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Assume that $\frac{x}{0}$ (and therefore $\frac{0}{0}$) is undefined and $\mathbb{Q}$ is our universum (when I tried to do it for $\mathbb{R}$ things just got way too complicated; if anyone knows a generalisation akin to mine — and possibly more correct — please add it to your answer).

We write

$$x^0 = x^{1} \cdot x^{-1} = \frac{x}{x} = 1; \quad x \neq 0.$$

We know that

$$0^x = 0 \; \text{ because } \; \underbrace{0 \cdot 0 \cdot \dots \cdot 0}_{x} = 0 \; \text{ for } \; x \in \mathbb{N}, $$

and

$$0^{a/b} = \sqrt[b]{0^a} = \sqrt[b]{0} = 0 \; \text{ for } \; a,b \in \mathbb{N} \; \text{ or } \; a,b \in\mathbb{Z}^-$$.

Then we consider the case

$$x^y \cdot x^0 = x^{y + 0} = x^y,$$

and plug for $x = 0$.

$$0^y \cdot 0^0 = 0^y = 0 \cdot 0^0 = 0 \; \text{ for } y \in \mathbb{Q}^+,$$

and now we have a scenario $0x = 0$ (where $0^0$ is our $x$) and we can't divide by zero (and write the "solution" $x = \frac{0}{0}$). Because we know that $0 \cdot z = 0$ for every $z \in \mathbb{Q}$ the solution set is $\mathcal{S} = \mathbb{Q}$.

So I've heard that asserting $0^0 = 1$ is why binomial theorem (don't ask me what that is, I've just heard of it, I'm a high school student currently) even works in the first place. So the way I see it, the assertion of $0^0 =1$, $0^0 = 0$ or any other equivalence $0^0 = z; \; z \in \mathbb{Q}$ is valid because $0z = 0; \; \forall z \in\mathbb{Q}$ holds. So we essentially select whatever value "fits" for us in a certain context, right?

Please correct any of my errors and outline the whole "being able to decide what the value is" thing. It will help me to understand mathematical concepts much deeper.

P.S.: I've blindly assumed the equivalence $\sqrt[b]{0}; \; b \in \mathbb{N}$, but I have no idea how to prove it. If you included the proof of this statement, that'd be a nice bonus. Thanks!

God bless
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    We can define $0^0$ however we want, or leave it undefined. But certain theorem statements are simplified if we define $0^0 =1$, so that seems to be a convenient definition. Often when using this definition we'll make a remark like, "We are using the convention that $0^0 =1$". – littleO Jul 06 '18 at 13:06
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    You argue any definition is valid because the rule $\forall z \in \mathbb{Q}, 0 \cdot z = 0$ holds, but what about other rules? – Sambo Jul 06 '18 at 13:50

1 Answers1

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We can define $0^0$ any way we want, and there's a good chance some of the laws that hold in easier cases will continue to hold, even if we decide that $0^0$ should be 17. Furthermore, there's no way to define $0^0$ so that all the familiar laws continue to hold, because one of those laws says $x^0=1$ while another one says $0^y=0$. (The latter is only for $y>0$, not for negative $y$, so we have somewhat less justification for wanting it to hold also for $y=0$.) So the issue really becomes: Which laws are the important ones? Which laws are we willing to give up (for this particular case) in order that others can continue to hold?

For me, the most important property of exponents is that, when $x$ and $y$ are the cardinalities of sets $X$ and $Y$, respectively, then $x^y$ is the cardinality of the set of all functions from $Y$ into $X$. (For example, this property provides a good definition of exponentiation for infinite cardinal numbers.) Once I decide that I want this property to hold, it tells me that $0^0=1$ because there is exactly one function from the empty set to itself.

Andreas Blass
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