Proof
(related to Proposition: Uniqueness of Inverse Real Numbers With Respect to Multiplication)
 For a given real number \(x\in\mathbb R\), \(x\neq 0\), let \(y\) be any real number with $xy=1\label{E18322}\tag{1}.$
 Applying the existence of inverse real 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 real one and since the multiplication of real numbers is cancellative, it follows that $x^{1}(xy)=x^{1}.$
 Because the multiplication of real numbers is associative, it follows $(x^{1}x)y=x^{1}.$
 Applying the existence of inverse real numbers with respect to multiplication once again leads to $1\cdot y=x^{1}.$
 Using the existence of real one again, results in $y=x^{1}.$
 Thus, the there is only one such real number $y$ for which $x\cdot y=1$, namely $y=x^{1}$.
 Note that because the multiplication of real numbers is commutative, we would get the same result, if we multiplied both sides of the equation $\eqref{E18322}$ from the right side by \(x^{1}\).
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References
Bibliography
 Forster Otto: "Analysis 1, Differential und Integralrechnung einer VerĂ¤nderlichen", Vieweg Studium, 1983