Proof of Theorem pellexlem1
| Step | Hyp | Ref
| Expression |
| 1 | | nncn 11028 |
. . . . . . 7
⊢ (𝐴 ∈ ℕ → 𝐴 ∈
ℂ) |
| 2 | 1 | 3ad2ant2 1083 |
. . . . . 6
⊢ ((𝐷 ∈ ℕ ∧ 𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ) → 𝐴 ∈
ℂ) |
| 3 | 2 | sqcld 13006 |
. . . . 5
⊢ ((𝐷 ∈ ℕ ∧ 𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ) → (𝐴↑2) ∈
ℂ) |
| 4 | | nncn 11028 |
. . . . . . 7
⊢ (𝐷 ∈ ℕ → 𝐷 ∈
ℂ) |
| 5 | 4 | 3ad2ant1 1082 |
. . . . . 6
⊢ ((𝐷 ∈ ℕ ∧ 𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ) → 𝐷 ∈
ℂ) |
| 6 | | nncn 11028 |
. . . . . . . 8
⊢ (𝐵 ∈ ℕ → 𝐵 ∈
ℂ) |
| 7 | 6 | 3ad2ant3 1084 |
. . . . . . 7
⊢ ((𝐷 ∈ ℕ ∧ 𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ) → 𝐵 ∈
ℂ) |
| 8 | 7 | sqcld 13006 |
. . . . . 6
⊢ ((𝐷 ∈ ℕ ∧ 𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ) → (𝐵↑2) ∈
ℂ) |
| 9 | 5, 8 | mulcld 10060 |
. . . . 5
⊢ ((𝐷 ∈ ℕ ∧ 𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ) → (𝐷 · (𝐵↑2)) ∈ ℂ) |
| 10 | 3, 9 | subeq0ad 10402 |
. . . 4
⊢ ((𝐷 ∈ ℕ ∧ 𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ) → (((𝐴↑2) − (𝐷 · (𝐵↑2))) = 0 ↔ (𝐴↑2) = (𝐷 · (𝐵↑2)))) |
| 11 | | nnne0 11053 |
. . . . . . . 8
⊢ (𝐵 ∈ ℕ → 𝐵 ≠ 0) |
| 12 | 11 | 3ad2ant3 1084 |
. . . . . . 7
⊢ ((𝐷 ∈ ℕ ∧ 𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ) → 𝐵 ≠ 0) |
| 13 | | sqne0 12930 |
. . . . . . . 8
⊢ (𝐵 ∈ ℂ → ((𝐵↑2) ≠ 0 ↔ 𝐵 ≠ 0)) |
| 14 | 7, 13 | syl 17 |
. . . . . . 7
⊢ ((𝐷 ∈ ℕ ∧ 𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ) → ((𝐵↑2) ≠ 0 ↔ 𝐵 ≠ 0)) |
| 15 | 12, 14 | mpbird 247 |
. . . . . 6
⊢ ((𝐷 ∈ ℕ ∧ 𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ) → (𝐵↑2) ≠
0) |
| 16 | 3, 5, 8, 15 | divmul3d 10835 |
. . . . 5
⊢ ((𝐷 ∈ ℕ ∧ 𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ) → (((𝐴↑2) / (𝐵↑2)) = 𝐷 ↔ (𝐴↑2) = (𝐷 · (𝐵↑2)))) |
| 17 | | sqdiv 12928 |
. . . . . . . . . 10
⊢ ((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐵 ≠ 0) → ((𝐴 / 𝐵)↑2) = ((𝐴↑2) / (𝐵↑2))) |
| 18 | 17 | fveq2d 6195 |
. . . . . . . . 9
⊢ ((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐵 ≠ 0) →
(√‘((𝐴 / 𝐵)↑2)) =
(√‘((𝐴↑2)
/ (𝐵↑2)))) |
| 19 | 2, 7, 12, 18 | syl3anc 1326 |
. . . . . . . 8
⊢ ((𝐷 ∈ ℕ ∧ 𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ) →
(√‘((𝐴 / 𝐵)↑2)) =
(√‘((𝐴↑2)
/ (𝐵↑2)))) |
| 20 | | nnre 11027 |
. . . . . . . . . . 11
⊢ (𝐴 ∈ ℕ → 𝐴 ∈
ℝ) |
| 21 | 20 | 3ad2ant2 1083 |
. . . . . . . . . 10
⊢ ((𝐷 ∈ ℕ ∧ 𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ) → 𝐴 ∈
ℝ) |
| 22 | | nnre 11027 |
. . . . . . . . . . 11
⊢ (𝐵 ∈ ℕ → 𝐵 ∈
ℝ) |
| 23 | 22 | 3ad2ant3 1084 |
. . . . . . . . . 10
⊢ ((𝐷 ∈ ℕ ∧ 𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ) → 𝐵 ∈
ℝ) |
| 24 | 21, 23, 12 | redivcld 10853 |
. . . . . . . . 9
