Inverse of Field Product with Inverse

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Theorem

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

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


Then:

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

This can also be expressed using the notation of division as:

$1 / \paren {a / b} = b / a$


Proof 1

By definition, a field is a non-trivial division ring whose ring product is commutative.

By definition, a division ring is a ring with unity such that every non-zero element is a unit.

Hence we can use Inverse of Division Product:

$\paren {\dfrac a b}^{-1} = \dfrac {1_R} {\paren {a / b}} = \dfrac b a$

which applies to the group of units of a comutative ring with unity.

$\blacksquare$


Proof 2

\(\ds \paren {a \times b^{-1} }^{-1}\) \(=\) \(\ds \paren {b^{-1} }^{-1} \times a^{-1}\) Inverse of Field Product
\(\ds \) \(=\) \(\ds b \times a^{-1}\) Inverse of Multiplicative Inverse

$\blacksquare$


Sources

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