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

Solution

True. To show that $\equiv$ is reflexive we need to show that $\forall x\in\mathbb{R} :x\equiv x$. Let $x\in\mathbb{R} , x\equiv x \mbox{ if } x^{2}=x^{2}$, which is obvious.

$[x] ={y\in\mathbb{R} |x\equiv y} =[0]={y\in\mathbb{R}|0\equiv y}={0}$. Hence $y^{2} =0^{2} =0$ which implies that $y=0$.

Can anyone please give me correct answer to this question.

Dennis Gulko
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user153017
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  • I edited your question a bit so that it starts being readable. You must wrap latex command with dollar signs (e.g., type \equiv between dollar signs to get $\equiv$). When you post a question here, you must make sure you can easily read it, so at the very least all latex command should render properly. If you can't do that, then nobody else would like to try to help you. Make and effort, and you'll see good results. – Ittay Weiss May 25 '14 at 09:44

2 Answers2

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Proof of symmetry and transitivity:

Symmetry:
Let $x,y\in\Bbb{R}$ with $x\equiv y$, so we have $x²=y²$ and trivially we get $y²=x²$, hence $y\equiv x$.

Transitivity:
Let $x,y,z\in\Bbb{R}$ with $x\equiv y$ and $y\equiv z$. We now show that $x\equiv z$: Since $x²=y²$ and $y²=z²$ we get by transitivity of equality $x²=z²$ and therefore $x\equiv z$.

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$$x\equiv y \iff x=y \lor x=-y$$ Classes are $\{0\}$ and $\{-x, x\}$ for all $x>0$.