Let S be the set of all real numbers and let

Solution:

In order to show $R$ is an equivalence relation we need to show $R$ is Reflexive, Symmetric and Transitive.

Given that, $\forall a, b \in S, R=\{(a, b): a=\pm b\}$

Now,

$\underline{R}$ is Reflexive if $(a, a) \in \underline{R} \underline{\forall} \underline{a} \in \underline{S}$

For any $a \in S$, we have

$a=\pm a$

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

Thus, $R$ is reflexive.

$\underline{R}$ is Symmetric if $(a, b) \in \underline{R} \Rightarrow(b, a) \in \underline{R} \forall \underline{a}, b \in \underline{S}$

$(a, b) \in R$

$\Rightarrow a=\pm b$

$\Rightarrow b=\pm a$

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

Thus, $R$ is symmetric.

$\underline{R}$ is Transitive if $(a, b) \in \underline{R}$ and $(b, c) \in \underline{R} \Rightarrow(a, c) \in \underline{R} \underline{\forall} a, b, c \in \underline{S}$

Let $(a, b) \in R$ and $(b, c) \in R \forall a, b, c \in S$

$\Rightarrow \mathrm{a}=\pm \mathrm{b}$ and $\mathrm{b}=\pm \mathrm{c}$

$\Rightarrow \mathrm{a}=\pm \mathrm{c}$

$\Rightarrow(\mathrm{a}, \mathrm{c}) \in \mathrm{R}$

Thus, $R$ is transitive.

Hence, $R$ is an equivalence relation.