Why rationalize the denominator?

Question

In grade school we learn to rationalize denominators of fractions when possible. We are taught that $\frac{\sqrt{2}}{2}$ is simpler than $\frac{1}{\sqrt{2}}$. An answer on this site says that "there is a bias against roots in the denominator of a fraction". But such fractions are well-defined and I'm failing to see anything wrong with $\frac{1}{\sqrt{2}}$ - in fact, IMO it is simpler than $\frac{\sqrt{2}}{2}$ because 1 is simpler than 2 (or similarly, because the former can trivially be rewritten without a fraction).

So why does this bias against roots in the denominator exist and what is its justification? The only reason I can think of is that the bias is a relic of a time before the reals were understood well enough for mathematicians to be comfortable dividing by irrationals, but I have been unable to find a source to corroborate or contradict this guess.

Answer 1

This was very important before computers in problems where you had to do something else after computing an answer.

One simple example is the following: When you calculate the angle between two vectors, often you get a fraction containing roots. In order to recognize the angle, whenever when possible, it is good to have a standard form for these fractions [side note, I saw often students not being able to find the angle $\theta$ so that $\cos(\theta)=\frac{1}{\sqrt{2}}$]. The simplest way to define a standard form is by making the denominator or numerator integer.

If you wonder why the denominator is the choice, it is the natural choice: As I said often you need to make computations with fractions. What is easier to add: $$\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{6}+\sqrt{3}} \, \mbox{ or }\, \frac{\sqrt{3}}{3}+\frac{\sqrt{6}-\sqrt{3}}{3} \,?$$

Note that bringing fractions to the same denominator is usually easier if the denominator is an integer. And keep in mind that in many problems you start with quantities which need to be replaced by fractions in standard form [for example in trigonometry, problems are set in terms of $\cos(\theta)$ where $\theta$ is some angle].

But at the end of the day, it is just a convention. And while you think that $\frac{1}{\sqrt{2}}$ looks simpler, and you are right, the key with conventions is that they need to be consistent for the cases where you need recognition. The one which looks simpler is often relative...

Answer 2

The main reason I'd guess our math teacher culture tells us to require rationalizing the denominator is so that there is one set universal nomenclature among students about what a standard form means. In school, teachers have a lot of answers to check so if they have to keep seeing things like $\frac{1}{\sqrt{3}}$, $\frac{\sqrt{3}}{3}$, and $\sqrt{\frac{1}{3}}$ (just as a very simple example) floating around, it slows down checking slightly and is, also, to some extent, annoying.

Historical thing: before calculators, you had to do things by hand (duh). In this scenario, dividing $1$ by $\sqrt{3}$ is a lot harder than dividing $\sqrt{3}$ by $3$, so rationalizing the denominator (rather than rationalizing the numerator) seems logical.

Answer 3

can you compute this? $$ \left \lfloor \frac{1}{\sqrt{25}-\sqrt{24}} \right \rfloor$$without calculator !!!
can you compute $$ \left \lfloor \frac{1}{\sqrt{25}-\sqrt{24}}\frac{\sqrt{25}+\sqrt{24}}{\sqrt{25}+\sqrt{24}} \right \rfloor=\\ \left \lfloor \frac{\sqrt{25}+\sqrt{24}}{25-24} \right \rfloor=\\\left \lfloor \frac{\sqrt{25}+\sqrt{24}}{1} \right \rfloor=\left \lfloor \sqrt{25}+\sqrt{24}\right \rfloor=\\\left \lfloor 5+\sqrt{24} \right \rfloor=9$$now which one is easy to understand ?

Answer 4

Adding up two fractions with irrational denominators looks like less roots.

See

$ \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} = \frac{\sqrt{3}+\sqrt{2}}{\sqrt{6}} $

vs

$ \frac{\sqrt{2}}{2} + \frac{\sqrt{3}}{3} = \frac{2\sqrt{3}+3\sqrt{2}}{6} $

Answer 5

The historical reason for rationalizing the denominator is that before calculators were invented, square roots had to be approximated by hand.

