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I am mostly a lazy person and so if I can reduce what I write, all the better. Obviously ultimately it is for the professor to decide whether something is or not an abuse in notation, but I want to know whether this is, at least, acceptable in the general case.

I have made the following statement:

Assume $n\equiv 2 \pmod5$:

Then $n^5\equiv2^5\pmod5 \implies n^5-n\equiv 2^5-2=30\equiv0 \pmod5 \implies n^5-n=5k$

for some $k\in\mathbb{N}$

The part I am unsure about is the equals sign followed by a congruency symbol. It seems to me readable enough, but I am very lax with notation.

Edit:

I know this is not the whole proof to my problem, there are 4 other cases to consider, but it's pretty much copy paste this assumption and change the numbers, it's explicetly the notation that I am conerned about.

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    $\begingroup$ What are you trying to prove? And yes, that is proper notation. $\endgroup$ – user_194421 Jan 19 '18 at 4:06
  • $\begingroup$ That for all $n$, $n^5-n$ = $5k$ $\endgroup$ – Makogan Jan 19 '18 at 4:07
  • $\begingroup$ You could just replace the $=$ sign with the $\equiv$ symbol with no effect on your result. $\endgroup$ – Riley Jan 19 '18 at 4:07
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    $\begingroup$ Looks fine to me. Though the proof is not complete, because you assumed $n \equiv 2 \pmod 5$ $\endgroup$ – Landuros Jan 19 '18 at 4:09
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    $\begingroup$ Are you required to use modular arithmetic? Because there is a simple proof by induction that just uses a binomial expansion. $$\begin{split}0^5-0 & =0 \\ (n+1)^5-(n+1) & = (n^5-n)+5(n^4+2n^3+2n^2+n)\end{split}$$ $\endgroup$ – Graham Kemp Jan 19 '18 at 4:33

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