Digging into Rational Exponents

Mr. Laiche

So we’ve worked with exponents. We know the basic exponent properties. However, the lingering question is; What do we do with a rational exponent? We will derive a Rational Exponents/Radicals Formula that will help us easily convert between radicals and exponents.

What happens when our exponent is not a whole number? What shall we do if our exponent is a rational number in the form of a fraction?

First, let’s take a look at an example. What happens when we decompose a whole number into the sum of two or more fractions.

\(\large{ \mathbf{ 5 = \frac{5}{1} }}\)

Before we get started let’s review briefly;

\(\large{ \mathbf{ x^{a} \cdot x^{b}=x^{a+b}} }\)\(\large{ \mathbf{ (x^{a})^{b}=x^{a \cdot b} }}\)\(\large{ \mathbf{ 1=\frac{n}{n} }}\)
\(\large{ \mathbf{ \frac{x^a}{x^b} = x^{a-b} }}\)\(\large{ \mathbf{ \sqrt{x}=\sqrt[2]{x} }}\)\(\large{ }\)

To the right, we start with \(\large{4^{1} }\) and decompose the exponent into the sum, \(\large{\frac{1}{2}+\frac{1}{2} }\). Using the product property of exponents we can turn a power with a sum for an exponent into a square product. And finally, taking the square root, we are left with;

\(\LARGE{\boxed{ \sqrt[2]{4^{1}} = 4^{\frac{1}{2}} }}\)

 

\(\Large{ \mathbf{ \begin{aligned} 4 &= 4^{1} \\ 4^{1} &=4^{\frac{1}{2}+\frac{1}{2}} \\ 4^{1} &= 4^{\frac{1}{2}} \cdot 4^{\frac{1}{2}} \\ 4^{1} &= \sqrt[2]{4^{\frac{1}{2}} \cdot 4^{\frac{1}{2}}} \\ \sqrt[2]{4^{1}} &= 4^{\frac{1}{2}} \\ \sqrt[2]{x} &= x^{\frac{1}{2}} \end{aligned}} }\)

Generalizing

We need to generalize this process so that, independent of values, we have a identity relating radicals and exponents. We will start by applying this concept, of decomposing the exponent into a sum, to powers whose base is a variable.

Square Roots

\( \Large{ \mathbf{ \begin{aligned} x &= x^{1} \\ x &= x^{\frac{2}{2}} \\ x &= x^{\frac{2}{2}} \\ x &= x^{(\frac{1}{2}+\frac{1}{2})} \\ x &= x^{\frac{1}{2}}+x^{\frac{1}{2}} \\ \sqrt[2]{x} &= \sqrt[2]{x^{\frac{1}{2}}+x^{\frac{1}{2}}} \\ \sqrt[2]{x} &= x^{\frac{1}{2}} \end{aligned} }}\)

Cube Roots

\( \Large{ \mathbf{ \begin{aligned} x &= x^{1} \\ x &= x^{\frac{2}{2}} \\ x &= x^{\frac{3}{3}} \\ x &= x^{(\frac{1}{3}+\frac{1}{3}+\frac{1}{3})} \\ x &= x^{\frac{1}{3}+x^\frac{1}{3}+x^\frac{1}{3}} \\ \sqrt[3]{x} &= \sqrt[3]{x^{\frac{1}{3}+x^\frac{1}{3}+x^\frac{1}{3}}} \\ \sqrt[3]{x} &= x^{\frac{1}{3}} \end{aligned} } }\)

bth roots

\( \Large{ \mathbf{ \begin{aligned} x &= x^{1} \\ x &= x^{\frac{b}{b}} \\ x &= x^{(\frac{1}{b}+\frac{1}{b}+\frac{1}{b}+\dots)} \\ x &= x^{\frac{1}{b}} \cdot x^{\frac{1}{b}} \cdot x^{\frac{1}{b}}\dots \\ \sqrt[b]{x} &= \sqrt[b]{x^{\frac{1}{b}}+x^{\frac{1}{b}}+x^{\frac{1}{b}}+\dots} \\ \sqrt[b]{x} &= x^{\frac{1}{b}} \end{aligned} } }\)

One More Step

Now, let’s take this and generalize it further for a fraction of the form \(\Large{ \frac{a}{b} }\).

\( \Large{ \mathbf{ \begin{aligned} x^{\frac{a}{b}} &= x^{a\cdot\frac{1}{b}} \\ &= (x^{a})^{\frac{1}{b}} \\ &= \boxed{ \sqrt[b]{x^{a}}} \end{aligned} }} \)
\( \Large{ \mathbf{ \begin{aligned} x^{\frac{a}{b}} &= x^{\frac{1}{b}\cdot a} \\ &= (x^{\frac{1}{b}})^{a} \\ &= \boxed{\left( \sqrt[b]{x} \right) ^{a}} \end{aligned} }} \)

The Fractional Exponent/Radical Formula

With these two conclusions, we are left with the final rule that we use for converting from exponential form to radical form and vice versa:

\( \mathbf{ \huge { x^{ \frac{a}{b}} = \left( \sqrt[b]{x} \right) ^{a}} }\)