Let R be a relation defined on the set of natural numbers N as

Question:

Let R be a relation defined on the set of natural numbers N as
R = {(xy) : x N, 2x + y = 41}
Find the domain and range of R. Also, verify whether R is (i) reflexive, (ii) symmetric (iii) transitive.

Solution:

Domain of is the values of x and range of R is the values of that together should satisfy 2x+y = 41.
So,
Domain of R = {1, 2, 3, 4, ... , 20}
Range of R = {1, 3, 5, ... , 37, 39}

Reflexivity: Let x be an arbitrary element of R. Then,

$x \in R$

$\Rightarrow 2 x+x=41$ cannot be true.

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

So, $R$ is not reflexive.

Symmetry:

Let $(x, y) \in R$. Then,

$2 x+y=41$

$\not \Rightarrow 2 y+x=41$

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

So, $R$ is not symmetric.

Transitivity:

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

$\Rightarrow 2 x+y=41$ and $2 y+z=41$

$\Rightarrow 2 x+z=2 x+41-2 y 41-y-2 y=41-3 y$

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

Thus, $R$ is not transitive.

Leave a comment

Close

Click here to get exam-ready with eSaral

For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.

Download Now