)) → ((√‘𝐴) # (√‘𝐵) ↔ 𝐴 # 𝐵))
Theorem	sqrt11 11732	The square root function is one-to-one. Also see sqrt11ap 11731 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))
Theorem	sqrt00 11733	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))
Theorem	rpsqrtcl 11734	The square root of a positive real is a positive real. (Contributed by NM, 22-Feb-2008.)
⊢ (𝐴 ∈ ℝ+ → (√‘𝐴) ∈ ℝ+)
Theorem	sqrtdiv 11735	Square root distributes over division. (Contributed by Mario Carneiro, 5-May-2016.)
⊢ (((𝐴 ∈ ℝ ∧ 0 ≤ 𝐴) ∧ 𝐵 ∈ ℝ+) → (√‘(𝐴 / 𝐵)) = ((√‘𝐴) / (√‘𝐵)))
Theorem	sqrtsq2 11736	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) = 𝐵 ↔ 𝐴 = (𝐵↑2)))
Theorem	sqrtsq 11737	Square root of square. (Contributed by NM, 14-Jan-2006.) (Revised by Mario Carneiro, 29-May-2016.)
⊢ ((𝐴 ∈ ℝ ∧ 0 ≤ 𝐴) → (√‘(𝐴↑2)) = 𝐴)
Theorem	sqrtmsq 11738	Square root of square. (Contributed by NM, 2-Aug-1999.) (Revised by Mario Carneiro, 29-May-2016.)
⊢ ((𝐴 ∈ ℝ ∧ 0 ≤ 𝐴) → (√‘(𝐴 · 𝐴)) = 𝐴)
Theorem	sqrt1 11739	The square root of 1 is 1. (Contributed by NM, 31-Jul-1999.)
⊢ (√‘1) = 1
Theorem	sqrt4 11740	The square root of 4 is 2. (Contributed by NM, 3-Aug-1999.)
⊢ (√‘4) = 2
Theorem	sqrt9 11741	The square root of 9 is 3. (Contributed by NM, 11-May-2004.)
⊢ (√‘9) = 3
Theorem	sqrt2gt1lt2 11742	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)
Theorem	absneg 11743	Absolute value of negative. (Contributed by NM, 27-Feb-2005.)
⊢ (𝐴 ∈ ℂ → (abs‘-𝐴) = (abs‘𝐴))
Theorem	abscl 11744	Real closure of absolute value. (Contributed by NM, 3-Oct-1999.)
⊢ (𝐴 ∈ ℂ → (abs‘𝐴) ∈ ℝ)
Theorem	abscj 11745	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))
Theorem	absvalsq 11746	Square of value of absolute value function. (Contributed by NM, 16-Jan-2006.)
⊢ (𝐴 ∈ ℂ → ((abs‘𝐴)↑2) = (𝐴 · (∗‘𝐴)))
Theorem	absvalsq2 11747	Square of value of absolute value function. (Contributed by NM, 1-Feb-2007.)
⊢ (𝐴 ∈ ℂ → ((abs‘𝐴)↑2) = (((ℜ‘𝐴)↑2) + ((ℑ‘𝐴)↑2)))
Theorem	sqabsadd 11748	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))))))
Theorem	sqabssub 11749	Square of absolute value of difference. (Contributed by NM, 21-Jan-2007.)
⊢ ((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ) → ((abs‘(𝐴 − 𝐵))↑2) = ((((abs‘𝐴)↑2) + ((abs‘𝐵)↑2)) − (2 · (ℜ‘(𝐴 · (∗‘𝐵))))))
Theorem	absval2 11750	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))))
Theorem	abs0 11751	The absolute value of 0. (Contributed by NM, 26-Mar-2005.) (Revised by Mario Carneiro, 29-May-2016.)
⊢ (abs‘0) = 0
Theorem	absi 11752	The absolute value of the imaginary unit. (Contributed by NM, 26-Mar-2005.)
⊢ (abs‘i) = 1
Theorem	absge0 11753	Absolute value is nonnegative. (Contributed by NM, 20-Nov-2004.) (Revised by Mario Carneiro, 29-May-2016.)
⊢ (𝐴 ∈ ℂ → 0 ≤ (abs‘𝐴))
Theorem	absrpclap 11754	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+)
Theorem	abs00ap 11755	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))
Theorem	absext 11756	Strong extensionality for absolute value. (Contributed by Jim Kingdon, 12-Aug-2021.)
⊢ ((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ) → ((abs‘𝐴) # (abs‘𝐵) → 𝐴 # 𝐵))
Theorem	abs00 11757	The absolute value of a number is zero iff the number is zero. Also see abs00ap 11755 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))
Theorem	abs00ad 11758	A complex number is zero iff its absolute value is zero. Deduction form of abs00 11757. (Contributed by David Moews, 28-Feb-2017.)
