There are two ways of looking at the mod operator (on integers): 1. As a map from the integers Z to Z/pZ. Then n mod p is defined as: n mod p = { k | k in Z, k = n + ip for some i in Z } Instead of the set, we ususally write its smallest nonnegative element. And yes, in that sense, Z/0Z gives: n mod 0 = { k | k in Z, k = n } = { k } =~ k 2. As the remainder under division by p. Since n mod 0 would be the remainder under division by 0, this correctly gives a division by zero error. I used to think that the definitions were equivalent... apparently not. Thomas On Wed, Jul 22, 2009 at 10:05, Chris Kuklewicz<haskell@list.mightyreason.com> wrote:
Nathan Bloomfield wrote:
Hello haskell-cafe;
I'm fiddling with this <http://cdsmith.wordpress.com/2009/07/20/calculating-multiplicative-inverses-in-modular-arithmetic/> blog post about inverting elements of Z/(p), trying to write the inversion function in pointfree style. This led me to try executing statements like
n `mod` 0
which in the ring theoretic sense should be n, at least for integers*. (MathWorld agrees. <http://mathworld.wolfram.com/Congruence.html>)
I agree that (n `mod` 0) ought to be n. Specifically
divMod n 0 = (0,n)
and
quotRem n 0 = (0,n)
In (divMod n m) the sign of the remainder is always the same as the sign of m, unless n or m is zero. In (quotRem n m) the sign of the quotient is the product of the signs of n and m, unless n or m is zero.
-- Chris
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