
Re: On Multiplicative Inverse Notation
Posted:
May 21, 2000 11:08 PM


Pagadala wrote in message <8g9rug$uan$1@nnrp1.deja.com>... >Well, I've started study of abstract algebra as advised by the forum >gurus. I was reading Abstarct Algebra by Beachy/Blair, 2nd ed. Chapter >4, Polynomials. > >I could't agree some notations. Following is one of them. > ><quote> >Roots; unique factorization >4.1.1. Definition. ............... >(vi) Inverse elements: > >For each a in set F, the equations >a + x = 0 and x + a = 0 >have a solution x in set F, called an additive inverse of a, and >denoted by a. > >For each nonzero element a in set F, the equations >a*x = 1 and x*a = 1 >have a solution x in set F, called a multiplicative inverse of a, and >denoted by a^(1). ><unquote> > >in other words, >additive inverse of a is 0a >multiplicative inverse of a is a^(1) > >Thats strange ! > >The minus sign is prefixed to "a" but not to the additive identity(0) >in case of additive inverse, while the multiplicative identity (1) gets >the minus sign in case of multiplicative inverse.
a^(1) is just a notation meaning the multiplicative inverse of a. The 1 being used here is the integer 1, not 1 in the ring which contains a. I think that is the source of your confusion, but I'm not sure. This notation is convenient because one can define a^n for any integer n (assuming a is invertible): a^0=1, a^2=a*a, a^(2)=a^(1)*a^(1), etc. Then one has the relation a^(m+n) = a^m * a^n for any integers m,n.

