Question

Let S be the set of all real numbers except
—1. Define an operation + on S by
x+y=x+y+xy; x,y belongs to S. Show that
< S, +> is an abelian group. Find a solution
of the equation 1+x = 2 in S.

Answer 1

$S=R/\{-1\}$

$x+y=x+y+xy, x,y\in S$

1.

$\forall x,y\in S\implies (x+y+xy)\in R/\{1\}$

2.

$\forall x,y,z\in S\\ (x+y)+z=x+(y+z)\\ (x+y)+z=(x+y+xy)+z=x+y+xy+\\ +z+(x+y+xy)z=\\=x+y+z+xy+xz+yz+xyz\\ x+(y+z)=x+(y+z+yz)=x+y+z+\\ +yz+x(y+z+yz)=\\=x+y+z+xy+xz+yz+xyz$

3.

$\exist0\in S\forall x\in S:x+0=x\\ x+0=x+0+x\cdot0=x$

4.

$\forall x\in S\exist y\in S: x+y=0\\ x+y+xy=0\\ y(1+x)=-x\\ y=\frac{-x}{1-x}\in S (x\neq 0)$

5.

$\forall x,y\in S\\ x+y=y+x\\ x+y=x+y+xy\\ y+x=y+x+yx$

$<S,+>$ is an abelian group.

$1+x=2\\ 1+x=1+x+1\cdot x\\ 1+x+x=2\\ 2x=1\\ x=\frac{1}{2}\in S$