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Metamath Proof Explorer

Theorem divreczi

Description: Relationship between division and reciprocal. Theorem I.9 of Apostol p. 18. (Contributed by NM, 11-Oct-1999)

Ref Expression
Hypotheses divclz.1 ⊢ 𝐴 ∈ ℂ
divclz.2 ⊢ 𝐵 ∈ ℂ
Assertion divreczi ( 𝐵 ≠ 0 → ( 𝐴 / 𝐵 ) = ( 𝐴 · ( 1 / 𝐵 ) ) )

Proof

Step Hyp Ref Expression
1 divclz.1 ⊢ 𝐴 ∈ ℂ
2 divclz.2 ⊢ 𝐵 ∈ ℂ
3 divrec ⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐵 ≠ 0 ) → ( 𝐴 / 𝐵 ) = ( 𝐴 · ( 1 / 𝐵 ) ) )
4 1 2 3 mp3an12 ⊢ ( 𝐵 ≠ 0 → ( 𝐴 / 𝐵 ) = ( 𝐴 · ( 1 / 𝐵 ) ) )