Associative binary operations and the
Pythagorean Theorem
Denis Bell
April 22, 2016
The Pythagorean Theorem
a c b a2 +b2 = c2 Babylonians, Egyptians, Greeks,… 2000 BC – The approach of Lucio Berrone
Denote the length of the hypotenuse of the right triangle with legs
a and b, by a ◦ b.
Berrone’s approach (American Math Monthly, 2009) is to prove
the theorem from the following qualitative properties of ◦:
(ii) Homogeneity: (λa) ◦ (λb) = λ(a ◦ b).
(iii) Associativity: (a ◦ b) ◦ c = a ◦ (b ◦ c).
(iii) Continuity: (x, y ) 7→ x ◦ y is continuous.
(iv) Reducibility: x ◦ a = y ◦ a =⇒ x = y ,
a ◦ x = a ◦ y =⇒ x = y .
Berrone uses a theorem of J. Aczel to prove the following result.
Theorem
Properties (i) - (iv) above of ◦ imply that ◦ has the form
x ◦ y = (x p + y p )1/p
for some p ∈ R.
The value of p is found from the construction 1o1 1o1 1 1 1 2 = 1o1o1o1 = (1o1)(1o1). 1o1 =
2
. Setting x = y = 1 in the previous formula gives p = 2. Hence xoy = (x 2 + y 2 )1/2 Berrone’s approach is interesting because it brings together two
totally different areas of mathematics, Euclidean geometry and
functional equations.
However, the proof of Aczel’s theorem, on which the argument is
based is rather difficult (occupies 10 pages of his book!).
Aczel’s theorem does not assume homogeneity (which Berrone
anyway needs to use later).
The main result
We provide an elementary treatment of Berrone’s theorem. We
prove the following result.
Theorem
Suppose ◦ : (0, ∞) × (0, ∞) 7→ (0, ∞) is a binary operation that is
homogeneous and associative, and monotone, i.e.
x ≥ α, y ≥ β =⇒ x ◦ y ≥ α ◦ β.
Suppose furthermore that 1 ◦ 1 6= 1. Then
x ◦ y = (x p + y p )1/p
where p =
1
log2 1◦1 .
Remark: Above theorem replaces continuity and reducibility
assumptions in Berrone’s theorem by monotonicity. A minor
modification of our argument proves Berrone’s original theorem
(Math Intelligencer, 2011).
Observation: Berrone’s hypotheses imply 1 ◦ 1 6= 1. Argue by
contradiction. Suppose 1 ◦ 1 = 1. Then
x ◦ x = x(1 ◦ 1) = x, ∀x.
In particular
1 ◦ 1 ◦ 2 = 1 ◦ 2 = 1 ◦ 2 ◦ 2.
Since ◦ is reducible, we can cancel 1 on the left and 2 on the right
to obtain 1 = 2.
About the hypotheses...
Monotonicty seems self evident in the Pythagorean setting.
Homogeneity follows from similar triangles.
There is a simple geometric proof of associativity that will be
discussed at the end of the talk.
The result was proved by F. Bohnenblust in 1940, assuming also
that ◦ is commutative.
Proof of the Theorem
Suppose that 1 ◦ 1 > 1. Define f : N 7→ R by
f (n) = 1 ◦ 1 ◦ . . . ◦ 1,
where ◦ is applied n − 1 times. Then by monotonicity
1 ◦ 1 > 1 =⇒ 1 ◦ 1 ◦ 1 ≥ 1 ◦ 1 =⇒ 1 ◦ 1 ◦ 1 ◦ 1 ≥ 1 ◦ 1 ◦ 1 ◦ 1, etc.
i.e. f is non-decreasing. Furthermore, f satisfies
f (n) ◦ f (m) = f (n + m).
(1)
and
f (nm) = f (n) ◦ ... ◦ f (n) = f (n)(1 ◦ 1 ◦ ... ◦ 1) = f (n)f (m). (2)
We extend f to Q, then to (0, ∞) so that these properties
continue to hold.
