Associative BinaryOperations and thePythagoreanTheoremDENIS BELL

IIn a recent article [2], L. Berrone presented a newapproach to the Pythagorean Theorem (PT). The idea isto derive the geometric theorem from analytic and

algebraic properties, by methods of functional equations.(So we are not dealing with a method that was an optionfor the ancients!) I thought about Berrone’s ideas, withinhis context of functional equations. Some pleasant sur-prises fell out. Then a surprising gift of functional equationsback to geometry closed the circle for me – and will closethis article.

Let a � b denote the hypotenuse of a right triangle withlegs a and b. The operation � : ð0;1Þ2 7!ð0;1Þ is evidentlycontinuous. It is also(a) Homogeneous (of degree one):

ðkxÞ � ðkyÞ ¼ kðx � yÞ; 8k[ 0: ð1Þ

(b) Reducible: either of the equations a � x ¼ a � y; x �a ¼y �a implies x = y.(c) Associative.

The homogeneity condition states that if the two legs ofa right triangle are scaled by a common factor, then thehypotenuse is scaled by the same factor. This is proved inBook VI of Euclid’s Elements. The reducibility of � is easy toprove within the framework of Euclidean geometry. Asso-ciativity will be discussed later.

Berrone proves the following result in [2]. His argumentis based on a deep theorem of J. Aczel [1, page 256]1

THEOREM 1 Suppose � is a continuous binary operation

on (0,?) that satisfies conditions (a)-(c) above. Then there

exists p such that

x � y ¼ ðx p þ y pÞ1=p: ð2Þ

The construction in Figure 1 shows that, for the partic-ular operation � in the Pythagorean case,

ð1 � 1Þ2 ¼ 1 � 1 � 1 � 1 ¼ 2:

Hence p = 2 in the representation (2), and the PythagoreanTheorem follows.

I was intrigued by the wayBerrone’s work brings togetherthe ancient subject of Euclidean geometry and the very dif-ferent tradition of functional equations. At the same time Iwas disappointed to see a sophisticated result in the latterarea (Aczel’s theorem) invoked to prove a basic result in theformer. It ought to be possible to provide a direct and self-contained proof of Theorem 1, and so to place Berrone’sapproach to PT on an elementary basis. And so it turned out.

This search led me to a more general study of associa-tive, homogeneous, binary operations. I will sketch a fewof these by-products within the field of functional

1This theorem implies that � admits a representation x � y ¼ f �1 f ðxÞ þ f ðyÞð Þ:

92 THE MATHEMATICAL INTELLIGENCER � 2010 Springer Science+Business Media, LLC

equations. Theorem 2 below is an alternative to Theorem 1in which continuity and reducibility are replaced by amonotonicity hypothesis that appeared in an earlier work ofBohnenblust [3]. In Theorem 3, I study monotonic binaryoperations satisfying the condition 1 � 1 ¼ 1 and characterizethese operations according to four possible ‘‘boundaryconditions’’ that they can have.

But this has left a gap. In order to prove PT from The-orem 1, it is necessary to demonstrate the associativity ofthe Pythagorean operation. This question, which could nothave occurred to Pythagoras or Euclid, is addressed at theend of the paper in terms they would have appreciated.

First let me prove Theorem 1. This requires the follow-ing preliminary result.

LEMMA Under the hypotheses of Theorem 1, one of the

following two conditions holds:

x�1 [ x; 8x [ 0; ð3Þ

x�1\x; 8x [ 0: ð4Þ

PROOF. We can argue by contradiction that 1 � 1 6¼ 1. If

this is false, then (1) yields

x � x ¼ xð1 � 1Þ ¼ x; 8x:

In particular, 1 � 1 � 2 ¼ 1 � 2 � 2: Since � is reducible, wecan cancel 1 on the left and 2 on the right to obtain 1 = 2,an absurd conclusion. Hence 1 � 1 6¼ 1 as claimed. It followsthat there exists no a such that a � 1 ¼ a, as this wouldimply a � 1 � 1 ¼ a � 1) 1 � 1 ¼ 1: Because the functionx 7! x � 1 is continuous, the Intermediate-Value Theoremimplies that either (3) or (4) holds.

1o1 1o11

1 1

1 o 1 o 1 o 1

Figure 1. Evaluation of �.

.........................................................................

AU

TH

OR DENIS BELL was born in London and

earned his doctorate from the University of

Warwick. His area of research ordinarily is

stochastic analysis. Aside from mathematics,

his occupations are spending time with hisfamily, listening to music, and surfing – the

web, that is.

