# A^0 definition
**URL:** https://boards.straightdope.com/t/a-0-definition/976776
**Category:** Factual Questions
**Created:** [December 19, 2022, 10:19am UTC](https://boards.straightdope.com/t/a-0-definition/976776 "2022-12-19T10:19:13Z")
**Posts on this page:** 4
**Page:** 2
### Author:  [@Dr.Strangelove](https://boards.straightdope.com/u/Dr.Strangelove)
#### Post date: [December 21, 2022, 12:55am UTC](https://boards.straightdope.com/t/a-0-definition/976776/21 "2022-12-21T00:55:50Z")
I’m not saying it’s impossible to find functions with different limits. Just that they’re less common. If you have g(x)^{f(x)}, you need to make f(x) approach 0 logarithmically more slowly than g(x) to dilute its influence.
Saying it’s undefined is perfectly legitimate. But if we _want_ to define it, 1 is probably the only sensible answer.
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### Author:  [@glowacks](https://boards.straightdope.com/u/glowacks)
#### Post date: [December 21, 2022, 1:16am UTC](https://boards.straightdope.com/t/a-0-definition/976776/22 "2022-12-21T01:16:44Z")
If you’re straight up asking what 0 ^ 0 equals as an expression devoid of context, it’s best left undefined because of the tension between considering it 0 (from 0^a) or 1 (from a^0).
If you have 0^0 in a limit, you might rearrange it into e ^ (0 \* log (0)), in which case you then end up with a 0 \* infinity indeterminate form in the exponent, but in that case it might be easier to see how to evaluate it.
If you have 0 ^ 0 as part of the output of a combinatorial formula, it usually means that you’re effectively counting the number of ways to put 0 items into 0 boxes, as stated earlier, and there’s exactly one way of doing that, because doing nothing is a thing that is always counted. There might be other areas of study in which you have formulas that might give you 0 ^ 0 for certain data sets, and in those fields it might be a bit more abstract what exactly it’s determining, but it’s probably safer to have it equal to 1 following from combinatorics, because that’s probably how the exponential that could be 0 got into your formula.
Unambiguously though, 0! = 1.
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### Author:  [@Hari\_Seldon](https://boards.straightdope.com/u/Hari_Seldon)
#### Post date: [December 21, 2022, 1:36am UTC](https://boards.straightdope.com/t/a-0-definition/976776/23 "2022-12-21T01:36:00Z")
> [@Dr.Strangelove](#):
>
> aba^b could be seen as the number of ways you can pack b bb items into a aa boxes, assuming a box can contain multiple items. Given that, it’s obvious that a^0=1 a0=1a^0=1 , since there’s just one way to pack no items into some number of boxes. It also implies that 0^0=1 00=10^0=1 , though as stated, that is more definitional.
Another version of the same thing is that for (finite) sets A and B with a and b elements, resp., the number of functions A\to B is a^b. The number of functions of the empty set to itself is 1. The definition, if any, of 0^0 is going to depend on the context. In the context of set theory, 0^0=1. In the context of real variables, it is undefined.
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### Author:  [@Chronos](https://boards.straightdope.com/u/Chronos)
#### Post date: [December 21, 2022, 3:06pm UTC](https://boards.straightdope.com/t/a-0-definition/976776/24 "2022-12-21T15:06:48Z")
No matter what you do, the function has a discontinuity at (0,0). A hole in a domain is a kind of discontinuity, after all. So it’s not enough to say “0^0 is undefined so we don’t have a discontinuity there”. We’re not deciding whether to have a discontinuity; we’re just deciding what kind of discontinuity to have.
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