Relay-Version: version B 2.10 5/3/83; site utzoo.UUCP Posting-Version: version B 2.10 5/3/83 based; site homxa.UUCP Path: utzoo!watmath!clyde!cbosgd!cbdkc1!desoto!packard!edsel!bentley!hoxna!houxm!homxa!young From: young@homxa.UUCP (Y.HUH) Newsgroups: net.math Subject: RE: Re: Beyond Exponention Message-ID: <696@homxa.UUCP> Date: Thu, 7-Feb-85 16:21:42 EST Article-I.D.: homxa.696 Posted: Thu Feb 7 16:21:42 1985 Date-Received: Sat, 9-Feb-85 07:44:17 EST Organization: AT&T Bell Labs, Holmdel NJ Lines: 59 >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$a. > e e e e >Any comments/flames? >"And you can do the algebra yourself." - Dr. D. > Mike Moroney > > ..decwrl!rhea!jon!moroney VERY INTERESTING! Note that "$" could be defined slightly differently while maintaining the algebraic properties, because the logarithm used in the definition can be of any base, not necessarily e. Suppose that we choose to use e as the base, as before. Then the identity element for this operator is e, that is, e$a = a = a$e . And the $-inverse of a is e^(1/log.e(a)) . Let R = the real numbers, S = the real numbers except 0, P = the positive real numbers, T = the positive real numbers except 1. Then: (R,+) is a group. [1] (S,*) is a group. [2] (P,*) is a group. [3] (T,$) is a group. [4] (R,+,*) is a ring. (from [1] and [2]) (P,*,$) is a ring. (from [3] and [4]) Now: Define an algebraic system with three operators, (+,*,$) ??? Can we keep going on doing this? +, *, $, ... what next??? groups, rings, ... -- Young Huh AT&T Bell Laboratories homxa!young