Prove 2 groups made up of rational numbers are isomorphic. Given groups are - 
$Q*$ multiplicative group of non zero rational numbers
$G$ defined as $G= \{ x : x\in\mathbb{Q}|x\neq1\}$;
$a*b=a+b-ab$
Identity in $G$ is $0$ and inverse of any element $a$ is $−a/(1-a)$
How do I go about it?
 A: We are looking for a bijection $f: G \to  \Bbb Q^\times$ such that $f(a * b)=f(a)f(b)$ for all $a,b$. The first thing to do is look at "special" elements of $\Bbb Q$.
As you correctly noted, $0$ is the identity element in $G$, so we must have $f(0)=1$.
Another special element of $\Bbb Q$ is $-1$: it is the only element of order $2$. So we must have $f(2) = -1$ since $2$ is the only element of order $2$ in $G$.
This could be enough for a first guess, but let's make a final observation. If we extend the operation of $G$ to all the rational numbers (using the same formula), we see that $1*a=1$. That is, $1$ behaves with $*$ like $0$ does with ordinary multiplication in $\Bbb Q$. That is just a hint that looking for an $f:\Bbb Q \to \Bbb Q$ such that $f(1)=0$ might not be a bad idea.
If you plot the three above points $(a,f(a))$ for $a\in\{0,1,2\}$, you'll see that they lie on the same line $y=1-x$. So $f(x)=1-x$ is our first guess. And it works, as you can easily verify that it respects the group operation and it is clearly a bijection between $G$ and $\Bbb Q^\times$.
A: The map $f\colon x\mapsto \frac{x-1}x$ is a group isomorphism $\mathbb{Q}^\times\to G$ 
Proof: $f$ is bijective with inverse $y\mapsto\frac 1{1-y}$. We have to see that it is a group homomorphism. This is a straightforward calculation:
$$f(x)*f(y)=\frac{x-1}x+\frac{y-1}y-(\frac{x-1}x)(\frac{y-1}y) =\frac{(xy-y)+(xy-x)-(xy-y-x+1)}{xy} = f(xy)$$
