Metamath Proof Explorer


Theorem goldratmolem4

Description: Lemma 4 for determining the value of golden ratio. (Contributed by Ender Ting, 24-Jul-2026)

Ref Expression
Hypothesis goldra.val ⊢ F = 2 ⁢ cos ⁡ π 5
Assertion goldratmolem4 ⊢ F 2 - F - 1 = 0

Proof

Step Hyp Ref Expression
1 goldra.val ⊢ F = 2 ⁢ cos ⁡ π 5
2 1 goldrarr ⊢ F ∈ ℝ
3 2re ⊢ 2 ∈ ℝ
4 2 3 readdcli ⊢ F + 2 ∈ ℝ
5 1 goldrapos ⊢ 0 < F
6 2pos ⊢ 0 < 2
7 2 3 5 6 addgt0ii ⊢ 0 < F + 2
8 4 7 gt0ne0ii ⊢ F + 2 ≠ 0
9 8 neii ⊢ ¬ F + 2 = 0
10 2 recni ⊢ F ∈ ℂ
11 10 goldpolyfactor ⊢ F 2 - F - 1 ⁢ F 2 - F - 1 ⁢ F + 2 = F 5 − 5 ⁢ F 3 + 5 ⁢ F + 2
12 1 goldratmolem3 ⊢ F 5 − 5 ⁢ F 3 + 5 ⁢ F + 2 = 0
13 11 12 eqtri ⊢ F 2 - F - 1 ⁢ F 2 - F - 1 ⁢ F + 2 = 0
14 10 sqcli ⊢ F 2 ∈ ℂ
15 14 10 subcli ⊢ F 2 − F ∈ ℂ
16 ax-1cn ⊢ 1 ∈ ℂ
17 15 16 subcli ⊢ F 2 - F - 1 ∈ ℂ
18 17 17 mulcli ⊢ F 2 - F - 1 ⁢ F 2 - F - 1 ∈ ℂ
19 2cn ⊢ 2 ∈ ℂ
20 10 19 addcli ⊢ F + 2 ∈ ℂ
21 18 20 mul0ori ⊢ F 2 - F - 1 ⁢ F 2 - F - 1 ⁢ F + 2 = 0 ↔ F 2 - F - 1 ⁢ F 2 - F - 1 = 0 ∨ F + 2 = 0
22 13 21 mpbi ⊢ F 2 - F - 1 ⁢ F 2 - F - 1 = 0 ∨ F + 2 = 0
23 orcom ⊢ F 2 - F - 1 ⁢ F 2 - F - 1 = 0 ∨ F + 2 = 0 ↔ F + 2 = 0 ∨ F 2 - F - 1 ⁢ F 2 - F - 1 = 0
24 22 23 mpbi ⊢ F + 2 = 0 ∨ F 2 - F - 1 ⁢ F 2 - F - 1 = 0
25 9 24 mtpor ⊢ F 2 - F - 1 ⁢ F 2 - F - 1 = 0
26 17 msq0i ⊢ F 2 - F - 1 ⁢ F 2 - F - 1 = 0 ↔ F 2 - F - 1 = 0
27 25 26 mpbi ⊢ F 2 - F - 1 = 0