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.

  • $\begingroup$ Reflexivity and symmetry are straightforward. Prove transitivity. $\endgroup$
    – Berci
    Sep 29, 2020 at 9:45
  • $\begingroup$ @Berci I thought symmetry and reflexivity were straightforward. I've seen the proofs for some examples. I just never know if it's "enough" sometimes since I don't think I'm really proving anything most of the time, just rearranging some variables. $\endgroup$
    – 23408924
    Sep 29, 2020 at 9:47
  • 1
    $\begingroup$ Yes, often rearranging the variables and exchanging the two sides of an equation does give a valid proof. $\endgroup$
    – Berci
    Sep 29, 2020 at 9:49

1 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.

  • $\begingroup$ Thanks for the answer! That helped a lot! Just a quick question about the second exercise. What do you mean with Q having three equivalence classes? $\endgroup$
    – 23408924
    Sep 29, 2020 at 12:39
  • $\begingroup$ Note that $e \mathcal Q \pi$ because $e(\sqrt{\frac {\pi}{e}})^2=\pi$. Does that help you see when two real numbers are $\mathcal Q$-equivalent to one another? $\endgroup$ Sep 30, 2020 at 0:05
  • $\begingroup$ So since they cancel out , this works for any real number in $\mathbb {R}$ $\endgroup$
    – 23408924
    Sep 30, 2020 at 8:45
  • $\begingroup$ @23408924 Try it for $e$ and $- \pi$. $\endgroup$ Sep 30, 2020 at 9:13

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