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.

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- Apr 4th 2006, 06:05 AMTexasGirlGroup 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. - Apr 4th 2006, 02:13 PMThePerfectHackerQuote:

Originally Posted by**TexasGirl**

**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, $\displaystyle -\frac{a}{a+1}\in G$