🚀 go-pugleaf

lori@rt7.cs.wisc.edu (Lori Lehman) writes:



>Everyone will agree that  

>    (a)  5 div 3 = 1,
>    (b)  5 mod 3 = 2,

>and most people seem to agree that  

>    (c)  -5 div 3 = -2,
>    (d)  -5 mod 3 = 1.

>Note that the divisor in these examples, 3, is positive.  What happens when the
>divisor is _negative_?  That is, how do we define the following?

>    (e)   5 div -3,
>    (f)   5 mod -3,

>    (g)  -5 div -3,
>    (h)  -5 mod -3.
>     


>ONE POSSIBLE SOLUTION
>---------------------

>Generally speaking, given numbers m and n, we want to find the quotient and the
>remainder when m is divided by n.  The quotient is "m div n" and the remainder
>is "m mod n."  Mathematically speaking, given integers m and n (n <> 0), we can 
>solve for q and r in the equation

>    m = qn + r,

>where q and r are integers.  If we force 0 <= r < n, we can get a unique 
>solution.


I am delighted that "most" people are now agreed about the -ve dividends, but
I do not like the proposed solution for -ve divisors.

ANOTHER POSSIBLE SOLUTION
-------------------------

I have no gut feeling for what 5 mod -3 should do, so let us fix div first.
The consensus for -ve divisors implies that m div n = floor(m/n). Let us try
to keep this for the new case.

	(e)  5 div -3 =  -2	( = -5 div 3 )
	(g) -5 div -3 =   1

But we should like to preserve
    m = qn + r,
which inevitably leads to

	(f)  5 mod -3 =  -1
	(h) -5 mod -3 =  -2

Now try this for gut feeling, and we have, for all m and -ve n, n < m mod n <= 0
just as we had,                            for all m and +ve n, 0 <= m mod n < n
The whole thing feels quite comfortable for me, and it just remains for some Group
Theorist to report whether it is now a respectable algebra.

To summarize:
	(a)  5 div 3 =  1      (e)  5 div -3 = -2
	(b)  5 mod 3 =  2      (f)  5 mod -3 = -1
	(c) -5 div 3 = -2      (g) -5 div -3 =  1
	(d) -5 mod 3 =  1      (h) -5 mod -3 = -2