# Inverse of Field Product

## Theorem

Let $\struct {F, +, \times}$ be a field whose zero is $0_F$ and whose unity is $1_F$.

Let $a, b \in F$ such that $a \ne 0$ and $b \ne 0$.

Then:

$\paren {a \times b}^{-1} = b^{-1} \times a^{-1}$

## Proof

We are given that $a \ne 0$ and $b \ne 0$.

From Field has no Proper Zero Divisors and Rule of Transposition, we have:

$a \times b \ne 0$

By Field Axiom $\text M4$: Inverses for Product we have that $\paren {a \times b}^{-1}$ exists.

Then we have:

 $\ds \paren {b^{-1} \times a^{-1} } \times \paren {a \times b}$ $=$ $\ds b^{-1} \times \paren {a^{-1} \times a} \times b$ Field Axiom $\text M1$: Associativity of Product $\ds$ $=$ $\ds b^{-1} \times 1_F \times b$ Field Axiom $\text M4$: Inverses for Product $\ds$ $=$ $\ds b^{-1} \times b$ Field Axiom $\text M3$: Identity for Product $\ds$ $=$ $\ds 1_F$ Field Axiom $\text M4$: Inverses for Product

Hence the result by definition of multiplicative inverse.

$\blacksquare$