Republic of Mathematics blog

Is rationalizing rational?

Posted by: Gary Ernest Davis on: May 3, 2011

Rationalize or get it wrong?

Text books and test setters regularly mark students wrong if they give an answer of \frac{1}{\sqrt{2}} to a problem.

A student is supposed to know, as part of mathematical convention, that \frac{1}{\sqrt{2}} should be rationalized to \frac{\sqrt{2}}{2}.

This issue has been discussed before by @suburbanlion at SuburbanLion’s Blog and @jamestanton at Rationalising the Denominator, among others.

Marking a student wrong for not carrying out a rationalization like this is just BS (and I don’t mean “Bachelor of Science”).

And there is no such mathematical convention – it’s made up by people who write text books and set tests.

This issue came up again recently in #mathchat when @davidwees wrote: “I tell my students that people used to rationalize denominators b/c it made calculations easier” and “1.414…/2 is much easier to do than 1/1.414… without a calculator.”

My issue with what David wrote (and indeed with part of what James Tanton says in his video) is how, without a calculator, do we know that \sqrt{2}\approx 1.414?

Of course if someone told us that \sqrt{2}\approx 1.414 and we figured that \frac{1}{\sqrt{2}}=\frac{\sqrt{2}}{2} then we could probably be adept enough to calculate \frac{1}{\sqrt{2}}\approx 1.414/2=0.707 by hand.

The field \mathbb{Q}[\sqrt{2}]

Numbers of the form a+b\sqrt{2} where a \textrm{ and }b are rational numbers (fractions) collectively have a remarkable structural property:

not only can we add, subtract and multiply such numbers and still get numbers of the same form (e.g. (1+2\sqrt{2})\times (2-3\sqrt{2})=-10+\sqrt{2}), we can also divide such numbers and still get a number of the same form:

e.g. \frac{1+\sqrt{2}}{2-3\sqrt{2}}= \frac{1+\sqrt{2}}{2-3\sqrt{2}}\times \frac{2+3\sqrt{2}}{2+3\sqrt{2}}=\frac{14+7\sqrt{2}}{-14}=-1-\frac{1}{2}\sqrt{2}

Collectively, numbers of the form a+b\sqrt{2} constitute a mathematical field, denoted by \mathbb{Q}[\sqrt{2}] – a structure in which addition, subtraction, multiplication and division (by non-zero numbers) is always possible.

From this perspective, rationalizing \frac{1}{\sqrt{2}}=\frac{1}{2}\sqrt{2} is just expressing the fact that the reciprocal of \sqrt{2} is again in the field \mathbb{Q}[\sqrt{2}].

What about rational approximations?

There are technical senses in which the continued fraction of a real number gives the “best” rational approximation to that number.

So how do we find the continued fraction for \sqrt{2}?

We are looking for an (infinite) expression of the form :

\sqrt{2}=a_0+\frac{1}{a_1+\frac{1}{a_2+\ldots}}

where the a_n are whole numbers.

Because the piece \frac{1}{a_1+\frac{1}{a_2+\ldots}} after a_0 is less than 1, we see that a_0 is the largest integer less than \sqrt{2}, which is 1 (we know 1< \sqrt{2} < 2 because 1=1^2 < 2=\sqrt{2}^2 < 4=2^2).

This leaves \frac{1}{a_1+\frac{1}{a_2+\ldots}}= \sqrt{2}-1 so taking reciprocals we see that a_1+\frac{1}{a_2+\ldots} =\frac{1}{\sqrt{2}-1}.

This means that a_1 is the largest integer less than \frac{1}{\sqrt{2}-1}.

In other words, a_1<\frac{1}{\sqrt{2}-1}<a_1+1, which means that 1+\frac{1}{a_1+1}<\sqrt{2} < 1+\frac{1}{a_1}.

We know 1+\frac{1}{3} <\sqrt{2} < 1+ \frac{1}{2} because (1+\frac{1}{3})^2=\frac{16}{9} < 2 <\frac{9}{4}= (1+ \frac{1}{2})^2, so a_1=2 .

Continuing in this vein gets trickier, because next we have to find the largest integer less than \frac{1}{\frac{1}{\sqrt{2}-1}-2}.

Because \sqrt{2} satisfies a quadratic equation we can use a cute trick to find the continued fraction for \sqrt{2}.

Let’s write \alpha = \sqrt{2}+1 so that \alpha^2=(\sqrt{2}+1)^2=2+2\sqrt{2}+1=2\alpha+1.

Dividing by \alpha gives:

\alpha = 2+\frac{1}{\alpha} = 2+\frac{1}{2+\frac{1}{\alpha}}=2+\frac{1}{2+\frac{1}{2+\frac{1}{\alpha}}}

Taking this to a limit we get \alpha = 2+\frac{1}{2+\frac{1}{2+\ldots}}.

Since \sqrt{2}=\alpha - 1 we get \alpha = 1+\frac{1}{2+\frac{1}{2+\ldots}}.

By successively terminating this continued fraction we get the following rational approximations to \sqrt{2}: 1, \frac{3}{2}, \frac{7}{5}, \frac{17}{12},\frac{41}{29}.

This immediately gives us the following rational approximations to \frac{1}{\sqrt{2}}: \frac{2}{3}, \frac{5}{7}, \frac{12}{17},\frac{29}{41}.

