## A proof with a hole solution

Submitted by plusadmin on September 1, 2006## A proof with a hole: equals 2

Forget all that business about being an irrational number with infinitely many decimal places: I can prove conclusively that is equal to 2.

First of all, let's recall that is defined to be the ratio between the circumference and the diameter of a circle, which is the same regardless of the size of the circle. So, using the notation of the diagram below, we have * = c/d.*

Now let's start with a circle of circumference 2, and only consider one half of it, as shown in the figure. Since it's exactly one half of the circle, the length of this semi-circle is 1. Now let's divide in half the diameter *d* of the circle, and draw a new, smaller semi-circle on each of the two halves. Since the ratio between diameter and circumference is the same for *any*
circle, you can work out that the two smaller semi-circles — which are built on half the diameter of the larger one — have circumference half that of the larger one. In other words, the length of each of the two smaller semi-circles is 1/2.

Now continue in the same manner: divide the original diameter *d* into 4 equal pieces and draw on each of them a semi-circle of length 1/4; then divide it into 8 equal pieces and draw on each of them a semi-circle of length 1/8, etc, etc. After *n* steps you have *2 ^{n}* semi-circles, each of length

*1/2*.

^{n}Obviously, the semi-circles get smaller and smaller at each stage, and after a great number of steps, your string of semi-circles will hardly be distinguishable from the straight line which forms the diameter of the largest circle. The string of semi-circles *approximates* the diameter *d*, and the approximation gets better and better the more steps you take. This means that the
lengths of the semi-circles all added up approximate *d*. In fact, *d* is the *limit* of this sum as the number of steps *n* tends to infinity:

*d = lim*

_{n→∞}2^{n}×1/2^{n}= 1.*c*of the large circle is 2, so

*= c/d = 2/1 = 2*, which proves my claim. Or have I made a mistake?

## The solution

What's wrong here? Well, it is definitely true that after *n* steps there will be *2 ^{n}* semi-circles, each of length

*1/2*. It is also true that the diameter of the largest circle is in some sense a

^{n}*limit*of the strings of semi-circles: by making

*n*large enough, you can ensure that the

*nth*string of semi-circles squeezes as closely to the diameter

*d*as you like.

The mistake lies in the next step: the assumption that the lengths of the strings of semi-circles tend to the length of the diameter. This is false! What happens in fact is that while at each stage we replace the semi-circles involved by smaller ones, these also become more numerous, so that the replacement makes no difference at all to the overall length. *Every* string of semi-circles
has length 1, since

*2 ^{n}×1/2^{n} = 1,*

*n*. We've simply replaced a number of big wriggles by a greater number of smaller wriggles without changing the length.

The lesson is, when it comes to things like length, don't be fooled by appearance. A curve may look like a straight line, but it may be so wriggly at a small scale that in fact it's a lot longer than the line. Even worse, a set may look like a decent curve that should have finite length, but may in fact be infinitely long. And what exactly do we mean by "length" anyway? For an example of a
curve with infinite length, check out the *Koch curve* in *Plus* article Jackon's fractals. And to learn about notions of length read *Plus* article Measure for measure.

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