Metamath Proof Explorer


Theorem pell1qrge1

Description: A Pell solution in the first quadrant is at least 1. (Contributed by Stefan O'Rear, 17-Sep-2014)

Ref Expression
Assertion pell1qrge1 ⊢ D ∈ ℕ ∖ ◻ ℕ ∧ A ∈ Pell1QR ⁡ D → 1 ≤ A

Proof

Step Hyp Ref Expression
1 elpell1qr ⊢ D ∈ ℕ ∖ ◻ ℕ → A ∈ Pell1QR ⁡ D ↔ A ∈ ℝ ∧ ∃ a ∈ ℕ 0 ∃ b ∈ ℕ 0 A = a + D ⁢ b ∧ a 2 − D ⁢ b 2 = 1
2 1red ⊢ D ∈ ℕ ∖ ◻ ℕ ∧ A ∈ ℝ ∧ a ∈ ℕ 0 ∧ b ∈ ℕ 0 ∧ a 2 − D ⁢ b 2 = 1 → 1 ∈ ℝ
3 simplrl ⊢ D ∈ ℕ ∖ ◻ ℕ ∧ A ∈ ℝ ∧ a ∈ ℕ 0 ∧ b ∈ ℕ 0 ∧ a 2 − D ⁢ b 2 = 1 → a ∈ ℕ 0
4 3 nn0red ⊢ D ∈ ℕ ∖ ◻ ℕ ∧ A ∈ ℝ ∧ a ∈ ℕ 0 ∧ b ∈ ℕ 0 ∧ a 2 − D ⁢ b 2 = 1 → a ∈ ℝ
5 eldifi ⊢ D ∈ ℕ ∖ ◻ ℕ → D ∈ ℕ
6 5 ad3antrrr ⊢ D ∈ ℕ ∖ ◻ ℕ ∧ A ∈ ℝ ∧ a ∈ ℕ 0 ∧ b ∈ ℕ 0 ∧ a 2 − D ⁢ b 2 = 1 → D ∈ ℕ
7 6 nnnn0d ⊢ D ∈ ℕ ∖ ◻ ℕ ∧ A ∈ ℝ ∧ a ∈ ℕ 0 ∧ b ∈ ℕ 0 ∧ a 2 − D ⁢ b 2 = 1 → D ∈ ℕ 0
8 7 nn0red ⊢ D ∈ ℕ ∖ ◻ ℕ ∧ A ∈ ℝ ∧ a ∈ ℕ 0 ∧ b ∈ ℕ 0 ∧ a 2 − D ⁢ b 2 = 1 → D ∈ ℝ
9 7 nn0ge0d ⊢ D ∈ ℕ ∖ ◻ ℕ ∧ A ∈ ℝ ∧ a ∈ ℕ 0 ∧ b ∈ ℕ 0 ∧ a 2 − D ⁢ b 2 = 1 → 0 ≤ D
10 8 9 resqrtcld ⊢ D ∈ ℕ ∖ ◻ ℕ ∧ A ∈ ℝ ∧ a ∈ ℕ 0 ∧ b ∈ ℕ 0 ∧ a 2 − D ⁢ b 2 = 1 → D ∈ ℝ
11 simplrr ⊢ D ∈ ℕ ∖ ◻ ℕ ∧ A ∈ ℝ ∧ a ∈ ℕ 0 ∧ b ∈ ℕ 0 ∧ a 2 − D ⁢ b 2 = 1 → b ∈ ℕ 0
12 11 nn0red ⊢ D ∈ ℕ ∖ ◻ ℕ ∧ A ∈ ℝ ∧ a ∈ ℕ 0 ∧ b ∈ ℕ 0 ∧ a 2 − D ⁢ b 2 = 1 → b ∈ ℝ
