# Why rationalize the denominator?

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.

• by this time such bias is largely restricted to school level. Furthermore, it is restricted to cases with very short terms; if I need to estimate $$\frac{\sqrt{n+1} - \sqrt n}{2}$$ my quickest trick is to rationalize the numerator Commented Dec 29, 2014 at 19:31
• See this thread in MathEducators.SE. Many answers and explanations there. The discussion is pedagogical rather than mathematical, but anyway. A quick summary: comparing answers is easier in standard form, b4 computers yadda yadda, nice to learn how to do this so that you can when it really matters. Commented Dec 29, 2014 at 19:54
• @JyrkiLahtonen, I think it is three yadda's. Hmmm; usually three but varies youtube.com/watch?v=O6kRqnfsBEc Commented Dec 29, 2014 at 19:58
• At least it's a surd number of yaddas, rather than an absurd number! Commented Dec 30, 2014 at 8:33
• This is part of the moral relativism and moral dissonance problems. We try to rationalize everything! :-) Commented Dec 30, 2014 at 10:33

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...

• While 1/sqrt(2) might look simpler than sqrt(2)/2, I personally think it is not. sqrt(2)/2 is half of sqrt(2). By contrast, 1/sqrt(2) is the inverse of sqrt(2). I think "half" is simpler than "inverse". I know that sqrt(2) is approximately 1.414, so I can immediately calculate its half sqrt(2)/2 = 0.707. However, calculating 1/sqrt(2) requires taking the inverse of 1.414, which I don't know how to do.
– Stef
Commented Apr 28, 2022 at 15:11

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 $2$.

• Minor nitpick: I think you meant "square roots had to be approximated [by hand]." Commented Jan 1, 2015 at 23:04
• What would be the problem of using the approximation in the denominator? Is it because the errors would be 'expanded', while in the numerator they would not? Commented Aug 25, 2016 at 17:42
• @Anoldmaninthesea. Before calculators were invented, approximating square roots was difficult. However, once a particular square root had been calculated, it was easier to rationalize the denominator and use a known approximation rather than calculate a new square root and verify that calculation. In short, rationalizing the denominator was a labor saving device. Commented Aug 25, 2016 at 17:50
• @N.F.Taussig Thanks. I understood what you wrote in your answer. My doubt was specific to the last part, namely the example. We have a proximation for the square root of 2. Why should I use $\sqrt{2}/2$ instead of $1/\sqrt{2}$? the approximation used in both is the same... Commented Aug 25, 2016 at 17:55
• @Anoldmaninthesea. Sorry, I did not initially understand what you were asking. The reason that it makes more sense to use $\sqrt{2}/2$ than $1/\sqrt{2}$ is that it is easier to divide the approximation $\sqrt{2} \approx 1.414214$ by $2$ than it is to divide $1$ by $1.414214$. Simpler calculations lead to fewer errors. Commented Aug 25, 2016 at 18:24

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.

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.

• I can buy the rationale that rationalizing the denominator makes fractions easier to compute by hand. But in that case we should prefer series representations of certain expressions to expressions containing certain common functions like exponentials. I think that terseness, symmetry, and forms that provide insight into the properties of a number are more important than ease of hand-computation, especially considering that we have calculators. Commented Dec 29, 2014 at 19:47
• @Ahaan - +1 for the answer. Nonetheless, a good teacher, at any level, should have no trouble realizing that your three radical expressions represent the same number. However, having said that, and being a US resident, maybe not. Commented Dec 29, 2014 at 20:06
• @ChrisLeary Thanks for the +1! Well, often times, there are standard forms. Anyhow, the more traditional reason, I think, is the historical reason I mention. Commented Dec 29, 2014 at 20:08
• @Solomonoff'sSecret: Traditionally, when computing by hand, the way you'd deal with exponential or trig functions would be to look them up in a table (or use a slide rule). That's easier than using a series expansion, although of course you could do that, if you didn't have a suitable lookup table available. Commented Dec 30, 2014 at 2:05
• +1 for the historical thing - after discussing standard-forms for so long one tends to forget that for some people the numerical results do matter... Commented Dec 30, 2014 at 7:45

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.

