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Topic: On Multiplicative Inverse Notation
Replies: 8   Last Post: May 25, 2000 1:13 AM

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Chuck Cadman

Posts: 16
Registered: 12/12/04
Re: On Multiplicative Inverse Notation
Posted: May 21, 2000 11:08 PM
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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 0-a
>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.








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