A Beautiful Geometric Proof

Tennenbaum’s Proof that Square Root 2 is Irrational

Brett Berry
Oct 3, 2018 · 3 min read

Proofs are the bedrock of mathematics. It’s how we know that every rule and theorem we use holds true. Without the logical rigor of proofs, math would be a bunch of wishy-washy assumptions. Proofs come in all shapes and sizes. Some are long and arduous discernible by very few, others stand on such fundamental logic most anyone could reproduce them with a little motivation.

One such classic proof of number theory and analysis is demonstrating that irrational numbers exist, most commonly that the square root of 2 is irrational. Now there are many ways to prove this result, and I’ve talked about this before…but that was before I encountered Tennenbaum’s proof. A magnificently clever, yet straight-forward proof of irrationality. Using a basic understanding of geometry we can logically show that square root 2 is irrational. It’s quite a fun little proof!

Interestingly, this proof has been largely overlooked historically. Perhaps that’s because our problem is an ancient problem that’s been solved many times over, yet this proof wasn’t concocted until the mid-twentieth century by American mathematician Stanley Tennenbaum. The proof was overlooked until the 1990’s when a past student of Tennenbaum’s published it in a book on the power of mathematics. Finally, the proof began to be appreciated for its beauty and simplicity. Watch or read on!!

the proof.

Much like the popular square root two proof, we begin with a proof by contradiction assuming that the square root of two is rational and therefore can be written as the ratio of two positive integers: p/q.

We then do a little rearranging to get the result that p-squared is equal to 2q-squared.

Here is where a little geometry love comes into play. Instead of taking an algebraic interpretation of our statement, we’ll look at it geometrically.

So we have a square with positive integer side lengths of p, whose area is equal to the combined area of two smaller q-by-q squares who also have positive integer side lengths.

At this point, we can make the assumption that our p-by-p square is the smallest such square we can make that is equal to two smaller q-by-q squares with positive integer side lengths.

Next, we’ll take our smaller squares and neatly place them inside the larger p-by-p square.

This results in two things:

  1. We have a square area where the two q-by-q squares overlap.
  2. We have two smaller square regions where the q-by-q squares do not cover the larger p-by-p square.

Because the area of the two q-by-q squares equals the area of the p-by-p squares, the area of the overlap (1) must equal the area of the uncovered squares (2).

Lastly, to make our contradiction, we deduce that the overlapping square (1) and the uncovered squares (2) also all have positive integer side lengths (I explain this in more detail in the video above).

Therefore, we have found our contradiction because we stated that the p-by-p square we started with was the smallest we could make of this kind, yet we have discovered a smaller square (the overlapping square) who also has integer side lengths and has an area equal to two smaller squares with positive integer side lengths.

Hence, the square root of 2 must be irrational.


❤️ STAY CONNECTED❤️

Instagram | Facebook | Twitter

Math Hacks

Tutorials with a fresh perspective.

Brett Berry

Written by

Check out my YouTube channel “Math Hacks” for hands-on math tutorials and lots of math love ♥️

Math Hacks

Tutorials with a fresh perspective.