0

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.

user142943
  • 53
  • Hint: $(-1)^2=(1)^2 \implies -1\equiv 1$ – Graham Kemp May 24 '14 at 03:01
  • What did you try? What is the doubt? – Swapnil Tripathi May 24 '14 at 03:35
  • i don't know how to do this question – user153017 May 24 '14 at 03:37
  • Do you know the definition of an equivalence relation? – Nishant May 24 '14 at 03:38
  • Why do you use both $X$ and $x$? They're supposed to mean the same number, right? Font/case usually means that this is not the case. Also, the last line is confused. You found a single equivalence containing one number. Why would that say anything about the number of equivalance classes? Actually you cannot even begin to talk about equivalence classes before you have shown that $\equiv$ is an equivalence relation. – Jyrki Lahtonen May 24 '14 at 07:07

2 Answers2

2

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 $y = -x$.

marty cohen
  • 110,450
  • Is this answer correct.True. To show that \equiv is reflexive we need to show that \forall x\in\mathbb{R} :x=x. Let x\in\mathbb{R} , x\equiv\mbox{\ensuremath{x}} 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 .

     – user153017 May 24 '14 at 04:19
  • 1
    @user153017 Tip: You need to place $ symbols before and after MathJax text for it to display in human readable form. Also, yes, marty cohen's answer is correct. ${0}$ is the only equivalence class with a single member; the rest of the uncountably many equivalence classes each contain two members: a positive real and the corresponding negative real. – Graham Kemp May 24 '14 at 04:52
1

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: $$[a]=\{x\in X\mid f(x)=f(a)\}$$

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.

drhab
  • 153,781