Showing a relation is an equivalence relation on $\mathbb {N}$ or $\mathbb {R}$

Question

I have an exercise that consists of 2 parts that I don't really know how to prove.

Consider the relation $\mathcal R$ on $\mathbb {N}$ defined as follows: for all $a, b$ ∈ $\mathbb {N}$, a$\mathcal R$b if there exist $m, n$ ∈ $\mathbb {N}$\{0} such that $am^2$ = $bn^2$ . Show that $\mathcal R$ is an equivalence relation. Show that the set $\mathbb {N}$/$\mathcal R$ is infinite.

Consider the relation $\mathcal Q$ on $\mathbb {R}$ defined as follows: for all $x, y$ ∈ $\mathbb {R}$, x$\mathcal Q$y if there exist $r$ ∈ $\mathbb {R}$\{0} such that $x = yr^2$ . Show that $\mathcal Q$ is an equivalence relation. How many elements are there in $\mathbb {R}$/$\mathcal Q$?

So I've been reading my script and I know that an equivalence relation is a relation that is reflexive, symmetric and transitive. For the first task, it seems somewhat intuitive that this is an equivalence relation. But I'm struggling with proving this mathematically correct.

Answer 1

Note that $a \cdot 1^2 = a \cdot 1^2$, so $a \mathcal R a$.

If $a \mathcal R b$, then $\exists m, n \in \Bbb N$ such that $am^2=bn^2$, so $bn^2 = am^2$ and $b \mathcal R a$.

If $a \mathcal R b, b \mathcal R c$, then $\exists m, n, k, t \in \Bbb N$ such that $am^2=bn^2, bk^2=ct^2$, so $a(mk)^2=b(nk)^2 = c(nt)^2$ and $a \mathcal R c$.

Note that if $p, q$ are distinct primes, $pm^2 = qn^2$ is not possible (because an odd power of $p$ divides the left-hand side but an even power of $p$ divides the right-hand side), so $\mathcal R$ has infinitely many different equivalence classes.

A similar proof shows that $\mathcal Q$ is an equivalence relation on $\Bbb R$. Any positive number is a square in $\Bbb R$, so $x \mathcal Q y \iff x \text{ and } y \text{ have the same sign}$ and $\mathcal Q$ has three equivalence classes.