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Metamath Proof Explorer

Theorem divrecd

Description: Relationship between division and reciprocal. Theorem I.9 of Apostol p. 18. (Contributed by Mario Carneiro, 27-May-2016)

Ref Expression
Hypotheses div1d.1 ⊢ ( 𝜑 → 𝐴 ∈ ℂ )
divcld.2 ⊢ ( 𝜑 → 𝐵 ∈ ℂ )
divcld.3 ⊢ ( 𝜑 → 𝐵 ≠ 0 )
Assertion divrecd ( 𝜑 → ( 𝐴 / 𝐵 ) = ( 𝐴 · ( 1 / 𝐵 ) ) )

Proof

Step Hyp Ref Expression
1 div1d.1 ⊢ ( 𝜑 → 𝐴 ∈ ℂ )
2 divcld.2 ⊢ ( 𝜑 → 𝐵 ∈ ℂ )
3 divcld.3 ⊢ ( 𝜑 → 𝐵 ≠ 0 )
4 divrec ⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐵 ≠ 0 ) → ( 𝐴 / 𝐵 ) = ( 𝐴 · ( 1 / 𝐵 ) ) )
5 1 2 3 4 syl3anc ⊢ ( 𝜑 → ( 𝐴 / 𝐵 ) = ( 𝐴 · ( 1 / 𝐵 ) ) )