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Say I want to calculate say 12 (mod 13).

Can I do the following:

swap 12 and 13 and say that since 13 (mod 12) = 1, I can therefore conclude that 12 (mod 13)= -1?

If not, could someone please explain the correct procedure were I can get negative remainders?

Thanks very much.

  • In this case it works, but in general for a positive integer $;a;$ , you'll get $;-a=b\pmod m\iff a+b=0\pmod m;$ , or what is the same: iff $;a+b;$ is a multiple of $;m;$ . – DonAntonio Mar 26 '17 at 12:25
  • Thanks for that. I'm still wondering how negative remainders are calculated though. I have seen this when a is less than b. Could you please explain? –  Mar 26 '17 at 12:53
  • Exactly as I told you. For example, if we work $;\pmod {24};$ , say, then $;-5=19\pmod{24};$ , because $;5+19=24=0\pmod{24};$ , or also $;-88=8\pmod{24};$ , because $;88+8=96=24\cdot4=0\pmod{24};$ , etc. ... – DonAntonio Mar 26 '17 at 13:40
  • With the -88 = 8 (mod 24) example, doesn't that mean there are endless answers that would work? I'm sorry if my questions are annoying, I'm still trying to get my head around the negative remainders and I'm afraid I'm still not getting it :( –  Mar 26 '17 at 14:01
  • @u Yes, of course there are infinite possible answers, but it you choose and fix one set of representatives of the equivalence classes modulo $;m;$ , then the answer is unique in any case. For example, $;{...-5,19,43,67,...};$ are all equal to $;19\pmod{24};$ , and this is an infinite set. Yet if you choose the usually standard representatives set, namely $;{0,1,2,3,...,23};$ , then the only one that fits in is $;19;$ . – DonAntonio Mar 26 '17 at 14:11
  • Oh I see. Thank you very much for your explanation. –  Mar 27 '17 at 00:30

1 Answers1

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This does not work in general. Notice that $14$ is $4 \bmod 5$, but $5$ is not $-4 \bmod 14$.

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