⊢ (𝜑 → 𝐴 ∈ ℂ)    ⇒   ⊢ (𝜑 → ((abs‘𝐴) = 0 ↔ 𝐴 = 0))
Theorem	abs00bd 11759	If a complex number is zero, its absolute value is zero. (Contributed by David Moews, 28-Feb-2017.)
⊢ (𝜑 → 𝐴 = 0)    ⇒   ⊢ (𝜑 → (abs‘𝐴) = 0)
Theorem	absreimsq 11760	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)))
Theorem	absreim 11761	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))))
Theorem	absmul 11762	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)))
Theorem	absdivap 11763	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)))
Theorem	absid 11764	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)
Theorem	abs1 11765	The absolute value of 1. Common special case. (Contributed by David A. Wheeler, 16-Jul-2016.)
⊢ (abs‘1) = 1
Theorem	absnid 11766	A negative number is the negative of its own absolute value. (Contributed by NM, 27-Feb-2005.)
⊢ ((𝐴 ∈ ℝ ∧ 𝐴 ≤ 0) → (abs‘𝐴) = -𝐴)
Theorem	leabs 11767	A real number is less than or equal to its absolute value. (Contributed by NM, 27-Feb-2005.)
⊢ (𝐴 ∈ ℝ → 𝐴 ≤ (abs‘𝐴))
Theorem	qabsor 11768	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))
Theorem	qabsord 11769	The absolute value of a rational number is either that number or its negative. (Contributed by Jim Kingdon, 8-Nov-2021.)
⊢ (𝜑 → 𝐴 ∈ ℚ)    ⇒   ⊢ (𝜑 → ((abs‘𝐴) = 𝐴 ∨ (abs‘𝐴) = -𝐴))
Theorem	absre 11770	Absolute value of a real number. (Contributed by NM, 17-Mar-2005.)
⊢ (𝐴 ∈ ℝ → (abs‘𝐴) = (√‘(𝐴↑2)))
Theorem	absresq 11771	Square of the absolute value of a real number. (Contributed by NM, 16-Jan-2006.)
⊢ (𝐴 ∈ ℝ → ((abs‘𝐴)↑2) = (𝐴↑2))
Theorem	absexp 11772	Absolute value of positive integer exponentiation. (Contributed by NM, 5-Jan-2006.)
⊢ ((𝐴 ∈ ℂ ∧ 𝑁 ∈ ℕ0) → (abs‘(𝐴↑𝑁)) = ((abs‘𝐴)↑𝑁))
Theorem	absexpzap 11773	Absolute value of integer exponentiation. (Contributed by Jim Kingdon, 11-Aug-2021.)
⊢ ((𝐴 ∈ ℂ ∧ 𝐴 # 0 ∧ 𝑁 ∈ ℤ) → (abs‘(𝐴↑𝑁)) = ((abs‘𝐴)↑𝑁))
Theorem	abssq 11774	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)))
Theorem	sqabs 11775	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)))
Theorem	absrele 11776	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))
Theorem	absimle 11777	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))
Theorem	nn0abscl 11778	The absolute value of an integer is a nonnegative integer. (Contributed by NM, 27-Feb-2005.)
⊢ (𝐴 ∈ ℤ → (abs‘𝐴) ∈ ℕ0)
Theorem	zabscl 11779	The absolute value of an integer is an integer. (Contributed by Stefan O'Rear, 24-Sep-2014.)
⊢ (𝐴 ∈ ℤ → (abs‘𝐴) ∈ ℤ)
Theorem	ltabs 11780	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)
Theorem	abslt 11781	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 < 𝐴 ∧ 𝐴 < 𝐵)))
Theorem	absle 11782	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 ↔ (-𝐵 ≤ 𝐴 ∧ 𝐴 ≤ 𝐵)))
Theorem	abssubap0 11783	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) → (𝐵 − 𝐴) # 0)
Theorem	abssubne0 11784	If the absolute value of a complex number is less than a real, its difference from the real is nonzero. See also abssubap0 11783 which is the same with not equal changed to apart. (Contributed by NM, 2-Nov-2007.)
⊢ ((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℝ ∧ (abs‘𝐴) < 𝐵) → (𝐵 − 𝐴) ≠ 0)
Theorem	absdiflt 11785	The absolute value of a difference and 'less than' relation. (Contributed by Paul Chapman, 18-Sep-2007.)
