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Metamath Proof Explorer

Theorem qdivcl

Description: Closure of division of rationals. (Contributed by NM, 3-Aug-2004)

Ref Expression
Assertion qdivcl ( ( 𝐴 ∈ ℚ ∧ 𝐵 ∈ ℚ ∧ 𝐵 ≠ 0 ) → ( 𝐴 / 𝐵 ) ∈ ℚ )

Proof

Step Hyp Ref Expression
1 qcn ⊢ ( 𝐴 ∈ ℚ → 𝐴 ∈ ℂ )
2 qcn ⊢ ( 𝐵 ∈ ℚ → 𝐵 ∈ ℂ )
3 id ⊢ ( 𝐵 ≠ 0 → 𝐵 ≠ 0 )
4 divrec ⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐵 ≠ 0 ) → ( 𝐴 / 𝐵 ) = ( 𝐴 · ( 1 / 𝐵 ) ) )
5 1 2 3 4 syl3an ⊢ ( ( 𝐴 ∈ ℚ ∧ 𝐵 ∈ ℚ ∧ 𝐵 ≠ 0 ) → ( 𝐴 / 𝐵 ) = ( 𝐴 · ( 1 / 𝐵 ) ) )
6 qreccl ⊢ ( ( 𝐵 ∈ ℚ ∧ 𝐵 ≠ 0 ) → ( 1 / 𝐵 ) ∈ ℚ )
7 qmulcl ⊢ ( ( 𝐴 ∈ ℚ ∧ ( 1 / 𝐵 ) ∈ ℚ ) → ( 𝐴 · ( 1 / 𝐵 ) ) ∈ ℚ )
8 6 7 sylan2 ⊢ ( ( 𝐴 ∈ ℚ ∧ ( 𝐵 ∈ ℚ ∧ 𝐵 ≠ 0 ) ) → ( 𝐴 · ( 1 / 𝐵 ) ) ∈ ℚ )
9 8 3impb ⊢ ( ( 𝐴 ∈ ℚ ∧ 𝐵 ∈ ℚ ∧ 𝐵 ≠ 0 ) → ( 𝐴 · ( 1 / 𝐵 ) ) ∈ ℚ )
10 5 9 eqeltrd ⊢ ( ( 𝐴 ∈ ℚ ∧ 𝐵 ∈ ℚ ∧ 𝐵 ≠ 0 ) → ( 𝐴 / 𝐵 ) ∈ ℚ )