1. ## Group Structure

Show that a*b=a+b+ab defines a group structure on G=R\{-1} using that 1+a*b=(1+a)(1+b).

not that G=R\{-1} means that G is all real numbers excluding -1.

2. Originally Posted by TexasGirl
Show that a*b=a+b+ab defines a group structure on G=R\{-1} using that 1+a*b=(1+a)(1+b).

not that G=R\{-1} means that G is all real numbers excluding -1.
Binary Operation is Closed: The only case where it would not be closed is when, $\displaystyle a,b \not = -1$
$\displaystyle a*b=-1$
Thus,
$\displaystyle a+b+ab=-1$
Thus,
$\displaystyle b(a+1)=-a-1=-(a+1)$
Thus,
$\displaystyle b=-1$
Impossible, thus the binary operation is closed.

Associative: We have,
$\displaystyle (a*b)*c=(a+b+ab)*c=a+b+ab+c+ac+bc+abc$
And,
$\displaystyle a*(b*c)=a*(b+c+bc)=a+b+c+bc+ab+ac+abc$
Thus, $\displaystyle a*(b*c)=(a*b)*c$

Identity: We have, that,
$\displaystyle 0*a=0+a+a0=a$ and,
$\displaystyle a*0=a+0+00=a$.

Inverse: Notice that,
$\displaystyle a*-\frac{a}{a+1}=-\frac{a}{a+1}*a=0$
Finally, we need to show if, $\displaystyle a\not = -1$ then, the element $\displaystyle -\frac{a}{a+1}\not = -1$
Assume, it were then,
$\displaystyle -\frac{a}{a+1}=-1$
Thus,
$\displaystyle \frac{a}{a+1}=1$
Thus,
$\displaystyle a=a+1$
Thus,
$\displaystyle 1=0$
Impossible, thus, $\displaystyle -\frac{a}{a+1}\in G$