```Ok, I think I understand now.  But to clarify, when you mention
the range as [0,1) you are referring to the remainder with respect
to the denominator, right?

So given

n % d = m

then

0 <= m < d

Is that it? But this doesn't hold if d is negative: 10 % -9 = -8
Argh!  Maybe I still don't quite get it.

" It depends.
"
" If you promise that the remainder will always be in the range
" [0,1), it is what you have to do.  If you are a speed demon
" you may prefer saying that the sign of the modulus is not
" defined and leave this out.  If you intend to use the modulus
" it is more useful to know that it is in that range.  But the
" vast majority of the time, people are not using the modulus.
"
" So -2 is theoretically a "better" answer.  But it is slower to
" produce.
"
" It is therefore probably not a coincidence that of the 6
" languages that you tried, the ones which are more concerned
" with speed all gave -1 for an answer, and the ones which are
" trying to be high-level languages gave -2.
"
" BTW to add a data point just to be confusing, Perl does it 3
" ways. :-)
"
" 1. By default Perl gives an answer in floating point.
" 2. If you use integer (saying you want speed) you get -1.
" 3. If you use the % (ie modulus) operator by default you
"    get 8 (reflecting the "careful" rounding) but if you
"    have "use integer" in place you get -1 for an answer.
"
" Note with item 3 that the answer that Perl gives *changes*
" depending on whether you are hinting that you care about
" speed.
"
" Cheers,
" Ben
"
Thanks,
Mike
```