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Is a relation that is pruely reflexive also symmetric?

For example, say you have a relation defined as $R = \{(a,a),(b,b)\}$. This is purely reflexive, but is it also symmetric? The typical symmetric definition is $aRb \Leftrightarrow bRa$, which is kinda shown in this as $aRa \Leftrightarrow aRa$, but I am unsure. Sorry if this is a trivial question, I am just learning about this stuff in a proof course and am slightly confused.

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  • $\begingroup$ Yes, it’s symmetric, because it is certainly true that $a\,R\,a\leftrightarrow a\,R\,a$. $\endgroup$ – Brian M. Scott Jul 26 '20 at 2:10
  • $\begingroup$ Yes. You need to show that, if $x \, R \, y$, then $y \, R \, x$ to show symmetry. In such relations, if $x \, R \, y$, then we must have $x = y$, and so $x \, R \, y$ means the same as $x \, R \, x, y \, R \, x,$ and $y \, R \, y$. $\endgroup$ – user810049 Jul 26 '20 at 2:29
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I've seen a useful method for getting a more intuitive understanding of some of these properties of relations, namely using a table to indicate a relation and then visually deduce the property. In the case you describe one obtains:

enter image description here

For example, if a relation is reflexive, the diagonal elements will all be populated with 1's. Note that a "1" indicates that the corresponding row entry and column entry is in the relation, and a "0" means that combination in not in the relation. For example, $\left(a,a\right) \in R$, but $\left(a,b\right) \notin R$, etc. in the above table.

If the relation is symmetric, then the table will identical if you reflect it about the diagonal. Therefore, for the relation you have described in the question, this is clearly true.

Here is another question in Stack Exchange that illustrates this idea further: Checking the binary relations, symmetric, antisymmetric and etc

I hope this helps.

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  • $\begingroup$ That's wonderful! Thank you! $\endgroup$ – Paul J Jul 26 '20 at 2:35

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