Rational Exponents and Radicals

The meaning is forced, not chosen

The exponent rules were built for whole numbers. One of them is:

$$(x^a)^b = x^{ab}$$

Now ask what $x^{1/2}$ could possibly mean, insisting that rule keeps working.

$$\left(x^{1/2}\right)^2 = x^{\frac{1}{2} \times 2} = x^1 = x$$

So $x^{1/2}$ is the number that gives $x$ when squared. That is the definition of the square root.

$$x^{1/2} = \sqrt{x}$$

Nobody decided this. Once the rules are required to hold for fractions, the meaning is the only one available. The same argument works for any denominator:

$$\left(x^{1/3}\right)^3 = x \quad\Rightarrow\quad x^{1/3} = \sqrt[3]{x}$$ $$x^{1/n} = \sqrt[n]{x}$$

Numerator and denominator

A general fraction does both jobs:

$$x^{m/n} = \sqrt[n]{x^m} = \left(\sqrt[n]{x}\right)^m$$

Part	Job
$n$, the denominator	which root to take
$m$, the numerator	which power to raise to

$$8^{2/3} = \left(\sqrt[3]{8}\right)^2 = 2^2 = 4$$

Take the root first. Both orders give the same answer, but rooting first keeps the numbers small: the other route is $\sqrt[3]{64}$, which needs a bigger cube to recognize.

Expression	Root first	Power first
$8^{2/3}$	$2^2 = 4$	$\sqrt[3]{64} = 4$
$16^{3/4}$	$2^3 = 8$	$\sqrt[4]{4096} = 8$
$27^{4/3}$	$3^4 = 81$	$\sqrt[3]{531441} = 81$

The third row is the argument for root-first on its own.

Negative fractional exponents

A minus sign in the exponent still means reciprocal, exactly as it does for whole numbers:

$$x^{-m/n} = \frac{1}{x^{m/n}}$$ $$16^{-1/2} = \frac{1}{16^{1/2}} = \frac{1}{4}$$

Handle the fraction and the sign separately. The sign flips the expression over; the fraction says which root and power. Doing them in either order gives the same result.

Why exponent form is worth the switch

Radical notation hides the arithmetic. Exponent notation exposes it.

Simplify $\sqrt{x} \cdot \sqrt[3]{x}$.

In radical form there is no move to make — the indexes differ. In exponent form it is a fraction addition:

$$x^{1/2} \cdot x^{1/3} = x^{\frac{1}{2} + \frac{1}{3}} = x^{5/6} = \sqrt[6]{x^5}$$

Every exponent rule now applies to fractions unchanged:

Rule	With fractional exponents
$x^a \cdot x^b = x^{a+b}$	$x^{1/2} \cdot x^{1/4} = x^{3/4}$
$\tfrac{x^a}{x^b} = x^{a-b}$	$\tfrac{x^{3/4}}{x^{1/4}} = x^{1/2}$
$(x^a)^b = x^{ab}$	$\left(x^{2/3}\right)^{3} = x^{2}$

One set of rules covers powers and roots together, which is the whole reason this notation exists.

Simplifying a radical using exponents

$$\sqrt[4]{x^{12}} = \left(x^{12}\right)^{1/4} = x^{3}$$

That is one line. Doing it in radical form means splitting $x^{12}$ into four equal groups and arguing about why each group leaves the root.

$$\frac{\sqrt[3]{x^5}}{\sqrt{x}} = \frac{x^{5/3}}{x^{1/2}} = x^{\frac{5}{3} - \frac{1}{2}} = x^{7/6}$$

Worked examples

Common mistakes

Practice problems