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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 17:18

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.RTitleUserPersonal
Name		DateLines
289.1ALIEN::POSTPISCHILThu May 23 1985 10:1142
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 +).
289.2TURTLE::GILBERTTue May 28 1985 12:1713
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.