No radical in the denominator — why?

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  • Why rationalize the denominator?

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I have had students seem surprised when I told them that I have no preference for $\tan\left(\frac{\pi}{6}\right)$ to be “simplified” to $\frac{\sqrt 3}{3}$, for example. A high school teacher whom I told this was even more surprised, but no reason was given in defense of requiring the practice. Nonetheless, to some extent I am glad that high school teachers often tell students that they have to rationalize denominators, because it means that many of them will at least be familiar with the idea when the time comes that “rationalizing” (both denominators and numerators) does serve a purpose in simplifying algebraic expressions, in particular when finding limits in calculus.

So maybe $\frac{1}{4+\sqrt 3}$ is no less simple than $\frac{4-\sqrt 3}{13}$, but $\frac{h}{\sqrt{x+h}-\sqrt{x}}$ is harder to deal with than $\sqrt{x+h}+\sqrt{x}$ when $h$ goes to $0$, and this comes more easily if you already know how to rationalize denominators with numbers.

The usual reason I’ve heard is that dividing by integers is computationally easier — it’s easier to find, say, $(5\sqrt{3})/3$ by computing $5 \times \sqrt{3} \approx 8.66 $ and then dividing by $3$ to get $2.89$ then to find $5/\sqrt{3}$ by dividing $5/1.73$ directly.

The slightly cynical teacher in me wants to say that the reason for demanding no radicals in the denominator is so there is only one right answer to each question, which simplifies grading.

I heard from a high school math teacher once that this was done back in the day (before calculators) so that people could look up the values of these expressions in tables. If this is indeed the case, there is no good reason for this practice now.

This transformation is known as $ $ rationalizing the denominator. $ $ As the terminology suggests, $\ $it serves to simplify by transforming an irrational divisor into a rational divisor. This can lead to all sorts of simplifications. Below are a couple examples.

In this prior question is an example where RTD transforms a limit of indeterminate form into a simple determinate limit by way of cancelling an apparent singularity at $\rm\ x = a\ $

$$\rm \frac{x^2-a\sqrt{ax}}{\sqrt{ax}-a}\ =\ \frac{x^2-a\sqrt{ax}}{\sqrt{ax}-a} \ \frac{\sqrt{ax}+a}{\sqrt{ax}+a}\ =\ \frac{ax\:(x-a)+\sqrt{ax}\ (x^2-a^2) }{a\:(x-a) }\ =\ x+(x+a)\sqrt{\frac{x}{a}}$$

Here’s another example from number theory showing how RTD serves to reduce divisibility of algebraic integers to rational integers. Consider the Gaussian integers $\rm\ \mathbb I = \{ m + n\ i\ : \ m,n\in \mathbb Z \}\ $. As in any ring we define divisibilty by $\rm\ a\ |\ b\ in\ \mathbb I \iff b/a \in \mathbb I\:.\ $ Suppose we wish to know if $\rm\ 2+3\ i\ |\ 91\ in\ \mathbb I\:,\:$ i.e. is $\rm\ w = 91/(2+3\ i)\in \mathbb I\ ?\ $ Now in fact $\rm\:\mathbb I\:$ happens to have a division algorithm which we could apply. But it is simpler to RTD yielding $\rm\ w = 91\ (2-3\ i)/(2^2+3^2) = 7\ (2-3\ i)\ $ so, indeed, $\rm\: w\in \mathbb I\:.\ $ More generally we can often reduce problems about algebraic numbers to problems about rational numbers by taking norms, traces, etc. In fact this is how Kronecker constructed his divisor theory for algebraic integers, see e.g. Harold Edwards: Divisor Theory.

The form with neither denominators in radicals nor radicals in denominators and with only squarefree expressions under square-root signs, etc., is a canonical form, and two expressions are equal precisely if they’re the same when put into canonical form.

When are two fractions equal? How do you know that $\dfrac{51}{68}$ is the same as $\dfrac{39}{52}$? They’re both the same when reduced to lowest terms.

How do you know that $\dfrac{1}{3+\sqrt{5}}$ is the same as $\dfrac{3\sqrt{5}-5}{4\sqrt{5}}$? Again, they’re the same when put into canonical form.

How do you know that $\dfrac{13+i}{1+2i}$ is the same as $\dfrac{61}{6+5i}$? Same thing.

Don’t blame your high school teacher

They are right

$\frac{1}{\sqrt{3}}$ is correctly “simplified” to $\frac {\sqrt{3}}{3}$.

Its difficult to find value $\frac{1}{\sqrt{3}}$ as $\sqrt{3}$ is irrational number

Its easy to find value of $\frac {\sqrt{3}}{3}$ as here we are dividing it with 3