# Abstract Algebra any help is appreciated!

1. Feb 12, 2008

### lostinmath08

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.

3. since we know x*e=x, e*x=x, so i attempted:
using e as y, because it would just mean y.

so
2(x+e)-xe-2= 2x+2e-xe-2 = 2x+e(2-x)-2 = 2

so wouldnt my e be 2?

i think i am right, but am not too sure, any input would be appreciated!

2. Feb 12, 2008

### Dick

Try that again. You should solve e*y=y, or 2*(e+y)-ey-2=y. Solve that for e.

3. Feb 12, 2008

### lostinmath08

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!

4. Feb 12, 2008

### HallsofIvy

Staff Emeritus
Since this is not symbolic logic, that $\rightarrow$ is not "implies". It simply indicates that the function if "from" R x R "to" R and changes (x, y) into 2(x+ y)- xy- 2.

5. Feb 12, 2008

### Hurkyl

Staff Emeritus

This step is sketchy -- this step is only permissible if 2-y is known to be (globally) nonzero. Some methods for getting around this problem by handling 2-y=0 as a separate case, making an explicit assumption that 2-y is not zero, managing to translate the problem appropriately into univariate polynomials, or my favorite: factoring rather than dividing.

That's not what you've shown! You've shown that if there is a neutral element, it has to be 1. You have not actually shown there exists a neutral element. Fortunately, it's easy to prove that 1 is, in fact, a neutral element.

Last edited: Feb 12, 2008
6. Feb 12, 2008

thank you!