Department of Mathematics

University of North Florida

Jacksonville, FL 32224

USA

e-mail: dbell@unf.edu

PROOF OF THEOREM 1 Assume (3) holds (the argu-

ment will work equally well under (4)). Consider the

function f : N 7!R defined by

f ðnÞ ¼ 1 � 1 � � � � � 1; ð5Þ

where � is applied n - 1 times. Then f is strictly increasing

by (3) and satisfies

f ðnÞ � f ðmÞ ¼ f ðnþmÞ: ð6Þ

Furthermore, (1) yields

f ðnmÞ ¼ f ðnÞ � � � � � f ðnÞ ¼ f ðnÞð1 � 1 � � � � � 1Þ¼ f ðnÞf ðmÞ: ð7Þ

I will show how to extend the domain of f first to the set ofrational numbers, then to the reals, in such a way that theabove properties continue to hold. Set

f ðn=mÞ ¼ f ðnÞf ðmÞ ð8Þ

and observe that f is well-defined by (7). Then (6) and (7)extend to Q. Indeed, the extension of (7) is immediate.Using (6)–(8), we have

fa

b

� �� f

c

d

� �¼ f ðaÞ

f ðbÞ

� �� f ðcÞ

f ðdÞ

� �¼ f ðaÞf ðdÞ

f ðbÞf ðdÞ

� �� f ðbÞf ðcÞ

f ðbÞf ðdÞ

� �

¼ 1

f ðbÞf ðdÞ f ðaÞf ðdÞ½ � � f ðbÞf ðcÞ½ �

¼ 1

f ðbdÞ f ðadÞ � f ðbcÞ

¼ f ðad þ bcÞf ðbdÞ ¼ f

a

bþ c

d

� �;

so (6) also holds at rational points. The function f isincreasing on Q, since a/b \ c/d implies

f ðaÞf ðdÞ ¼ f ðadÞ\f ðbcÞ ¼ f ðbÞf ðcÞ;

hence

fa

b

� �¼ f ðaÞ

f ðbÞ\f ðcÞf ðdÞ ¼ f

c

d

� �:

To extend f to (0,?), define f ðxÞ ¼ supf f ðrÞ : r 2Q; r � xg: Then f has the multiplicative property

f ðxÞf ðyÞ ¼ f ðxyÞ; 8x; y [ 0: ð9Þ

To see this, choose sequences of rationals rn " x and sn " ysuch that f ðxÞ ¼ lim f ðrnÞ and f ðyÞ ¼ lim f ðsnÞ. Thenf (xy) C f (rnsn) = f (rn)f (sn)?f (x)f (y), thus f (xy) C

f (x)f (y). Conversely, let {tn} be a sequence of rationals

such that tn " xy and f ðxyÞ ¼ lim f ðtnÞ. Writing tn = rnsn

where rn and sn are rationals with rn \ x and sn \ y, we

have f (x)f (y) C f (rn)f (sn) = f (tn)?f (xy), implying f (x)f (y)

C f (xy). Thus (9) holds. A similar argument shows that f is

everywhere nondecreasing. Now it is well-known that con-

dition (9) implies one of the following two situations: either

f is a power function x 7! xr or f is everywhere discontinuous.

The latter case can be ruled out because a monotone function

� 2010 Springer Science+Business Media, LLC, Volume 33, Number 1, 2011 93

cannot be discontinuous on an uncountable set of points. It

follows that f (x) = xr for some r[0, hence

xr � yr ¼ ðx þ yÞr ; 8x; y 2 Q: ð10Þ

The continuity of � now implies that (10) holds for all

x and y. Replacing xr by x and yr by y in (10), yields (2)

with p = 1/r.A minor modification of the proof of Theorem 1 (the

details will be omitted) gives the same conclusion underalternative hypotheses.

THEOREM 2 Suppose � is associative, homogeneous, and

satisfies the monotonicity condition

x� a and y� b) x � y� a �b: ð11Þ

Suppose further that 1 � 1 6¼ 1. Then

x � y ¼ ðx p þ y pÞ1=p

where p ¼ 1log2ð1 � 1Þ :

Now this result was proved by Bohnenblust [3] in 1940under the extra assumption that � is commutative. This isjust the point that made Berrone’s approach appealing tome initially: Theorem 1 or Theorem 2 seems to workfrom hypotheses not related to commutativity and endwith a conclusion that the relation is commutative.

Obviously, the condition 1 � 1 6¼ 1 is necessary for theconclusion of Theorem 2. What can be said if this condi-tion fails? Let me now address this question. Assume as inTheorem 2 that � is associative and homogeneous, andsatisfies (11). It may be extended by standard arguments to[0, ?) 9 [0, ?) preserving these properties, and I will statethe answer to the question for the extended operation.

The homogeneity of � yields

ð1 � 0Þ2 ¼ 1 � 0 � 0 ¼ 1 � 0:

Hence 1�0 is either 0 or 1, and similarly for 0�1. So we aredown to four cases.

