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From: moroney@jon.DEC
Newsgroups: net.math
Subject: Re: Beyond Exponention
Message-ID: <460@decwrl.UUCP>
Date: Mon, 4-Feb-85 16:10:06 EST
Article-I.D.: decwrl.460
Posted: Mon Feb  4 16:10:06 1985
Date-Received: Thu, 7-Feb-85 07:32:07 EST
Sender: daemon@decwrl.UUCP
Organization: DEC Engineering Network
Lines: 17

I contend that the third function in the series {a+b, a*b, a^b} is NOT the
exponention function a^b ("^" is the exponentiation function) but rather
a variation, a^(log (b))  Before you all think I am mad, let me explain
                   e
my reasoning.  First of all I will define a$b as a^(log (b)).  Unlike a^b,
                                                       e
but like the 2 preceding functions a+b and a*b, a$b is both communitive
and associative, that is a$b==b$a and a$(b$c)==(a$b)$c.  It also obeys
the distributive law, that is a$(b*c)==(a$b)*(a$c).  The exponention function
obeys none of these.  For those who don't see how I got this far, remember,
a$b==a^(log (b))==e^(log (a)*log (b))==b^(log (a))==b$a.
           e            e       e            e
Any comments/flames?
"And you can do the algebra yourself." - Dr. D.
							Mike Moroney
						..decwrl!rhea!jon!moroney
Sun 3-Feb-1985 23:44 EST