To approximate $\sqrt{n}$, where $n \in \mathbb{N}$, the ancient Babylonians used the following method:

1. Make an initial guess, $x_0$.

2. Let $$x_{k + 1} = \frac{x_k + \dfrac{n}{x_k}}{2}$$

If you use this method, which is equivalent to applying Newton's Method to the function $f(x) = x^2 - n$, to approximate the square root of $2$ by hand with $x_0 = 3/2$, you will see that while the sequence converges quickly, the calculations become onerous after a few steps. However, once an approximation was known, it was easy to calculate $$\frac{1}{\sqrt{2}}$$ quickly by rationalizing the denominator to obtain $$\frac{\sqrt{2}}{2}$$ then dividing the approximation by $2$.

Answer 6

Calculate $\frac1{\sqrt{3.0000001}-\sqrt3}$ and compare with the answer obtained from the rationalized form $\frac{\sqrt{3.0000001}+\sqrt3}{0.0000001}$.

Adjust the number of $0$'s to the precision of the calculator or software used.

In Maple with Digits := 10, the first expression gives $3.571428571\cdot10^7$, while the second gives $3.464101644\cdot10^7$.

Answer 7

The reason I've seen is historical, because the Greeks could easily construct a fraction with radicals on the numerator by constructing all the radicals and then dividing as necessary, but to divide by an irrational amount is very difficult. This problem is changed to simply dividing up rationals which is easily done with straightedge and compass

Answer 8

Anecdotally it shows that the inverse of $a+b\sqrt n$ can also be written as $c+d\sqrt n$ (with $a, b, c, d \in \Bbb Q$), which is key in showing that $\Bbb Q[\sqrt n]$ (the set $\{ P(\sqrt n) \mid P \text{ is a rational polynomial} \}$) is a field.

Answer 9

Like so many things it is nothing to get particularly obsessed about, but knowing how to rationalise denominators is quite a useful tool to have at one's disposition. In fact more so in the general context of manipulating expressions than just for simplifying numbers. It is based on a small trick that is easy to understand, but which most people would probably not have thought of if it were not taught to them.

Notably, I would not like to do without this method when trying to decide whether a rational expression involving a single square root is equal to$~0$.

Also I think that a similar method (though maybe better called realising than rationalising) is used in the most striaghtforward proof of the fact that the complex numbers are a field.

Answer 10

I have a fairly clear memory of doing some high-school math problem and ending up with $\frac3{\sqrt3}$, and thinking "I'll go ahead and get the root out of the denominator, even though it never makes the result any more useful." I was surprised to discover that in that case the rationalized result is more useful (it's just $\sqrt3$, of course).

Nowadays I have the facility to make good decisions about whether it's more parsimonious to put the root in the numerator or the denominator — I, like you, prefer $\frac1{\sqrt2}$ to $\frac{\sqrt2}2$. But I developed that facility after rationalizing a bunch of denominators, which makes me think that it's perfectly useful as a pedagogical bias.

Answer 11

This is quite related (but not identical) to making the denominator real for complex valued fractions such as

\begin{align} \frac1{1+3i} &= \frac{1-3i}{10} \\ &= \frac1{10} - \frac3{10}i \end{align}

which is necessary in order to separate the fraction into its real and imaginary part. Of course that can be intermingled with non-rational numbers, e.g.

\begin{align} \frac1{\sqrt3-\sqrt7i} &= \frac{\sqrt3+\sqrt7i}{10} \\ &= \frac{\sqrt3}{10} + \frac{\sqrt7}{10}i. \end{align}

Expressions get of course more complicated once $\sqrt[3]{\ }$ and the like occurs, and it gets even funnier when you wonder about the real and imaginary parts of something like $\sqrt{3+7i}$...

Answer 12

Rationalizing the denominator (RTD) (a special case of the method of simpler multiples) is useful because it often serves to simplify problems, e.g. by transforming an irrational denominator (or divisor) into a simpler rational one. This can lead to all sorts of simplifications, e.g. below.

