On the theorem of Pythagoras

For the theorem of Pythagoras, I start from Coxeter's formulation ("Introduction to Geometry", p.8)

"In a right-angled triangle, the square on the hypotenuse is equal to the sum of the squares on the two catheti."

Let us play a little bit with that formulation. In a triangle with sides a, b, and c —different from 0 so as to make its angles well-defined— we introdue the usual nomenclature α, β, and γ for their respective opposite angles. (We introduced one angle name so as to be able to express right-angledness, and the other two for reasons of symmetry.)

A formal expression of Coxeter's formulation is

γ = π/2     ⇒     a² + b² = c²     .

Besides the nomenclature we introduced, this formulation contains the (transcendantal!) constant π. Fortunately we can eliminate it thanks to

π = α + β +γ     .

Elementary arithmetic yields the equivalent formulation

α + β = γ     ⇒     a² + b² = c²     .

Isn't that nicely symmetric? It immediately suggests —at least to me— the strengthening


(This will turn out to be a theorem.) We get an equivalent formulation by negating both sides:

α + β ≠ γ     ≡     a² + b² ≠ c²     .

But x ≠ y   ≡   x < y   ∨   x > y , and the latter disjuncts are mutually exclusive. Remembering that the larger angle is opposite to the larger side, is it bold to guess

(1)      α + β < γ     ≡     a² + b² < c²                               and
(2)      α + β > γ     ≡     a² + b² > c²                  ?

Note that (0), (1) and (2) are not independent: from any two of them, the third can be derived. They can be jointly formulated in terms of the function sgn —read "signum"— given by

sgn.0 = 0     ∧     (sgn.x =1   ≡   x > 0)     ∧     (sgn.x = -1   &equiv   x < 0) ,
viz.        sgn.(α + β - γ)    =    sgn.(a² + b² - c²)     .

Consider now the following figure. We have drawn


the case α + β < γ, in which the triangles ΔCKB and ΔAHC, of disjoint areas, don't cover the whole of ΔACB; denoting the area of ΔXYZ by "XYZ" we have in this case

CKB + AHC < ACB     .

In the case α + β = γ, H and K coincide and we have

CKB + AHC = ACB     ,

and in the case α + β > γ, the two triangles overlap and we have

CKB + AHC > ACB     .

In summary

sgn.(α + β - γ) = sgn.(CKB + AHC - ACB)     .

The three areas of the right-hand side are those of similar triangles and hence have the same ratios as the squares of corresponding lines, in particular

CKB/a² = AHC/b² = ACB/c² > 0     ;

hence

sgn.(CKB + AHC - ACB) = sgn.(a² + b² - c²)     ,

Hence we have proved

sgn.(α + β - γ) = sgn.(a² + b² - c²)     ,

a theorem, say, 4 times as rich as the one we quoted from Coxeter.

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The title of this note could make one wonder why I would waste my time flogging a horse as dead as Pythagoras's Theorem. So let us try to summarize what we could learn from this exercise.

•    Three cheers for formalization! Instead of setting out to prove a² + b² = c² for a right-angled triangle, we included the antecedent γ = π/2 in the formal statement of what was to be proved. It was only after the introduction of π that we could eliminate it and met the "nicely symmetric" formulation.

•   Three cheers for the equivalence! It makes quite clear that the theorem is not about right-angled triangles, but about triangles in general.

•   Three cheers for the notational device captured in sgn. If we had not been careful, we would have ended up proving

(α + β) R γ    ≡    (a² + b²) R c²

for R any of the six relations =, ≠, <, ≤, >, and ≥

•   No cheers at all for that stage of the argument in which lack of axiomatization forced us to resort to a picture. Pictures are almost unavoidably overspecific and therby often force a case analysis upon you. Note that I carefully avoided the pictures for α + β > γ; there are 9 of them: K to the right of A, coincident with A, and to the left of A, and similarly for the pair H and B. For the argument these distinctions are irrelevant but, when drawing a picture, you can hardly avoid making them.

•   One of these days I would like to find a convincing explanation of the circumstance that youngsters continue to be educated with the theorem of Pythagoras in its diluted form as quoted from Coxeter. Notice that the 9 figures could have been avoided by also proving

CKB + AHC < ACB      ⇒      α + β < γ          and
CKB + AHC = ACB      ⇒      &alpha + β = γ     ,

i.e. proving (0) and (1) in full.

•   Notice that our figure was not pulled out of a magicians hat! As soon as sgn.(α+β–γ) occurs in the demonstrandum, it is sweetly reasonable to construct that difference. In order not to destroy the symmetry between α and β, one starts with γ and subtracts α at the one side and β at the other:

and this is the germ of the figure we drew.

Epilogue. I am in a paradoxical situation. I am convinced that of the people knowing the theorem of Pythagoras, almost no one can read the above without being surprised at least once. Furthermore, I think that all those surprises relevant (because telling about their education in reasoning). Yet I don't know of a single respectable journal in which I could flog this dead horse. prof.dr.Edsger W.Dijkstra
Department of Computer Sciences
The University of Texas at Austin
Austin, TX 78712-1188, USA

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Transcription by Joel Hockey
Last revised on Thu, 27 May 2010.