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I'm interested in knowing whether $a^0 = 1$ ('$a$' not zero) is a definition. If not, can anyone please help me with proving this?

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    $\begingroup$ It depends on the context. In some cases, exponentiation is defined recursively by $a^0=1$ and $a^{x+1}=a^x\cdot a.$ If it is defined in a different way, it may be something provable. How is it defined for you, and in what context? $\endgroup$ – Cameron Buie Apr 5 '15 at 15:58
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    $\begingroup$ @EnjoysMath Are you being serious? $\endgroup$ – Jack M Apr 5 '15 at 16:09
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    $\begingroup$ @JackM That's what I was thinking...hardly seems like a useful question to ask for a question such as this. $\endgroup$ – Daniel W. Farlow Apr 5 '15 at 16:17
  • $\begingroup$ I would just say that $a^0=a^{x-x}=a^x a^{-x}=aa\cdots a a^{-1}a^{-1}\cdots a{-1}=ee\cdots e = e$. In certain groups (multiplicative ones, in particular) the identity element $e$ is simply 1 $\endgroup$ – Patrick Shambayati Apr 5 '15 at 16:37
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It's a definition. A convenient definition.

We know from our early encounter with mathematics that $a^m\times a^n=a^{m+n}$ if $m,n\in \mathbb{N}$ (not including zero) because it's very natural: "Multiplying $m$ times and multiplying $n$ times, and then multiplying those values should be the same as multiplying $m+n$ times".

What if $m$ or $n$ is $0$? We'd like that formula to still be true, i.e. we'd like

$$a^0\times a^n = a^{0+n}=a^n$$

However, the only number that satisfies this is $1$ so it's necessary to define $a^0:=1$ in order to keep this property.

This is exactly the same reason why we define

$$a^n:=\frac{1}{a^{-n}}$$

if $n$ is a negative integer, because we want laws as

$$\frac{a^m}{a^n}=a^{m-n}$$

to hold even if $m$ is not greater than $n$.

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    $\begingroup$ +1 for noting that it's a definition. Just like $0!=1$. That's the important thing here in my opinion. $\endgroup$ – Daniel W. Farlow Apr 5 '15 at 16:05
  • $\begingroup$ In a way, it's obvious that it's a definition. If $a^n=aaa...a$, $n$ times, then $a^0$ just flat out doesn't make sense, so it clearly needs to be given a special meaning before it can be said to have any value at all. $\endgroup$ – Jack M Apr 5 '15 at 16:11
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    $\begingroup$ It certainly can be a definition. In some cases, though, it can (and must) be proved! For example, consider cardinal exponentiation. It's easy to prove, but it's still not part of the definition. $\endgroup$ – Cameron Buie Apr 5 '15 at 16:12
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    $\begingroup$ It is possible to define $a^n$ as the product of the $n$-tuple $\left(a,a,\ldots,a\right)$ instead of defining it recursively. Then, $a^0 = 1$ is not a definition, but rather a consequence of the definition saying that the product of the empty $0$-tuple is $1$. The difference between these approaches, of course, is negligible; I just wanted to point out that you don't have to use recursion to define powers if you already have used recursion to define products of tuples (which are a more important notion). $\endgroup$ – darij grinberg Apr 5 '15 at 16:33
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$${a^b\over a^c} = a^{b-c} $$and vice versa. {Index rule}

Therefore, $$a^0 = a^{x-x}$$ for any x

$$= {a^x\over a^x} ,$$ using the aforementioned index rule.

$$= 1 $${since any thing divided by itself is 1 except 0)

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