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.

$\blacksquare$

Sources

• 1973: C.R.J. Clapham: Introduction to Mathematical Analysis ... (previous) ... (next): Chapter $1$: Axioms for the Real Numbers: $2$. Fields: Theorem $3 \ \text {(vii)}$