TITLE: Find a binary operation $\ast$ on $\mathbb{Q}$ such that $f:(\mathbb{Q},+) \to (\mathbb{Q},\ast)$ is an isomorphism.
QUESTION [5 upvotes]: Let $f: \mathbb{Q} \to \mathbb{Q}$ be defined by $f(x) = 3x - 1$.

(a) Find a binary operation $\ast$ on $\mathbb{Q}$ such that $f:(\mathbb{Q},+) \to (\mathbb{Q},\ast)$ is an isomorphism.

Here's my work. Is it correct?
Since $f(x)$ is bijective, we just need to find a binary operation $\ast$ such that $f: (\mathbb{Q},+) \to (\mathbb{Q},\ast)$ is a homomorphism.
Define $\ast:\mathbb{Q} \times \mathbb{Q} \to \mathbb{Q}$ as $(a,b) \mapsto a + b + 1$. Then $$ \begin{align} f(a + b) = 3(a+b)-1 &= 3a + 3b - 1 \\&=3a + 3b -2 + 1\\&=(3a-1)+(3b-1) + 1\\&=(3a-1) \ast (3b -1)\\&=f(a) \ast f(b).  \end{align}$$ Since $f: (\mathbb{Q},+) \to (\mathbb{Q},\ast)$ satisfies $f(a + b) = f(a) \ast f(b)$ for all $a,b \in \mathbb{Q}$ we have $(\mathbb{Q},+) \cong (\mathbb{Q},\ast)$ on $\ast$.

(b) Find a binary operation $\ast$ such that $f:(\mathbb{Q},\ast) \to (\mathbb{Q},+)$ is an isomorphism.

Define $\ast$ by $(a,b) \mapsto a + b - \frac{1}{3}$. Then $$\begin{align} f(a \ast b) = 3(a + b - \frac{1}{3}) - 1 &= 3a + 3b - 2 \\&= (3a - 1) + (3b - 1) \\&=f(a)+f(b). \end{align}$$ Since for $f:(\mathbb{Q},\ast) \to (\mathbb{Q},+)$ we have $f(a \ast b) = f(a) + f(b)$ for all $a,b \in \mathbb{Q}$ it follows that  $(\mathbb{Q}, \ast) \cong (\mathbb{Q},+)$ on $\ast$.

REPLY [4 votes]: As Meelo says, your proof is involved, but correct. However, it all boils down to the following general statement.
In your case, $f(x)=3x-1$ is a bijection and you don't need to actually use it. The new operation is obtained by doing the original one on the preimages of the elements and then going back with the map $f$.
Theorem. Let $X$ be a set and let $\cdot$ be an operation on $X$. If $f\colon X\to X$ is a bijection, define a new operation $*$ on $X$ by
$$
a*b=f(f^{-1}(a)\cdot f^{-1}(b))
$$
for all $a,b\in X$. Then $f$ is an isomorphism between $(X,\cdot)$ and $(X,*)$.
Proof. Let $x,y\in X$; then, by definition,
$$
f(x)*f(y)=f\bigl(f^{-1}(f(x))\cdot f^{-1}(f(x))\bigr)=f(x\cdot y)
\qquad\text{QED}
$$
Motivation. Basically, you need $f(x\cdot y)=f(x)*f(y)$; set $a=f(x)$ and $b=f(y)$, so that $x=f^{-1}(a)$ and $y=f^{-1}(b)$; then
$$
a*b=f(x\cdot y)=f(f^{-1}(a)\cdot f^{-1}(b)).
$$
We can note that $a$ and $b$ are arbitrary elements of $X$, because $f$ is bijective.
For the operation $*$ such that $f$ is an isomorphism of $(X,*)$ onto $(X,\cdot)$, you want that
$$
f(x*y)=f(x)\cdot f(y)
$$
so the idea is to define
$$
x*y=f^{-1}(f(x)\cdot f(y)).
$$