Mitar wrote:
Why is 0/0 (which is NaN) > 1 == False and at the same time 0/0 < 1 == False. This means that 0/0 == 1? No, because also 0/0 == 1 == False. I understand that proper mathematical behavior would be that as 0/0 is mathematically undefined that 0/0 cannot be even compared to 1. There is probably an implementation reason behind it, but do we really want such "hidden" behavior? Would not it be better to throw some kind of an error?
Like nearly all programming languages, Haskell implements
the standard IEEE behavior for floating point numbers.
That leads to some mathematical infelicities that are
especially irking to us in Haskell, but the consensus was
that it is best to follow the standard.
That is why NaN==NaN, NaN
I think it's a bug. Here is why:
let f = (\x -> x/0) in f 0 == f 0
Referential transparency say that f 0 must equal to f 0, but in this case it is not. :-)
This does not violate referential transparency. f 0 is always the same value. (==) is a function like any other; in this case, it does not satisfy the mathematical laws we would like it to in order to conform to standard floating point behavior.