0
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ 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
1
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.