Associative Binary
Operations and the
Pythagorean
Theorem
DENIS BELL
n a recent article [2], L. Berrone presented a new
approach to the Pythagorean Theorem (PT). The idea is
to 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 option
for the ancients!) I thought about Berrone’s ideas, within
his context of functional equations. Some pleasant surprises fell out. Then a surprising gift of functional equations
back to geometry closed the circle for me – and will close
this article.
Let a b denote the hypotenuse of a right triangle with
legs a and b. The operation : ð0; 1Þ2 7!ð0; 1Þ is evidently
continuous. It is also
(a) Homogeneous (of degree one):
I
ðkxÞ ðkyÞ ¼ kðx yÞ;
8k [ 0:
ð1Þ
(b) Reducible: either of the equations a x ¼ a y; x a ¼
y a implies x = y.
(c) Associative.
The homogeneity condition states that if the two legs of
a right triangle are scaled by a common factor, then the
hypotenuse is scaled by the same factor. This is proved in
Book VI of Euclid’s Elements. The reducibility of is easy to
prove within the framework of Euclidean geometry. Associativity will be discussed later.
1
Berrone proves the following result in [2]. His argument
is based on a deep theorem of J. Aczél [1, page 256]1
T HEOREM 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 p Þ1=p :
ð2Þ
The construction in Figure 1 shows that, for the particular operation in the Pythagorean case,
ð1 1Þ2 ¼ 1 1 1 1 ¼ 2:
Hence p = 2 in the representation (2), and the Pythagorean
Theorem follows.
I was intrigued by the way Berrone’s work brings together
the ancient subject of Euclidean geometry and the very different tradition of functional equations. At the same time I
was disappointed to see a sophisticated result in the latter
area (Aczél’s theorem) invoked to prove a basic result in the
former. It ought to be possible to provide a direct and selfcontained proof of Theorem 1, and so to place Berrone’s
approach to PT on an elementary basis. And so it turned out.
This search led me to a more general study of associative, homogeneous, binary operations. I will sketch a few
of these by-products within the field of functional
This theorem implies that admits a representation x y ¼ f 1 ð f ðxÞ þ f ðyÞÞ:
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P ROOF OF T HEOREM 1 Assume (3) holds (the argument will work equally well under (4)). Consider the
function f : N 7! R defined by
1 o1
1o1
f ðnÞ ¼ 1 1 1;
1
1
where is applied n - 1 times. Then f is strictly increasing
by (3) and satisfies
1
1o1 o1o1
f ðnÞ f ðmÞ ¼ f ðn þ mÞ:
equations. Theorem 2 below is an alternative to Theorem 1
in which continuity and reducibility are replaced by a
monotonicity hypothesis that appeared in an earlier work of
Bohnenblust [3]. In Theorem 3, I study monotonic binary
operations satisfying the condition 1 1 ¼ 1 and characterize
these operations according to four possible ‘‘boundary
conditions’’ that they can have.
But this has left a gap. In order to prove PT from Theorem 1, it is necessary to demonstrate the associativity of
the Pythagorean operation. This question, which could not
have occurred to Pythagoras or Euclid, is addressed at the
end of the paper in terms they would have appreciated.
First let me prove Theorem 1. This requires the following preliminary result.
f ðnmÞ ¼ f ðnÞ f ðnÞ ¼ f ðnÞð1 1 1Þ
¼ f ðnÞf ðmÞ:
f ðn=mÞ ¼
8x [ 0;
ð3Þ
x1\x;
8x [ 0:
ð4Þ
PROOF. We can argue by contradiction that 1 1 6¼ 1. If
¼
1
½ f ðaÞf ðdÞ ½ f ðbÞf ðcÞ
f ðbÞf ðdÞ
¼
1
f ðadÞ f ðbcÞ
f ðbdÞ
¼
a c f ðad þ bcÞ
¼f
þ ;
f ðbdÞ
b d
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, we
can 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 follows
that there exists no a such that a 1 ¼ a, as this would
imply a 1 1 ¼ a 1 ) 1 1 ¼ 1: Because the function
x 7! x 1 is continuous, the Intermediate-Value Theorem
implies that either (3) or (4) holds.
.........................................................................
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 his
family, listening to music, and surfing – the
web, that is.
DENIS BELL
Department of Mathematics
University of North Florida
Jacksonville, FL 32224
USA
e-mail: [email protected]
f ðnÞ
f 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
a c f ðaÞ f ðcÞ f ðaÞf ðdÞ f ðbÞf ðcÞ ¼
f
f
¼
b
d
f ðbÞ
f ðdÞ
f ðbÞf ðdÞ
f ðbÞf ðdÞ
following two conditions holds:
x1 [ x;
ð7Þ
I will show how to extend the domain of f first to the set of
rational numbers, then to the reals, in such a way that the
above properties continue to hold. Set
L EMMA Under the hypotheses of Theorem 1, one of the
AUTHOR
ð6Þ
Furthermore, (1) yields
Figure 1. Evaluation of .
