# Each of the following defines a relation on N:

Question:

Each of the following defines a relation on N:

(i) $x>y, x, y \in \mathbf{N}$

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

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

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

Determine which of the above relations are reflexive, symmetric and transitive.  [NCERT EXEMPLAR]

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