Show that the relation

Solution:

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

Given that, $R$ be the relation in $N \times N$ defined by $(a, b) R(c, d)$ if $a+d=b+c$ for $(a, b),(c, d)$ in $N \times N$.

$\underline{R}$ is Reflexive if $(a, b) R(a, b)$ for $(a, b)$ in $N \times N$

Let $(a, b) R(a, b)$

$\Rightarrow a+b=b+a$

which is true since addition is commutative on $\mathrm{N}$.

$\Rightarrow \mathrm{R}$ is reflexive.

$R$ is Symmetric if $(a, b) R(c, d) \Rightarrow(c, d) R(a, b)$ for $(a, b),(c, d)$ in $N \times N$

Let $(a, b) R(c, d)$

$\Rightarrow \mathrm{a}+\mathrm{d}=\mathrm{b}+\mathrm{c}$

$\Rightarrow \mathrm{b}+\mathrm{c}=\mathrm{a}+\mathrm{d}$

$\Rightarrow \mathrm{c}+\mathrm{b}=\mathrm{d}+\mathrm{a}$ [since addition is commutative on $\mathrm{N}$ ]

$\Rightarrow(c, d) R(a, b)$

$\Rightarrow \mathrm{R}$ is symmetric.

$\underline{R}$ is Transitive if $(a, b) R(c, d)$ and $(c, d) R(e, f) \Rightarrow(a, b) R(e, f)$ for $(a, b),(c, d),(e, f)$ in $N \times N$

Let $(a, b) R(c, d)$ and $(c, d) R(e, f)$

$\Rightarrow \mathrm{a}+\mathrm{d}=\mathrm{b}+\mathrm{c}$ and $\mathrm{c}+\mathrm{f}=\mathrm{d}+\mathrm{e}$

$\Rightarrow(\mathrm{a}+\mathrm{d})-(\mathrm{d}+\mathrm{e})=(\mathrm{b}+\mathrm{c})-(\mathrm{c}+\mathrm{f})$

$\Rightarrow \mathrm{a}-\mathrm{e}=\mathrm{b}-\mathrm{f}$

$\Rightarrow \mathrm{a}+\mathrm{f}=\mathrm{b}+\mathrm{e}$

$\Rightarrow(\mathrm{a}, \mathrm{b}) \mathrm{R}(\mathrm{e}, \mathrm{f})$

$\Rightarrow \mathrm{R}$ is transitive.

Hence, $R$ is an equivalence relation.