## April 5, 2011

### Tell Me Why I'm Wrong: A Geometric "Proof"

This little geometric riddle is something that I have thought about way too much, and it's driving me crazy. I know this is incorrect but I can't figure out why:
The diagonal of a square is twice the length of its sides.
Crazy, right? We all know it's not true. The Pythagorean Theorem states that for a triangle with sides a, b, and hypotenuse c,
a2 + b2 = c2
which means that for triangles formed by a diagonally bisected square with sides of length n,
n2 + n2 = c2
where c is the length of the diagonal. Simple algebra gives us
c = √2n
That's what we all learned in geometry, right? The length of the diagonal of a square is √2 times the length of its sides. Physical reality backs up this fact. Grab a ruler and see for yourself. Now then, why does the following "proof" make sense?

Assuming a square with sides of length a, the distance between opposite corners of the square traveling along the border of the square is 2a.

By dividing both sides into two equal segments and alternating directions, we create a new path of the same distance 2a.
2(a/2) + 2(a/2) = a + a = 2a

By continually dividing each side into n equal segments of length a/n and rearranging them in this manner, we create a path that is arbitrarily close to the diagonal without changing the total length of the path:
n(a/n) + n(a/n) = a + a = 2a

As the number of segments n approaches infinity, the total length remains unchanged.
lim(n → ∞) 2an/n = lim(n → ∞) 2a = 2a

Thus the length of the diagonal is 2a or twice the length of a side.

Please, someone tell me why I'm wrong.

#### 3 comments:

guitargirl said...

........................

Math

*runs away screaming*

Kazemi said...

I suspect the problem lies in taking the limit. If you take the limit of (2a_n)/n as n goes to infinity you get 0, since you wind up "dividing" by infinity. Thus, your limit does not equal 2a, but rather a nonsensical 0.

I think this is because each time you double your n, you are simply doubling the number of triangles you have. You don't actually create a diagonal line. In effect, you create an infinite number of infinitely small triangles. An infinitely small triangle can't have a "length" to its sides, since if it did we could make it smaller.

starcrashx said...

It's a great line of thinking. You're thinking outside the box and challenging the well-defined and tested notion. But as you already knew, you've gotten something wrong, and that would be your basic premise.

Given a wrong premise, we can prove almost anything is true. You started with the premise that the line is equal to 2A... but why? Because you actually started with your conclusion and worked it backwards (a logical fallacy known as "begging the question"). So it shouldn't be surprising that your math led you to the conclusion that you started with.

But I'm sure this isn't enough to convince you, so let's see why it was faulty. Did you notice that your "line" as you went increasing smaller had more than a length or width, and was actually made up of both? Your line -- at every point in the process -- had area, and the equation to figure out its area is L x W (or as you can easily see, A * A, or A squared) rather than L + W. You assumed that you final line had no length and width and thus no area, despite proving that it had an area at every step of the process... why did you assume that this part of the function ceased to be true? It didn't.

So your out-of-the-box thinking is actually another great way to prove that the diagonal is A squared when done correctly.