How does an exponent work when it's less than one?

Solution 1:

You're right that there is something interesting going on. It's certainly true that $25^{1/2} = 5$ is a bit different than $3^2 = 3 \cdot 3 = 9$.

One of the reasons why we've chosen $25^{1/2}$ to mean $\sqrt{25}$ goes like this.

For normal, positive integer exponents, we have the really nice exponent laws $a^b \cdot a^c = a^{b+c}$. For instance, $3^3 \cdot 3^5 = 3^8$. What if we wanted this law to work even when we used numbers less than $1$, or maybe fractions between integers?

Then we would want $25^{1/2} \cdot 25^{1/2} = 25^{\frac{1}{2} + \frac{1}{2}} = 25^1 = 25$, and so we would want $25^{1/2} = \sqrt {25}$. This works with others, too. For instance, $27^{1/3}$ should satisfy $27^{1/3} \cdot 27^{1/3} \cdot 27^{1/3} = 27^1 = 27$, so that $27^{1/3} = \sqrt[3]{27} = 3$.

For this also works for fractions bigger than $1$. Thus $36^{3/2}$ should satisfy $36^{3/2} \cdot 36^{3/2} = 36^3$, so $36^{3/2} = \sqrt{36^3}$. Note that you could also rationalize this as thinking that $36^{3/2}$ should equal $(36^3)^{1/2}$, and indeed that works here!

More generally, choosing $25^{1/2}$ to mean $\sqrt {25}$ (and the related identities) agrees with all of our previous rules regarding exponents, so it seems like a very natural choice to make. And indeed, it is the choice we make.

As an aside: one can further extend exponents to decimal expansions instead of fractions, or go further still.

Solution 2:

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 $\sqrt[n]{b^m} = \sqrt[q]{\sqrt[n]{b^m}^q}=\sqrt[nq]{b^{mq}} = \sqrt[nq]{b^{np}} = \sqrt[q]{\sqrt[n]{(b^p)^n}} = \sqrt[q]{b^p}$