= Solution
\b[True.] Choose the <irrational number> $\alpha=\sqrt2$, and partition the <rational numbers> into
$$
C_n=\mathbb Q\cap(n+\alpha,n+1+\alpha),\qquad n\in\mathbb Z.
$$
No endpoint is <rational>, so these disjoint sets cover $\mathbb Q$. Each is nonempty by the <density of the rational numbers>, and each is <clopen> in the <subspace topology> on $\mathbb Q$.
Let two <rational numbers> be equivalent exactly when they belong to the same $C_n$. The <bijection>
$$
h:\mathbb Q/{\sim}\longrightarrow\mathbb Z,\qquad h([q])=n\text{ when }q\in C_n
$$
is well-defined. For every <subset> $B\subseteq\mathbb Z$, the <preimage> of $h^{-1}(B)$ under the <canonical projection of a quotient set> is $\bigcup_{n\in B}C_n$, which is <open> in $\mathbb Q$. Thus every <subset> of $\mathbb Q/{\sim}$ is <open>: its <quotient topology> is the <discrete topology>. The usual <subspace topology> on the <integers> is also the <discrete topology>, because $\{n\}=\mathbb Z\cap(n-1/2,n+1/2)$. Consequently $h$ is a <homeomorphism>.
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