[SOLVED] Proving 1 isn't in my group

Quote:

Originally Posted by beebe

$Let G=\mathbb{Q}-\{1\}$
$a*b=a+b-ab \forall a,b \in G$

I need to show my group is closed under *, and I'm happy with my proof that all of the possible outcomes are rational numbers. So now I need to show that 1 is not a possible outcome. Is my proof of this acceptable?

Assume $a+b-ab=1$

$a-ab=1-b$

$a(1-b)=1-b$

$a=\frac{1-b}{1-b}$

$a=1$

$1 \notin G$

Therefore, $a+b-ab \neq 1$

it is correct but I think you just need to modified it like this want to prove that for all $a,b\;\in \mathbb{Q} - \{1\}$
$a*b \ne 1$
your proof is by contradiction so we will say
suppose in the contrary that there exist $a,b \in \mathbb{Q} - \{1\}$ such that $a*b = 1$

$a+b - ab = 1$
... in the end we face a contradiction $a \in G$ and $a = 1$ this gives $1 \in G$ contradiction