# Symmetric Relations

Symmetric relations : A relation R on a set A is said to be a symmetric-relations if and if only**(a,b) $\in$ R $\Rightarrow $ (b,a) $\in$ R for all a, b $\in$ A**

aRb $\Rightarrow $ bRa for all a,b $\in$ A.

aRb $\Rightarrow $ bRa for all a,b $\in$ A.

For example, The identity and the universal relations on a non-void set are symmetric-relations.

## Examples on Symmetric Relations

**Example 1 :**Let A = {1,2,3,4} and let R1 be relations, R1= {(1,3),(1,4)(3,1),(2,2)(4,1)} and R2 be relations, R2={(1,1),(3,3)(3,1),(2,2)}

**Solution :**As we know that,

(a,b) $\in$ R $\Rightarrow $ (b,a) $\in$ R for all a, b $\in$ A

So R1 is symmetric-relation on set A.

However, R2 is not a symmetric-relations on set A

because (3,1) $\notin$ R2.

**Example 2 :**Prove that a relation R on a set A is symmetric if and only if R = R$^{-1}$

**Solution :**Let R be a symmetric-relation on set A. Then we have to prove that R = R$^{-1}$ .

In order to prove this we have to prove that R $\subseteq$ R$^{-1}$ and R$^{-1}$$\subseteq$ R.

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

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

**[ Since R is symmetric]**

$\Rightarrow$ (a, b) $\in$ R$^{-1}$

**[ By definition of inverse relation]**

Thus, (a,b)$\in$ R $\Rightarrow$ (a,b) $\in$ R$^{-1}$ for all a, b $\in$ A

So, R $\subseteq$ R$^{-1}$ --------------(1)

Now let (x,y) be an arbitary element of R$^{-1}$, then

(x,y) $\in$ R$^{-1}$

$\Rightarrow$ (x,y) $\in$ R$^{-1}$

**[ Since R is symmetric]**

$\Rightarrow$ (y, x) $\in$ R

**[ By definition of inverse relation]**

Thus, (x,y)$\in$ R$^{-1}$ $\Rightarrow$ (x,y) $\in$ R for all x, y $\in$ A

So, R$^{-1}$ $\subseteq$ R --------------(2)

From (1) and (2) we get,

R $\subseteq$ R$^{-1}$

**Converse is also true, which is R be a relation on set A such that**

R = R$^{-1}$, then R is a symmetric relation on set A.

R = R$^{-1}$, then R is a symmetric relation on set A.

**Example 3 :**Suppose R, S are relations on a set A. If R, S are both reflexive, then R$\cap$ S is reflexive.

**Solution :**Suppose a $\in$ A. Since R, S are both reflexive on A, (a, a) $\in$ R and (a, a) $\in$ S. Since (a, a) is in both R and S, (a, a) $\in$ R$\cap$ S, so R$\cap$ S is reflexive.

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