**moroney@jon.DEC** (02/05/85)

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

**gwyn@brl-tgr.ARPA (Doug Gwyn <gwyn>)** (02/06/85)

> 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). If you want to get algebraic, observe that a*b is not a group (no inverse for 0) and in general one expects to lose an algebraic characteristic for each operation in the progression. This is a lot like the progression real -> complex -> quaternion. I don't think there is a single "right" answer to the question, "What operation is the third in the sequence +, *, ...?" any more than there is a single right answer to "What is the fourth number in the sequence 3, 5, 7, ...?" George Terell told me once that he had investigated the problem of making a generalized series of +, *, ... operators when he was in high school. I never got any details, though. George was last seen by me at the Houston Medical Center and/or Rice campus.

**young@homxa.UUCP (Y.HUH)** (02/08/85)

>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