Reflexive and relation

Question

Please give me feedback on my answer to this question. Question: For all $x,y\in R$ define that $x\equiv y$ if $x^{2}=y^{2}$. Then $\equiv$ is an equivalence relation on $R$, there are infinitely many equivalence classes, one of them consists of one element and the rest consist of two elements. Answer: False, since;

$x\equiv y$ if $x^{2}=y^{2}$ on $R$

To show $\equiv$ is reflexive, we need to show that ,

$\forall x\in R:X\equiv x.$

Let $x\in R$, $X\equiv x$ if $X^{2}=x^{2}$ , which is obvious.

$[x]\triangleq:\{y\in R/x\equiv y\}$

$[0]=[y\in R/0\equiv y\},$ when $y=0$,

Then $y^{2}=0^{2}=0$

Thus; there are not many infinity many equivalence classes.

Answer 1

This is essentially the same as an earlier problem I answered that I can't find.

The "trick" is that $x^2 = y^2$ has two $y$'s for any $x$ (except for $x=0$): $y = x$ and $y = -x$.

Answer 2

General (to remember)

Whenever $f:X\rightarrow Y$ is a function the relation on $X$ defined by: $a\equiv b$ if $f(a)=f(b)$ is an equivalence relation. This is not difficult to check:

$f(a)=f(a)$ (reflexivity)

$f(a)=f(b)\Rightarrow f(b)=f(a)$ (symmetry)

$f(a)=f(b)\wedge f(b)=f(c)\Rightarrow f(a)=f(c)$ (transitivity)

The equivalence classes are exactly the fibers of $f$, i.e. the sets: $$[a]=\{x\in X\mid f(x)=f(a)\}$$

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Application

You are dealing with $f:\mathbb R\rightarrow \mathbb R$ prescribed by $x\mapsto x^2$ and can apply this right away.

For every $a\in\mathbb R$ we find equivalenceclass: $[a]=\{x\in\mathbb R\mid x^2=a^2\}=\{a,-a\}$

Evidently $[0]=\{0\}$ so has exactly one element. This in contrast with the other classes that all contain exactly two elements.

On base of $0\le a<b\Rightarrow[a]\ne[b]$ we conclude that there are infinitely many classes.