A Glowing Geometric Proof that Root-2 is Irrational

Tennenbaum-00It was a great shock to the Pythagoreans to discover that the diagonal of a unit square could not be expressed as a ratio of whole numbers. This discovery represented a fundamental fracture between the mathematical domains of Arithmetic and Geometry: since the Greeks recognized only whole numbers and ratios of whole numbers, the result meant that there was no number to describe the diagonal of a unit square.

Proving the Result

The usual way of proving the irrationality of 2 is to assume that it can be expressed as a ratio p/q and, by a simple arithmetical argument, show that this leads to a contradiction. Around 1950, the American mathematician Stanley Tennenbaum discovered an ingenious proof of the irrationality of 2. He never published this result, although he told many other people about it.

We begin by recalling the theorem of Pythagoras, that a² + b² = c², where a, b and c are the sides of a right-angled triangle, with c being the hypotenuse. It is sufficient for us to consider an isosceles triangle, as illustrated in the picture below (left panel). The theorem implies that the areas of the two smaller blue squares add up to the area of the larger red square.


Suppose that the side lengths of the triangle are whole numbers, m and n. Then the theorem implies that m² + m² = n², as illustrated in the right hand panel. But this means 2m² = n², giving 2 = n/m, the ratio of two whole numbers.

Now, if there are two such whole numbers m and n, then there must be two smallest numbers with the property that 2m² = n². Without any loss of generality, we may assume that m and n have already been chosen to be the smallest numbers having this property.

Tennenbaum’s clever idea was to fit the two smaller squares, each of area , into the larger square of area . This is shown in the figure below (left panel). Of course, the two smaller squares must overlap (if they did not, we would have m² + m² < n²). The overlapping region is a square of side 2m–n. So, since this region is covered twice, it must have the same area as the region omitted, the area of the two red squares.


But now we have produced two identical squares (of side n–m) whose areas add up to the area of the square of side 2m–n. The two red squares add up to the purple square:

2(n–m)² = (2m–n)² .

This contradicts the assumption that m and n are the minimal values. The inescapable conclusion is that no such numbers m and n exist.

Another Geometric Proof

Another geometric proof of the irrationality of √2 is illustrated in the figure below. Assume ABC is an isosceles right-angled triangle with integer sides m and hypotenuse n. As with Tennenbaum’s argument, assume these are the smallest integers for which such a triangle exists.

Draw a circular arc centred at C with radius m to cut the side AC at D. Then draw a tangent at D cutting the side AB at E. It is easy to show that the smaller triangle ADE (coloured) is similar to ABC and has smaller integer sides as shown. This contradicts the hypothesis, proving that no such triangle can exist.



Conway, John H: The Power of Mathematics. PDF.

Miller, Steven and David Montague, 2009. Irrationality from the book. ArXiV: PDF.

David Richeson blog post: https://divisbyzero.com/2009/10/06/tennenbaums-proof-of-the-irrationality-of-the-square-root-of-2/

Stanley Tennenbaum: American Original.


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