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:

$\text{R}:\mathbb{N}\times \mathbb{N}\to \mathbb{N}\times \mathbb{N}$, 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

$\boxed{{\displaystyle x\text{R}x}}\mathrm{\forall }x$

(ii) Symmetric

$\boxed{{\displaystyle x\text{R}y}}\Rightarrow \boxed{{\displaystyle y\text{R}x}}\mathrm{\forall }x,y$

(iii) Transitive

$\boxed{{\displaystyle x\text{R}y}}\text{and}\boxed{{\displaystyle y\text{R}z}}\Rightarrow \boxed{{\displaystyle x\text{R}z}}\mathrm{\forall }x,y,z$

Proof:

(i) Condition for Reflexive Relation:

$(a,b)\text{R}(a,b)$

$\mathrm{\forall }(a,b)\in \mathbb{N}\times \mathbb{N}$ 

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

$a+b=a+b$

$a+b=b+a$

So, $(a,b)\text{R}(a,b)\mathrm{\forall }(a,b)\in \mathbb{N}\times \mathbb{N}$ 

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

(ii) Condition for Symmetric Relation:

$(a,b)\text{R}(c,d)\Rightarrow (c,d)\text{R}(a,b)$

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

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

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

$c+b=d+a$

So, $(c,d)\text{R}(a,b)\mathrm{\forall }(a,b),(c,d)\in \mathbb{N}\times \mathbb{N}$

Hence $\text{R}$ is symmetric.

(ii) Condition for Transitive Relation:

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

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

$\mathrm{\forall }(a,b),(c,d),(e,f)\in \mathbb{N}\times \mathbb{N}$

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

As $(a,b)\text{R}(c,d)$So, $a+d=b+c(1)$

As $(c,d)\text{R}(e,f)$So, $c+f=d+e(2)$

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

$a+d+c+f=b+c+d+e$

$a+f=b+e$

So, $(a,b)\text{R}(e,f)\mathrm{\forall }(a,b),(e,f)\in \mathbb{N}\times \mathbb{N}$

Hence $\text{R}$ is transitive.

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

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