<s>[INST] Prove that $\mathbb{Z}$ is isomorphic to $\mathbb{Z}_x$ Let $\mathbb{Z}_x$ denote the ring of integers with the operations $\odot, \oplus $ defined as $a \odot b$ = $ a+b -a b$ and $a \oplus b$ = $a+b -1$. Prove that $\mathbb{Z}$ is isomorphic to $\mathbb{Z}_x$.
Not sure how to start.
 [/INST] So we want to find a bijection $f \colon \def\Z{\mathbb Z}\Z \to \Z$ such that 
$$ f(a+b) = f(a) \oplus f(b) , \quad f(ab) = f(a) \odot f(b) $$
holds for all $a,b \in \Z$. Let's suppose we have such an $f$ and want to find $f(0)$, for all $a$ we must have 
\begin{align*}
  f(a) &= f(a+0)\\
       &= f(a) \oplus f(0)\\
       &= f(a) + f(0) - 1\\
\iff f(0) &= 1
\end{align*}
Using the same idea, we must have
\begin{align*}
  f(a) &= f(a\cdot 1)\\
      &= f(a) \odot f(1)\\
      &= f(a) + f(1) - f(a)f(1)\\
 \iff f(1) &= f(a)f(1)
\end{align*}
As $f$ is onto, we must have $f(1) = 0$. Now we can get an idea what to do for a general $n$: 
\begin{align*}
  f(n+1) &= f(n) \oplus f(1)\\
       &= f(n) \oplus 0\\
       &= f(n) - 1
\end{align*}
So if there is such an $f$, we must have: $f(1) = 0$, $f(n+1) = f(n)- 1$ for any $n$, so we must have $f(n) = 1-n$ for every $n$. Now check if this $f$ is the desired isomorphism.
</s>