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Why does $-14 \bmod 12 = 10$? I would be grateful if someone could explain this to me step by step, for I am but a novice in the field of modular arithmetic. [edit] I obtained this equation by playing around with values for (x,y,z) in: x mod y = z on the Google calculator. Thank you!

Edit: I would like to hit myself on the head- I see it now. Thanks to everyone who responded! Unless there is a discussion going on in this post right now, it can be marked as resolved.

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closed as off-topic by user296602, Shailesh, Leucippus, metamorphy, RRL Jan 31 at 6:34

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  • $\begingroup$ I think it's the latter, but that wouldn't make much sense here. I was playing around with the google calculator when I came up with this equation. $\endgroup$ – lєαf Jan 30 at 23:27
  • $\begingroup$ @GunnarSveinsson : I would downvote your comment if I could. There is one and only one sane definition of $a \mod n$ (where $n>0$). It is the unique value in $[0,n)$ which can be obtained by adding an integral multiple of $n$ to $a$. $\endgroup$ – MPW Jan 30 at 23:29
  • $\begingroup$ @MPW It is true that the first definition / notation is wrong, but it is not true that there is only one definition of operational mod, e.g. it might denote the canonical residue in a system of least magnitude reps, e.g. $\, -1,0,1\pmod{3}.\ $ Its definition is relative to the complete system of reps employed. $\endgroup$ – Bill Dubuque Jan 30 at 23:33
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    $\begingroup$ @MPW: ...that having been said, I'd be much happier if the definition were that $a\equiv b\pmod{n}$ if $n| (b - a)$, rather than writing it as function-like thing "$a\, \text{mod}\, n$". $\endgroup$ – anomaly Jan 31 at 0:09
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    $\begingroup$ @anomaly : I think of ‘mod’ as giving the representative in the base coset $\endgroup$ – MPW Jan 31 at 0:51
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$-14\equiv10\pmod{12}\,$ because $\,{-}14=10-12\cdot 2$. In other words, $-14$ and $10$ leaves the same remainder after dividing $12.$

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    $\begingroup$ The notation means not only they are congruent but, further that $10$ is the canonical rep from the equivalence class $\,-14+12\Bbb Z\,$ wrt the complete system of residues $\bmod 12\,$ that one is using. $\endgroup$ – Bill Dubuque Jan 31 at 0:14
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14 hours before midnight, it's 10 o'clock.

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-24 is the largest number less than -14 which is congruent to 0 mod 12. -24 + 10 = -14 so -14 = 10 mod 12.

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Because $10 -(-14) = 24$ which is a multiple of 12. In general, $ a\equiv b \pmod m $ means that $m$ divides $a-b$.

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If you have $a\text{ mod }b = c$ then there exists $k\in\mathbb Z$ such that $a=kb+c$. In your case, you have $a=-14$, $b=12$ and $c=10$ while $k=-2$: $$ -14 = -2\cdot 12+10. $$

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Whenever you get a positive number, you remove $12$ until you get a number between $0$ And $11$ to get the modulo.

Example: $\ 37\bmod{12}\equiv 37-12\equiv 25\equiv 25-12\equiv 13\equiv 13-12\equiv 1\pmod{12}$

For a negative number, in the same way just add $12$ until you get something positive.

Example: $\ {-}14\bmod{12}\equiv -14+12\equiv -2\equiv -2+12\equiv 10\pmod{12}$

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I know it's tempting to think of mod as an operation, but most often in modular arithmetic, it's a modifier of $\equiv$, specifying exactly what "congruence" means. Really, if the notation had been $-14\equiv_{12}10$ instead, this would've been so much clearer.

So with this interpretation in mind, $-14\mod 12$ doesn't mean anything. On the other hand, $-14\equiv10\mod12$ is a statement. We can check whether it's true by using the definition: $$ a\equiv b\mod c\iff c\mid a-b $$ In this case, we get $12\mid -14-10$, which is true. This means that $-14\equiv 10\mod 12$ is true.

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    $\begingroup$ I disagree that $-14 \mod 12$ doesn't mean anything. I have seen this been treated as the image of $-14$ under the function $\mod 12 : \mathbb{Z}\to \mathbb{Z}$ or even of the canonical function $ \mathbb{Z}\to \mathbb{Z}/12\mathbb{Z}$. In a computer science course I took this expression was treated as a number. $\endgroup$ – Gunnar Sveinsson Jan 30 at 23:09
  • $\begingroup$ Although if "modular aritmetic" is more well defined than I think, you may be right that it doesnt really mean anything in that "field". $\endgroup$ – Gunnar Sveinsson Jan 30 at 23:17
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    $\begingroup$ The operation $\bmod$ is certainly in wide use in mathematics, computer science, etc. So it is incorrect and misleading to claim the it "doesn't mean anything". $\endgroup$ – Bill Dubuque Jan 30 at 23:17
  • $\begingroup$ @BillDubuque I tried to make it clear that I'm talking in the context of modular arithmetic. $\endgroup$ – Arthur Jan 31 at 0:20
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    $\begingroup$ @Arthur And so am I. $\endgroup$ – Bill Dubuque Jan 31 at 0:23

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