Set
f (n/m) =
f (n)
.
f (m)
This is well defined since by (2)
f (kn/km) =
f (kn)
f (k)f (n)
f (n)
=
=
.
f (km)
f (k)f (m)
f (m)
It follows similarly from (1) that for rationals r and s
f (rs) = f (r )f (s).
f is non-decreasing on Q since a/b < c/d implies
f (a)f (d) = f (ad) ≤ f (bc) = f (b)f (c),
thus
f (a/b) =
f (c)
f (a)
≤
= f (c/d).
f (b)
f (d)
We extend f to (0, ∞) by defining
f (x) = sup{f (r ) : r ∈ Q, r ≤ x}.
Then f satisfies
f (xy ) = f (x)f (y ), x, y ∈ (0, ∞).
(3)
To show this, choose sequences of rationals rn ↑ x and sn ↑ y .
Then f (rn ) → f (x), f (sn ) → f (y ), and, since rn sn ↑ xy , we have
f (rn sn ) → f (xy ). Then (3) holds with x = rn , y = sn and taking
n → ∞, we get (3) for x and y .
Similarly, we see f is non-decreasing. For x < y , choose sequences
of rationals rn ↑ x and sn ↑ y with rn < sn , ∀n. Then
f (x) = lim f (rn ) ≤ lim f (sn ) = f (y ).
It follows from these two properties (multiplicative and
non-decreasing) that f has the form
f (x) = x r .
In particular, 2r = f (2) = 1 ◦ 1, so r = log2 1 ◦ 1 > 0.
The final step is to establish the relationship between ◦ and f .
Recall
f (n) ◦ f (m) = f (n + m), n, m ∈ N.
We need to establish this for all positive real values n and m.
Consider
ad + bc a
c
+
=f
f
b d
bd
f (ad) ◦ f (bc)
f (ad + bc)
=
=
f (bd)
f (bd)
h f (ad) i h f (bc) i
=
◦
f (bd)
f (bd)
a c =f
◦f
.
b
d
Finally, let x, y ∈ (0, ∞). Choose sequences of rationals
rn , sn , tn , un such that rn ↑ x, sn ↑ y , tn ↓ x, un ↓ y . Then by
monotonicity of ◦ and f , we have
f (rn + sn ) = f (rn ) ◦ f (sn ) ≤ f (x) ◦ f (y )
≤ f (tn )◦f (un ) = f (tn +un ).
Taking n → ∞, we have
f (x) ◦ f (y ) = f (x + y )
where f (x) = x r . Setting x r = a, y r = b, and p = 1/r , we have
a ◦ b = (ap + b p )1/p .
About the monotonicity condition
In the course of proving the theorem, we proved that a monotone
function f : N 7→ R with the multiplicative property
f (nm) = f (n)f (m) necessarily has the form f (x) = x r .
The most general class of functions satisfying f (nm) = f (n)f (m)
on N is obtained as follows. Define f arbitrarily on the primes.
Then define
f (p1r1 p2r2 . . . pnrn ) = f (p1 )r1 f (p2 )r2 . . . f (pn )rn .
f will not, in general, be monotone.
The nilpotency condition
Clearly 1 ◦ 1 6= 1 is necessary for the theorem. What can be said if
1 ◦ 1 = 1?
Note that by homogeneity (1 ◦ 0)(1 ◦ 0) = 1 ◦ 0 ◦ 0 = 1 ◦ 0, thus
1 ◦ 0 = 0 or 1. Similarly for 0 ◦ 1. This yields four scenarios and we
have the following result.
Theorem
Suppose 1 ◦ 1 = 1. Then
(i) If 1 ◦ 0 = 1 = 0 ◦ 1, then x ◦ y = max{x, y }.
(ii) If 1 ◦ 0 = 0 = 0 ◦ 1, then x ◦ y = min{x, y }.
(iii) If 1 ◦ 0 = 1 and 0 ◦ 1 = 0, then x ◦ y = x.
(iv) If 1 ◦ 0 = 0 and 0 ◦ 1 = 1, then x ◦ y = y .
Limiting behavior
Observe that
lim (x p + y p )1/p = max{x, y },
p→∞
lim (x p + y p )1/p = min{x, y }.
p→−∞
What happens as p → 0?
lim (x p + y p )1/p = ∞,
p→0+
lim (x p + y p )1/p = 0.
p→0−
Associativity of the Pythagorean law? c a aob (aob)oc b a ao(boc) b boc c Thinking inside the box boc a (aob)oc = ao(boc) b c aob