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:


 * $\forall \epsilon > 0 : \exists t \in \N : \forall i > t : \left|{x_i - x}\right| < \epsilon$

Let $\epsilon = \left|{x}\right|$. This is possible because $x \ne 0$, since it is required that $\epsilon > 0$.

Construct a sequence $\left\langle{y_n}\right\rangle$ as follows:


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

Let $\mu_n$ denote the Heaviside step function with parameter $t$.

We have:

Similarly for $\left[\!\left[{\left \langle{y_n}\right \rangle}\right]\!\right] \times \left[\!\left[{\left \langle{x_n}\right \rangle}\right]\!\right]$.

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