Metamath Proof Explorer


Theorem goldratval

Description: Value of the golden ratio. (Contributed by Ender Ting, 24-Jul-2026)

Ref Expression
Hypothesis goldra.val ⊢ F = 2 ⁢ cos ⁡ π 5
Assertion goldratval ⊢ F = 1 + 5 2

Proof

Step Hyp Ref Expression
1 goldra.val ⊢ F = 2 ⁢ cos ⁡ π 5
2 1lt5 ⊢ 1 < 5
3 0le1 ⊢ 0 ≤ 1
4 5nn0 ⊢ 5 ∈ ℕ 0
5 4 nn0ge0i ⊢ 0 ≤ 5
6 1re ⊢ 1 ∈ ℝ
7 5re ⊢ 5 ∈ ℝ
8 6 7 sqrtlti ⊢ 0 ≤ 1 ∧ 0 ≤ 5 → 1 < 5 ↔ 1 < 5
9 3 5 8 mp2an ⊢ 1 < 5 ↔ 1 < 5
10 2 9 mpbi ⊢ 1 < 5
11 negneg1e1 ⊢ − -1 = 1
12 sqrt1 ⊢ 1 = 1
13 11 12 eqtr4i ⊢ − -1 = 1
14 5pos ⊢ 0 < 5
15 7 14 sqrtpclii ⊢ 5 ∈ ℝ
16 15 recni ⊢ 5 ∈ ℂ
17 16 addridi ⊢ 5 + 0 = 5
18 10 13 17 3brtr4i ⊢ − -1 < 5 + 0
19 neg1rr ⊢ − 1 ∈ ℝ
20 19 renegcli ⊢ − -1 ∈ ℝ
21 0re ⊢ 0 ∈ ℝ
22 20 15 21 ltsubadd2i ⊢ - -1 - 5 < 0 ↔ − -1 < 5 + 0
23 18 22 mpbir ⊢ - -1 - 5 < 0
24 20 15 resubcli ⊢ - -1 - 5 ∈ ℝ
25 2re ⊢ 2 ∈ ℝ
26 25 6 remulcli ⊢ 2 ⋅ 1 ∈ ℝ
27 2pos ⊢ 0 < 2
28 2t1e2 ⊢ 2 ⋅ 1 = 2
29 27 28 breqtrri ⊢ 0 < 2 ⋅ 1
30 24 21 26 29 ltdiv1ii ⊢ - -1 - 5 < 0 ↔ - -1 - 5 2 ⋅ 1 < 0 2 ⋅ 1
31 23 30 mpbi ⊢ - -1 - 5 2 ⋅ 1 < 0 2 ⋅ 1
32 26 recni ⊢ 2 ⋅ 1 ∈ ℂ
33 21 29 gtneii ⊢ 2 ⋅ 1 ≠ 0
34 32 33 div0i ⊢ 0 2 ⋅ 1 = 0
35 31 34 breqtri ⊢ - -1 - 5 2 ⋅ 1 < 0
36 24 26 33 redivcli ⊢ - -1 - 5 2 ⋅ 1 ∈ ℝ
37 36 21 ltnsymi ⊢ - -1 - 5 2 ⋅ 1 < 0 → ¬ 0 < - -1 - 5 2 ⋅ 1
38 35 37 ax-mp ⊢ ¬ 0 < - -1 - 5 2 ⋅ 1
39 1 goldrapos ⊢ 0 < F
40 breq2 ⊢ F = - -1 - 5 2 ⋅ 1 → 0 < F ↔ 0 < - -1 - 5 2 ⋅ 1
41 39 40 mpbii ⊢ F = - -1 - 5 2 ⋅ 1 → 0 < - -1 - 5 2 ⋅ 1
42 38 41 mto ⊢ ¬ F = - -1 - 5 2 ⋅ 1
43 1 goldrarr ⊢ F ∈ ℝ
44 43 recni ⊢ F ∈ ℂ
45 44 sqcli ⊢ F 2 ∈ ℂ
46 ax-1cn ⊢ 1 ∈ ℂ
