Find a binary operation * on $\Bbb Q$ such that $f:(\Bbb Q,+)\rightarrow (\Bbb Q,*)$ is an isomorphism Find a binary operation * on $\Bbb Q$ such that  $f:(\Bbb Q,+)\rightarrow (\Bbb Q,*)$ is an isomorphism
$f(x)=3x-1$
Approach
I already proved that $f$ is bijective now I have to find *, so $f$ is homomorphic .
$f(a+b)=f(a)*f(b)$ for some $a,b \in \Bbb Q$
so $$3(a+b)-1=(3(a)-1)*(3(b)-1)$$
what would be a good one?
 A: Hint $f$ is a bijection with inverse $f^{-1}(x)=\frac{x+1}{3}$. Since you want $f$ to be an isomorphism, so is $f^{-1}$. 
Therefore 
$$f^{-1}(x*y)=f^{-1}(x)+f^{-1}(y) \\
f^{-1}(x*y)=\frac{x+y+2}{3}
$$
Now apply $f$ to this relation.
A: Assuming $*$ can be any binary operation, $a*b= a+b+1$ should do the trick.
A: Here's a way to find such an $\ast$ (and in fact show that there's only a single choice of $\ast$). You've already shown that $f$ is a bijection, and since the operation $+$ is already given, we can 'push forward' $+$ through $f$ to get the desired $\ast$.  
To understand this a bit more, you've come up with the equation 
$$3(a+b)-1=(3a-1)\ast (3b-1)$$
or (turning it around)
$$(3a-1)\ast (3b-1)=3(a+b)-1$$
We want to give a formula for $x\ast y$ for any $x,y$, and since you've shown that $f$ is a bijection, we know there are $a,b$ so that $f(a)=x=3a-1$ and $f(b)=y=3b-1$ (namely, $a=(x+1)/3$ and $b=(y+1)/3$).  Hence, we can rewrite this as
$$x\ast y=3(a+b)-1 = 3((x+1)/3+(y+1)/3)-1=x+y+1$$
A: For $a,b \in \mathbb Q$ let $a * b := a+b +1$. Then
$$
\begin{align*}
f(a) * f(b) &= (3a-1) * (3b-1) \\
&= (3a-1)+(3b-1) +1 \\
&= 3(a+b) -1 \\
&= f(a+b).
\end{align*}
$$
How does one obtain such a solution without black magic?
Well, let $a = f(\bar{a})$, $b = f(\bar{b})$, i.e. $\bar{a} = \frac{a+1}{3}$ and $\bar{b} = \frac{b+1}{3}$. Then
$$
\begin{align*}
a * b &= f(\bar{a}) * f(\bar{b}) \\
&= f(\bar{a} + \bar{b}) \\
&= 3(\bar{a} + \bar{b}) - 1 \\
&= 3( \frac{a+1}{3} + \frac{b+1}{3}) - 1 \\
&= a + b + 1.
\end{align*}
$$
The key point is the second line in the second equation. We set $f(\bar{a}) * f(\bar{b}) = f(\bar{a} + \bar{b})$, because this is the equation we want to fullfill. So by 'pretending' that this equation holds, we obtain the unique definition of $*$ that actually satisfies this identity.
