Answer to Question #269747 in Discrete Mathematics for Haider

Question

29. Prove or disprove that if m and n are integers such that mn = 1 then either m = 1 and n = 1 , or else m = - 1 and n = - 1 .

Answer 1

Let us prove that if $m$ and $n$ are integers such that $mn = 1$ then either $m = 1$ and $n = 1 ,$ or else $m = - 1$ and $n = - 1 .$

Let us consider the following three cases.

1) If $m=0$ then $mn=0\cdot n=0,$ and hence $mn\ne 1.$

2) If $|m|>1,$ then $mn = 1$ implies $|n|=\frac{1}{|m|}<1.$ Since $n\ne 0,$ we conclude that there is no such integer $n$ with $0<|n|<1.$

3) Finally, let $|m|=1$, then either $m=1$ or $m=-1.$ If $m=1,$ then $n=\frac{1}{m}=1\in\Z.$ If $m=-1,$ then $n=\frac{1}{m}=-1\in\Z.$

We conclude that only in the case 3 the equality $mn = 1$ is possible, and consequently, if $m$ and $n$ are integers such that $mn = 1$ then either $m = 1$ and $n = 1 ,$ or else $m = - 1$ and $n = - 1 .$