THE MATHEMATICAL INTELLIGENCER
ð5Þ
so (6) also holds at rational points. The function f is
increasing on Q, since a/b \ c/d implies
f ðaÞf ðdÞ ¼ f ðadÞ\f ðbcÞ ¼ f ðbÞf ðcÞ;
hence
f
a
b
¼
c
f ðaÞ f ðcÞ
:
\
¼f
f ðbÞ f ðdÞ
d
To extend f to (0,?), define f ðxÞ ¼ supf f ðrÞ : r 2
Q; 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 " y
such that f ðxÞ ¼ lim f ðrn Þ and f ðyÞ ¼ lim f ðsn Þ. Then
f (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 ðtn Þ. 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 condition (9) implies one of the following two situations: either
f is a power function x !
7 x r or f is everywhere discontinuous.
The latter case can be ruled out because a monotone function
cannot be discontinuous on an uncountable set of points. It
follows that f (x) = xr for some r [ 0, hence
x r y r 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 under
alternative hypotheses.
T HEOREM 2 Suppose is associative, homogeneous, and
satisfies the monotonicity condition
x a and
y b ) x y a b:
1
1
1
b
1 1 [ 0:
1
bn a
ba
a
a
This contradicts the fact that limn#0 1 n ¼ 1 0 ¼ 0:
A similar 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: It
follows that if x B y then
y
¼x
xy ¼ x 1
x
and if y B x then
xy ¼ y
x
1
y
¼ y:
Hence x y ¼ minfx; yg as claimed.
Consider the dual operation defined by
1 1 1
:
xy ¼
x y
p 1=p
x y ¼ ðx þ y Þ
where p ¼
1
ð11Þ
Suppose further that 1 1 6¼ 1. Then
p
Iterating (12), starting with x = a and successively replacing x by bx yields, for all n C 1
1
log2 ð1 1Þ :
Now this result was proved by Bohnenblust [3] in 1940
under the extra assumption that is commutative. This is
just the point that made Berrone’s approach appealing to
me initially: Theorem 1 or Theorem 2 seems to work
from hypotheses not related to commutativity and end
with a conclusion that the relation is commutative.
Obviously, the condition 1 1 6¼ 1 is necessary for the
conclusion of Theorem 2. What can be said if this condition fails? Let me now address this question. Assume as in
Theorem 2 that is associative and homogeneous, and
satisfies (11). It may be extended by standard arguments to
[0, ?) 9 [0, ?) preserving these properties, and I will state
the answer to the question for the extended operation.
The homogeneity of yields
ð1 0Þ2 ¼ 1 0 0 ¼ 1 0:
Hence 10 is either 0 or 1, and similarly for 01. So we are
down to four cases.
T HEOREM 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 maxfx; y 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) If 1 0 ¼ 0 ¼ 0 1, then x y ¼ minfx; y g.
Then it is easy to show that satisfies all the hypotheses
imposed on in the previous theorems with the exception
of its values at the points (1,0) and (0,1). The roles of 0 and
1 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) of
Theorem 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
0; x 6¼ y
xy ¼
x; x ¼ y
which is associative and homogeneous, shows that the
monotonicity assumption in Theorem 3 is necessary.
I conclude with the problem raised earlier concerning
the applicability of these results to Euclidean geometry. Let
ab denote the hypotenuse of a right triangle with legs
a and b. Recall that in order to prove the Pythagorean
Theorem via Theorem 1, one has to know that is
D
B
The proofs of (a) - (c) are simple and will be omitted.
The proof of (d) is as follows. I first prove by contradiction
that the function x 7! x 1 is bounded above by 1. Suppose
that a 1 ¼ b [ 1 for some a [ 0. Then
x 1 b;
2
z
(x o y) o z
x o (y o z)
8x a:
yoz
Dividing through by x, composing on the left-hand side
with 1, and using (1), (11) and the condition 1 1 ¼ 1 gives
1
b
1 1 ;
x
x
x
8x a:
ð12Þ
xoy
A
y
x
C
y
Figure 2. Plane thinking.
This case was established in [3].
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z
z
C
F
y
B
l
xo y
E
x
immediately that l ¼ ðx yÞ z ¼ x ðy zÞ, establishing
the associativity of . (This ‘‘box’’ proof of associativity is
also given in [2].) It appears that an excursion to
3-dimensional space is required to verify a 2-dimensional
law! Another example of this phenomenon is the
3-dimensional proof of Desargues’s Theorem in the plane
[4]. Are there other instances where the proof of a geometrical theorem requires a construction in higher dimensions?
It would be worthwhile to list some and examine their
independence.
yo z
A
D
Figure 3. Thinking inside the box.
REFERENCES
[1] J. Aczél, Lectures on Functional Equations and their Applications,
Academic Press, New York, 1966.
[2] L. Berrone, ‘‘The Associativity of the Pythagorean Law’’, Amer-
associative. I have not been able to find an independent
proof of this fact via planar figures.
One would need to show that the lengths AB and CD in
Figure 2 coincide, and it is not clear how to do this without
using PT. However, consider the 3-dimensional rectangular
box depicted in Figure 3. Since the diagonal AF is the
hypotenuse of both the right triangles ACF and AEF, we see
THE MATHEMATICAL INTELLIGENCER
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