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Exercise (The Last Element Of UN Is Always Its Own Inverse)
Pick an integer $n$ greater than 1. Prove that $(n-1)\in U(n)$ is always its own multiplicative inverse.
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Notice that $(n-1)\pmod n = (-1)\pmod n$, which means $(n-1)^2\pmod n = (-1)^\pmod n = 1\pmod n$. This shows that the order of $n-1$ is 2, and also that $n-1$ is its own inverse.