Describe equivalence classes from equivalence relations

Question

I don't really understand the way to do these.

Describe equivalence classes for the following equivalence relations on the given set $S$:

(i) $S$ is the set of all points in the plane, and $a\sim b$ means $a$ and $b$ have the same distance from the origin.

(ii) $S = \mathbb{N}$, and $a\sim b$ if and only if $ab$ is a square.

(iii) $S = \mathbb{R}$, and $a\sim b$ if and only if $a = b$.

Answer 1

This means describe the subsets $S_a=\{x\in S:x\sim a\}$ given the relation $\sim$ on $S$ in each case.

(i) If $a$ and $b$ have the same distance from the origin, this means they lie on the same circle centered at the origin. So choose an $a$, then $S_a$ is the set of all points on the same circle centered at the origin as $a$; in other words, the equivalence classes are just circles centered at the origin.

(ii) If the product $ab$ is a square, we can write $ab=c^2$ where $c$ is also a natural number. So the equivalence class $S_a$ is the set $\{n^2/a, n\in \Bbb N:a|n^2\}$ of all perfect squares evenly divided by $a$, divided by $a$. (A little confusing, but basically take $a$, find all the perfect squares which $a$ evenly divides, and divide them by $a$; that collection is $S_a$.)

(iii) This is the easiest one! Choose $a$, the set of all reals which are equal to $a$ is just $\{a\}$, so the equivalence classes are just sets which contain exactly one real number. You could consider the equivalence classes just as the real numbers themselves.

Answer 2

(i) Hint: Given a distance $r$, what is the set of all points in the place at distance $r$ from the origin?

(ii) Every integer $n$ has a unique decomposition as a product of powers of distinct primes: $$n = \prod_{p\text{ is prime}}p^{n_p}.$$ For integers $a,b$, $ab$ is a perfect square $\iff$ for every prime $p$, the exponent $n_p$ of $p$ is divisible by 2 — that is, if $n_p$ is even. With this representation, $$\begin{align} ab &= \prod_{p\text{ is prime}}p^{a_p}\prod_{p\text{ is prime}}p^{a_p} \\ &= \prod_{p\text{ is prime}}p^{a_p+b_p}. \end{align}$$ Thus, $ab$ is a perfect square if for every prime p, $a_p + b_p$, the exponent of $p$ in the decomposition of $ab$, is even — that is, if $a_p$ and $b_p$ are either both even and both odd.

It's not hard to see that if $a\sim c^2b$ then $a\sim b$. Let $e(k) = 1$ if $k$ is odd, $0$ if $k$ is even. Then for every $a$, $$ a = \prod_{p\text{ is prime}}p^{a_p} = \prod_{p\text{ is prime}}p^{e(a_p)}, $$ where the rightmost product is square-free — not divisible by any square (other than $1$). Square-free integers are just products of distinct primes, where all the exponents of primes are 1. So, for all $a$, there is a square-free $b$ such that $a\sim b$. It's also easy to see that if $a, b$ are square-free and $a\ne b$, then $a\not\sim b$. Finally, if $a\sim b$ then $a\sim c^2 b$. So the equivalence classes are the sets $$ \{n^2 a\mid n\in\Bbb N\}\text{, for each $a$ square-free}. $$

(iii) Exercise.