Start with a single paper square:

Fold from upper right to lower left (colors added to both sides of the paper for clarity):

Follow up with one more fold:

Voila, a proof that the square root of 2 is irrational.

(Mind you, there is some reasoning involved, but what’s the fun in giving that away? Start with “if the square root of 2 is rational, then there is some isosceles right triangle where the sides are the smallest possible integers.”)

This proof first appears in slightly different form in 1892. The paper folding version is from Conway & Guy in 1996.

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Filed under: Mathematics, Puzzles |

Dr. M, on August 2, 2013 at 4:43 pm said:Thanks for this. I think that I’ll write it up and give it as a challenge problem in my Honors classes.

I knew that a diagram like this was involved in the first proof of the irrationality of the square root of 2, but I’d never looked into the details. I’m quite pleased with myself at the moment. I figured it out.

Spoiler alert!

The second fold yields two right triangles. The one in the upper left is similar to the right triangle that results from the first fold. Thus the ratio of hypotenuse of leg is the same in each.

Let m and n be hypotenuse and leg respectively in the second-fold, upper-left right triangle. Let p and q be the hypotenuse and leg respectively of the first-fold right triangle. m/n = p/q. Thus since p/q equals the square root of 2, so too does m/n.

Now, assume that p and q are both integers. Indeed assume that they are integers with no common factor greater than 1. (It’s certainly possible.) This means that there are no integers x < p and y < q such that x/y equals the square root of 2.

Now, a bit of reflection reveals that n = p – q and m = 2q – p. Thus if p and q are integers, so too are m and n. But n is smaller than q and p is smaller than m! This is contrary to our assumption and so we conclude that p and q cannot both be integers.

This means of course that the square root of 2 isn't rational.