Show that the relation R on N × N, defined by (a, b) R (c, d) such that a + d = b + c is an equivalent relation.

Given:

$\text{R}: \mathbb{N}\times \mathbb{N}\rightarrow \mathbb{N}\times \mathbb{N}$, defined by

$(a, b)\; \text{R}\; (c, d)$ such that $a + d = b + c$

where $(a,b), (c,d) \in \mathbb{N}\times \mathbb{N}$

To Prove:

$\text{R}$ is an equivalence relation i.e.

(i) Reflexive

$\hspace{0.5cm}\boxed{x\;\text{R}\;x}\hspace{0.3cm}\forall\; x$

(ii) Symmetric

$\hspace{0.5cm}\boxed{x\;\text{R}\;y}\;\Rightarrow\; \boxed{y\;\text{R}\;x}\hspace{0.3cm}\forall\; x, y$

(iii) Transitive

$\hspace{0.5cm}\boxed{x\;\text{R}\;y} \;\text{and}\;\boxed{y\;\text{R}\;z}\; \Rightarrow\; \boxed{x\;\text{R}\;z}\hspace{0.3cm}\forall\; x, y, z$

Proof:

(i) Condition for Reflexive Relation:

$\hspace{0.6cm}(a,b)\;\text{R}\;(a,b)$

$\hspace{0.6cm}\forall\; (a,b)\in\mathbb{N}\times\mathbb{N}$ 

Let $(a,b)\in  \mathbb{N}\times \mathbb{N}$ be any element.

$\hspace{0.5cm}a+b=a+b$

$\hspace{0.5cm}a+b=b+a$

So, $(a,b)\;\text{R}\;(a,b)\hspace{0.5cm}\forall\; (a,b)\in\mathbb{N}\times\mathbb{N}$ 

Hence, $\text{R}$ is reflexive.

(ii) Condition for Symmetric Relation:

$\hspace{0.6cm}(a,b)\;\text{R}\;(c,d)\hspace{0.2cm}\Rightarrow\hspace{0.2cm} (c,d)\;\text{R}\;(a,b)$

$\hspace{0.6cm}\forall\; (a,b), (c,d) \in\mathbb{N}\times\mathbb{N}$

Let $(a,b), (c,d)\in  \mathbb{N}\times\mathbb{N}$ be any element.

As $(a,b)\;\text{R}\;(c,d)\hspace{0.2cm}$So, $\hspace{0.2cm}a+d=b+c$

$\hspace{4.3cm}c+b=d+a$

So, $ (c,d)\;\text{R}\;(a,b)\hspace{0.5cm}\forall\; (a,b), (c,d) \in\mathbb{N}\times\mathbb{N}$

Hence $\text{R}$ is symmetric.

(ii) Condition for Transitive Relation:

$\hspace{0.6cm}(a,b)\;\text{R}\;(c,d)$ and $(c,d)\;\text{R}\;(e,f)$

$\hspace{0.6cm}\Rightarrow\hspace{0.2cm} (a,b)\;\text{R}\;(e,f)$

$\hspace{0.6cm}\forall\; (a,b), (c,d), (e,f) \in\mathbb{N}\times \mathbb{N}$

Let $(a,b), (c,d), (e,f)\in  \mathbb{N}\times \mathbb{N}$ be any element.

As $(a,b)\;\text{R}\;(c,d)\hspace{0.2cm}$So, $\hspace{0.2cm}a+d=b+c\hspace{0.3cm}(1)$

As $(c,d)\;\text{R}\;(e,f)\hspace{0.2cm}$So, $\hspace{0.2cm}c+f=d+e\hspace{0.3cm}(2)$

Adding $(1)$ and $(2)$, we get

$\hspace{0.5cm}a+d+c+f=b+c+d+e$

$\hspace{2.1cm}a+f=b+e$

So, $ (a,b)\;\text{R}\;(e,f)\hspace{0.5cm}\forall\; (a,b), (e,f) \in\mathbb{N}\times \mathbb{N}$

Hence $\text{R}$ is transitive.

From (i), (ii) and (iii), we can say that

$\text{R}$ is an equivalence relation.