Fractional Indices

Fractional Indices with Numerator being 1

First, using the multiplication index law, we have:

$2^{\frac12}\cdot2^{\frac12}=2^{\frac12+\frac12}=2^1=2$

The definition of a square root is a value which multiplies by itself to give the original number. We know that $\sqrt2\cdot\sqrt2=2$.

Comparing these two statements we see that $2^{\frac12}=\sqrt2$.

We can confirm this by using the power of a power rule: $\left(2^{\frac12}\right)^2=2^{\frac12\cdot2}=2^1=2$.

In a similar way, we can look at the value $2^{\frac13}$. Using the multiplication law: $2^{\frac13}\cdot2^{\frac13}\cdot2^{\frac13}=2$.

From the definition of a cube root: $\sqrt[3]{2}\cdot\sqrt[3]{2}\cdot\sqrt[3]{2}=2$.

Once again, by comparison we see that $2^{\frac13}=\sqrt[3]{2}$.

These results can be generalised for any positive denominator. That is, $a^{\frac{1}{n}}=\sqrt[n]{a}$.

Simplifying Surds with Other Numerators

Generally, the fractional index law states: $a^{\frac{m}{n}}=\sqrt[n]{a^{m}}=\left(\sqrt[n]{a}\right)^{m}$ . In this equation, $m$ is the power and $n$ is the root.

When solving problems with fractional indices, it doesn’t matter whether you start with the powers or the roots (although it might be easier to do it one way for any particular expression).

Let’s simplify $16^{\frac32}$ in two ways:

Starting with the root:

$16^{\frac32}=\left(\sqrt{16}\right)^3$

=$4^3$

=$64$

Now let’s start with the powers:

$16^{\frac32}=\left(\sqrt{16^3}\right)$

=$\sqrt{4096}$

=$64$