0
$\begingroup$

In $\Bbb{Z}/m\Bbb{Z}$, show that $([a]_m)^{qd+r} = (([a]_m)^d)^q([a]_m)^r$.

My first attempt at this question was to use simple arithmetic properties to prove this true, however, this is incorrect.

What is the best way to prove this?

$\endgroup$
0
$\begingroup$

You could use $$([a]_m)^{qd+r}=\underbrace{[a]_m[a]_m\cdots[a]_m}_{qd+r\textrm{ times}}=_*[a^{qd+r}]_m=[(a^d)^qa^r]_m$$ where $(*)$ happens by definition.

$\endgroup$
  • $\begingroup$ I was thinking of doing this, but then wouldn't it entail proving $([a^d])([a^q]) = [a^{qd}]$? $\endgroup$ – flutterbug98 Sep 27 at 2:51
  • $\begingroup$ If so, then how would that be proven? $\endgroup$ – flutterbug98 Sep 27 at 2:51
0
$\begingroup$

Start by showing the following is well-defined: $$[x]_m \cdot [y]_m = [x\cdot y]_m$$

That is, if you take representatives $x_1, x_2, y_1, y_2 \in \mathbb{Z}$ such that $[x_1]_m = [x_2]_m$ and $[y_1]_m = [y_2]_m,$ then $$x_1 y_1 - x_2 y_2 = (x_1-x_2)y_1 + x_2(y_1-y_2)$$ is divisible by $m.$


This property allows us to write $$([a]_m)^n = [a^n]_m$$ and these two statements together allow you to use those simple arithmetic properties to prove your statement.

$\endgroup$

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.