Inverse for Real Multiplication

Theorem

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

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

Proof

By the definition of real number:

$\forall \epsilon \in \R_{>0}: \exists t \in \N: \forall i > t: \size {x_i - x} < \epsilon$

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Let $\epsilon = \size x$.

This is possible because $x \ne 0$, since it is required that $\epsilon > 0$.

\(\ds \exists t \in \N: \forall i > t: \, \)

\(\ds \size {x_i - x}\)

\(<\)

\(\ds \size x\)

\(\ds \leadsto \ \ \)

\(\ds \exists t \in \N: \forall i > t: \, \)

\(\ds \size {x_i - x}\)

\(<\)

\(\ds \size {x_i - x} + \size {x_i}\)

Triangle Inequality for Real Numbers

\(\ds \leadsto \ \ \)

\(\ds \exists t \in \N : \forall i > t: \, \)

\(\ds \size {x_i}\)

\(>\)

\(\ds 0\)

Construct a sequence $\sequence {y_n}$ as follows:

$y_n = \begin {cases} \dfrac 1 {x_n} & n > t \\ 0 & n \le t \end {cases}$

Let $u_n$ denote the Heaviside step function with parameter $t$.

We have:

\(\ds \eqclass {\sequence {x_n} } {} \times \eqclass {\sequence {y_n} } {}\)

\(=\)

\(\ds \eqclass {\sequence {x_n y_n} } {}\)

\(\ds \)

\(=\)

\(\ds \eqclass {\sequence {u_n} } {}\)

\(\ds \)

\(=\)

\(\ds 1\)

Similarly for $\eqclass {\sequence {y_n} } {} \times \eqclass {\sequence {x_n} } {}$.

So the inverse of $x \in \struct {\R_{\ne 0}, \times}$ is $x^{-1} = \dfrac 1 x = y$.

$\blacksquare$

Sources

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