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.
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.