13 10 12 remulcld ⊢ D ∈ ℕ ∖ ◻ ℕ ∧ A ∈ ℝ ∧ a ∈ ℕ 0 ∧ b ∈ ℕ 0 ∧ a 2 − D ⁢ b 2 = 1 → D ⁢ b ∈ ℝ
14 4 13 readdcld ⊢ D ∈ ℕ ∖ ◻ ℕ ∧ A ∈ ℝ ∧ a ∈ ℕ 0 ∧ b ∈ ℕ 0 ∧ a 2 − D ⁢ b 2 = 1 → a + D ⁢ b ∈ ℝ
15 2nn0 ⊢ 2 ∈ ℕ 0
16 15 a1i ⊢ D ∈ ℕ ∖ ◻ ℕ ∧ A ∈ ℝ ∧ a ∈ ℕ 0 ∧ b ∈ ℕ 0 ∧ a 2 − D ⁢ b 2 = 1 → 2 ∈ ℕ 0
17 11 16 nn0expcld ⊢ D ∈ ℕ ∖ ◻ ℕ ∧ A ∈ ℝ ∧ a ∈ ℕ 0 ∧ b ∈ ℕ 0 ∧ a 2 − D ⁢ b 2 = 1 → b 2 ∈ ℕ 0
18 7 17 nn0mulcld ⊢ D ∈ ℕ ∖ ◻ ℕ ∧ A ∈ ℝ ∧ a ∈ ℕ 0 ∧ b ∈ ℕ 0 ∧ a 2 − D ⁢ b 2 = 1 → D ⁢ b 2 ∈ ℕ 0
19 18 nn0ge0d ⊢ D ∈ ℕ ∖ ◻ ℕ ∧ A ∈ ℝ ∧ a ∈ ℕ 0 ∧ b ∈ ℕ 0 ∧ a 2 − D ⁢ b 2 = 1 → 0 ≤ D ⁢ b 2
20 18 nn0red ⊢ D ∈ ℕ ∖ ◻ ℕ ∧ A ∈ ℝ ∧ a ∈ ℕ 0 ∧ b ∈ ℕ 0 ∧ a 2 − D ⁢ b 2 = 1 → D ⁢ b 2 ∈ ℝ
21 2 20 addge02d ⊢ D ∈ ℕ ∖ ◻ ℕ ∧ A ∈ ℝ ∧ a ∈ ℕ 0 ∧ b ∈ ℕ 0 ∧ a 2 − D ⁢ b 2 = 1 → 0 ≤ D ⁢ b 2 ↔ 1 ≤ D ⁢ b 2 + 1
22 19 21 mpbid ⊢ D ∈ ℕ ∖ ◻ ℕ ∧ A ∈ ℝ ∧ a ∈ ℕ 0 ∧ b ∈ ℕ 0 ∧ a 2 − D ⁢ b 2 = 1 → 1 ≤ D ⁢ b 2 + 1
23 sq1 ⊢ 1 2 = 1
24 23 a1i ⊢ D ∈ ℕ ∖ ◻ ℕ ∧ A ∈ ℝ ∧ a ∈ ℕ 0 ∧ b ∈ ℕ 0 ∧ a 2 − D ⁢ b 2 = 1 → 1 2 = 1
25 nn0cn ⊢ a ∈ ℕ 0 → a ∈ ℂ
26 25 ad2antrl ⊢ D ∈ ℕ ∖ ◻ ℕ ∧ A ∈ ℝ ∧ a ∈ ℕ 0 ∧ b ∈ ℕ 0 → a ∈ ℂ
27 26 sqcld ⊢ D ∈ ℕ ∖ ◻ ℕ ∧ A ∈ ℝ ∧ a ∈ ℕ 0 ∧ b ∈ ℕ 0 → a 2 ∈ ℂ
28 5 ad2antrr ⊢ D ∈ ℕ ∖ ◻ ℕ ∧ A ∈ ℝ ∧ a ∈ ℕ 0 ∧ b ∈ ℕ 0 → D ∈ ℕ
