Answer to Question #157417 in Real Analysis for Vincent

Question

Show that x inverse is not equal to 0 and is unique

Answer 1

Here we have $x$ $\in\R$ (which is invertible).

Now, let $a\in\R$ be the multiplicative inverse of $x$ . So, by definition we get

$$a\cdot x=1_\R (identity\>element\>in\>\R)\\ \Rightarrow a\cdot x=1\\ \Rightarrow x=a^{-1}$$

Now, let us assume $a=x^{-1}=0$ , so we get

So, we have

$$x^{-1}=0\\ \Rightarrow x=\frac{1}{0}=undefined \> in \> \R$$

So, we have a contradiction, which arises from our assumption that $a=0$ , so we get $a\neq0$ .

Now, let us assume $b\in\R$ , such that $b$ is the multiplicative inverse of $x$.

So $b=x^{-1}$

But, we already know that $a=x^{-1}$ , so

$\therefore a =b=x^{-1}$

Hence the multiplicative inverse of $x$ is unique.