Math Core

So far exponents have been integers. But what could $9^{1/2}$ or $8^{2/3}$ mean? It turns out fraction exponents are just another way to write roots, and the exponent rules you already know tell you exactly how they must behave.

Square roots and cube roots

A square root of $a$ is a number that gives $a$ when squared. Both $5^2$ and $(-5)^2$ equal $25$ , but the symbol $\sqrt{25}$ always means the principal (nonnegative) root, so $\sqrt{25} = 5$ .

A cube root of $a$ is a number that gives $a$ when cubed. $\sqrt[3]{64} = 4$ because $4^3 = 64$ . Cube roots of negative numbers are fine: $\sqrt[3]{-8} = -2$ because $(-2)^3 = -8$ .

In general, $\sqrt[n]{a}$ is the $n$ th root of $a$ . The small $n$ is called the index; a square root has index $2$ , which is not written.

Simplifying square roots

A square root is simplified when the number under the root has no perfect-square factor other than $1$ . The key fact is that roots split over multiplication:

\sqrt{ab} = \sqrt{a} \cdot \sqrt{b} \qquad (a, b \ge 0)

To simplify $\sqrt{72}$ , find the largest perfect square that divides $72$ . That's $36$ :

\sqrt{72} = \sqrt{36 \cdot 2} = \sqrt{36} \cdot \sqrt{2} = 6\sqrt{2}.

If you pick a smaller perfect square first, like $\sqrt{72} = \sqrt{4} \cdot \sqrt{18} = 2\sqrt{18}$ , you're not done: $\sqrt{18} = 3\sqrt{2}$ , so you still reach $6\sqrt{2}$ .

Fraction exponents as roots

What should $9^{1/2}$ mean? Suppose the power-of-a-power rule still works. Then

\left(9^{1/2}\right)^2 = 9^{\frac{1}{2} \cdot 2} = 9^1 = 9.

So $9^{1/2}$ is a number whose square is $9$ . That's $\sqrt{9} = 3$ . In the same way, $\left(8^{1/3}\right)^3 = 8$ , so $8^{1/3} = \sqrt[3]{8} = 2$ .

Now build up to other numerators. Since $\dfrac{2}{3} = \dfrac{1}{3} \cdot 2$ ,

8^{2/3} = \left(8^{1/3}\right)^2 = 2^2 = 4.

Rational exponents

For a positive integer $n$ (and $a \ge 0$ when $n$ is even):

a^{1/n} = \sqrt[n]{a} \qquad\qquad a^{m/n} = \left(\sqrt[n]{a}\right)^m = \sqrt[n]{a^m}

The denominator is the root. The numerator is the power.

Both forms of $a^{m/n}$ give the same answer, but taking the root first keeps the numbers small. For $8^{2/3}$ : root first, $\sqrt[3]{8} = 2$ and $2^2 = 4$ . Power first, $8^2 = 64$ and $\sqrt[3]{64} = 4$ . Same result, but the first way needs no big numbers.

Worked example: Evaluating rational exponents

Evaluate each expression.

$49^{1/2}$
$125^{1/3}$
$16^{3/4}$
$27^{-2/3}$

Solutions.

$49^{1/2} = \sqrt{49} = 7$ .
$125^{1/3} = \sqrt[3]{125} = 5$ , since $5^3 = 125$ .
$16^{3/4} = \left(\sqrt[4]{16}\right)^3 = 2^3 = 8$ , since $2^4 = 16$ .
The negative sign means reciprocal: $27^{-2/3} = \dfrac{1}{27^{2/3}} = \dfrac{1}{\left(\sqrt[3]{27}\right)^2} = \dfrac{1}{3^2} = \dfrac{1}{9}$ .

Common mistake

Don't confuse the parts of a rational exponent.

$a^{1/2}$ is the square root of $a$ , not half of $a$ : $16^{1/2} = 4$ , not $8$ .
In $a^{m/n}$ the denominator is the root. $8^{2/3}$ means "cube root, then square," not "square root, then cube."
A negative exponent still means reciprocal, not a negative answer: $4^{-1/2} = \dfrac{1}{2}$ .

Converting between forms

You can switch between radical form and exponent form in either direction. Match the index with the denominator and the power with the numerator.

Worked example: Radical form and exponent form

Write $\sqrt[5]{x^3}$ with a rational exponent.
Write $y^{2/7}$ in radical form.

Solutions.

The index $5$ becomes the denominator and the power $3$ becomes the numerator: $\sqrt[5]{x^3} = x^{3/5}$ .
The denominator $7$ is the index and the numerator $2$ is the power: $y^{2/7} = \sqrt[7]{y^2}$ .

The exponent rules still work

Every property of exponents holds for rational exponents too. You just need to be comfortable with fraction arithmetic.

Worked example: Rules with fraction exponents

Simplify $x^{1/2} \cdot x^{1/3}$ and $\left(a^{3/4}\right)^{8}$ .

For the product, add the exponents with a common denominator:

x^{1/2} \cdot x^{1/3} = x^{\frac{3}{6} + \frac{2}{6}} = x^{5/6}.

For the power of a power, multiply:

\left(a^{3/4}\right)^8 = a^{\frac{3}{4} \cdot 8} = a^6.

Tip

When evaluating something like $32^{3/5}$ , rewrite the base as a power whose exponent matches the denominator: $32 = 2^5$ . Then $32^{3/5} = (2^5)^{3/5} = 2^3 = 8$ .