If $x^a$ is $x$ multiplied $a$ times, then how does $x^{-1}$ make sense?

Question

What is the meaning of $x$ raised to any non-positive value? We know that $x^{-a} = \dfrac{1}{x^a}$ and $x^0 = 1$, but where does that come from? What is the proof? Why is this true? What about $x$ raised to a fraction, like say $\frac{1}{3}$? How do you multiply $x$ $\frac{1}{3}$ times?

Answer 1

We know that $x^{a+b}=x^ax^b$ since an $(a+b)$ amount of $x$'s is the same as an $a$ amount of $x$'s times a $b$ amount of $x$'s.

From this, we have

$$x^0=x^{a-a}=x^ax^{-a}$$

Thus, we have

$$x^{-a}=\frac{x^0}{x^a}$$

and once you show $x^0=1$, your done.

As for fractions, show that

$$x^{\frac ab}=y$$

For some value of $y$ which we are trying to solve for.

We see this is the same as

$$x^{\frac ab}=\underbrace{x\times x\times x\times\dots x}_{a/b?}=y$$

However, we see that by multiplying both sides on itself $b$ times, we have

$$\underbrace{x\times x\times x\times\dots x}_{a/b}=y\implies\underbrace{x\times x\times x\times\dots x}_{a}=y^b$$

and so,

$$\underbrace{x\times x\times x\times\dots x}_{a/b}=\sqrt[b]{\underbrace{x\times x\times x\times\dots x}_{a}}$$

$$x^{\frac ab}=\sqrt[b]{x^a}$$

Answer 2

From that definition, it doesn't make sense. It's what we mathematicians call a generalisation. The original definition of powers only allow natural numbers to be exponents.

However, one can ask "Is it possible to come up with a definition of $x^{-a}$ that plays well along with the old definition?" By which we mean that $x^{b+c}=x^bx^c$ and $(x^b)^c=x^{bc}$ still holds. And it turns out that there is exactly one definition that works.

In the same way you get $x^{1/n}=\sqrt[n]x$. Not because the old definition says so, but because it's the only thing it can be if we want it to work.