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 reduces divisibility of algebraic integers to rational integers. Consider the Gaussian integers $\rm\ \mathbb I = \{ m + n\ i\ : \ m,n\in \mathbb Z \}.\, $ Suppose we wish to know if $\rm\ 2+3\ i\,\mid\, 91\ in\ \mathbb I,\,$ 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 more elementary to simply RTD, which quickly yields $\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 (roughly) how Kronecker constructed his divisor theory for algebraic integers, $ $ see e.g. Harold Edwards: Divisor Theory.

RTD can also "rationalize" to base fields in any algebraic extension, e.g. we can "realize" denominators of complex fractions, lifting "existence of inverses of elements $\ne 0\,$" from $\mathbb R$ to $\mathbb C.\:$ Namely, since $\mathbb R$ is a field, $\rm\ 0\ne r\in \mathbb R\ \Rightarrow\ r^{-1}\in \mathbb R,\:$ so with $\,\alpha' = $ conjugate of $\alpha,$

$$\rm 0\ne\alpha\in\mathbb C\ \ \Rightarrow\ \ 0\ne\alpha\alpha' = r\in \mathbb R\ \ \Rightarrow\ \frac{1}\alpha\, =\, \frac{\alpha'}{\alpha\:\alpha'}\, =\, \frac{\alpha'}r\in\mathbb C $$

Thus $ $ field $\mathbb R\, \Rightarrow\, $ field $\mathbb C\ $ by using the norm $\rm\:\alpha\to\alpha\!\ \alpha'\:$ to lift existence of inverses from $\mathbb R$ to $\mathbb C.$

Answer 13

I may have missed it, but there is an important reason that I think has been omitted from the other answers. (Ahaan Rungta mentioned it, but did not explain in detail.)

Recall how something like $\frac3{17}$ was calculated prior to around 1964:

$$\require{enclose} \begin{array}{rl} 17&\enclose{longdiv}{3.000\ldots} \end{array}$$

$$\begin{array}{rlll} & \ \ \ \,0.1\\ 17&\enclose{longdiv}{3.000\ldots} \\ & \ \ 1.7 \\ \hline & \ \ 1\ 3 \end{array}$$

$$\begin{array}{rlll} & \ \ \ \,0.17\\ 17&\enclose{longdiv}{3.000\ldots} \\ & \ \ 1.7 \\ \hline & \ \ 1\ 30 \\ & \ \ 1\ 19 \\ \hline & \ \ \ \ \ 11 \end{array}$$

And so on. The difficulty of the calculations depends only on the complexity of the divisor, which is 17. To extract a result with any required degree of precision one needs only continue the calculation until the required number of digits have been emitted. But the operations themselves are determined by the divisor.

Now let us take $\frac3{\sqrt2}$ as an example. To calculate this directly we need to evaluate:

$$1.4142\ldots \enclose{longdiv}{3.000\ldots} $$

which is quite onerous. Using an exact value for the divisor is impossible because of the way the algorithm works, so you must truncate the divisor. It's not clear how much error will be introduced by this truncation. And if you round off the divisor to $n$ digits of precision, you must perform many multiplications and subtractions of $n$-digit numbers.

In contrast, calculating $\frac{3\sqrt2}2$ is much easier. First calculate $3\times \sqrt2$ with a single multiplication, to obtain $4.242640\ldots$. (If you need more digits later you can easily produce them when you need them.)

Then perform the following division:

$$2 \enclose{longdiv}{4.242640\ldots} $$

which requires only trivial integer calculations throughout.

Answer 14

If you can visualize $ {1 \over \sqrt 2} $ as a number representing ratios of sides of an isosseles right angled triangle in a right of its own, fine, well and good.

But if you wish to find difference say between $ 1/(\sqrt p - \sqrt q) $ and $ 1/(\sqrt p + \sqrt q) $, you need to pay toll at the gate of denominator.

Answer 15

Given $\sqrt{2}\approx1.41421356237$, suppose you're challenged to approximately calculate $1/\sqrt{2}$. Now you will find that $1:1.41421356237=100000000000:141421356237=0.7?$, where the calculation of ? isn't easily done. Knowing that $1/\sqrt2=\sqrt2/2$ it becomes a piece of cake: $1.41421356237:2=0.707106781185$.