(related to Proposition: Uniqueness of Inverse Rational Numbers With Respect to Multiplication)

- For a given rational number \(x\in\mathbb R\), \(x\neq 0\), let \(y\) be
*any*rational number with $xy=1\label{E18324}\tag{1}.$ - Applying the existence of inverse rational numbers with respect to multiplication, we can multiply both sides of the equation by \(x^{-1}\)
*from the left side*, resulting in $x^{-1}(xy)=x^{-1}\cdot 1.$ - From the existence of rational one and since the multiplication of rational numbers is cancellative, it follows that $x^{-1}(xy)=x^{-1}.$
- Because the multiplication of rational numbers is associative, it follows $(x^{-1}x)y=x^{-1}.$
- Applying the existence of inverse rational numbers with respect to multiplication once again leads to $1\cdot y=x^{-1}.$
- Using the existence of rational one once again, results in $y=x^{-1}.$
- Thus, the there is only one such rational number $y$ for which $x\cdot y=1$, namely $y=x^{-1}$.
- Note that because the multiplication of rational numbers is commutative, we would get the same result, if we multiplied both sides of the equation $\eqref{E18324}$
*from the right side*by \(x^{-1}\).∎