Each of the following defines a relation on N:

Solution:

(i) We have,

$R=\{(x, y): x>y, x, y \in \mathbf{N}\}$

As, $x=x \forall x \in \mathbf{N}$

$\Rightarrow(x, x) \notin R$

So, $R$ is not a reflexive relation'

Let $(x, y) \in R$

$\Rightarrow x>y$

but $y

$\Rightarrow(y, x) \notin R$

So, $R$ is not a symmeteric relation

Let $(x, y) \in R$ and $(y, z) \in R$

$\Rightarrow x>y$ and $y>z$

$\Rightarrow x>z$

$\Rightarrow(x, z) \in R$

So, $R$ is a transitive relation

(ii) We have,

$R=\{(x, y): x+y=10, x, y \in \mathbf{N}\}$

$R=\{(1,9),(2,8),(3,7),(4,6),(5,5),(6,4),(7,3),(8,2),(9,1)\}$

As, $(1,1) \notin R$

So, $R$ is not a reflexive relation

Let $(x, y) \in R$

$\Rightarrow x+y=10$

$\Rightarrow y+x=10$

$\Rightarrow(y, x) \in R$

So, $R$ is a symmeteric relation

As, $(1,9) \in R$ and $(9,1) \in R$ but $(1,1) \notin R$

So, $R$ is not a transitive relation

(iii) We have,

$R=\{(x, y): x y$ is square of an integer, $x, y \in \mathbf{N}\}$

As, $x \times x=x^{2}$, which is a square of an integer $x$

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

So, $R$ is a reflexive relation

Let $(x, y) \in R$

$\Rightarrow x y$ is square of an integer

$\Rightarrow y x$ is also a square of an integer

$\Rightarrow(y, x) \in R$

So, $R$ is a symmeteric relation

Let $(x, y) \in R$ and $(y, z) \in R$

$\Rightarrow x y$ is square of an integer and $y z$ is also a square of an interger

$\Rightarrow x z$ must be a square of an integer

$\Rightarrow(x, z) \in R$

So, $R$ is a transitive relation

(iv) We have,

$R=\{(x, y): x+4 y=10, x, y \in \mathbf{N}\}$

$R=\{(2,4),(6,1)\}$

As, $(2,2) \notin R$

So, $R$ is not a reflexive relation

As, $(2,4) \in R$ but $(4,2) \notin R$

So, $R$ is not a symmeteric relation

As, $(2,4) \in R$ but 4 is not related to any natural number

So, $R$ is a transitive relation