⊢ ((𝐴 ∈ ℝ ∧ 𝐵 ∈ ℝ ∧ 𝐶 ∈ ℝ) → ((abs‘(𝐴 − 𝐵)) < 𝐶 ↔ ((𝐵 − 𝐶) < 𝐴 ∧ 𝐴 < (𝐵 + 𝐶))))
Theorem	absdifle 11786	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‘(𝐴 − 𝐵)) ≤ 𝐶 ↔ ((𝐵 − 𝐶) ≤ 𝐴 ∧ 𝐴 ≤ (𝐵 + 𝐶))))
Theorem	elicc4abs 11787	Membership in a symmetric closed real interval. (Contributed by Stefan O'Rear, 16-Nov-2014.)
⊢ ((𝐴 ∈ ℝ ∧ 𝐵 ∈ ℝ ∧ 𝐶 ∈ ℝ) → (𝐶 ∈ ((𝐴 − 𝐵)[,](𝐴 + 𝐵)) ↔ (abs‘(𝐶 − 𝐴)) ≤ 𝐵))
Theorem	lenegsq 11788	Comparison to a nonnegative number based on comparison to squares. (Contributed by NM, 16-Jan-2006.)
⊢ ((𝐴 ∈ ℝ ∧ 𝐵 ∈ ℝ ∧ 0 ≤ 𝐵) → ((𝐴 ≤ 𝐵 ∧ -𝐴 ≤ 𝐵) ↔ (𝐴↑2) ≤ (𝐵↑2)))
Theorem	releabs 11789	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))
Theorem	recvalap 11790	Reciprocal expressed with a real denominator. (Contributed by Jim Kingdon, 13-Aug-2021.)
⊢ ((𝐴 ∈ ℂ ∧ 𝐴 # 0) → (1 / 𝐴) = ((∗‘𝐴) / ((abs‘𝐴)↑2)))
Theorem	absidm 11791	The absolute value function is idempotent. (Contributed by NM, 20-Nov-2004.)
⊢ (𝐴 ∈ ℂ → (abs‘(abs‘𝐴)) = (abs‘𝐴))
Theorem	absgt0ap 11792	The absolute value of a number apart from zero is positive. (Contributed by Jim Kingdon, 13-Aug-2021.)
⊢ (𝐴 ∈ ℂ → (𝐴 # 0 ↔ 0 < (abs‘𝐴)))
Theorem	nnabscl 11793	The absolute value of a nonzero integer is a positive integer. (Contributed by Paul Chapman, 21-Mar-2011.) (Proof shortened by Andrew Salmon, 25-May-2011.)
⊢ ((𝑁 ∈ ℤ ∧ 𝑁 ≠ 0) → (abs‘𝑁) ∈ ℕ)
Theorem	abssub 11794	Swapping order of subtraction doesn't change the absolute value. (Contributed by NM, 1-Oct-1999.) (Proof shortened by Mario Carneiro, 29-May-2016.)
⊢ ((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ) → (abs‘(𝐴 − 𝐵)) = (abs‘(𝐵 − 𝐴)))
Theorem	abssubge0 11795	Absolute value of a nonnegative difference. (Contributed by NM, 14-Feb-2008.)
⊢ ((𝐴 ∈ ℝ ∧ 𝐵 ∈ ℝ ∧ 𝐴 ≤ 𝐵) → (abs‘(𝐵 − 𝐴)) = (𝐵 − 𝐴))
Theorem	abssuble0 11796	Absolute value of a nonpositive difference. (Contributed by FL, 3-Jan-2008.)
⊢ ((𝐴 ∈ ℝ ∧ 𝐵 ∈ ℝ ∧ 𝐴 ≤ 𝐵) → (abs‘(𝐴 − 𝐵)) = (𝐵 − 𝐴))
Theorem	abstri 11797	Triangle inequality for absolute value. Proposition 10-3.7(h) of [Gleason] p. 133. (Contributed by NM, 7-Mar-2005.) (Proof shortened by Mario Carneiro, 29-May-2016.)
⊢ ((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ) → (abs‘(𝐴 + 𝐵)) ≤ ((abs‘𝐴) + (abs‘𝐵)))
Theorem	abs3dif 11798	Absolute value of differences around common element. (Contributed by FL, 9-Oct-2006.)
⊢ ((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ) → (abs‘(𝐴 − 𝐵)) ≤ ((abs‘(𝐴 − 𝐶)) + (abs‘(𝐶 − 𝐵))))
Theorem	abs2dif 11799	Difference of absolute values. (Contributed by Paul Chapman, 7-Sep-2007.)
⊢ ((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ) → ((abs‘𝐴) − (abs‘𝐵)) ≤ (abs‘(𝐴 − 𝐵)))
Theorem	abs2dif2 11800	Difference of absolute values. (Contributed by Mario Carneiro, 14-Apr-2016.)
⊢ ((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ) → (abs‘(𝐴 − 𝐵)) ≤ ((abs‘𝐴) + (abs‘𝐵)))