Show that the relation $\mathrm{R}$ on $\mathrm{N} \times \mathrm{N}$, defined by $(a, b) R(c, d) \Leftrightarrow a+d=b+c$ is an equivalent relation.

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Show that the relation $\mathrm{R}$ on $\mathrm{N} \times \mathrm{N}$ , defined by $(a, b) R(c, d) \Leftrightarrow a+d=b+c$ is an equivalent relation.

Show that the relation $\mathrm{R}$ on $\mathrm{N} \times \mathrm{N}$ , defined by $(a, b) R(c, d) \Leftrightarrow a+d=b+c$ is an equivalent relation.

Solution:

If $R$ is Reflexive, Symmetric and Transitive, then $R$ is an equivalence relation.

Reflexivity:
Suppose $a$ and $\mathrm{b}$ be an arbitrary element of $\mathrm{N} \times \mathrm{N}$
$\begin{array}{l} \Rightarrow(\mathrm{a}, \mathrm{b}) \in \mathrm{N} \times \mathrm{N} \\ \Rightarrow(\mathrm{a}, \mathrm{b}) \in \mathrm{N} \\ \Rightarrow \mathrm{a}+\mathrm{b}=\mathrm{b}+\mathrm{a} \\ \Rightarrow(\mathrm{a}, \mathrm{b}) \mathrm{R}(\mathrm{b}, \mathrm{a}) \end{array}$
Therefore, $\mathrm{R}$ is reflexive.

Symmetric:
Suppose $(a, b)$ and $(c, d) \in N \times N$ such that $(a, b) R(c, d)$
$\begin{array}{l} \Rightarrow \mathrm{a}+\mathrm{d}=\mathrm{b}+\mathrm{c} \\ \Rightarrow \mathrm{b}+\mathrm{c}=\mathrm{a}+\mathrm{d} \\ \Rightarrow \mathrm{c}+\mathrm{b}=\mathrm{d}+\mathrm{a} \\ \Rightarrow(\mathrm{c}, \mathrm{d}) \mathrm{R}(\mathrm{a}, \mathrm{b}) \end{array}$
Therefore, $\mathrm{R}$ is symmetric.

Transitivity:
Suppose $(a, b),(c, d),(e, f) \in N \times N$ such that $(a, b) R(c, d)$ and $(c, d) R$ $(\mathrm{e}, \mathrm{f})$
$\Rightarrow \mathrm{a}+\mathrm{d}=\mathrm{b}+\mathrm{c}$ and $\mathrm{c}+\mathrm{f}=\mathrm{d}+\mathrm{e}$
On adding both the equations we obtain
$\begin{array}{l} \Rightarrow a+d+c+f=b+c+d+e \\ \Rightarrow a+f=b+e \\ \Rightarrow(a, b) R(e, f) \end{array}$
$\Rightarrow \mathrm{R}$ is transitive.
As a result, $\mathrm{R}$ is an equivalence relation.