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?
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...