Metamath Proof Explorer


Theorem recval

Description: Reciprocal expressed with a real denominator. (Contributed by Mario Carneiro, 1-Apr-2015)

Ref Expression
Assertion recval ⊢ A ∈ ℂ ∧ A ≠ 0 → 1 A = A ‾ A 2

Proof

Step Hyp Ref Expression
1 cjcl ⊢ A ∈ ℂ → A ‾ ∈ ℂ
2 1 adantr ⊢ A ∈ ℂ ∧ A ≠ 0 → A ‾ ∈ ℂ
3 simpl ⊢ A ∈ ℂ ∧ A ≠ 0 → A ∈ ℂ
4 2 3 mulcomd ⊢ A ∈ ℂ ∧ A ≠ 0 → A ‾ ⁢ A = A ⁢ A ‾
5 absvalsq ⊢ A ∈ ℂ → A 2 = A ⁢ A ‾
6 5 adantr ⊢ A ∈ ℂ ∧ A ≠ 0 → A 2 = A ⁢ A ‾
7 4 6 eqtr4d ⊢ A ∈ ℂ ∧ A ≠ 0 → A ‾ ⁢ A = A 2
8 abscl ⊢ A ∈ ℂ → A ∈ ℝ
9 8 adantr ⊢ A ∈ ℂ ∧ A ≠ 0 → A ∈ ℝ
10 9 recnd ⊢ A ∈ ℂ ∧ A ≠ 0 → A ∈ ℂ
11 10 sqcld ⊢ A ∈ ℂ ∧ A ≠ 0 → A 2 ∈ ℂ
12 cjne0 ⊢ A ∈ ℂ → A ≠ 0 ↔ A ‾ ≠ 0
13 12 biimpa ⊢ A ∈ ℂ ∧ A ≠ 0 → A ‾ ≠ 0
14 11 2 3 13 divmuld ⊢ A ∈ ℂ ∧ A ≠ 0 → A 2 A ‾ = A ↔ A ‾ ⁢ A = A 2
15 7 14 mpbird ⊢ A ∈ ℂ ∧ A ≠ 0 → A 2 A ‾ = A
16 15 oveq2d ⊢ A ∈ ℂ ∧ A ≠ 0 → 1 A 2 A ‾ = 1 A
17 abs00 ⊢ A ∈ ℂ → A = 0 ↔ A = 0
18 17 necon3bid ⊢ A ∈ ℂ → A ≠ 0 ↔ A ≠ 0
19 18 biimpar ⊢ A ∈ ℂ ∧ A ≠ 0 → A ≠ 0
20 sqne0 ⊢ A ∈ ℂ → A 2 ≠ 0 ↔ A ≠ 0
21 10 20 syl ⊢ A ∈ ℂ ∧ A ≠ 0 → A 2 ≠ 0 ↔ A ≠ 0
22 19 21 mpbird ⊢ A ∈ ℂ ∧ A ≠ 0 → A 2 ≠ 0
23 11 2 22 13 recdivd ⊢ A ∈ ℂ ∧ A ≠ 0 → 1 A 2 A ‾ = A ‾ A 2
24 16 23 eqtr3d ⊢ A ∈ ℂ ∧ A ≠ 0 → 1 A = A ‾ A 2