Proof
(related to Proposition: \((xy)^{-1}=x^{-1}y^{-1}\))
- From the existence of the reciprocal numbers it follows that $(xy)\cdot(xy)^{-1}=1$ for all $x,y\in\mathbb R.$
- Multiplying both sides of the equation by $x^{-1}$ results in $y\cdot(xy)^{-1}=x^{-1}.$
- Since $x=ba^{-1}$ is the only unique solution of $ax=b,$ it follows that the equation $yz=x^{-1}$ has the unique solution $z=x^{-1}y^{-1}.$
- This proofs the statement $(xy)^{-1}=x^{-1}y^{-1}.$
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References
Bibliography
- Forster Otto: "Analysis 1, Differential- und Integralrechnung einer Veränderlichen", Vieweg Studium, 1983