29 28 nncnd ⊢ D ∈ ℕ ∖ ◻ ℕ ∧ A ∈ ℝ ∧ a ∈ ℕ 0 ∧ b ∈ ℕ 0 → D ∈ ℂ
30 nn0cn ⊢ b ∈ ℕ 0 → b ∈ ℂ
31 30 ad2antll ⊢ D ∈ ℕ ∖ ◻ ℕ ∧ A ∈ ℝ ∧ a ∈ ℕ 0 ∧ b ∈ ℕ 0 → b ∈ ℂ
32 31 sqcld ⊢ D ∈ ℕ ∖ ◻ ℕ ∧ A ∈ ℝ ∧ a ∈ ℕ 0 ∧ b ∈ ℕ 0 → b 2 ∈ ℂ
33 29 32 mulcld ⊢ D ∈ ℕ ∖ ◻ ℕ ∧ A ∈ ℝ ∧ a ∈ ℕ 0 ∧ b ∈ ℕ 0 → D ⁢ b 2 ∈ ℂ
34 1cnd ⊢ D ∈ ℕ ∖ ◻ ℕ ∧ A ∈ ℝ ∧ a ∈ ℕ 0 ∧ b ∈ ℕ 0 → 1 ∈ ℂ
35 27 33 34 subaddd ⊢ D ∈ ℕ ∖ ◻ ℕ ∧ A ∈ ℝ ∧ a ∈ ℕ 0 ∧ b ∈ ℕ 0 → a 2 − D ⁢ b 2 = 1 ↔ D ⁢ b 2 + 1 = a 2
36 35 biimpa ⊢ D ∈ ℕ ∖ ◻ ℕ ∧ A ∈ ℝ ∧ a ∈ ℕ 0 ∧ b ∈ ℕ 0 ∧ a 2 − D ⁢ b 2 = 1 → D ⁢ b 2 + 1 = a 2
37 36 eqcomd ⊢ D ∈ ℕ ∖ ◻ ℕ ∧ A ∈ ℝ ∧ a ∈ ℕ 0 ∧ b ∈ ℕ 0 ∧ a 2 − D ⁢ b 2 = 1 → a 2 = D ⁢ b 2 + 1
38 22 24 37 3brtr4d ⊢ D ∈ ℕ ∖ ◻ ℕ ∧ A ∈ ℝ ∧ a ∈ ℕ 0 ∧ b ∈ ℕ 0 ∧ a 2 − D ⁢ b 2 = 1 → 1 2 ≤ a 2
39 0le1 ⊢ 0 ≤ 1
40 39 a1i ⊢ D ∈ ℕ ∖ ◻ ℕ ∧ A ∈ ℝ ∧ a ∈ ℕ 0 ∧ b ∈ ℕ 0 ∧ a 2 − D ⁢ b 2 = 1 → 0 ≤ 1
41 3 nn0ge0d ⊢ D ∈ ℕ ∖ ◻ ℕ ∧ A ∈ ℝ ∧ a ∈ ℕ 0 ∧ b ∈ ℕ 0 ∧ a 2 − D ⁢ b 2 = 1 → 0 ≤ a
42 2 4 40 41 le2sqd ⊢ D ∈ ℕ ∖ ◻ ℕ ∧ A ∈ ℝ ∧ a ∈ ℕ 0 ∧ b ∈ ℕ 0 ∧ a 2 − D ⁢ b 2 = 1 → 1 ≤ a ↔ 1 2 ≤ a 2
43 38 42 mpbird ⊢ D ∈ ℕ ∖ ◻ ℕ ∧ A ∈ ℝ ∧ a ∈ ℕ 0 ∧ b ∈ ℕ 0 ∧ a 2 − D ⁢ b 2 = 1 → 1 ≤ a
44 8 9 sqrtge0d ⊢ D ∈ ℕ ∖ ◻ ℕ ∧ A ∈ ℝ ∧ a ∈ ℕ 0 ∧ b ∈ ℕ 0 ∧ a 2 − D ⁢ b 2 = 1 → 0 ≤ D
