Q:

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

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 $\ast$.

A:

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
$$
x*y=f^{-1}(f(x)\cdot f(y)).
$$

A:

I found your proofs a bit hard to follow, though they're certainly correct; if you just remove where you restate the definition of a homomorphism and addition, it'd be far more readable - though your process might have done that first, it's generally helpful to the reader if you jump right in to the insight.
However, if you really wanted to make the proofs look good, you could phrase them in a way that makes clear why those are the answers; I don't know how you found those functions, but the argument I would use (which might be how you did it too) would be:

A homomorphism must take the identity element to the identity element. The identity of the operation $a*b=a+b-c$ is $c$ and $f$ takes $0$ to $-1$, so $-1$ is the identity of $*$, meaning the only possible operation of that form is $a*b=a+b+1$.

And then you can proceed to show that this is a homomorphism by your previous method. - Your proofs certainly work at the basic level of "convincing in a formal way", but it's good to push towards "offering insight into a problem's solution."

