Let $f: \mathbb{Z}_5 \rightarrow \mathbb{Z}_{10}$ and $g: \mathbb{Z}_5 \rightarrow \mathbb{Z}_{10}$ be defined by $f([x]_5)=[x]_{10}$ and $g([x]_5)=([x]_{10})^5$ Prove that $f=g$
I know that the domain and codomain are the same but I'm unsure how to go about showing that for every input both functions have the same output.
Best Answer
You could make a table... and show that $f = g$ for all $x$ in the domain.
But, let's do some algebra.
We want to show that: $(x^5 - x) \equiv 0 \pmod {10}$
$(x^5 - x) =(x)(x+1)(x-1)(x^2+1)$
Clearly for $x \equiv 0,1,-1\pmod5, (x^5 - x)\equiv 0$
So we still need to check, $2,3$
$(4^2+1)\equiv 0\pmod 5\\ (9^2+1)\equiv 0\pmod 5$
And since one of $(x)(x-1)(x+1)$ must be even. $(x^5-x) \equiv 0 \pmod 5 \ \implies (x^5-x)\equiv 0 \pmod {10}$
Using similar logic we can extend this....One of $(x)(x-1)(x+1)$ is divisible by $3$ $(x^5-x)\equiv 0 \pmod {10}\implies(x^5-x)\equiv 0 \pmod {30}$