Newton and the Babylonians

Image courtesy Bill Casselman

Isaac Newton’s method for finding roots of functions, applied to the polynomial x^2-2 leads directly to a method for approximating \sqrt{2} known to the Babylonians.

Using this method we start with an approximation x_0 – let’s start with x_0=1 because its the first approximation from the continued fraction for \sqrt{2}: it’s the integer part of \sqrt{2}.

We then average x_0 \textrm{ and } \frac{2}{x_0} to get the next approximation x_1.

We generate a sequence of rational number approximations x_n to \sqrt{2} where. at each step, x_{n+1}=\frac{1}{2}(x_n+\frac{2}{x_n}).

This gives us the following rational number approximations to \sqrt{2}: 1, \frac{3}{2}, \frac{17}{12},\frac{577}{408}.

\sqrt{2} as a decimal

So now we know how to approximate \sqrt{2} by rational numbers, we can use the rationalization \frac{1}{\sqrt{2}}=\frac{\sqrt{2}}{2} to get rational approximations to \frac{1}{\sqrt{2}}, but you see the absurdity – why bother? Just invert the rational approximations to \sqrt{2}.

How about the decimal expansion of \sqrt{2} – where does that come from?

Decimal approximations to \sqrt{2} are usually obtained from the rational approximation from the continued fraction, of from the more rapidly convergent Newton’s (= Babylonian) method.

The record at the time of writing this post is  is 1,000,000,000,000 decimal places due to S. Kondo & A.J. Yee  in 2010.

The empirical evidence is that \sqrt{2} is normal in base 10, meaning that, to date, each of the digits appears with proprtion \frac{1}{10}, each pair of digits with proportion \frac{1}{100}, each triple of digits with proportion \frac{1}{1000} and so on.

However no one yet has a proof of this.

 

 

 

 

 

 

 

 

8 Responses to "Is rationalizing rational?"

Nice derivation using the fact that \sqrt{2} satisfies a quadratic, and I like the rational approximations both to \sqrt{2} and \frac{1}{\sqrt{2}}.

I’m not a rationalizer either, preferring to write \frac{1}{\sqrt{2}} primarily because it looks more like what it is: namely, the square root of \frac{1}{2}.

I always assumed that rationalizing had something to do with the monks of yore having tables of square roots at hand; if you can look up \sqrt{2}, it’s easier to divide that approximation by 2 than 1 by that approximation. (I guess monks = medieval calculators). Either way, not a sensible reason nowadays.

I think I was making the assumption that in typical calculations, one has a table in which they can look up an approximation to the square root of 2 to whatever accuracy has been previously calculated, and then use that approximation in their further calculations. Is it not true that engineers would have relied on tables of values in their calculations and that rationalizing the denominator of a fraction would have made those calculations easier for them to do?

Davis,

i think that’s right if one had look up tables. I appreciate your tweets which got me thinking about this again.

In researching our high school curriculum, I believe the two main sources of this rationalizing behavior are

(1) The ability to perform calculations by hand using lookup tables
(2) The “requirement” of having a single right answer to a question

(1) is no longer directly relevant, although the ability for a student to approximate the answer to something like sqrt(2) / 2 is good: I think this approximation is much easier for students to perform than approximating 1 / sqrt(2).

(2) is unacceptable, born out of test-giving, perhaps. There is some value in it: two students come up with different-looking answers to the same problem. Are sqrt(2)/2 and 1/sqrt(2) the same number? That sort of thing is pretty valuable, since it can focus the discussion on the meaning of number and representations. There are several situations where 1/sqrt(2) is more useful, notably the fundamental trigonometric identity cos^2(x) + sin^2(x) = 1, so the decision comes down to finding the most useful representation of what you’re working with. This same lesson can apply to fractions, decimals, radicals, polynomial expressions, and more, a general concept that applies through many topics and courses.

Our texts teach rationalizing and radical simplification (sqrt(12) + sqrt(75) = ?), but notes that they are conventions and not nearly as important as, say, the distributive property.

Thanks and keep writing!
– Bowen

Interesting post. I’d always get frustrated at school when the teacher couldn’t say why giving the final answers in rationalized form was so important. Now I have a wider view of things.

I have a question about your division example for numbers in Q[sqrt(2)]. Surely the numerator in the middle step should be 8 + 5(sqrt(2))? Which makes the rest of it incorrect? If I’m wrong, please excuse me, and if I’m right, please excuse my pedantry.

I want to marry you.

Okay, wait. Maybe that’s not the best comment to make on a blog. Do-over.

Thanks so much for this, Gary. I’ve been a non-rationalizer for years. So much so that I’ve informed students of my willingness to beat up any subsequent math instructor they have that makes them rationalize. Nobody’s taken me up on it. (bummer)

This reminds me of the Pythagorean’s murder of Hippasos for proving root 2 as irrational. I love math-based cults!

I have gone back and forth on this – but usually come back to teaching students to rationalize denominators mainly because they will run into it so often “in future math classes”. When asked why it is necessary, I show them since an irrational number has an infinitely long decimal representation, dividing 1 (or any rational) by that irrational number is “impossible”. (First make the dividend a whole number by moving the decimal point an infinite number of places, then …).

It has really become a requirement for all mathematics students in West African Examinations to write answers in fractions sucth that the divisor must be greater than the dividend.

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