# Thread: Prove or disprove that the complement of a relation is symmetric

1. ## Prove or disprove that the complement of a relation is symmetric

Consider a relation $R$ on a set $A$. Suppose that you are told that $R$ is symmetric. Prove or disprove that $\overline{R}$ is also symmetric.

I only really know this much:
$R\subset A\times A$ and $\overline{R}\subset A\times A-R$

2. In fact, $\bar{R}$ is not only a subset, but is equal to $(A\times A)-R$ (by definition).

To get a better intuition, consider an example, such as $xRy$ iff $x-y$ is even ( $x,y$ are integers), or $A\,R\,B$ iff $A\cap B\ne\emptyset$ ( $A,B$ are sets). Take a pair $(x,y)$ from $\bar{R}$ and see if $(y,x)\in\bar{R}$. Do this several times and then try to have a different outcome for whether $(y,x)\in\bar{R}$.

3. Originally Posted by emakarov
In fact, $\bar{R}$ is not only a subset, but is equal to $(A\times A)-R$ (by definition).

To get a better intuition, consider an example, such as $xRy$ iff $x-y$ is even ( $x,y$ are integers), or $A\,R\,B$ iff $A\cap B\ne\emptyset$ ( $A,B$ are sets). Take a pair $(x,y)$ from $\bar{R}$ and see if $(y,x)\in\bar{R}$. Do this several times and then try to have a different outcome for whether $(y,x)\in\bar{R}$.
I'm afraid that didn't really help me. Would it be possible for you to show me?

4. If $(x,y)\in \overline{R}$ then $(x,y)\notin R$.
But $R$ is symmetric so $(y,x)\notin R$.
That means that $(y,x)\in \overline{R}$. DONE.