Inverse for Rational Multiplication

Theorem

Each element $x$ of the set of non-zero rational numbers $\Q_{\ne 0}$ has an inverse element $\dfrac 1 x$ under the operation of rational number multiplication:

$\forall x \in \Q_{\ne 0}: \exists \dfrac 1 x \in \Q_{\ne 0}: x \times \dfrac 1 x = 1 = \dfrac 1 x \times x$

From the definition, the field $\struct {\Q, +, \times}$ of rational numbers is the field of quotients of the integral domain $\struct {\Z, +, \times}$ of integers.

From Rational Multiplication Identity is One, the identity for $\struct {\Q, \times}$ is $1 = \dfrac 1 1 = \dfrac p p$ where $p \in \Z$ and $p \ne 0$.

$\paren {\dfrac p q}^{-1} = \dfrac q p$

which can be demonstrated directly:

$\ds \frac p q \times \frac q p$

$=$

$\ds \frac {p \times q} {q \times p}$

$\ds $

$=$

$\ds \frac {p \times q} {p \times q}$

$\ds $

$=$

$\ds \frac 1 1$

Note that this is defined only when $p \ne 0$ and $q \ne 0$.

Now let $x = \dfrac p q \in \Q$.

We define $\dfrac 1 x$ as $1 \times \paren {\dfrac p q}^{-1}$.

Hence:

$\ds \frac 1 x$

$=$

$\ds \frac 1 1 \times \paren {\frac p q}^{-1}$

$\ds $

$=$

$\ds \frac 1 1 \times \frac q p$

$\ds $

$=$

$\ds \frac {1 \times q} {1 \times p}$

$\ds $

$=$

$\ds \frac q p$

$\blacksquare$

1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text I$: Algebraic Structures: $\S 4$: Neutral Elements and Inverses