# How do I subtract times?

I have an embarrassingly basic modular arithmetic question. I understand that I can subtract, for example, 1 hour from 10 o'clock to get 9 o'clock, or even 2 hours from 1 o'clock to get 11 o'clock; but how do I subtract 2 o'clock from 11 o'clock to get 3 hours.

That is, I want a function that takes two times on the face of a clock, and gives me the interval between them.

For example: I have two angular values measured in "hours". One value is the right ascension of the apparent sun, $\alpha(t)$, while the other is the right ascension of the "mean sun", $\langle\alpha\rangle(t)$; both are mod 24. I'm interested in the angular difference between these two values $$E(t) = \langle\alpha\rangle(t) - \alpha(t)$$ Is this just inherently ambiguous, so that I need to impose some additional constraint based on information about the system (e.g., here that the values can be positive or negative, and are always small, so that, say $\pm 22$ should be interpreted as $\mp 2$), or is there some simple systematic way to ensure that I get the correct values?

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math.stackexchange.com/questions/388361/… check this answer – iostream007 May 11 '13 at 16:50

Just subtract. Your answer will only be well-defined mod 12 (because maybe 15 hours passed between 11 and 2 o clock). If you want a "canonical" answer, you can always take the least residue mod 12. In this case, we get $(2 - 11) = -9 = 3.$

In fancy language, the times on a clock are a torsor for $\frac{\mathbb{Z}}{12}$. John Baez has a great blog entry on this.

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@Jean-Sébastien: I've added an example of my application to make things a bit more concrete. – raxacoricofallapatorius Nov 19 '12 at 20:01

2 o'clock - 11 o'clock = -9 = 3 mod 12 ?

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$$2-11=-9\equiv 3 \mod 12$$ you can take the absolute value of that, or work with the $24$ hours format and use the fact that $2$ is also $14$ and then have $$14-11=3$$

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I think I need a nap... :-S – amWhy Nov 19 '12 at 19:17
@amWhy you only needed to add the fact that we work mod $12$ – Jean-Sébastien Nov 19 '12 at 19:17
I saw that others already covered that...I knew what I was thinking, but that's hardly the same as being clear and explicit in the first place! (+1) by the way! – amWhy Nov 19 '12 at 19:36
@amWhy I'd be down for a nap as well, great idea ;). I like the 24 hours format for that kind of thing. Obviously you could argue that the problem comes back if you want ot know the time difference between say $23$ hours and $2$ but that is equivalent to $11$ and $14$ anywya so – Jean-Sébastien Nov 19 '12 at 19:46

If you just have times with no dates attached, you can't tell +2 from -22. You can choose to have the result be in the range $[-12,+12)$ or any other range of length $24$ like $(-5,19]$ that you like. Based on what you day, I would opt for one centered at zero.

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What you're doing is also a problem in directional statistics with measuring angles.

If the times are $t_1$ and $t_2$, then I think you want

\begin{cases} |t_1 - t_2| & \text{If}\; |t_1 - t_2| \le 12\\ 24 - |t_1 - t_2| & \text{If}\; |t_1 - t_2|>12 \\ \end{cases}

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