Is Python's == an equivalence relation on the floats?

Question

In native Python, without using NumPy (for which numpy.nan != numpy.nan ) there is no NaN, so am I right in thinking that Python's floating point == is reflexive? Then since it is symmetric ( a == b implies b == a ) and transitive (if a==b and b==c then a==c ), can we say that Python's == is an equivalence relation on the float s?

EDIT: OK, so I learned that there is a NaN: float('nan') (thanks @unutbu) which will propagate through various operations, but does any native Python method return it (rather than raising an Exception) without me introducing it by this assignment?

Answer 1

== is reflexive for all numbers, zero, -zero, ininity, and -infinity, but not for nan.

You can get inf , -inf , and nan in native Python just by arithmetic operations on literals, like below.

These behave correctly, as in IEEE 754 and without math domain exception:

>>> 1e1000 == 1e1000
True
>>> 1e1000/1e1000 == 1e1000/1e1000
False

1e1000 is a very big number, so float and double represent it as an infinity.

• infinity is equal to infinity
• infinity divided by infinity is not a number
• not a number != not a number

Floating-point arithmetic in Python also works OK for infinity minus infinity etc.:

>>> x = 1e1000
>>> x
inf
>>> x+x
inf
>>> x-x
nan
>>> x*2
inf
>>> x == x
True
>>> x-x == x-x
False
>>> 

And for the zero and minus zero case:

>>> inf = float("inf")
>>> 1/inf
0.0
>>> -1/inf
-0.0
>>> -1/inf == 1/inf
True
>>> 

Answer 2

float('nan') exists in native Python and float('nan') != float('nan') . So no, == is not an equivalence relation since it lacks reflexivity:

In [40]: float('nan') == float('nan')
Out[40]: False