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Here is my problem. The equivalence class of say $5$, denoted $[5]$, is a subset of $A$ containing the elements $\left\{1,5\right\}$. And $[3]$, for example, is a subset containing the elements $\left\{1,3\right\}$. In fact every equivalence class of an element in set A will contain "$1$" as an element, so none of the subsets created by these equivalence classes are disjoint, which is a requirement for a partition.

However, I do believe that the relation $R$ meets the requirements for an equivalence relation, meaning that it is reflexive, symmetric, and transitive. And since every equivalence relation determines a partition, the equivalence classes of R should create a partition. But seeing as none of the equivalence relations are disjoint, this does not appear to be the case.

Where has my reasoning gone wrong?

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  • $\begingroup$ Equivalence relations must be reflexive, symmetric, and transitive. $\endgroup$ – Graham Kemp Oct 19 '16 at 23:42
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Definition of equivalence relations should include symmetry and exclude antisymmetric.

$1$ divides $2$ but $2$ doesn't divide $1$. Hence it is not an equivalence relation.

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Equivalence relations must be reflexive, symmetric, and transitive.

It's the lack of symmetry (and the antisymmetry) which means $1$ divides every other element but none of them divide $1$, and so forth, thus leading you to have non-disjoint classes.

${[5]}_{\rm R}=\{1,5\}, {[10]}_{\rm R}=\{1,5,10\}$ et cetera.

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