# Why does ${-}14 \bmod 12 = 10$? [closed]

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.  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.

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

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Community, Shailesh, Leucippus, metamorphy, RRL
If this question can be reworded to fit the rules in the help center, please edit the question.

• 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. – lєαf Jan 30 at 23:27
• @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 $a$. – MPW Jan 30 at 23:29
• @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. – Bill Dubuque Jan 30 at 23:33
• @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 "$a\, \text{mod}\, n$". – anomaly Jan 31 at 0:09
• @anomaly : I think of ‘mod’ as giving the representative in the base coset – MPW Jan 31 at 0:51

$$-14\equiv10\pmod{12}\,$$ because $$\,{-}14=10-12\cdot 2$$. In other words, $$-14$$ and $$10$$ leaves the same remainder after dividing $$12.$$

• 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. – Bill Dubuque Jan 31 at 0:14

14 hours before midnight, it's 10 o'clock.

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

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

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 $$k=-2$$: $$-14 = -2\cdot 12+10.$$

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

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.

• 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. – Gunnar Sveinsson Jan 30 at 23:09
• Although if "modular aritmetic" is more well defined than I think, you may be right that it doesnt really mean anything in that "field". – Gunnar Sveinsson Jan 30 at 23:17
• 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". – Bill Dubuque Jan 30 at 23:17
• @BillDubuque I tried to make it clear that I'm talking in the context of modular arithmetic. – Arthur Jan 31 at 0:20
• @Arthur And so am I. – Bill Dubuque Jan 31 at 0:23