Proving 1 isn't in my group

$\displaystyle Let G=\mathbb{Q}-\{1\}$

$\displaystyle a*b=a+b-ab \forall a,b \in G$

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 $\displaystyle a+b-ab=1$

$\displaystyle a-ab=1-b$

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

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

$\displaystyle a=1$

$\displaystyle 1 \notin G$

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

Re: Proving 1 isn't in my group

Quote:

Originally Posted by

**beebe** $\displaystyle Let G=\mathbb{Q}-\{1\}$

$\displaystyle a*b=a+b-ab \forall a,b \in G$

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 $\displaystyle a+b-ab=1$

$\displaystyle a-ab=1-b$

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

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

$\displaystyle a=1$

$\displaystyle 1 \notin G$

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

it is correct but I think you just need to modified it like this want to prove that for all $\displaystyle a,b\;\in \mathbb{Q} - \{1\} $

$\displaystyle a*b \ne 1 $

your proof is by contradiction so we will say

suppose in the contrary that there exist $\displaystyle a,b \in \mathbb{Q} - \{1\} $ such that $\displaystyle a*b = 1 $

$\displaystyle a+b - ab = 1 $

... in the end we face a contradiction $\displaystyle a \in G $ and $\displaystyle a = 1 $ this gives $\displaystyle 1 \in G $ contradiction