Pythagoras: two favourite proofs

Recently in class when I asked students to name five greatest mathematicians (Newton, Leibniz, Euler, Gauss, Poincaré, Hilbert — and yes, I do know how to count), someone bravely shouted Pythagoras. Pythagoras gets points for being a vegetarian, if he indeed was a vegetarian, and for having a golden thigh, but if his claim to mathematical fame rests on having proved the Pythagoras theorem, then it should be more widely known that he did not, and that his proof is known from earlier Babylonian sources.

There are many proofs of the Pythagoras theorem, which as you know says that in a right angle triangle with hypotenuse c and sides a and b we necessarily have that a^2+b^2=c^2.  My favourite simple proof is given below: just work out the length of the side of the small square in the middle, and get the theorem from the fact that the area of the big square must equal the area of the small square together with the four triangles.

But I also have a favourite complicated proof. This one comes from a book by Gregory Barenblatt (Scaling, CUP, Cambridge 2003) which I recommend. This proof is guaranteed to break the ice at parties. Why should one bother with a complicated proof of a simple fact? Because it introduces in a reasonably straightforward way a very useful idea which can be used for harder facts.

First let us introduce some notation. We will use [x] to denote the units of a quantity x. So for example [mass] is kg, or some multiple thereof.

Let us see one application of thinking in units, or as it is put more frequently, “dimensionally”. Consider a pendulum with a bob of mass m, and arm of length L. Clearly, it swings about because there is the force of gravity, so the acceleration of free fall g must also enter the picture. I claim that the period T of oscillations of the pendulum must be proportional to \sqrt{L/g}. How do I know?

[T]=\hbox{sec}, \; [m]=\hbox{kg}, \; [L]=\hbox{metre}, \;[g]=\displaystyle\frac{\hbox{metre}}{\hbox{sec}^2}.

So the only way to get seconds out of the ingredients of the problem (m, L, g) is to disregard the mass of the bob (there is nothing that will cancel out kilograms!) and take

T \sim \sqrt{\displaystyle\frac{L}{g}}


\left[ \sqrt{\displaystyle\frac{L}{g}} \right] = \left[\sqrt{\frac{\hbox{metre}}{\hbox{metre}/\hbox{sec}^2}} \right] = [\hbox{sec}].

This type of analysis, “dimensional analysis”, is powerful stuff. Let us see how to prove Pythagoras’ theorem by dimensional analysis.

First of all note that there are many ways of uniquely specifying a right-angle triangle. For example, it can be specified by the two perpendicular sides, a and b. A better way to do it, is to specify it by the hypotenuse c and the angle \phi at its foot (see Figure 2).

Now let us consider this triangle ABC. Its area A_{\small ABC} is clearly a function of the hypotenuse and the angle,

A_{\small ABC}= f(c,\phi).

But [A_{\small ABC}]=\hbox{metre}^2, while [c]=\hbox{metre} and \phi has no units. (This perhaps will come as a surprise, but if you think about it, you specify angles by ratios of length, so [\phi]=\hbox{metre}/\hbox{metre}. If you are unsure, think about any trigonometric function!) To get the units to match, we must have

A_{\small ABC}= c^2 g(\phi),

where g(\phi) is some irrelevantly complicated function of the angle which you are welcome to compute at leisure later. Note that g(\phi)>0.

If we now drop a perpendicular from B onto AC, by the same argument as before we have

A_{\small ABD}= b^2 g(\phi) \hbox{ and } A_{\small BDC}= a^2 g(\phi).


A_{\small ABC} = A_{\small ABD} + A_{\small BDC}.


c^2 g(\phi)=a^2 g(\phi)+ b^2 g(\phi).

So now cancel g(\phi) and you have Pythagoras’ theorem.


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3 Responses to Pythagoras: two favourite proofs

  1. Hmmm…. So let me see if I understand…

    You/Barenblatt are saying that if dimensional analysis is correct, than a surface is of dimension 2 (c^2), a length is of dimension 1 (c) and an angle must have no dimension (because c/c has no dimension). Than you are saying that if the pythagorean theorem is true, then A_{ABC} is a function of c and \phi, A_{ABD} is a function of b and \phi and A_{BDC} is a function of a and \phi. Lastly, you are saying that if the pythagorean theorem is correct, then A_{ABC} = A_{ABD}+A_{BDC}. And so, since the pythagorean theorem is c^2 = a^2 + b^2, it must follow that g(\phi) must be of dimension 1 (a scalar); we can divide both sides of equation A_{ABC} = A_{ABD}+A_{BDC} by g(\phi) without changing the structure of the equation.

    It gives me two thoughts: Firstly, this doesn’t look like a proof of the correctness of either Pythagoras or dimensional analysis because all statements are made assuming the correctness of either Pythagoras or dimensional analysis. Secondly, if we take the proof as a proof that Pythagoras and dimensional analysis concur with each other even if we change the angle, then it’s quite easy to proof that they do not! If g(\phi) is a scalar and Pythagoras is correct, than the lengths of the legs and base of the triangle must be given by the functions L_a = a \sqrt{g(\phi)}, L_b=b \sqrt{g(\phi)} and L_c=c \sqrt{g(\phi)}. So if \phi changes, the shape of a triangle does not change, but only the size of the base and legs of the triangle change which contradicts with what we sought out to proof.

    I would be interested to learn about your thoughts on my thoughts… 🙂

  2. Ah! I see what I did wrong. I thought the areas were the squares on the sides… oops! sorry.

  3. strathmaths says:

    You’re ahead of me here! But I think you are onto something when you say that “all statements are made assuming the correctness of… dimensional analysis”. If we were trying to build up geometry from first principles a la Euclid then it’s not obvious (or not to me) that the ideas of geometrical similarity would be introduced earlier than the classical proof of Pythagoras — so although Barenblatt’s proof is a lovely argument you’re quite right to say that it depends on our accepting another big result first…

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