Metamath Proof Explorer


Theorem numdensq

Description: Squaring a rational squares its canonical components. (Contributed by Stefan O'Rear, 15-Sep-2014)

Ref Expression
Assertion numdensq ⊢ A ∈ ℚ → numer ⁡ A 2 = numer ⁡ A 2 ∧ denom ⁡ A 2 = denom ⁡ A 2

Proof

Step Hyp Ref Expression
1 qnumdencoprm ⊢ A ∈ ℚ → numer ⁡ A gcd denom ⁡ A = 1
2 1 oveq1d ⊢ A ∈ ℚ → numer ⁡ A gcd denom ⁡ A 2 = 1 2
3 qnumcl ⊢ A ∈ ℚ → numer ⁡ A ∈ ℤ
4 qdencl ⊢ A ∈ ℚ → denom ⁡ A ∈ ℕ
5 4 nnzd ⊢ A ∈ ℚ → denom ⁡ A ∈ ℤ
6 zgcdsq ⊢ numer ⁡ A ∈ ℤ ∧ denom ⁡ A ∈ ℤ → numer ⁡ A gcd denom ⁡ A 2 = numer ⁡ A 2 gcd denom ⁡ A 2
7 3 5 6 syl2anc ⊢ A ∈ ℚ → numer ⁡ A gcd denom ⁡ A 2 = numer ⁡ A 2 gcd denom ⁡ A 2
8 sq1 ⊢ 1 2 = 1
9 8 a1i ⊢ A ∈ ℚ → 1 2 = 1
10 2 7 9 3eqtr3d ⊢ A ∈ ℚ → numer ⁡ A 2 gcd denom ⁡ A 2 = 1
11 qeqnumdivden ⊢ A ∈ ℚ → A = numer ⁡ A denom ⁡ A
12 11 oveq1d ⊢ A ∈ ℚ → A 2 = numer ⁡ A denom ⁡ A 2
13 3 zcnd ⊢ A ∈ ℚ → numer ⁡ A ∈ ℂ
14 4 nncnd ⊢ A ∈ ℚ → denom ⁡ A ∈ ℂ
15 4 nnne0d ⊢ A ∈ ℚ → denom ⁡ A ≠ 0
16 13 14 15 sqdivd ⊢ A ∈ ℚ → numer ⁡ A denom ⁡ A 2 = numer ⁡ A 2 denom ⁡ A 2
17 12 16 eqtrd ⊢ A ∈ ℚ → A 2 = numer ⁡ A 2 denom ⁡ A 2
18 qsqcl ⊢ A ∈ ℚ → A 2 ∈ ℚ
19 zsqcl ⊢ numer ⁡ A ∈ ℤ → numer ⁡ A 2 ∈ ℤ
20 3 19 syl ⊢ A ∈ ℚ → numer ⁡ A 2 ∈ ℤ
21 4 nnsqcld ⊢ A ∈ ℚ → denom ⁡ A 2 ∈ ℕ
22 qnumdenbi ⊢ A 2 ∈ ℚ ∧ numer ⁡ A 2 ∈ ℤ ∧ denom ⁡ A 2 ∈ ℕ → numer ⁡ A 2 gcd denom ⁡ A 2 = 1 ∧ A 2 = numer ⁡ A 2 denom ⁡ A 2 ↔ numer ⁡ A 2 = numer ⁡ A 2 ∧ denom ⁡ A 2 = denom ⁡ A 2
23 18 20 21 22 syl3anc ⊢ A ∈ ℚ → numer ⁡ A 2 gcd denom ⁡ A 2 = 1 ∧ A 2 = numer ⁡ A 2 denom ⁡ A 2 ↔ numer ⁡ A 2 = numer ⁡ A 2 ∧ denom ⁡ A 2 = denom ⁡ A 2
24 10 17 23 mpbi2and ⊢ A ∈ ℚ → numer ⁡ A 2 = numer ⁡ A 2 ∧ denom ⁡ A 2 = denom ⁡ A 2