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.

• This would be more pertinent if it had mentioned the inverse of $a+b\sqrt n$ with $a,b\in\Bbb Q$. Commented Dec 30, 2014 at 12:35
• True, laziness on my part. I'll change it. Commented Dec 30, 2014 at 12:38
• This is a very important conceptual reason, and becomes even more compelling when you want to show expressions $a + b\sqrt[3]{2} + c\sqrt[3]{4}$ with rational $a, b, c$ form a field. The only nonobvious aspect of being a field is that such expressions are preserved under inversion, since the trick taught in school for the square root case does not work anymore.
– KCd
Commented Dec 31, 2014 at 4:53
• @KCd: So what is a trick that works for $\mathbb{Q}(\sqrt[3]{2})$? Commented Jan 1, 2015 at 13:33
• Well, the point is that there is no trick. Maybe you just meant to ask how it is done at all? To invert a specific number, say $5 + \sqrt[3]{2} - 8\sqrt[3]{4}$, you could multiply it by an unknown $x + y\sqrt[3]{2} + z\sqrt[3]{4}$, set the product equal to $1 + 0\sqrt[3]{2} + 0\sqrt[3]{4}$ by equating coefficients, and solve the resulting set of 3 linear equations in 3 unknowns: it is a linear algebra problem. Another method is to solve the polynomial equation $(5 + t - 8t^2)u(t) + (t^3 - 2)v(t) = 1$ in $\mathbf Q[t]$ and use $u(\sqrt[3]{2})$.
– KCd
Commented Jan 1, 2015 at 17:12

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

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

• True, although dividing the numerator and denominator of $\frac{\sqrt{3}+\sqrt{2}}{\sqrt{6}}$ by $\sqrt{3}$ would produce a fraction with just two square roots. Nonetheless, your method generalizes better. Commented Dec 29, 2014 at 19:51

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 $\sqrt{3+7i}$...

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.

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.$$

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 $3.464101644\cdot10^7$.

• This really has little to do with rationalisation of denominators. Take the inverses of those fractions, and the argument backfires. What it really says is avoid differences of near-equal values in multiplicative formulas, and one can sometimes obtain that by methods similar to those used to rationalise denominators Commented Dec 30, 2014 at 13:16
• @MarcvanLeeuwen. The question was "So why does this bias against roots in the denominator exist and what is its justification?" My answer says: "to have a better precision in calculations". Of course, sometimes, it will be more intereting to rationalise the numerator, but I think we may extend the question a bit.
– MasB
Commented Dec 30, 2014 at 16:43
• @MarcvanLeeuwen Maybe this example is not the best, but I am quite convinced that the precision of numerical approximations is at the heart of rationalizing the denominator. Which is easier to compute $\sqrt{2}/10$ or $1/(5 \sqrt{2})$? If you computed their decimal approximations naïvely, which would give a more reliable result? Commented Dec 30, 2014 at 20:48
• This example is a numerical illustration of the way rationalizing denominators is used to explain the formula for the derivative of $\sqrt{x}$ from the limit definition of the derivative.
– KCd
Commented Dec 31, 2014 at 4:56

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: $1.41421356237:2=0.707106781185$.

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 ?

• This isn't a fair comparison because rationalizing the denominator eliminates the fraction. Commented Dec 29, 2014 at 19:46
• Sqrt of 24 is 4? Commented Dec 30, 2014 at 6:46
• @Ooker: Note the "floor" brackets.
– Dan
Commented Dec 30, 2014 at 7:00
• Now I understand Commented Dec 30, 2014 at 7:20

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

• I don't think this is correct. The method for dividing numbers with straightedge and compass is exactly the same whether the lengths are rational or irrational.
– MJD
Commented Jan 2, 2015 at 2:18

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.