## A geometric fallacy

A geometric fallacy is basically a geometric construction that leads to a false result.  I once found an obscure russian treatise on this matter (published by мир), that kept me entertained for weeks: Ya.S.Dubnov’s “Errores en las demostraciones geométricas.”  The edition I have is a decent translation into Spanish, made by K.Medkov in 1993, although unfortunately some of the translated statements had to be treated with a grain of salt.  Nonetheless, I did enjoy greatly browsing through the different geometric problems, and learned quite a bit about mathematical reasoning.

Most of the geometric fallacies rely on the use of correct reasoning on the incorrect figures. But I would like to present here one of the most amusing ones, not only for the fact that it didn’t use a wrong figure, but also because it takes us to deliver a contradiction to the Pythagorean Theorem!  It goes like this:

In all right triangles, the sum of the lengths of the legs equals the length of the hypothenuse.  That is, the sides $a$, $b$ and $c$ satisfy $a+b=c$, where the largest side is conventionally denoted $c$ and is called the hypotenuse.  The other two sides, of length $a$ and $b$ are called the legs (or the catheti)

The “proof” of this absurd result is done by taking to the limit a clever construction.  Let us start with a given right triangle, and consider its two legs alone:

Starting in the vertex $A$, consider a staircase of segments that goes through the midpoints of the sides of the triangle, until reaching the vertex $B$. Notice that the lengths of segments of this staircase add up to $a+b$ by construction.

Repeat the same staircase construction starting in the vertex $A$, and using the midpoints of the previous staircase, until reaching vertex $B$. The lengths of the segments of this new staircase still add up to $a+b$.

By taking this construction to the limit, new staircases of length $a+b$ are constructed that tend to the hypotenuse (in the sense that, after some iterations, the staircase and the hypotenuse will be as close to each other as desired).  We infere that the limit of these staircases is indeed the hypotenuse, and since the length of each staircase remained constant $a+b$, the length of the hypotenuse must also be $a+b$.

QED

This is obviously not true. What is wrong with this proof?

1. October 14, 2010 at 3:25 pm

Two curves being “as close to each other as desired” doesn’t mean they have the same length… With a straight line, and an epsilon of wiggle room, you can fit a line that’s twice as long (even a continuous one if you want).

• November 29, 2010 at 12:58 am

Spot on! So, why will this limiting process work in certain cases, and not in others? Consider the example of the computation of the length of a circle by approximation with regular polygons (see my reply to the comment of Daniel Maxin)

2. October 14, 2010 at 6:51 pm

The fallacy is due to the fact that there is no definition of what it means for the staircase to be as close as possible to the hypotenuse. The division process does not change at all the setting of the problem. At step n, you simply have 2^n triangles with sides a/2^n, b/2^n and c/2^n. As n goes to infinity, these sides converge to zero so you do have lim(a/2^n+b/2^n)=lim(c/2^n)=0 but of course this does not imply a+b=c. In other words this limiting process does not offer a bound on a+b-c

• November 29, 2010 at 12:52 am

The definition of “being as close to the hypotenuse” is technically a uniform approximation. As Aaron points out in his post, think of “an epsilon of wiggle room” around the curve. This is a standard definition and is used, for example, to compute the length of the circle:

In a circle of radius one, consider a sequence of circumscribed regular polygons with $n$ sides. The sides are smaller as $n$ grows (they are actually of size $2\sin \tfrac{\pi}{n}).$ As in the case of the hypotenuse, they tend to zero! Since there are $n$ segments of that size, the length of the polygon is $2n\sin\tfrac{\pi}{n}.$ Note that this tends to $2\pi$ as $n \to \infty.$

As you can see, the limiting process does indeed offer the proper answer. Why does this procedure works in this case, and not in the case of the hypotenuse of a right triangle?

3. January 20, 2011 at 8:52 pm

Along the same lines: you can “prove” that Pi=4 by repeatedly removing corners from a circumscribed square to the unit circle.

• January 24, 2011 at 1:25 pm

Yup, same idea: isn’t it grand? So the key question is: why does this fail?

4. April 9, 2014 at 7:02 am

A variation on this is one that my dad taught me. Begin with a circle of diameter 2. The circumference is therefore 2Pi, and the circumference of the semicircle is therefore Pi. Now draw 2 “half-size” semicircles spanning the original diameter. The circumference of each smaller semicircle is now one-half Pi, and they sum up to Pi. Repeat this process with 4 semicircles, then 8. As each sum is still Pi, one can “take this to the limit” and declare that ultimately that the line approaches the original diameter…equating Pi with 2.

• April 10, 2014 at 4:48 pm

Very nice!