[Python-ideas] checking for identity before comparing built-in objects

[Ethan Furman]

Steven D'Aprano wrote:
> 1) It is not the case that NaN <comp> NaN is always false.

Huh -- well, apparently NaN != Nan --> True.

However, borrowing Steven's earlier example, and modifying slightly:

sqr(-1) != sqr(-1)

Shouldn't this be False?

Or, to look at it another way, surely somewhere out in the Real World 
(tm) it is the case that two NaNs are indeed equal.

~Ethan~