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Conference rusure::math

Title:	Mathematics at DEC

Moderator:	RUSURE::EDP

Created:	Mon Feb 03 1986
Last Modified:	Fri Jun 06 1997
Last Successful Update:	Fri Jun 06 1997
Number of topics:	2083
Total number of notes:	14613

289.0. "Non Commutative Multiplication" by TOOLS::STAN () Wed May 22 1985 16:18

A problem proposed by Canada for a past International Math. Olympiad:

You are given an algebraic system with an addition and a multiplication
for which all the laws of ordinary arithmetic are valid except
commutativity of multiplication.  Show that


		       -1  -1          -1     -1
		(a + ab  a)   + (a + b)    = a   ,


       -1                            -1      -1
where x   is that element for which x  x = xx   = e, the multiplicative
identity.

T.R	Title	User	Date	Lines
289.1		ALIEN::POSTPISCHIL	`Thu May 23 1985 09:11`	42
	By "laws of ordinary . . .", I hope you mean associativity of + and , commutativity of +, and left and right distributivity of over +. My proof follows. To indicate the inverse, I am using a prime, "'", so x'' is the inverse of the inverse of x. Lemma 0: (ba)' = a'b'. Proof: (ba)(ba)' = e (definition of inverse). ba(ba)' = e (associativity of ). b'ba(ba)' = b'e (x = y -> zx = zy, because is a function). ea(ba)' = b'e (definition of inverse). a(ba)'= b' (definition of identity). a'a(ba)' = a'b' (x = y -> zx = zy). e(ba)' = a'b' (definition of inverse). (ba)' = a'b' (definition of identity). Lemma 1: a = a''. Proof: a'a = e (definition of inverse). a''a'a = a'' (x = y -> zx = zy). ea = a'' (definition of inverse). a = a'' (definition of identity). Theorem: (a+ab'a)'+(a+b)' = a'. Proof: (b+a)(b+a)' = e (definition of inverse). b(b+a)'+a(b+a)' = e ((b+a) was distributed). b''(b+a)'+a(b+a)' = e (Lemma 1). ((b+a)b')'+a(b+a)' = e (Lemma 0). (bb'+ab')'+a(b+a)' = e (b' was distributed). (e+ab')'+a(b+a)' = e (definition of inverse). a'((e+ab')'+a(b+a)') = a'e (x = y -> zx = zy). a'(e+ab')'+a'a(b+a)' = a'e (a' was distributed). a'(e+ab')'+e(b+a)' = a'e (definition of inverse). a'(e+ab')'+(b+a)' = a' (definition of identity). ((e+ab')a)'+(b+a)' = a' (Lemma 0). (ea+ab'a)'+(b+a)' = a' (a was distributed). (a+ab'a)'+(b+a)' = a' (definition of identity). (a+ab'a)'+(a+b)' = a' (commutativity of +).
289.2		TURTLE::GILBERT	`Tue May 28 1985 11:17`	13
	Left-multiplying each side of the identity b + a = (b + a) by (a+b)', and right multiplying by a' gives: (a+b)'ba' + (a+b)'aa' = (a+b)'(b+a)a' Noting that addition is commutative, and that x'y = (y'x)', we get: ((ba')'(a+b))' + (a+b)' = a' (ab'(a+b))' + (a+b)' = a' (ab'a + ab'b)' + (a+b)' = a' (a + ab'a)' + (a+b)' = a' Which is the desired result.