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The relation $R$ on the set $A=\{2,4,6,8,10\}$ is defined by $$R=\{(2,2),(2,6),(2,10),(4,4),(4,8),(6,2),(6,6),(6,10),(8,4),(8,8),(10,2),(10,6),(10,10)\}$$

Question 1

Verify if $R$ is an equivalence relation on $A$.

Answer:

Reflexive, because $(2,2),(4,4)$

Symmetric, because $(2,6),(6,2)\in A$

Transitive, because $(2,6),(6,2)$ is $(2,2)$

Yes, it is equivalence relation

Question 2

Find the corresponding partition, $A/R$.

What does it mean by corresponding partition?

Sorry I can't get the idea to start this.

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    $\begingroup$ On question 1: with this kind of "proofs" it can be shown that any real number is rational: "$1\in\mathbb R$ and $\frac{2}{3}\in\mathbb R$ are rational so..." $\endgroup$ – drhab Jan 8 '15 at 10:00
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Your proof of the first part is lacking as well. You only show that there is an example for the transitivity. You don't show that for every two pairs $(a,b)$ and $(b,c)$ also $(a,c)\in R$. Similarly for reflexivity and symmetry.

So you did not, in fact, verify that this is an equivalence relation.

As for the second part, given an equivalence relation, the corresponding partition, also sometimes called "the set of equivalence classes" is the set $A/R$ whose elements are subsets of $A$ satisfying:

  1. For every $a\in A$ there is some $B\in A/R$ such that $a\in B$.
  2. For every $a,b\in A$, and $B\in A/R$, $a\in B$ and $b\in B$ if and only if $(a,b)\in R$.

Namely, each set in $A/R$ is composed of elements equivalent in $R$, and if $B$ is such set and $a\in B$, then all the elements in $B$ are equivalent in $R$ to $a$.

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  • $\begingroup$ thanks, can i know what is the way to verify it? $\endgroup$ – Bilis Jan 8 '15 at 11:50
  • $\begingroup$ With these sort of relations? Go through the relation by hand and verify. Of course you can "cut corners", for example if $(2,6)$ and $(6,2)$ are in $R$, there's no need to check that $(6,2)$ and $(2,6)$ are in $R$, etc. $\endgroup$ – Asaf Karagila Jan 8 '15 at 11:55
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Hint on finding the partition.

Pick out an element $a\in A$ and define: $[a]=\{x\in A\mid \langle a,x\rangle\in R\}$. E.g. for $a=2$ you will find: $[2]=\{2,6,10\}$. Note here that $[2]=[6]=[10]$. The collection of distinct sets $[a]$ form a partition of $A$ and: $$A/R=\{[a]\mid a\in A\}$$

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