Does identity relation satisfies as an equivalence relation.

Question

Let $A = \{1,2\}$ be a set and $R = \{(1,1),(2,2)\}$ be a binary relation on $A$. Is $\mathcal{R}$ an equivalence relation on $A$?

Answer 1

Yes. It is

• symmetric, $ R=Transpose[R] $
• $ Id \subset R $
• transitive

Answer 2

Let $A = \{1,2\}$, and $R$ be a relation on $A$ defined as $R = \{(1, 1), (2, 2)\}$.

Reflexive: $R$ is reflexive iff $xRx$ for all $x \in A$.

Note that $(\forall x \in A): (x,x) \in R$. So $(\forall x \in A): xRx$. Therefore $R$ is reflexive.

Symmetric: $R$ is symmetric iff $xRy$ implies $yRx$ for all $x,y \in A$.

Since $(1,1)$ and $(2,2)$ are in $R$, then if we switch the digits in both ordered pairs, we see that $(1,1)$ and $(2,2)$ are also in $R$. Therefore $R$ is symmetric.

Transitive: $R$ is transitive iff $xRy$ and $yRz$ implies $xRz$ for all $x,y,z \in A$.

You can easily check that, since $(1,1) \in R$ and $(1,1) \in R$, then $(1,1) \in R$ (this is pretty obvious). The same goes for $(2,2)$. Therefore $R$ is transitive.

By definition, a relation is said to be an equivalence relation iff it is reflexive, symmetric and transitive. Since $R$ satisfies these three properties, we conclude that $R$ is an equivalence relation.