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I know this has been addressed here, but I confess to not fully understanding that, so I'm hoping someone can chime here.

First, is there a canonical formula for this? In programming language languages, different ones lead to varying results, which would seem to indicate not?

Anyway, in layman's terms, I've always understood modulo to mean something like:

  1. Fractional remainder * quotient = x
  2. ceiling(x) = result

So an example with 5 mod 3:

  1. 5/3 = 1.66666666...
  2. fractional = 0.666666 * 3 = 1.998
  3. result = ceiling(1.998) = 2

applying that to a negative number though, like -1 mod 18:

  1. -1/18 = -0.0555555
  2. fractional = -0.055555 * 18 = -0.99999
  3. result = ceiling(-0.99999) = 0
  4. This seems wrong, so maybe floor() makes sense for negatives, which would yield -1 as the result

(#)4 seems intuitively correct as positive 1 mod 18 = 1. But that's wrong, as it seems -1 mod 18 = 17, and I just don't see how to get to that.

Thanks!

Odj fourth
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  • the fractional part of a negative number is defined as it was positive, i.e. $x-\lfloor x\rfloor$, but not the fractional part of the absolute –  Dec 03 '19 at 23:11
  • You take the greatest integer less than or equal to $x$, and subtract it from $x$ to get the fractional part. The greatest integer less than or equal to $-1/18$ is $-1$, so the integral part of $-1/18$ is $-1$, and the remainder is $.9444444...$. Multiply that by $18$ and you get $16.99999992$, whose ceiling is $17$. – Arturo Magidin Dec 03 '19 at 23:30

1 Answers1

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You are computing the fractional part of a negative number wrong.

The fractional part of a negative number is not the negative of the fractional part of its absolute value: that is, the fractional part of, for example, $-1.4$, is not $-0.4$. The fractional part is $0.6$.

This is because the fractional part of $x$ is defined to be $x-\mathrm{floor}(x)$. For $x=-1.4$, $\mathrm{floor}(x) = -2$, so the fractional part of $-1.4$ is $-1.4 - (-2) = 2-1.4 = 0.6$.

Once you do this adjustment, your process will work. You have $$\begin{align*} -\frac{1}{18} &\approx -0.055555\\ \mathrm{floor}\left(-\frac{1}{18}\right) &= -1\\ \mathrm{fractional}\left(-\frac{1}{18}\right) &\approx -0.055555+1\\ \mathrm{fractional}\left(-\frac{1}{18}\right) &\approx 0.944445\\ \mathrm{fractional}\left(-\frac{1}{18}\right)*18 &\approx 16.999992\\ \mathrm{ceiling}\left(\mathrm{fractional}\left(-\frac{1}{18}\right)*18\right) &= 17. \end{align*}$$ But you have to be careful with approximations, lest you get a bit more than $17$ in the final step, for example.

That said, it is better to think of this as quotient with remainder, rather than fractional part. The remainder when dividing by $q>0$ must be an integer greater than or equal to $0$ and strictly less than $18$. So when you divide $-1$ by $18$, you get a quotient of $-1$ and a remainder of $17$: more generally, if $m\gt 0$, then to find $a$ modulo $m$, you can do division with remainder to get $$a = qm + r,\qquad 0\leq r\lt m$$ and then the modulo is $r$.

Arturo Magidin
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  • Using that method makes sense. But why does it seem that it isn't the universal rule, e.g., programming languages with different implementations of modulo that yield different results? Are they incorrect in not adhering a "remainders are positive" stricture of number theory. (Incorrect in a loose sense, assuming they have a practical reason for pretending mod math works differently.) – Odj fourth Dec 04 '19 at 14:31
  • @Jimthefourth: Don't confuse the remainder with the modular relation with the modulo binary operator. The modulo binary operator yields the unique element that is congruent to the input from a set of chosen representatives. There are two standard choices: either the integers $0,1,\ldots,m-1$, or the integers of smallest absolute value, $0, 1, -1, 2, -2,$ etc. (e.g., modulo $7$ you can either ask "modulo 7" to return an integer from among $0,1,\ldots,6$, or from amont $-3,-2,\ldots,3$). Different applications prefer one or the other. They are both correct: they are different functions. – Arturo Magidin Dec 04 '19 at 15:18
  • @Jimthefourth See also this and this. – Arturo Magidin Dec 04 '19 at 15:25