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.

 

 

 

 

 

 

 

 

Mathematics is an empirical science

Posted by: Gary Ernest Davis on: May 2, 2011

Many people – including some of my colleagues – like to argue that mathematics is an art, not a science, and that even if it is a science then it’s not like physics, chemistry or biology because it’s a deductive, not an empirical, science.

In the five videos below Doron Zeilberger argues a contrary point of view.