Lesson Rationalizing The Denominator Containing a Radical

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Rationalizing Denominators Containing Radicals

Numbers that have rational denominators are considered to be in simpler form than numbers with irrational denominators.  For example:

*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{\sqrt{2}}{2}]

is considered to be simpler form than

*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{1}{\sqrt{2}}]

even though, as we shall see, these two numbers are equal. Hence, the goal of the process of rationalizing the denominator is to change the irrational denominator to a rational one without changing the value of the overall fraction.

There are two forms that we will deal with in this lesson but both of them depend on the idea of the multiplicative identity, one of the first concepts we learn in algebra.

For any number a we know that:

*[tex \LARGE \ \ \ \ \ \ \ \ \ \ a\,\cdot\,1\ =\ a]: Multiplicative Identity

From which it follows that:

*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{a}{a}\ =\ 1]

Which is to say, you don't change the value of anything if you multiply it by 1, and anything divided by itself is 1.

The first form we will deal with is, like the example provided above, an expression in the form of a fraction that contains single term radical in the denominator.  For example, consider the fraction:

*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{2}{\sqrt{3}}]

In order to eliminate the radical in the denominator we will make use of the following definition of a radical:

*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sqrt{a}\cdot\sqrt{a} = a]

From which we can see that it would be helpful if we could multiply the denominator of the example fraction by *[tex \LARGE \sqrt{3}].  However, simply multiplying the denominator by something other than 1 would change the value of the fraction and, as previously stated, that is something we want to avoid.

But using the idea of the multiplicative identity, we can multiply the entire fraction by 1 without changing its value, and we can use the idea that anything divided by itself is equal to 1 to develop an appropriate multiplier that is equal to 1.

For this first example, we will use:

*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{\sqrt{3}}{\sqrt{3}} = 1]

as the multiplier, thus:

*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(\frac{2}{\sqrt{3}}\right)\left(\frac{\sqrt{3}}{\sqrt{3}}\right)]

Now just perform the multiplication operation just like multiplying any other fractions, i.e. numerator times numerator and denominator times denominator:

*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{2\sqrt{3}}{3}]

Which result is fraction with a rational denominator that is numerically equivalent to the expression we started with.

That's fine for square roots, but what about higher order radical indexes -- cube, 4th, or 5th roots for example?

Let's look at:

*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{3}{\sqrt[4]{2}}]

Here we need to realize that

*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(\sqrt[4]{a}\right)^4 = \sqrt[4]{a}\cdot\sqrt[4]{a}\cdot\sqrt[4]{a}\cdot\sqrt[4]{a}\ =\ a\ =\ \sqrt[4]{a}\left(\sqrt[4]{a}\right)^3]

From which we can see that the desired multiplier for the denominator in the example is:

*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(\sqrt[4]{2}\right)^3]

and the multiplier for the entire fraction is then:

*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{ \left(\sqrt[4]{2}\right)^3 }{\left(\sqrt[4]{2}\right)^3}]

and then we have:

*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{3}{\sqrt[4]{2}}\frac{ \left(\sqrt[4]{2}\right)^3 }{\left(\sqrt[4]{2}\right)^3}\ =\ \frac{3\left(\sqrt[4]{2}\right)^3}{2}]

***********************************************************************

The second form we will deal with is when we have a binomial expression in the denominator where one or both of the terms is a radical.  For example:

*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{1}{1 + 2\sqrt{2}}]

Multiplying the denominator by itself will not work in this case because:

*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(1 + 2\sqrt{2}\right)\left(1 + 2\sqrt{2}\right)\ =\ 1 + 4\sqrt{2} + 8\ =\ 9 + 4\sqrt{2}]

And we are no better off than when we started.  Fortunately there is a simple technique that eliminates this problem.  What we need to do is multiply by the conjugate.  Please recall the Difference of Two Squares factorization:

*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (a^2 - b^2) = (a + b)(a - b)]

The two factors shown above are called conjugates. You obtain the conjugate of a binomial by changing the sign in the middle. This is important to us here because, as you can see from the Difference of Two Squares factorization, multiplying a binomial by its conjugate results in the difference of two squares.

Here's how the technique is applied.  Again, we must multiply the entire fraction by 1, but this time in the form of the conjugate of the denominator divided by itself, thus:

*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(\frac{1}{1 + 2\sqrt{2}}\right)\left(\frac{1 - 2\sqrt{2}}{1 - 2\sqrt{2}}\right) ]

So just multiply:

*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{1 - 2\sqrt{2}}{1 - 8} = \frac{1 - 2\sqrt{2}}{-7}]

These techniques should allow you to rationalize denominators containing radicals for any problems you are likely to encounter in Algebra II through Calculus.

If you have questions on this lesson, feel free to write: reecejo@gmail.com

John
*[tex \LARGE e^{i\pi} + 1 = 0]