Rationalize Denominator in Rational Expressions

Fractions show parts of wholes. The numerator tells how many pieces a whole is cut into and the numerator tells how many of those pieces you have.

Because you can't divide a whole into an irrational number of pieces, fractions may not have an irrational number (a radical ($\sqrt{\text{ }}$ or an $i$) in the denominator. So, if you have a fraction (or rational expression) with a radical or imaginary number in the denominator, you need to "rationalize the denominator," which means getting the radical or imaginary number out of the denominator.

Why must imaginary numbers be removed, along with radicals? Imaginary numbers are basically radicals, because $i$ equals $\sqrt{-1}$.  Because $i=\sqrt{-1}$, $i$ is also a radical and must be rationalized out of the denominator. 

How do you rationalize a denominator?  The process of rationalizing a denominator relies on three basic math rules:

1. You can multiply any number by 1, without changing the value of the number (remember the identity property?).
2. Any fraction that has the same number in the numerator and the denominator is equal to 1: $\dfrac{x}{x}=1$.
3. Any square root times itself is equal to the number under the radical: $(\sqrt{x})\times (\sqrt{x})=x$.

So, if you have a fraction with a radical in the denominator, multiply that fraction by a fraction that is equal to 1, but also involves multiplying the square root in the denominator by itself, so that it cancels out the radical. 

Let's see what that looks like:

Example:

Simplify the rational expression: $\require{cancel}\dfrac{4}{\sqrt{3}}$

What we want to do is multiply the rational expression by a fraction that is equal to one, but also multiplies the denominator ($\sqrt{3}$) by itself: $\dfrac{\sqrt{3}}{\sqrt{3}}=1$

Let's see what happens when we do the multiplication: $\dfrac{4}{\sqrt{3}} \times \dfrac{\sqrt{3}}{\sqrt{3}}=\dfrac{4 \times \sqrt{3}}{\sqrt{3}\times \sqrt{3}}=\dfrac{4 \sqrt{3}}{3}$

And, you're done.  Fractions may have radicals in the numerator, so this fraction is fully rationalized. 

This also works with imagnary numbers!

Example:

Simplify the rational expression: $\dfrac{2}{i}$

What we want to do is multiply the rational expression by a fraction that is equal to one, but also multiplies the denominator ($i$) by itself: $\dfrac{i}{i}=1$

Let's see what happens when we do the multiplication: $\dfrac{2}{i} \times \dfrac{i}{i}=\dfrac{2i}{i \times i}=\dfrac{2i}{\sqrt{-1}\times \sqrt{-1}}=\dfrac{2i}{-1}=2i$

This fraction is also fully rationalized.

Sometimes, you have to rationalize out a radical or imaginary number that is part of a binomial. In order to do that, you have to use the special products rule. Remember: $(x-y)(x+y)=x^2 -y^2$. The special products rule is a great way to square part of a binomial without being left with more radicals or $i$s than you started with.

Example:

Simplify the rational expression: $\dfrac{3}{2+\sqrt{5}}$

What we want to do is multiply the rational expression by a fraction that is equal to one, but also multiplies the denominator ($(2+\sqrt{5}$) by its complementary special product ($2-\sqrt{5}$): $\dfrac{(2-\sqrt{5})}{(2-\sqrt{5})}=1$

Let's do the multiplication:

$\dfrac{3}{2+\sqrt{5}}\times \dfrac{(2-\sqrt{5})}{(2-\sqrt{5})}=\dfrac{3 \times (2-\sqrt{5})}{(2+\sqrt{5})(2-\sqrt{5})}=\dfrac{6-3\sqrt{5}}{2^2-\sqrt{5}^2}=\dfrac{6-3\sqrt{5}}{4-5}=\dfrac{6-3\sqrt{5}}{-1}$

This fraction is also fully rationalized (but can be further simplified to: $-6+3\sqrt{5}$)

Whenever you see a radical or an imaginary number in a denominator, rationalize it out!