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Given a binary relation $R$ over set $A$ ,prove the following statement:

$R$ is symmetric if and only if it is equal to its converse.

$\implies$

$R$ symmetric iff $\forall a,b \in A$:

$$(a,b) \in R \iff (b,a) \in R$$

But how to show that it is equal to its converse?

$\Longleftarrow$

$R$ is equal to its converse iff $$R=R^T \iff \left\{\left(a,b\right)\mid aRb\right\}=\left\{\left(b,a\right)\mid aRb\right\} \iff {\left(a,b\right)}={\left(b,a\right)}$$

How does this imply the relation is symmetric?

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    That final equivalence makes no sense. – José Carlos Santos Jul 31 '20 at 07:59
  • What are $a$ and $b$ at the RHS of$${(a,b)\mid a\mathrel Rb}={(b,a)\mid a\mathrel Rb}\iff(a,b)=(b,a)?$$ – José Carlos Santos Jul 31 '20 at 08:40
  • It still makes no sense. – José Carlos Santos Jul 31 '20 at 08:41
  • For every $(a,b) \in R$, by definition of the converse, we have $(b,a) \in R^T$. But $R=R^T$ implies that $(b,a) \in R$ as well. This means, for every $(a,b) \in R$, we will have $(b,a) \in R$. Thus $R$ is symmetric. NOTE: your conclusion about $(a,b)=(b,a)$ is definitely not true. – Anurag A Jul 31 '20 at 08:43
  • Are you saying that those assertions are equivalent for every $a$ and every $b$? That is false. Are you saying that they are equivalent for some $a$ and some $b$? That is false too. Or are you saying something else? Anyway, it makes no sense because the equality $(a,b)=(b,a)$ alone is neither true nor false, and therefore it cannot possibly be equivalent to an assertion which is either true or false. – José Carlos Santos Jul 31 '20 at 08:46

1 Answers1

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Look at the very definition ov converse relation:

$$(x,y)\in R^T\iff (y,x)\in R$$

So

$$R\;\;\text{is symmetric}\;\iff \left[(a,b)\in R\iff(b,a)\in R\right]\iff R=R^T\;$$

Another, slower, way: 1) Suppose $\;R\;$ is a symmetric relation and let $\;(a,b) \in R\;$, then also

$\;(b,a)\in R\implies (a,b)\in R^;$, and skipping intermediate steps, we actually got here$\;(a,b)\in R\implies (a,b)\in R^T\implies \color{red}{R\subset R^T}\;$ .

But also $\;(x,y)\in R^T\implies (y,x)\in R\;$, and since $\;R\;$ is symmentric then also $\;(x,y)\in R\;$ , and skipping intermediate steps we got

$$(x,y)\in R^T\implies (x,y)\in R\implies \color{red}{R^T\subset R}$$

Both opposite inclusions in red above yield $\;R=R^T\;$

(2) Suppose now $\;R=R^T\;$, and let $\;(a,b)\in R\;$. then also $\;(a,b)\in R^T\;$ and from here $\;(b,a)\in R\;$. Hoping over intermediate steps we actually got here that $\;(a,b)\in R\implies (b,a)\in R\implies R\;$ is symmetric.

And we're done with both directions...

DonAntonio
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  • @45465Because of the definition of $;R^T;$, which I wrote in the second line...in fact, to write the condition for symmetry is exactly the same as writing the definition for $;R=R^T;$ ...! – DonAntonio Jul 31 '20 at 08:18
  • @45465 I added a new proof for you, more specific and lengthy but, hopefully, clearer, too. – DonAntonio Jul 31 '20 at 13:24
  • Well, yes...that's what I wrote in the first part of my answer... – DonAntonio Aug 06 '20 at 11:11