Define ⊕ on RXR by setting (a,b)⊕(c,d)=(ac−bd,ad+bc)Show that (RXR,⊕) is an algebraic system

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- Aug 16th 2012, 01:24 PMmathrldAlgebraic system
Define ⊕ on R

*X*R by setting (*a*,*b*)⊕(*c*,*d*)=(*a**c*−*b**d*,*a**d*+*b**c*)Show that (R*X*R,⊕) is an algebraic system - Aug 16th 2012, 02:13 PMModusPonensRe: Algebraic system
What do you mean by algebraic system? A ring? A field?

- Aug 16th 2012, 04:20 PMmathrldRe: Algebraic system
Well I havent learnt abt any algebraic structures except a group. I am wondering what defines an algebraic structure

- Aug 16th 2012, 05:31 PMDevenoRe: 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.