## Rationalizing the Denominator

Imagine you have never seen a root sign, $\sqrt {}$, in your life.Consider the problem: $Rationalize \frac{1}{5^\frac{1}{2}}$.  This means that for some reason you want the denominator to have only integer exponents.  So $\frac{1}{5^\frac{1}{2}}=\frac{1}{5^\frac{1}{2}}\frac{5^\frac{1}{2}}{5^\frac{1}{2}}=\frac{5^\frac{1}{2}}{5}$.  If we start with $Rationalize \frac{1}{5^\frac{2}{3}}$, then $\frac{1}{5^\frac{2}{3}}=\frac{1}{5^\frac{2}{3}}\frac{5^\frac{1}{3}}{5^\frac{1}{3}}=\frac{5^\frac{1}{3}}{5}$ and even if we get the problem, $Rationalize \frac{1}{5^\frac{7}{9}}$, then $\frac{1}{5^\frac{7}{9}}=\frac{1}{5^\frac{7}{9}}\frac{5^\frac{2}{9}}{5^\frac{2}{9}}=\frac{5^\frac{2}{9}}{5}$.

What if we were using roots? The above problems would be transformed as follows. $Rationalize \frac{1}{\sqrt{5}}$.  This means that for some reason you want the denominator to have no roots. So $\frac{1}{\sqrt{5}}=\frac{1}{5^\frac{1}{2}}\frac{\sqrt{5}}{\sqrt{5}}=\frac{\sqrt{5}}{5}$.  If we start with $Rationalize \frac{1}{\sqrt[3]{5^2}}$, then $\frac{1}{\sqrt[3]{5^2}}=\frac{1}{\sqrt[3]{5^2}}\frac{\sqrt[3]{5}}{\sqrt[3]{5}}=\frac{\sqrt[3]{5}}{5}$ and even if we get the problem, $Rationalize \frac{1}{\sqrt[9]{5^7}}$, then $\frac{1}{\sqrt[9]{5^7}}=\frac{1}{\sqrt[9]{5^7}}\frac{\sqrt[9]{5^2}}{\sqrt[9]{5^2}}=\frac{\sqrt[9]{5^2}}{5}$

Which of the two techniques would be easier to explain?  Or to teach?  I think the first one.  This is one of the reasons I think that radical parts of an algebra curriculum can be dispensed with.  By the way the root method may not look to bad but that is only because I presented it second and your brain was already primed to see the pattern.