# Proving 1 isn't in my group

• Dec 13th 2011, 04:17 PM
beebe
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, $\displaystyle a+b-ab \neq 1$
• Dec 13th 2011, 08:11 PM
Amer
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 $\displaystyle a,b\;\in \mathbb{Q} - \{1\}$
$\displaystyle a*b \ne 1$
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