# Discrete Math - Relations Proofs.

Let $R$ be a relation on a group $A$, how to prove that if $R$ is symmetric and transitive relation then $R$ is also reflexive relation?

attempt:

$R$ is symmetric relation, so for all $x,y ∈ A$ if $xRy$ then $yRx$.

$R$ is transitive relation, so for all $x,y,z ∈ A$ if $xRy$ and also $yRz$ then $xRz$.

By the definitions we get that:

for all $x,y,z ∈ A$ if $xRy$ and also $yRz$ then $yRx$ , $xRz$ , $zRy$ , $zRx$.

I have no idea how to continue from here to prove that $R$ is also reflexive relation, i.e all $x ∈ A$ are in the relation $R$: $xRx$.

I tried this:

Because $xRy$ and $yRx$, (as we saw above by definitions), I thougt that $xRx$, because $y$ exists in both terms. but its not like that because $xR∘Rx$, i.e $(x,x) ∈ R∘R$.

Can someone help with an idea?

Thanks!!

Modulo @Alan Wang's objection, $x R y$ and $y R x$ because it is symmetric. Also apply the transitive property on the above. You will get $x R x$.

Your question is not true in general.
Let $R$ be an empty relation.
Then clearly $R$ is symmetric and transitive but not reflexive.