45 11 nn0ge0d ⊢ D ∈ ℕ ∖ ◻ ℕ ∧ A ∈ ℝ ∧ a ∈ ℕ 0 ∧ b ∈ ℕ 0 ∧ a 2 − D ⁢ b 2 = 1 → 0 ≤ b
46 10 12 44 45 mulge0d ⊢ D ∈ ℕ ∖ ◻ ℕ ∧ A ∈ ℝ ∧ a ∈ ℕ 0 ∧ b ∈ ℕ 0 ∧ a 2 − D ⁢ b 2 = 1 → 0 ≤ D ⁢ b
47 4 13 addge01d ⊢ D ∈ ℕ ∖ ◻ ℕ ∧ A ∈ ℝ ∧ a ∈ ℕ 0 ∧ b ∈ ℕ 0 ∧ a 2 − D ⁢ b 2 = 1 → 0 ≤ D ⁢ b ↔ a ≤ a + D ⁢ b
48 46 47 mpbid ⊢ D ∈ ℕ ∖ ◻ ℕ ∧ A ∈ ℝ ∧ a ∈ ℕ 0 ∧ b ∈ ℕ 0 ∧ a 2 − D ⁢ b 2 = 1 → a ≤ a + D ⁢ b
49 2 4 14 43 48 letrd ⊢ D ∈ ℕ ∖ ◻ ℕ ∧ A ∈ ℝ ∧ a ∈ ℕ 0 ∧ b ∈ ℕ 0 ∧ a 2 − D ⁢ b 2 = 1 → 1 ≤ a + D ⁢ b
50 49 adantrl ⊢ D ∈ ℕ ∖ ◻ ℕ ∧ A ∈ ℝ ∧ a ∈ ℕ 0 ∧ b ∈ ℕ 0 ∧ A = a + D ⁢ b ∧ a 2 − D ⁢ b 2 = 1 → 1 ≤ a + D ⁢ b
51 simprl ⊢ D ∈ ℕ ∖ ◻ ℕ ∧ A ∈ ℝ ∧ a ∈ ℕ 0 ∧ b ∈ ℕ 0 ∧ A = a + D ⁢ b ∧ a 2 − D ⁢ b 2 = 1 → A = a + D ⁢ b
52 50 51 breqtrrd ⊢ D ∈ ℕ ∖ ◻ ℕ ∧ A ∈ ℝ ∧ a ∈ ℕ 0 ∧ b ∈ ℕ 0 ∧ A = a + D ⁢ b ∧ a 2 − D ⁢ b 2 = 1 → 1 ≤ A
53 52 ex ⊢ D ∈ ℕ ∖ ◻ ℕ ∧ A ∈ ℝ ∧ a ∈ ℕ 0 ∧ b ∈ ℕ 0 → A = a + D ⁢ b ∧ a 2 − D ⁢ b 2 = 1 → 1 ≤ A
54 53 rexlimdvva ⊢ D ∈ ℕ ∖ ◻ ℕ ∧ A ∈ ℝ → ∃ a ∈ ℕ 0 ∃ b ∈ ℕ 0 A = a + D ⁢ b ∧ a 2 − D ⁢ b 2 = 1 → 1 ≤ A
55 54 expimpd ⊢ D ∈ ℕ ∖ ◻ ℕ → A ∈ ℝ ∧ ∃ a ∈ ℕ 0 ∃ b ∈ ℕ 0 A = a + D ⁢ b ∧ a 2 − D ⁢ b 2 = 1 → 1 ≤ A
56 1 55 sylbid ⊢ D ∈ ℕ ∖ ◻ ℕ → A ∈ Pell1QR ⁡ D → 1 ≤ A
57 56 imp ⊢ D ∈ ℕ ∖ ◻ ℕ ∧ A ∈ Pell1QR ⁡ D → 1 ≤ A