Actually, (x+y)(x+z)-(x+yz) = x(x+y+z-1), so the solution set is the union {<0,y,z>} u {<x,y,z> | x+y+z=1} --ms -----Original Message----- From: math-fun-bounces@mailman.xmission.com [mailto:math-fun-bounces@mailman.xmission.com]On Behalf Of Jason Sent: Sunday, September 07, 2008 14:01 To: math-fun Subject: Re: [math-fun] Distributivity of addition over multiplication On Sun, 7 Sep 2008, Dan Asimov wrote:
Of course
x + yz = (x + y)(x + z)
is not true in general for (x,y,z) in R^3.
[...] That was 17 days ago. So as Gertrude Stein is reputed to have said, "What is the question?"
You asked if there was an intuitive explanation as to why the satisficing set is 1 = x + y + z. _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun