## Return to Purple Squares

10 October 2009

In the last post, we show that the following simple diagram provides all the information needed to prove that $\sqrt{2}$ is irrational.

But as it true so often in mathematics, there is much more to see beyond the surface-level observations, and this time I cannot resist going back to this picture to say more.

Our main points last time were:

1. the big square has the same area as the two light squares together if and only if the dark square has the same area as the two white squares together
2. a square of side $m$ has the same area as two squares of side $n$ if and only if $m/n = \sqrt 2$.

We know we can’t get equality in either case though, which motivates the following approximate version.

1. the big square has almost the same area as the two light squares together if and only if the dark square has almost the same area as the two white squares together
2. a square of side $m$ has about the same area as two squares of side $n$ if and only if $m/n \approx \sqrt 2$.

In other words, if the ratio of the  sides of the large and light squares is “about” $\sqrt 2$, then the ratio of the sides of dark and white squares is also “about” $\sqrt 2$.  Which one is a better approximation?  The one involving the larger squares.  The absolute discrepancy in area between the squares is the same, but the relative discrepancy will be smaller if the areas are larger.  (The same reason I was so much more dramatically older than my sister when I was 6 and she was 2 than I will be when I’m 82 and she’s 78.)

Let’s add some letters to simplify the statements.  If  $m$ is the side of the dark square and $n$ is the side of the white square, then $m+2n$ is the side of the big square, and $m+n$ is the side of the light square.  (Make sure you can see this in the picture.)  Then our claim is that if $m/n$ is a reasonable approximation to $\sqrt 2$, then $(m+2n)/(m+n)$ will be a better one.

It’s too much to hope for $m^2=2m^2$, but if we take $m=1,n=1$, then they’re only off by 1.  So we can take $1/1$ as a starting point.  Then we expect $3/2=1.5$ to be a better approximation.  But why stop here?  Taking $m=3,n=2$, $7/5=1.4$ is a better estimate.  We can keep this up forever, giving the following sequence of increasingly good rational approximations to $\sqrt 2$:
$1/1, 3/2, 7/5, 17/12, 41/29, 99/70, 239/169, 577/408, 1393/985, \cdots$
These approximations are getting very close very fast (the last one is right to six decimal places, enough for any practical application I can think of), and we’re not working very hard to get them!

Actually, more still is true.  If we start with any fraction $m/n$, even one which is nowhere near $\sqrt 2$, repeatedly applying the rule $m/n \mapsto (m+2n)/(m+n)$ will give us a sequence of numbers that, in the long run, will converge to $\sqrt 2$.  Since the absolute area discrepancy doesn’t change, but the squares get larger and larger, the approximation is eventually as close as we might like.  The sequence from the previous paragraph is still the best one, though, because there the area discrepancy is 1, which is the best we can hope for since we proved last time that 0 is impossible.

Actually, it can be proven, without anything fancier than stuff we’ve already said, that all the solutions in positive integers of the equation $m^2-2n^2=\pm 1$ come from the sequence two paragraphs back. Can you see how?

It can also be proven that the approximations will be alternately overestimates and underestimates.  Can you see why?

There is a rich theory of  rationally approximating irrational numbers, including a method based on continued fractions for finding optimal approximating fractions to any real number.  What is amazing is that in this case we can get exactly the same answers predicted by the general theory without knowing anything sophisticated.  We don’t need continued fractions or even a precise definition of “good rational approximation”.  All we need is the picture.

(In case  you either don’t like pictures or really like algebra, then the corresponding algebra fact is $(m+2n)^2 - 2(m+n)^2 = -(m^2-2n^2)$, but that’s so much less colorful…)

P.S.

I am aware that the triangle diagram in the previous post somehow got removed from my WordPress uploads.  I can’t fix this until I get back in my office on Monday, but I will do it at that time.

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