Show its not a group for # where a#b=a+b-ab in the set of all real numbers

[brandy]

Homework Statement

In set theory, i have a two part question, the first is showing that the system S={set of all real numbers( $\Re$ )}, #}
where a#b=a+b-ab
we have to show that it's not a group.
and then find what c is so that the system = { $\Re$ $\cap$$\overline{c}$, # } is a group.

Homework Equations

the definition i have of a group is closed, associative, has an identity and each element has an inverse.

The Attempt at a Solution

i figured it has to be closed, as in a,b,a#b $\in$ set of all real numbers
identity: i did a#e-ae=0 ->e(1-a)=0,e=0
inverse: a#b=e=a+b-ab, a+b-ab=0 ->a=-b/(1-b), then you have b-b/(1-b)+b^2/(1-b) which is equal to 0 so it works.
so far so good.
associative: i checked it and unless i made a mistake (a#b)#c=a#(b#c), it looks commutive so i tried (a#b)#(c#d) verses a#(b#c)#d but unless i made a mistake this wasn't equal. The system didnt look like it should be associative but by using the definition we were given, it was.
so I'm confused. can someone help point me in the right direction?

 

[Office_Shredder]

a=-b/(1-b),

I can think of a value of b for which this doesn't work

 

[brandy]

*facepalm*
thanks!