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)
