### Question :

So Python has positive and negative infinity:

```
float("inf"), float("-inf")
```

This just seems like the type of feature that has to have some caveat. Is there anything I should be aware of?

##
Answer #1:

You can still get not-a-number (NaN) values from simple arithmetic involving `inf`

:

```
>>> 0 * float("inf")
nan
```

Note that you will normally *not* get an `inf`

value through usual arithmetic calculations:

```
>>> 2.0**2
4.0
>>> _**2
16.0
>>> _**2
256.0
>>> _**2
65536.0
>>> _**2
4294967296.0
>>> _**2
1.8446744073709552e+19
>>> _**2
3.4028236692093846e+38
>>> _**2
1.157920892373162e+77
>>> _**2
1.3407807929942597e+154
>>> _**2
Traceback (most recent call last):
File "<stdin>", line 1, in ?
OverflowError: (34, 'Numerical result out of range')
```

The `inf`

value is considered a very special value with unusual semantics, so it’s better to know about an `OverflowError`

straight away through an exception, rather than having an `inf`

value silently injected into your calculations.

##
Answer #2:

Python’s implementation follows the IEEE-754 standard pretty well, which you can use as a guidance, but it relies on the underlying system it was compiled on, so platform differences may occur. Recently¹, a fix has been applied that allows “infinity” as well as “inf”, but that’s of minor importance here.

The following sections equally well apply to any language that implements IEEE floating point arithmetic correctly, it is not specific to just Python.

### Comparison for inequality

When dealing with infinity and greater-than `>`

or less-than `<`

operators, the following counts:

- any number including
`+inf`

is higher than`-inf`

- any number including
`-inf`

is lower than`+inf`

`+inf`

is neither higher nor lower than`+inf`

`-inf`

is neither higher nor lower than`-inf`

- any comparison involving
`NaN`

is false (`inf`

is neither higher, nor lower than`NaN`

)

### Comparison for equality

When compared for equality, `+inf`

and `+inf`

are equal, as are `-inf`

and `-inf`

. This is a much debated issue and may sound controversial to you, but it’s in the IEEE standard and Python behaves just like that.

Of course, `+inf`

is unequal to `-inf`

and everything, including `NaN`

itself, is unequal to `NaN`

.

### Calculations with infinity

Most calculations with infinity will yield infinity, unless both operands are infinity, when the operation division or modulo, or with multiplication with zero, there are some special rules to keep in mind:

- when multiplied by zero, for which the result is undefined, it yields
`NaN`

- when dividing any number (except infinity itself) by infinity, which yields
`0.0`

or`-0.0`

². - when dividing (including modulo) positive or negative infinity by positive or negative infinity, the result is undefined, so
`NaN`

. - when subtracting, the results may be surprising, but follow common math sense:
- when doing
`inf - inf`

, the result is undefined:`NaN`

; - when doing
`inf - -inf`

, the result is`inf`

; - when doing
`-inf - inf`

, the result is`-inf`

; - when doing
`-inf - -inf`

, the result is undefined:`NaN`

.

- when doing
- when adding, it can be similarly surprising too:
- when doing
`inf + inf`

, the result is`inf`

; - when doing
`inf + -inf`

, the result is undefined:`NaN`

; - when doing
`-inf + inf`

, the result is undefined:`NaN`

; - when doing
`-inf + -inf`

, the result is`-inf`

.

- when doing
- using
`math.pow`

,`pow`

or`**`

is tricky, as it doesn’t behave as it should. It throws an overflow exception when the result with two real numbers is too high to fit a double precision float (it should return infinity), but when the input is`inf`

or`-inf`

, it behaves correctly and returns either`inf`

or`0.0`

. When the second argument is`NaN`

, it returns`NaN`

, unless the first argument is`1.0`

. There are more issues, not all covered in the docs. -
`math.exp`

suffers the same issues as`math.pow`

. A solution to fix this for overflow is to use code similar to this:`try: res = math.exp(420000) except OverflowError: res = float('inf')`

### Notes

*Note 1:* as an additional caveat, that as defined by the IEEE standard, if your calculation result under-or overflows, the result will not be an under- or overflow error, but positive or negative infinity: `1e308 * 10.0`

yields `inf`

.

*Note 2:* because any calculation with `NaN`

returns `NaN`

and any comparison to `NaN`

, including `NaN`

itself is `false`

, you should use the `math.isnan`

function to determine if a number is indeed `NaN`

.

*Note 3:* though Python supports writing `float('-NaN')`

, the sign is ignored, because there exists no sign on `NaN`

internally. If you divide `-inf / +inf`

, the result is `NaN`

, not `-NaN`

(there is no such thing).

*Note 4:* be careful to rely on any of the above, as Python relies on the C or Java library it was compiled for and not all underlying systems implement all this behavior correctly. If you want to be sure, test for infinity prior to doing your calculations.

¹) Recently means since version 3.2.

²) Floating points support positive and negative zero, so: `x / float('inf')`

keeps its sign and `-1 / float('inf')`

yields `-0.0`

, `1 / float(-inf)`

yields `-0.0`

, `1 / float('inf')`

yields `0.0`

and `-1/ float(-inf)`

yields `0.0`

. In addition, `0.0 == -0.0`

is `true`

, you have to manually check the sign if you don’t want it to be true.

##
Answer #3:

So does C99.

The IEEE 754 floating point representation used by all modern processors has several special bit patterns reserved for positive infinity (sign=0, exp=~0, frac=0), negative infinity (sign=1, exp=~0, frac=0), and many NaN (Not a Number: exp=~0, frac?0).

All you need to worry about: some arithmetic may cause floating point exceptions/traps, but those aren’t limited to only these “interesting” constants.

##
Answer #4:

I found a caveat that no one so far has mentioned. I don’t know if it will come up often in practical situations, but here it is for the sake of completeness.

Usually, calculating a number modulo infinity returns itself as a float, but a fraction modulo infinity returns `nan`

(not a number). Here is an example:

```
>>> from fractions import Fraction
>>> from math import inf
>>> 3 % inf
3.0
>>> 3.5 % inf
3.5
>>> Fraction('1/3') % inf
nan
```

I filed an issue on the Python bug tracker. It can be seen at https://bugs.python.org/issue32968.

Update: this will be fixed in Python 3.8.

##
Answer #5:

A VERY BAD CAVEAT :Division by Zero

in a `1/x`

fraction, up to `x = 1e-323`

it is `inf`

but when `x = 1e-324`

or little it throws `ZeroDivisionError`

```
>>> 1/1e-323
inf
>>> 1/1e-324
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
ZeroDivisionError: float division by zero
```

so be cautious!