⊢ ((𝐷 ∈ ℕ ∧ 𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ) → (𝐴 / 𝐵) ∈ ℝ) |
| 25 | | nnnn0 11299 |
. . . . . . . . . . . 12
⊢ (𝐴 ∈ ℕ → 𝐴 ∈
ℕ0) |
| 26 | 25 | nn0ge0d 11354 |
. . . . . . . . . . 11
⊢ (𝐴 ∈ ℕ → 0 ≤
𝐴) |
| 27 | 26 | 3ad2ant2 1083 |
. . . . . . . . . 10
⊢ ((𝐷 ∈ ℕ ∧ 𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ) → 0 ≤
𝐴) |
| 28 | | nngt0 11049 |
. . . . . . . . . . 11
⊢ (𝐵 ∈ ℕ → 0 <
𝐵) |
| 29 | 28 | 3ad2ant3 1084 |
. . . . . . . . . 10
⊢ ((𝐷 ∈ ℕ ∧ 𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ) → 0 <
𝐵) |
| 30 | | divge0 10892 |
. . . . . . . . . 10
⊢ (((𝐴 ∈ ℝ ∧ 0 ≤
𝐴) ∧ (𝐵 ∈ ℝ ∧ 0 < 𝐵)) → 0 ≤ (𝐴 / 𝐵)) |
| 31 | 21, 27, 23, 29, 30 | syl22anc 1327 |
. . . . . . . . 9
⊢ ((𝐷 ∈ ℕ ∧ 𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ) → 0 ≤
(𝐴 / 𝐵)) |
| 32 | 24, 31 | sqrtsqd 14158 |
. . . . . . . 8
⊢ ((𝐷 ∈ ℕ ∧ 𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ) →
(√‘((𝐴 / 𝐵)↑2)) = (𝐴 / 𝐵)) |
| 33 | 19, 32 | eqtr3d 2658 |
. . . . . . 7
⊢ ((𝐷 ∈ ℕ ∧ 𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ) →
(√‘((𝐴↑2)
/ (𝐵↑2))) = (𝐴 / 𝐵)) |
| 34 | | nnq 11801 |
. . . . . . . . 9
⊢ (𝐴 ∈ ℕ → 𝐴 ∈
ℚ) |
| 35 | 34 | 3ad2ant2 1083 |
. . . . . . . 8
⊢ ((𝐷 ∈ ℕ ∧ 𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ) → 𝐴 ∈
ℚ) |
| 36 | | nnq 11801 |
. . . . . . . . 9
⊢ (𝐵 ∈ ℕ → 𝐵 ∈
ℚ) |
| 37 | 36 | 3ad2ant3 1084 |
. . . . . . . 8
⊢ ((𝐷 ∈ ℕ ∧ 𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ) → 𝐵 ∈
ℚ) |
| 38 | | qdivcl 11809 |
. . . . . . . 8
⊢ ((𝐴 ∈ ℚ ∧ 𝐵 ∈ ℚ ∧ 𝐵 ≠ 0) → (𝐴 / 𝐵) ∈ ℚ) |
| 39 | 35, 37, 12, 38 | syl3anc 1326 |
. . . . . . 7
⊢ ((𝐷 ∈ ℕ ∧ 𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ) → (𝐴 / 𝐵) ∈ ℚ) |
| 40 | 33, 39 | eqeltrd 2701 |
. . . . . 6
⊢ ((𝐷 ∈ ℕ ∧ 𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ) →
(√‘((𝐴↑2)
/ (𝐵↑2))) ∈
ℚ) |
| 41 | | fveq2 6191 |
. . . . . . 7
⊢ (((𝐴↑2) / (𝐵↑2)) = 𝐷 → (√‘((𝐴↑2) / (𝐵↑2))) = (√‘𝐷)) |
| 42 | 41 | eleq1d 2686 |
. . . . . 6
⊢ (((𝐴↑2) / (𝐵↑2)) = 𝐷 → ((√‘((𝐴↑2) / (𝐵↑2))) ∈ ℚ ↔
(√‘𝐷) ∈
ℚ)) |
| 43 | 40, 42 | syl5ibcom 235 |
. . . . 5
⊢ ((𝐷 ∈ ℕ ∧ 𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ) → (((𝐴↑2) / (𝐵↑2)) = 𝐷 → (√‘𝐷) ∈ ℚ)) |
| 44 | 16, 43 | sylbird 250 |
. . . 4
⊢ ((𝐷 ∈ ℕ ∧ 𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ) → ((𝐴↑2) = (𝐷 · (𝐵↑2)) → (√‘𝐷) ∈
ℚ)) |
| 45 | 10, 44 | sylbid 230 |
. . 3
⊢ ((𝐷 ∈ ℕ ∧ 𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ) → (((𝐴↑2) − (𝐷 · (𝐵↑2))) = 0 → (√‘𝐷) ∈
ℚ)) |
| 46 | 45 | necon3bd 2808 |
. 2
⊢ ((𝐷 ∈ ℕ ∧ 𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ) → (¬
(√‘𝐷) ∈
ℚ → ((𝐴↑2)
− (𝐷 · (𝐵↑2))) ≠
0)) |
| 47 | 46 | imp 445 |
1
⊢ (((𝐷 ∈ ℕ ∧ 𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ) ∧ ¬
(√‘𝐷) ∈
ℚ) → ((𝐴↑2)
− (𝐷 · (𝐵↑2))) ≠
0) |