0

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

Alexander Gruber
  • 28,037
user130602
  • 89
  • http://math.stackexchange.com/questions/807330/reflexive-and-relation – Graham Kemp May 24 '14 at 04:14
  • Hint: $(-x)^2=(x)^2 \implies -x\equiv x$ – Graham Kemp May 24 '14 at 04:20
  • if I can say x=-1,1, do I have to go through the same as I have for 0,0 and my proof will be completed?? – user130602 May 24 '14 at 04:23
  • An equivalence class should be a subset of $\mathbb{R}$. So it's $[0]$, the equivalence class of $0$ (all the numbers that are equivalent to 0), and $[1]$, the equivalent class of $1$ (all the numbers that are equivalent to 1). – Braindead May 24 '14 at 04:29
  • Also, the wording of your statements are weird. For instance you want to show that $\forall x\in \mathbb{R}$, you have $ x \equiv x$. So write a sentence (or many) explaining why $x \equiv x$ is true for any $x\in\mathbb{R}$. Then, write a sentence (or many) explaining why does assuming $x\equiv y$ allow you to conclude $y\equiv x$. Proof is about explaining why something is true. Your current format doesn't really do that. – Braindead May 24 '14 at 04:38
  • I am trying to show that this is an equivalence relation if it satisfies the three points I have on my proof. How can I write a short description on my proof to show that the statement is an equivalence relations before I am actually showing the truth about my proof. Please can you write me my starting statement. Thank you so much. – user130602 May 24 '14 at 05:11
  • It helps if you state what the three points are (ie: name them and give their general expression) before trying to demonstrate that they hold for your specific relation. – Graham Kemp May 24 '14 at 06:02

1 Answers1

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.

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.

Graham Kemp
  • 133,231