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.

  • $\begingroup$ You can just calculate it. f.e. $f([3]_5) = [3]_{10}$ and $g([3]_5) = [3]*[3]*[3]*[3]*[3] = [7]*[3]*[3] = [1]*[3] = [3] $ (in subscript $10$). $\endgroup$ – bob Dec 12 '16 at 19:04
  • $\begingroup$ What are the definition of the functions? Given $[x]_5$, does one take a representative $x$ there and then modulo 10 to get $[x]_{10}$? Is this a well-defined? If that is the case, you need to show that $x^5-x = x(x^4-1)=x(x-1)(x+1)(x^2+1)$ is divisible by 10 for every $x \in \mathbb{Z}.$ Or, as another suggestion goes, once you know is well defined, just calculate the five values of both functions and compare. $\endgroup$ – Will M. Dec 12 '16 at 19:07

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 $3$ $(x^5-x)\equiv 0 \pmod {10}\implies(x^5-x)\equiv 0 \pmod {30}$


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