1
$\begingroup$

The relation $P$ on $ℝ$ is defined by $xPy$ iff $x^2=y^2.$

(a) Prove that the the relation $P$ is an equivalence relation.

(b) Describe the equivalence class of $3$ and $0$.

In order to solve this proof one must first understand what an equivalence relation is which means that a relation must be reflexive, symmetric and transitive. In order to do part (a) does one have to define what reflexive symmetric and transitive is in order to receive the right result? Does one prove this through example?

For part (b), I know that an equivalence class, the set of equivalence classes for the set relation $\Rightarrow $ m is denoted $Z_m$. which is (by substituting) we get, $3P-3$ and $-3P3$ so $[3] \Rightarrow $ $\{-3,3\}$ and $[0] = \{0\}$.

$\endgroup$
4
  • 1
    $\begingroup$ For (a), yes, it's necessary to show that $P$ is reflexive, symmetric, and transitive and no, you need to show this holds in general, so examples, while helpful for understanding, are not sufficient to qualify as proofs. $\endgroup$
    – Rick Decker
    Commented Jun 2, 2016 at 0:17
  • $\begingroup$ What would be a good tag to classify this as? $\endgroup$
    – Jon
    Commented Jun 2, 2016 at 0:27
  • 1
    $\begingroup$ We have a whole tag devoted to equivalence-relations . That seems like a good one for this. $\endgroup$
    – Graham Kemp
    Commented Jun 2, 2016 at 0:35
  • $\begingroup$ It is necessary to KNOW what the conditions reflexive, symmetric, and transitive mean. For the proof, show, one at a time, that the relation satisfies each of these 3 conditions. Always work from the definitions. Otherwise you cannot begin. $\endgroup$
    – DanielWainfleet
    Commented Jun 2, 2016 at 1:17

1 Answer 1

Reset to default
1
$\begingroup$

Yes, for part (a) to show that $P$ is an equivalence relation on $\mathbb{R}$ you need to check that it is reflexive, symmetric and transitive. Here is how you get started.

Reflexivity: Take $a\in \mathbb{R}$ and clearly $aPa$ holds, since $a^2=a^2$ is true for all real numbers $a$. Hence $P$ is reflexive.

Symmetry: Take $a,b\in \mathbb{R}$. Now if you know that $aPb$, does that mean $bPa$? If you can show that, then $P$ is symmetric.

Transitivity: Take $a,b,c \in \mathbb{R}$ and assume that $aPb$ and $bPc$ are both true. Can you show that also $aPc$ is true?

Your answer for part (b) is correct. You are asked to describe the equivalence class of the values $3$ and $0$ in $\mathbb{R}$. Basically, which values $a\in \mathbb{R}$ satisfy $aP3$, i.e. $a^2=3^2$? The resulting set $\{-3,3\}$ is your equivalence class for $3$. Similarly which values $b\in \mathbb{R}$ satisfy $bP0$, i.e. $0^2=b^2$? The resulting set $\{0\}$ is your equivalence class for $0$.

$\endgroup$
3
  • $\begingroup$ So then R is symmetric iff for all x and y∈A if $xRy$ then $yRx.$ R is transitive iff for all x, y and z ∈ A if $xRy$ and $yRz$ then $xRz.$ What I wonder is what I need to include in order to finish it? $\endgroup$
    – Jon
    Commented Jun 2, 2016 at 0:48
  • 1
    $\begingroup$ @LittleJon Just note what $xRy$ means in this case and substitute into your definitions. Then symmetry would mean the statement, $\forall x\in\Bbb R~\forall y\in\Bbb R~((x^2=y^2) \to (y^2=x^2))$, is true. Is it? $\endgroup$
    – Graham Kemp
    Commented Jun 2, 2016 at 0:53
  • 1
    $\begingroup$ @LittleJon And to check transitivity you need to show that: if $a^2=b^2$ and $b^2=c^2$, then $a^2=c^2$. Thats the exact same as showing: if $aPb$ and $bPc$, then $aPc$. $\endgroup$
    – M47145
    Commented Jun 2, 2016 at 1:50

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .