Fractional exponents are roots

If the exponent laws are to keep working, there is only one thing $9^{1/2}$ can possibly mean.

So far the little number on top has been a whole number, and it counted copies. Half a copy of something makes no sense, so $9^{1/2}$ looks like nonsense too. It is not, and the reason is that one of the laws already decides what it has to be.

The law $(a^{m})^{n} = a^{mn}$ says that if $9^{1/2}$ means anything at all, then squaring it must give $9^{(1/2)\cdot 2} = 9^{1} = 9$. So $9^{1/2}$ is the number whose square is 9. Ask the machine what it thinks, and check.

Roots, and the words for their parts

An $n$th root of a number $a$ is a number whose $n$th power is $a$. It is written with a radical sign, $\sqrt[n]{a}$. The small $n$ tucked into the crook is the index, saying which power is being undone, and the $a$ underneath is the radicand. An index of 2 is never written: a bare $\sqrt{\phantom{x}}$ always means a square root.

For an even index the radical sign always returns the non-negative root, called the principal root. Both 3 and $-3$ square to 9, but $\sqrt{9}$ is 3 and only 3. For an odd index there is exactly one real root and it keeps the sign of the radicand, so $\sqrt[3]{-8} = -2$.

Exponents that are any fraction

Once $a^{1/n}$ is settled, $a^{m/n}$ follows from the same law, because $\frac{m}{n} = m \cdot \frac{1}{n}$ and also $\frac{1}{n} \cdot m$. Both orders are legal, so both readings must give the same value.

Taking the root first keeps the numbers small. For $8^{2/3}$, the root-first route is $\sqrt[3]{8} = 2$ and then $2^{2} = 4$. The power-first route is $8^{2} = 64$ and then $\sqrt[3]{64} = 4$. Same answer, but you had to handle 64.

A negative fractional exponent stacks the two rules you already have: flip, then take the root, then take the power. $16^{-3/4}$ is $\frac{1}{16^{3/4}}$, and $16^{3/4} = \left(\sqrt[4]{16}\right)^{3} = 2^{3} = 8$, so the answer is $\frac{1}{8}$.

Where the real numbers run out

An even root of a negative number is not a real number, because no real number squares to something negative. An odd root of a negative number is perfectly ordinary. Machines are fussy about this, so it is worth watching one refuse.

The lesson from that is a working habit: take the root of the size, then put the sign back yourself, and only when the index is odd.

Translating both ways, fluently

Calculus asks for this translation constantly, in both directions, because the rules for differentiating are stated for powers and the answers are usually presented as roots.

The five exponent laws work on fractional exponents with no modification at all, which is the practical payoff of this lesson. Simplify $\dfrac{x^{3/4}\cdot x^{1/2}}{x^{1/4}}$: multiplying adds the exponents, so the top is $x^{3/4+2/4} = x^{5/4}$, and dividing subtracts, so the whole thing is $x^{5/4-1/4} = x^{1} = x$. Putting $\frac{1}{2}$ over a denominator of 4 is the only arithmetic involved.

One restriction is worth stating plainly. When the denominator of the exponent is even, the base must be zero or positive, because an even root of a negative number is not a real number. So $(-4)^{1/2}$ has no value here, and neither does $(-4)^{3/2}$. When the denominator is odd there is no restriction on the base.

A worked example, every step

Write $\dfrac{1}{x^{2}\sqrt[3]{x}}$ as a single power of $x$.

1. Put the root into exponent form first: $\sqrt[3]{x} = x^{1/3}$.
2. The bottom is now $x^{2} \cdot x^{1/3}$. Multiplying adds the exponents: $2 + \frac{1}{3}$.
3. Give 2 the same denominator: $2 = \frac{6}{3}$, so $\frac{6}{3} + \frac{1}{3} = \frac{7}{3}$.
4. The expression is $\dfrac{1}{x^{7/3}}$.
5. A power on the bottom moves to the top with its sign changed, so the answer is $x^{-7/3}$.
6. Check with a number. At $x = 8$: the original is $\frac{1}{64 \cdot 2} = \frac{1}{128}$, and $8^{-7/3} = \frac{1}{\left(\sqrt[3]{8}\right)^{7}} = \frac{1}{2^{7}} = \frac{1}{128}$.