How does an exponent work when it's less than one? I'm rather familiar with exponents, I know that $y^x = y_1 \cdot y_2 \cdot y_3 .... y_x$, but what if the exponent is less than one, how would that work? 
I put in my computer $25^{1/2}$ anyway, expecting it to give me an error, and I got an answer!! And even more surprising, when I did this with more numbers and a little research, I found that $$x^{1/y} = \sqrt[y] x$$ Is it just me, or can exponents take on the role of square roots?

One thing we notice right away about $b^n; n \in N$ is that $b^nb^m = b^{n+m}$ (This is obvious because $b^n = \underbrace{b\cdot b\cdots b}_{n\text{ times}}$ and $b^m = \underbrace{b\cdot b\cdots b}_{m\text{ times}}$, so $b^nb^m = \underbrace{b\cdot b\cdots b}_{n\text{ times}} \cdot \underbrace{b\cdot b\cdots b}_{m\text{ times}} = \underbrace{b\cdot b\cdots b}_{n+m\text{ times}} = b^{n+m}$). 
And that $(b^n)^m = \underbrace{b^n\cdot b^n\cdots b^n}_{m \text{ times}} = b^{n+n+...+n} = b^{n\cdot m}$.
So if we want to extend the definition of $b^n$ so that $n$ is not just a natural number but maybe $n = 0$ or $n < 0$ or $n \in \mathbb Z$ we realize that we want to define it so that $b^0b^n = b^{0+n} = b^n$.  That means we must define $b^0 = b^n/b^n = 1$.  (We really have not choice).
We also want it so that if $0 < n < m$ then $b^{m-n}=b^mb^{-n}$ so $b^{-n} = \frac{b^{m-n}}{b^m}= \frac{b^{m-n}}{b^{(m-n)+n}} = \frac {b^{m-n}}{b^{m-n}b^n} = \frac 1 {b^n}$.
So we must define $b^{-n} = 1/b^n$.  (We really have no choice.)
Now we also have $(b^n)^m = b^{nm}$ this means $ \sqrt[m]{b^{nm}} = b^n$.  This really isn't that surprising. After all $\sqrt[m]{b^{nm}} =\sqrt[m]{(b^{n})^m} = b^n$, after all.
But what if we aren't talking about whole integers? What if $\sqrt[m]{b^n} = \sqrt[m]{(b^{nm/m})}=\sqrt[m]{(b^{n/m})^m} = b^{n/m}$.  Does that make any sense at all?
Well, it makes perfect sense if $m$ divides $n$ and $n/m$ is an integer. 
But if $n/m$ then .... we haven't defined what $b^{n/m}$ means if $n/m$ isn't an integer. 
But why shouldn't we define $b^{n/m}$ if $n/m$ isn't an integer?  If we define $b^{n/m} = \sqrt[m]{b^n}$ that is a fine definition[@].  And because of our rules $(b^r)^n = b^{rn}$ we really have no choice.  We must define it that way.
So $a^{1/2} = \sqrt{a}$.  This is because if $a^{1/2} = x$ then $x^2 = (a^{1/2})^2 = a^{\frac 12 * 2} = a^1 = a$.  So $x = \sqrt{a}$.
[@] Actually, we have to show that if $r = m/n = p/q$ then $\sqrt[n]{b^m} = \sqrt[q]{b^p}$ as it turns out that is true.  $m/n = p/q$ mean $mq= np$ and $\sqrt[n]{b^m} = \sqrt[q]{\sqrt[n]{b^m}^q}=\sqrt[nq]{b^{mq}} = \sqrt[nq]{b^{np}} = \sqrt[q]{\sqrt[n]{(b^p)^n}} = \sqrt[q]{b^p}$
