Define a relation $R$ on $\mathbb{N}$ by declaring that $x R y$ if and only if $x = y\alpha^2$ for some $\alpha \in \mathbb{Q}$. Show that each equivalence class has the same cardinality as $\mathbb{N}$.
Not really sure how to proceed working through this one. Any help would be greatly appreciated.
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The number theory part of this is that two natural numbers are in the same squareclass if and only if their product is a square.
The set theory part is that any set of natural numbers has a minimum, call it $m$ for some class of interest. Furthermore, $m$ is squarefree. The class is the set of all $m n^2$ for $n = 1,2,3,4,5,6,7,...$
Notice that since the cardinality of equivalence classes is equal, it is enough to prove that the class containing $1$ has cardinality equal to that of $\mathbb{N}$. Now, $\forall\ x \in$ the equivalence class of $1$, then $\exists\ a \in \mathbb{Q}$ such that $x = a^{2}$.
$$\implies |\mathbb{R}_1| = |\{a^{2}: a \in\ \mathbb{Q}\}|$$ $$\implies |\mathbb{R}_1| = |\mathbb{Q}| = |\mathbb{N}|$$
Hence, proved.
Here's my approach for the relation on $\mathbb{N}$.
Claim: $\forall\ n \in \mathbb{N}, a \in \mathbb{Q}\ \land a^2 = n \implies a \in \mathbb{N}$
Proof: Let $\exists\ n \in \mathbb{N}$ such that $a \in \mathbb{Q}\ \land a \notin \mathbb{N}\ \land a^2 = n$
$\implies \exists\ p, q \in \mathbb{N}$ such that gcd($p, q$) = 1 & $n = \frac{p^2}{q^2} \implies n \notin \mathbb{N} \implies \impliedby$
$$\implies |\mathbb{N}_1| = |\{a^2: a\in \mathbb{N}\}| = |\mathbb{N}|$$
Hence, proved.