There are actuall two usefull definition for %. The first which Ada calls 'mod' always returns a value 0<=X<N and yes it has no working value that is an identity. The other which Ada calls 'rem' defined as follows:
Signed integer division and remainder are defined by the relation: A = (A/B)*B + (A rem B) where (A rem B) has the sign of A and an absolute value less than the absolute value of B. Signed integer division satisfies the identity: (-A)/B = -(A/B) = A/(-B) It does have a right side identity of +INF. -- Mark Biggar [EMAIL PROTECTED] [EMAIL PROTECTED] [EMAIL PROTECTED] > HaloO Mark, > > please don't regard the following as obtrusive. > > you wrote: > > If as usual the definition of a right identity value e is that a op e = a > > for > all a, > > then only +inf works. Besdies you example should have been; > > Or actually $n % any( abs($n)+1 .. Inf ) to really exclude 0 > from the junction. > > > $n % any (($n+1)..Inf), $n % $n = 0. > > That depends on the definition of % and the sign of $n. > With the euclidean definition 0 <= ($n % $N == $n % -$N) < abs($N) > and for $n < 0 there's no identity at all. The identity element > has to be an element of the set, which +Inf isn't. It's a type. > > BTW, is % defined as truncation in Perl6? > That would be a bit unfortunate. Simple but not well thought out. > -- > TSa (Thomas Sandlaß) >
