This is admittedly a homework question, but it seems erroneous to me.
Define the rational numbers $\mathbb{Q}$ as the set of equivalence classes on $\mathbb{Z} \times (\mathbb{Z} \backslash \{0\})$ corresponding to the equivalence relation $\sim$, where
$$(a,b)\sim (c,d) \Leftrightarrow ad = bc$$
Show that $f: \mathbb{Q} \rightarrow \mathbb{Q}$ given by $f([(a,b)]_\sim) = [(a+b,b)]_\sim$ is not well-defined.
But here is a proof (I think) that $f$ is well-defined:
$[(a,b)]_\sim = [(c,d)]_\sim \Rightarrow ad = bc$ by definition of $\sim$
$\Rightarrow ad = cb$, by commutativity of multiplication in $\mathbb{Z}$
$\Rightarrow ad + bd = cb+db$, by an addition property of equality and commutativity of multiplication
$\Rightarrow (a+b)d = (c+d)b$, by right distributivity of multiplication over addition
$\Rightarrow [(a+b,b)]_\sim=[(c+d,d)]_\sim$, by definition of $\sim$
$\Rightarrow f([(a,b)]_\sim) = f([(c,d)]_\sim)$, by definition of $f$.
Therefore $f$ is well-defined by the definition of a well-defined function. $\Box$
Is my proof incorrect?