# Why `divmod(float('inf'), 1) == (float('nan'), float('nan'))`

cool-RR ram.rachum at gmail.com
Wed Sep 17 16:55:32 CEST 2014

```Terry, that doesn't really answer the question "why", it just pushes it back to the documentation. Is there a real answer why? Why return NaN when Inf would make mathematical sense?

On Wednesday, September 17, 2014 4:13:38 AM UTC+3, Terry Reedy wrote:
> On 9/16/2014 5:40 PM, cool-RR wrote:
>
> > While debugging my code I found that the bug was because I assumed
>
> > that something like `divmod(float('inf'), 1)` would be equal to
>
> > `(float('inf'), float('nan'))`,  while Python returns `(float('nan'),
>
> > float('nan'))`. Why does Python make the division result be NaN in
>
> > this case? `float('inf') / 1` is still `float('inf')`.
>
>
>
> For integers, divmod(x, y) is defined as (x // y, x % y) == ((x - x%y)
>
> // y, x % y) == ((x - x%y) / y, x % y).
>
>
>
> For floats, Python is documented as using the third expression.
>
>
>
> "Help on built-in function divmod in module builtins:
>
> divmod(...)
>
>      divmod(x, y) -> (div, mod)
>
>      Return the tuple ((x-x%y)/y, x%y).  Invariant: div*y + mod == x."
>
>
>
> It does not really matter, however, as infinity cannot be 'floored' as
>
> required for //
>
>
>
>  >>> math.floor(float('inf'))
>
> Traceback (most recent call last):
>
>    File "<pyshell#21>", line 1, in <module>
>
>      math.floor(float('inf'))
>
> OverflowError: cannot convert float infinity to integer
>
>
>
> and hence
>
>
>
>  >>> float('inf') // 1.0
>
> nan
>
>
>
> --
>
> Terry Jan Reedy

```