Metamath Proof Explorer


Theorem numdenneg

Description: Numerator and denominator of the negative. (Contributed by Thierry Arnoux, 27-Oct-2017)

Ref Expression
Assertion numdenneg ( 𝑄 ∈ ℚ → ( ( numer ‘ - 𝑄 ) = - ( numer ‘ 𝑄 ) ∧ ( denom ‘ - 𝑄 ) = ( denom ‘ 𝑄 ) ) )

Proof

Step Hyp Ref Expression
1 qnegcl ⊢ ( 𝑄 ∈ ℚ → - 𝑄 ∈ ℚ )
2 qnumcl ⊢ ( 𝑄 ∈ ℚ → ( numer ‘ 𝑄 ) ∈ ℤ )
3 2 znegcld ⊢ ( 𝑄 ∈ ℚ → - ( numer ‘ 𝑄 ) ∈ ℤ )
4 qdencl ⊢ ( 𝑄 ∈ ℚ → ( denom ‘ 𝑄 ) ∈ ℕ )
5 4 nnzd ⊢ ( 𝑄 ∈ ℚ → ( denom ‘ 𝑄 ) ∈ ℤ )
6 neggcd ⊢ ( ( ( numer ‘ 𝑄 ) ∈ ℤ ∧ ( denom ‘ 𝑄 ) ∈ ℤ ) → ( - ( numer ‘ 𝑄 ) gcd ( denom ‘ 𝑄 ) ) = ( ( numer ‘ 𝑄 ) gcd ( denom ‘ 𝑄 ) ) )
7 2 5 6 syl2anc ⊢ ( 𝑄 ∈ ℚ → ( - ( numer ‘ 𝑄 ) gcd ( denom ‘ 𝑄 ) ) = ( ( numer ‘ 𝑄 ) gcd ( denom ‘ 𝑄 ) ) )
8 qnumdencoprm ⊢ ( 𝑄 ∈ ℚ → ( ( numer ‘ 𝑄 ) gcd ( denom ‘ 𝑄 ) ) = 1 )
9 7 8 eqtrd ⊢ ( 𝑄 ∈ ℚ → ( - ( numer ‘ 𝑄 ) gcd ( denom ‘ 𝑄 ) ) = 1 )
10 qeqnumdivden ⊢ ( 𝑄 ∈ ℚ → 𝑄 = ( ( numer ‘ 𝑄 ) / ( denom ‘ 𝑄 ) ) )
11 10 negeqd ⊢ ( 𝑄 ∈ ℚ → - 𝑄 = - ( ( numer ‘ 𝑄 ) / ( denom ‘ 𝑄 ) ) )
12 2 zcnd ⊢ ( 𝑄 ∈ ℚ → ( numer ‘ 𝑄 ) ∈ ℂ )
13 4 nncnd ⊢ ( 𝑄 ∈ ℚ → ( denom ‘ 𝑄 ) ∈ ℂ )
14 4 nnne0d ⊢ ( 𝑄 ∈ ℚ → ( denom ‘ 𝑄 ) ≠ 0 )
15 12 13 14 divnegd ⊢ ( 𝑄 ∈ ℚ → - ( ( numer ‘ 𝑄 ) / ( denom ‘ 𝑄 ) ) = ( - ( numer ‘ 𝑄 ) / ( denom ‘ 𝑄 ) ) )
16 11 15 eqtrd ⊢ ( 𝑄 ∈ ℚ → - 𝑄 = ( - ( numer ‘ 𝑄 ) / ( denom ‘ 𝑄 ) ) )
17 qnumdenbi ⊢ ( ( - 𝑄 ∈ ℚ ∧ - ( numer ‘ 𝑄 ) ∈ ℤ ∧ ( denom ‘ 𝑄 ) ∈ ℕ ) → ( ( ( - ( numer ‘ 𝑄 ) gcd ( denom ‘ 𝑄 ) ) = 1 ∧ - 𝑄 = ( - ( numer ‘ 𝑄 ) / ( denom ‘ 𝑄 ) ) ) ↔ ( ( numer ‘ - 𝑄 ) = - ( numer ‘ 𝑄 ) ∧ ( denom ‘ - 𝑄 ) = ( denom ‘ 𝑄 ) ) ) )
18 17 biimpa ⊢ ( ( ( - 𝑄 ∈ ℚ ∧ - ( numer ‘ 𝑄 ) ∈ ℤ ∧ ( denom ‘ 𝑄 ) ∈ ℕ ) ∧ ( ( - ( numer ‘ 𝑄 ) gcd ( denom ‘ 𝑄 ) ) = 1 ∧ - 𝑄 = ( - ( numer ‘ 𝑄 ) / ( denom ‘ 𝑄 ) ) ) ) → ( ( numer ‘ - 𝑄 ) = - ( numer ‘ 𝑄 ) ∧ ( denom ‘ - 𝑄 ) = ( denom ‘ 𝑄 ) ) )
19 1 3 4 9 16 18 syl32anc ⊢ ( 𝑄 ∈ ℚ → ( ( numer ‘ - 𝑄 ) = - ( numer ‘ 𝑄 ) ∧ ( denom ‘ - 𝑄 ) = ( denom ‘ 𝑄 ) ) )