__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 ⇒a^{2}+b^{2}=c^{2 }.

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

π =Elementary arithmetic yields the equivalent formulationα+β+γ.

Isn't that nicely symmetric? It immediately suggests —at least to me— the strengtheningα+β=γ⇒a^{2}+b^{2}=c^{2 }.

`α`+

`β`=

`γ`≡

`a`

^{2}+

`b`

^{2}=

`c`

^{2 }.

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

α+β≠γ≡a^{2}+b^{2}≠c^{2 }.

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

`α`+

`β`<

`γ`≡

`a`

^{2}+

`b`

^{2}<

`c`

^{2 }and

(2)

`α`+

`β`>

`γ`≡

`a`

^{2}+

`b`

^{2}>

`c`

^{2}?

Bold perhaps, but not unreasonable.

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.(

`a`

^{2}+

`b`

^{2}-

`c`

^{2}) .

Consider now the following figure. We have drawn

the case `α` + `β` < `γ`, in which the triangles Δ`CKB` and Δ`AHC`, of disjoint areas, don't cover the whole of Δ`ACB`; denoting the area of Δ`XYZ` by "`XYZ`" we have in this case

CKB+AHC<ACB .

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

and in the caseCKB+AHC=ACB,

`α`+

`β`>

`γ`, the two triangles overlap and we have

In summaryCKB+AHC>ACB.

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

henceCKB/a^{2}=AHC/b^{2}=ACB/c^{2}> 0 ;

sgn.(Hence we have provedCKB+AHC-ACB) = sgn.(a^{2}+b^{2}-c^{2}) ,

sgn.(`α` + `β` - `γ`) = sgn.(`a`^{2} + `b`^{2} - `c`^{2}) ,

*

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`a`

^{2}+

`b`

^{2}=

`c`

^{2}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

(forα+β)Rγ≡ (a^{2}+b^{2})Rc^{2}

__any of the six relations =, ≠, <, ≤, >, and ≥__

`R`• 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

i.e. proving (0) and (1) in full.CKB+AHC<ACB⇒α+β<γand

CKB+AHC=ACB⇒&alpha+β=γ,

• 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.

Austin, 7 September 1986

Department of Computer Sciences

The University of Texas at Austin

Austin, TX 78712-1188, USA

Transcription by Joel Hockey

Last revised on