Inverses 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:

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

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 $x^{-1} = \dfrac 1 x = y$.

$\blacksquare$