Listing equivalence relations class for $x\sim y \Leftrightarrow x^2=y^2$

Question

Can someone help me on the right track for my proof for the statement below. I started and got stuck but I need help. Please guide me to answer what this statement requires and how to word it out correctly.

For all $x,y\in\mathbb R$ define that $x\equiv y$ if $x^{2}=y^{2}.$ Then $\equiv$ is an equivalence relation on $\mathbb R$, there are infinitely many equivalence classes, one of them consists of one element and the rest consist of two elements.

True. Let $p=x,y,z\in\mathbb{R}$. For $x\equiv y$ if $x^{2}\equiv y^{2}$ on $\mathbb{R}.$ To be an equivalence relation on $\mathbb{R}$ we need to show that: $$\text{i).}\ \forall x\in\mathbb{R}; x\equiv x\ \text{such that}\ x^{2}=x^{2}$$ $$\text{ii).}\ \forall x,y\in\mathbb{R}\ \text{if}\ x\equiv y\ \text{then}\ y\equiv x\ \text{such that}\ x^{2}=y^{2}\ \text{but}\ y^{2}=x^{2}$$ $$\text{iii).}\ \forall x,y,z\in\mathbb{R}\ \text{if}\ x\equiv y\ \text{and}\ y\equiv z, \text{then}\ x\equiv z\ \text{such that}\ x^{2}=y^{2}\ \text{and}\ y^{2}=z^{2}\Rightarrow x^{2}=z^{2}$$

Therefore the equivalence class of $[0,0]=\{p=(x,y,z\in\mathbb{R}|p\equiv(0,0)\}=\{(0,0)\}$ then $p\equiv(0,0)\Leftrightarrow x^{2}=y^{2}$ which is equal to $0^{2}=0^{2}=0$ if and only if $x=y=0.$

Comment: I am stuck here I dont know where to continue on. Please help..thank you

Answer 1

To demonstrate that a given relationship, $\circ: \mathscr{R}\times \mathscr{R}$, is an equivalence relation you need to show that:

• Reflexivity holds: $\forall x\in \mathscr{R}: (x\circ x)$
• Symmetry holds: $\forall x,y\in \mathscr{R}: ((x\circ y)\iff(y\circ x))$
• Transitivity holds: $\forall x,y,z\in \mathscr{R}: ((x\circ y)\land(y\circ z)\implies(x\circ z))$

You have shown reflexivity: $$\begin{align}\because \forall x \in \mathscr{R} &: (x^2=x^2) \\ \therefore \forall x \in \mathscr{R} &: (x\equiv x)\end{align}$$

Yours to do: Repeat for the other properties.

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Next, construct equivalence classes, given that $x\equiv y \iff x^2=y^2$

$$\{[x]\mid \forall x \in\mathscr{R}\}=\{\{y\mid \forall y\in \mathscr{R}, y \equiv x\}\mid \forall x\in \mathscr{R}\}$$

Our hint is that $y^2=x^2 \implies y=\pm x$, so the equivalence classes are: $$\{[x]\mid \forall x \in\mathscr{R}\}=\{\{y\mid \forall y\in \mathscr{R}, y=\pm x\}\mid \forall x\in \mathscr{R}\}$$

Yours To Do: Show that this consists of 1 class with 1 member and uncountably many classes each with two members.