Inverses for Real Multiplication

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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:

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



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

\(\displaystyle \exists t \in \N : \forall i > t : \ \ \) \(\displaystyle \size {x_i - x}\) \(<\) \(\displaystyle \size x\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle \exists t \in \N : \forall i > t : \ \ \) \(\displaystyle \size {x_i - x}\) \(<\) \(\displaystyle \size {x_i - x} + \size {x_i}\) Triangle Inequality for Real Numbers
\(\displaystyle \leadsto \ \ \) \(\displaystyle \exists t \in \N : \forall i > t : \ \ \) \(\displaystyle \size {x_i}\) \(>\) \(\displaystyle 0\)


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:

\(\displaystyle \eqclass {\sequence {x_n} } {} \times \eqclass {\sequence {y_n} } {}\) \(=\) \(\displaystyle \eqclass {\sequence {x_n y_n} } {}\)
\(\displaystyle \) \(=\) \(\displaystyle \eqclass {\sequence {u_n} } {}\)
\(\displaystyle \) \(=\) \(\displaystyle 1\)

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

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

$\blacksquare$