THEOREM 3 Suppose � is associative, homogeneous,

monotonic (11), and idempotent 1 � 1 ¼ 1. (a) If

1 � 0 ¼ 1 ¼ 0 � 1, then x � y ¼ max x; yf g.2 (b) If 1 � 0 ¼ 1

and 0 � 1 ¼ 0, then x � y ¼ x. (c) If 1 � 0 ¼ 0 and 0 � 1 ¼ 1,

then x � y ¼ y. (d) If1 � 0 ¼ 0 ¼ 0 � 1, then x � y ¼ min x; yf g.

The proofs of (a) - (c) are simple and will be omitted.The proof of (d) is as follows. I first prove by contradictionthat the function x 7! x � 1 is bounded above by 1. Supposethat a � 1 ¼ b [ 1 for some a [ 0. Then

x � 1� b; 8x� a:

Dividing through by x, composing on the left-hand sidewith 1, and using (1), (11) and the condition 1 � 1 ¼ 1 gives

1 � 1

x� 1 � b

x; 8x� a: ð12Þ

Iterating (12), starting with x = a and successively replac-ing x by bx yields, for all n C 1

1 � 1

bna� � � � � 1 � 1

ba� 1 � 1

a� 1 � b

a[ 0:

This contradicts the fact that limn#0 1 � n ¼ 1 � 0 ¼ 0:Asimilar argument shows that 1 � y� 1; 8y: Since 1 � 1 ¼ 1

and � is monotone, this implies 1 � n ¼ 1 ¼ n � 1; 8n� 1: Itfollows that if x B y then

x � y ¼ x 1 � y

x

� �¼ x

and if y B x then

x � y ¼ yx

y� 1

� �¼ y:

Hence x � y ¼ minfx; yg as claimed.Consider the dual operation � defined by

x � y ¼ 1

x� 1

y

� ��1

:

Then it is easy to show that � satisfies all the hypothesesimposed on � in the previous theorems with the exceptionof its values at the points (1,0) and (0,1). The roles of 0 and1 at these points are interchanged in passing to the dual,that is

1 � 0 ¼ 1() 0 � 1 ¼ 0; 0 � 1 ¼ 1() 1 � 0 ¼ 0:

This observation yields an alternative proof of part (d) ofTheorem 3 since, if � satisfies the hypotheses of (d) then �satisfies the hypotheses of (a). Hence x � y = max{x, y}and this implies x � y ¼ minfx; yg.

Another side remark: The example

x�y ¼ 0; x 6¼ yx; x ¼ y

�

which is associative and homogeneous, shows that themonotonicity assumption in Theorem 3 is necessary.

I conclude with the problem raised earlier concerningthe applicability of these results to Euclidean geometry. Leta�b denote the hypotenuse of a right triangle with legsa and b. Recall that in order to prove the PythagoreanTheorem via Theorem 1, one has to know that � is

(x o y) o z

y

x

z

y

z

x

x o (y o z)

A

B

C

D

xoyyoz

Figure 2. Plane thinking.

2This case was established in [3].

94 THE MATHEMATICAL INTELLIGENCER

associative. I have not been able to find an independentproof of this fact via planar figures.

One would need to show that the lengths AB and CD inFigure 2 coincide, and it is not clear how to do this withoutusing PT. However, consider the 3-dimensional rectangularbox depicted in Figure 3. Since the diagonal AF is thehypotenuse of both the right triangles ACF and AEF, we see

immediately that l ¼ ðx � yÞ � z ¼ x � ðy � zÞ, establishingthe associativity of �. (This ‘‘box’’ proof of associativity isalso given in [2].) It appears that an excursion to3-dimensional space is required to verify a 2-dimensionallaw! Another example of this phenomenon is the3-dimensional proof of Desargues’s Theorem in the plane[4]. Are there other instances where the proof of a geomet-rical theorem requires a construction in higher dimensions?It would be worthwhile to list some and examine theirindependence.

REFERENCES

[1] J. Aczel, Lectures on Functional Equations and their Applications,

Academic Press, New York, 1966.

[2] L. Berrone, ‘‘The Associativity of the Pythagorean Law’’, Amer-

ican Mathematical Monthly 116 (2010), 936-939.

[3] F. Bohnenblust, ‘‘An Axiomatic Characterization of Lp-spaces’’.

Duke Math. J. 6 (1940), 627–640.

[4] ‘‘Desargues’ Theorem’’, Wikipedia article http://en.wikipedia.

org/wiki/Desargues’_theorem

C F

B

E

A D

x

y

z

xoy

yoz

l

Figure 3. Thinking inside the box.

� 2010 Springer Science+Business Media, LLC, Volume 33, Number 1, 2011 95