47 44 46 addcli ⊢ F + 1 ∈ ℂ
48 45 47 negsubi ⊢ F 2 + − F + 1 = F 2 − F + 1
49 45 mullidi ⊢ 1 ⁢ F 2 = F 2
50 44 mulm1i ⊢ -1 ⁢ F = − F
51 50 oveq1i ⊢ -1 ⁢ F + -1 = - F + -1
52 44 46 negdii ⊢ − F + 1 = - F + -1
53 51 52 eqtr4i ⊢ -1 ⁢ F + -1 = − F + 1
54 49 53 oveq12i ⊢ 1 ⁢ F 2 + -1 ⁢ F + -1 = F 2 + − F + 1
55 subsub4 ⊢ F 2 ∈ ℂ ∧ F ∈ ℂ ∧ 1 ∈ ℂ → F 2 - F - 1 = F 2 − F + 1
56 45 44 46 55 mp3an ⊢ F 2 - F - 1 = F 2 − F + 1
57 48 54 56 3eqtr4ri ⊢ F 2 - F - 1 = 1 ⁢ F 2 + -1 ⁢ F + -1
58 1 goldratmolem4 ⊢ F 2 - F - 1 = 0
59 57 58 eqtr3i ⊢ 1 ⁢ F 2 + -1 ⁢ F + -1 = 0
60 1cnd ⊢ ⊤ → 1 ∈ ℂ
61 ax-1ne0 ⊢ 1 ≠ 0
62 61 a1i ⊢ ⊤ → 1 ≠ 0
63 neg1cn ⊢ − 1 ∈ ℂ
64 63 a1i ⊢ ⊤ → − 1 ∈ ℂ
65 44 a1i ⊢ ⊤ → F ∈ ℂ
66 4cn ⊢ 4 ∈ ℂ
67 46 66 subnegi ⊢ 1 − -4 = 1 + 4
68 neg1sqe1 ⊢ − 1 2 = 1
69 63 mullidi ⊢ 1 ⁢ -1 = − 1
70 69 oveq2i ⊢ 4 ⁢ 1 ⁢ -1 = 4 ⁢ -1
71 66 46 mulneg2i ⊢ 4 ⁢ -1 = − 4 ⋅ 1
72 66 mulridi ⊢ 4 ⋅ 1 = 4
73 72 negeqi ⊢ − 4 ⋅ 1 = − 4
74 70 71 73 3eqtri ⊢ 4 ⁢ 1 ⁢ -1 = − 4
75 68 74 oveq12i ⊢ − 1 2 − 4 ⁢ 1 ⁢ -1 = 1 − -4
76 df-5 ⊢ 5 = 4 + 1
77 66 46 76 comraddi ⊢ 5 = 1 + 4
78 67 75 77 3eqtr4ri ⊢ 5 = − 1 2 − 4 ⁢ 1 ⁢ -1
79 78 a1i ⊢ ⊤ → 5 = − 1 2 − 4 ⁢ 1 ⁢ -1
80 60 62 64 64 65 79 quad ⊢ ⊤ → 1 ⁢ F 2 + -1 ⁢ F + -1 = 0 ↔ F = - -1 + 5 2 ⋅ 1 ∨ F = - -1 - 5 2 ⋅ 1
81 80 mptru ⊢ 1 ⁢ F 2 + -1 ⁢ F + -1 = 0 ↔ F = - -1 + 5 2 ⋅ 1 ∨ F = - -1 - 5 2 ⋅ 1
82 59 81 mpbi ⊢ F = - -1 + 5 2 ⋅ 1 ∨ F = - -1 - 5 2 ⋅ 1
83 82 ori ⊢ ¬ F = - -1 + 5 2 ⋅ 1 → F = - -1 - 5 2 ⋅ 1
84 42 83 mt3 ⊢ F = - -1 + 5 2 ⋅ 1
85 11 oveq1i ⊢ - -1 + 5 = 1 + 5
86 85 28 oveq12i ⊢ - -1 + 5 2 ⋅ 1 = 1 + 5 2
87 84 86 eqtri ⊢ F = 1 + 5 2