i) (for all or any) x,y E R implies x+(-y) E R
ii) (for all or any) x,y E R implies xy E R ( R is closed under mulitplication)
so using the requirements of a subring...this is what i came up with:
x-y=y-x
-y-y=-x-x
-2y=-2x
y=x
and vice versa.
The above is just to satisfy the...
The following axioms must be satisfied
a) (for all or any) x,y E R implies x+(-y) E R
b) (for all or any) x,y E R implies xy E R ( R is closed under mulitplication)
The above are the requirements for a subring to be valid.
This is something i got from wikipedia:
Let R be a ring. Any...
1. [B]The problem statement
Let R be a ring. The center of R is defines as follows:
Z(R)= {x E R where xy = yx for all y E R}
Show that Z(R) is a subring of R
The Attempt at a Solution
I know that rings have to follow 4 axioms
a) its an abelian group under addition
b)...
1. [B]The Problem
If S and T are subrings of a ring R, show that S intersects T, is a subring of R.
The Attempt at a Solution
I don't know how to go about answering this question.
Homework Statement
On the set G=R-{1/3} the following operation is defined:
*G: GxG arrow G
(x,y) arrow x*y=x+y-3xy
Show that (G,*) is an abelian group.
Homework Equations
To proove something is an abelian group:
The Associative Law need to hold true x*(y*x)=(x*y)*x...
Associative Law...help please..thanks!
b1. Homework Statement [/b]
On the set of real numbers R, the following is defined *:RxR arrow R
(x,y) arrow x*y=a(x+y)-xy
find all the values of the real parameter a such that the operation is associative
Homework Equations
associative...
i know where i made my mistake the first time, so e=1...im thinking.
this is how i got to my second conclusion:
2(e+y)-ey-2=y
2e+2y-ey-2=y
2e-ey+2y-2=y
e(2-y)=y-2y+2
e(2-y)=(2-y)
e=(2-y)/(2-y)
e=1
so my neutral element would be 1.
thanks for the help!
[b]1. On the set of real numbers, R the following operation is defined:
*RxR implies (arrow) R, (x,y) implies (arrow) x*y=2(x+y)-xy-2
Find the neutral element of this operation.
[b]3. since we know x*e=x, e*x=x, so i attempted:
using e as y, because it would just mean y...