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Everyone knows that anything to the power of 0 equals 1. Why is this? Are there exceptions? I have looked all over the internet for a comprehendible answer but I have yet to find one. Sorry ahead of time if this question is not suited to this forum. I just really think that understanding this will be a big leap towards my reaching of mathematical maturity.

Unbothered
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  • $1=a^m/a^m=a^{m-m}=a^0$ – Z Ahmed Jan 08 '24 at 06:14
  • It’s mostly just a convention. We want to ensure that $a^m \cdot a^n = a^{m + n}$ for all $m$ and $n$, and the only way to do so is to set $a^0 = 1$. – David Gao Jan 08 '24 at 06:23
  • (I should point out that this argument does not apply when $a = 0$, and indeed there are different conventions on what is $0^0$ or whether it is even defined.) – David Gao Jan 08 '24 at 06:25
  • I've found this explanation to click with a number of people: $2^5 = 32$, $2^4 = 16$ (divide by 2), $2^3 = 8$, $2^2 = 4$, $2^1 = 2$, $2^0 = 1$ (divide by 2 again). – Jean Abou Samra Jan 08 '24 at 06:42
  • @DavidGao A few people are unhappy with defining $0^0=1$, even though it is the sensible value to use in polynomials and set theory and combinatorics and probability: see the various answers to https://math.stackexchange.com/questions/11150/zero-to-the-zero-power-is-00-1 – Henry Jan 08 '24 at 07:12
  • @Henry I am aware - I’m only informing the OP of the fact that $0^0$ can be somewhat trickier. To add on to what you said, it also makes sense if one thinks of $\mathbb{R}$ as a multiplicative monoid, then the most sensible thing to do is to define the zeroth power to always be the multiplicative identity. I will agree that in any area of math that’s more “discrete”, it’s more sensible to say $0^0 = 1$. In analysis, that is a much different matter. There it is some time better to left it undefined or even set $0^0 = 0$. – David Gao Jan 08 '24 at 07:39
  • @Henry Quite honestly, as long as it is clear from the context what is the intended value of $0^0$, I don’t have a problem with an author choosing any specific convention over others. It is, in the end, just a convention that makes it easier to state things in a given context, after all. – David Gao Jan 08 '24 at 07:40
  • @DavidGao I have never come across a case where anything other than $0^0=1$ is useful. You would need to want something peculiar like requiring $f(x)=0^x$ to be continuous on $[0,\infty)$ and I have never seen that: even with $0^0=1$ you can say $\int_0^1 0^x, dx =0$. – Henry Jan 08 '24 at 07:47
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    @Henry Well, I’m an operator algebraist. For example, if you want the range of functional calculus to never venture outside the support projection of a given operator (which is necessary, for example, if you need the result of functional calculus to stay inside a direct summand of a larger algebra and therefore not containing the identity of the larger algebra), the easiest way is to let $0^0 = 0$ by convention. I’ll admit situations in which it is useful to define $0^0$ as anything other than 1 are rare, but they definitely exist. – David Gao Jan 08 '24 at 07:54

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