# Algebraic system

• Aug 16th 2012, 12:24 PM
mathrld
Algebraic system
Define on RXR by setting (a,b)(c,d)=(acbd,ad+bc)Show that (RXR,) is an algebraic system
• Aug 16th 2012, 01:13 PM
ModusPonens
Re: Algebraic system
What do you mean by algebraic system? A ring? A field?
• Aug 16th 2012, 03:20 PM
mathrld
Re: Algebraic system
Well I havent learnt abt any algebraic structures except a group. I am wondering what defines an algebraic structure
• Aug 16th 2012, 04:31 PM
Deveno
Re: Algebraic system
the algebraic system formed by (RxR,⊕) is called a commutative monoid (which is almost like a group, except some elements don't have inverses). it's "almost" an abelian group. in fact, it would be an abelian group if you left out (0,0). so what you can prove:

⊕ is associative.
⊕ is commutative.
⊕ has an identity (what is it?)
every element except (0,0) has an inverse.

EDIT: formally, an "algebra" is:

a vector space V, together with an operation * such that * is bilinear, and distributive over + (vector sums).

in other words:

(V,+) is an abelian group
there is a map FxV → V (usually called scalar multiplication, where F is a field) with (for all a,b in F and u,v in V):

a(u+v) = au + av
(a+b)u = au + bu
a(bu) = (ab)u
1u = u

and (V,+,*) is a ring, with a(u*v) = (au)*v = u*(av).

it turns out that (RxR,+,⊕) is actually an algebra under this definition, as well, if we define the vector addition and scalar multiplication like so:

(a,b) + (c,d) = (a+c,b+d)
r(a,b) = (ra,rb). i doubt you are being asked to show this, but it's possible.