Associative binary operations and the Pythagorean Theorem

Transcription

1 Associative binary operations and the Pythagorean Theorem Denis Bell April 22, 2016

2 The Pythagorean Theorem a c b a 2 +b 2 = c 2 Babylonians, Egyptians, Greeks, 2000 BC

3 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) x y is continuous. (iv) Reducibility: x a = y a = x = y, a x = a y = x = y.

4 Berrone uses a theorem of J. Aczel to prove the following result. Theorem Properties (i) - (iv) above of imply that has the form for some p R. x y = (x p + y p ) 1/p

5 The value of p is found from the construction 1o1 1 1o = 1o1o1o1 = (1o1)(1o1). 1o1 = 2. Setting x = y = 1 in the previous formula gives p = 2. Hence xoy = (x 2 + y 2 ) 1/2

6 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).

7 The main result We provide an elementary treatment of Berrone s theorem. We prove the following result. Theorem Suppose : (0, ) (0, ) (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 Then where p = 1 log x y = (x p + y p ) 1/p

8 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 Argue by contradiction. Suppose 1 1 = 1. Then In particular x x = x(1 1) = x, x = 1 2 = Since is reducible, we can cancel 1 on the left and 2 on the right to obtain 1 = 2.

9 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.

10 Proof of the Theorem Suppose that 1 1 > 1. Define f : N R by f (n) = , where is applied n 1 times. Then by monotonicity 1 1 > 1 = = , etc. i.e. f is non-decreasing. Furthermore, f satisfies and f (n) f (m) = f (n + m). (1) f (nm) = f (n)... f (n) = f (n)( ) = f (n)f (m). (2) We extend f to Q, then to (0, ) so that these properties continue to hold.

11 Set This is well defined since by (2) f (n/m) = f (n) f (m). f (kn/km) = f (kn) f (k)f (n) = f (km) f (k)f (m) = f (n) f (m). It follows similarly from (1) that for rationals r and s f (rs) = f (r)f (s).

12 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 (a) f (b) f (c) f (d) = f (c/d).

13 We extend f to (0, ) by defining Then f satisfies f (x) = sup{f (r) : r Q, r x}. f (xy) = f (x)f (y), x, y (0, ). (3) To show this, choose sequences of rationals r n x and s n y. Then f (r n ) f (x), f (s n ) f (y), and, since r n s n xy, we have f (r n s n ) f (xy). Then (3) holds with x = r n, y = s n and taking n, we get (3) for x and y.

14 Similarly, we see f is non-decreasing. For x < y, choose sequences of rationals r n x and s n y with r n < s n, n. Then f (x) = lim f (r n ) lim f (s n ) = f (y). It follows from these two properties (multiplicative and non-decreasing) that f has the form f (x) = x r. In particular, 2 r = f (2) = 1 1, so r = log > 0.

15 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 ( a f b + c ) ( ad + bc ) = f d bd f (ad + bc) f (ad) f (bc) = = f (bd) f (bd) [ f (ad) ] [ f (bc) ] = f (bd) f (bd) ( a ) ( c ) = f f. b d

16 Finally, let x, y (0, ). Choose sequences of rationals r n, s n, t n, u n such that r n x, s n y, t n x, u n y. Then by monotonicity of and f, we have f (r n + s n ) = f (r n ) f (s n ) f (x) f (y) Taking n, we have f (x) f (y) = f (x + y) f (t n ) f (u n ) = f (t n +u n ). where f (x) = x r. Setting x r = a, y r = b, and p = 1/r, we have a b = (a p + b p ) 1/p.

17 About the monotonicity condition In the course of proving the theorem, we proved that a monotone function f : N 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 (p r 1 1 pr prn n ) = f (p 1 ) r 1 f (p 2 ) r 2... f (p n ) rn. f will not, in general, be monotone.

18 The nilpotency condition Clearly is necessary for the theorem. What can be said if 1 1 = 1? Note that by homogeneity (1 0)(1 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.

19 Limiting behavior Observe that lim (x p + y p ) 1/p = max{x, y}, p What happens as p 0? lim (x p + y p ) 1/p = min{x, y}. p lim p 0 +(x p + y p ) 1/p =, lim p 0 (x p + y p ) 1/p = 0.

20 Associativity of the Pythagorean law? c a aob (aob)oc b a b boc ao(boc) c

21 Thinking inside the box boc c aob b a (aob)oc = ao(boc)