Practice Rationalizing Numerators/Denominators with Radicals

Especially in math contests, which would result in an incorrect answer, radicals in the denominator tend to be converted into a real number (preferably an integer for many intents and purposes) by removing the radical through various methods in a process called rationalizing the denominator.

Consider the $\cos{\frac{\pi}{4}} = \frac {1}{\sqrt{2}}$. As you can see, we have a radical in the denominator and now we want to get rid of it. So the question is, how can we remove it? This is one of the simpler cases where we just multiply by $\frac{\sqrt{2}}{\sqrt{2}}\rightarrow \frac {1}{\sqrt{2}}\ast \frac{\sqrt{2}}{\sqrt{2}} = \frac{\sqrt{2}}{2}$ Now, we have rationalized the denominator.

Let's look at another example. This one will resemble most other rationalize the denominator questions you might encounter. The solution will resemble familiar factoring techniques. Let's begin with $\frac{1}{\sqrt{3}-\sqrt{2}}$. Notice how our previous method doesn't apply well. Under what condition would the denominator result in an integer? One avenue of thought is wishing the individual terms are squared, thereby removing the radical. If you recall, the expression $x^{2}-y^{2} = (x+y)(x-y)$. Replacing $x$ with $\sqrt{3}$ and $y$ with $\sqrt{2}$, we see that multipling the fraction by $\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}+\sqrt{2}} \rightarrow \frac{1}{\sqrt{3}-\sqrt{2}} \ast \frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}+\sqrt{2}} = \frac{\sqrt{3}+\sqrt{2}}{(\sqrt{3})^{2}-(\sqrt{2})^2}=\frac{\sqrt{3}+\sqrt{2}}{1}=\sqrt{3}+\sqrt{2}$. By utilizing difference of squares, we were able to rationalize the denominator. If it was $+$ instead of $-$ in the denominator, then we would simply use the other factor $\sqrt{3}-\sqrt{2}$ and continue from there.

The last example is similar to the one above in its use of factoring techniques; however, this one based around cube roots. Recognize that the factor we multiply $\frac{1}{\sqrt[3]{2}}$ by $\frac{\sqrt[3]{4}}{\sqrt[3]{4}}$ and not by $\sqrt{2}$ in order to cancel out the $\sqrt[3]{}$. On to a more difficult example, we want to simpliy the fraction $\frac{1}{\sqrt[3]{2}-1}$. If want to rationalize this denominator, we want to cube both terms, so we can apply difference of cubes, whose factored form is as follows: $$x^{3}-y^{3} = (x-y)(x^{2} + xy + y^{2})$$

If we have $x\rightarrow \sqrt[3]{2}, y\rightarrow 1$, then we see we multiply by $\frac{\sqrt[3]{2^{2}}+\sqrt[3]{2}*1 +1^{2}}{\sqrt[3]{2^{2}}+\sqrt[3]{2}*1 +1^{2}} \rightarrow \frac{1}{\sqrt[3]{2}-1}\ast \frac{\sqrt[3]{2^{2}}+\sqrt[3]{2}*1 +1^{2}}{\sqrt[3]{2^{2}}+\sqrt[3]{2}*1 +1^{2}} = \frac{\sqrt[3]{2^{2}}+\sqrt[3]{2}*1 +1^{2}}{sqrt[3]{2^{3}}-1^{3}} = \sqrt[3]{4}+\sqrt[3]{2} +1$.

So as we have seen, there are many ways to rationalize the denominator, some are more direct while others may need some manipulation to recognize factoring analogs.