Question

Let R be a relation on the set N be defined by {(x, y): 3x + y = 33}. Then R is-
A. Reflexive
B. Symmetric
C. Transitive
D. None of the above

Answer 1

Any relation can be classified as reflexive, symmetric and transitive. If aRa exists in the relation, then it is said to be reflexive. If aRb and bRa both exist in the relation, then it is said to be symmetric. If aRb and bRc exist implies that aRc also exists, the relation is transitive.

Complete step by step answer:
For R to be reflexive, (x, x) should be an element of R, which can be written as-
3x + x = 33
4x = 33
x = 8.25
But it is given that R is a relation on the set on natural numbers, but x is not a natural number. Hence, R is not reflexive.
For R to be symmetric, if (x, y) is an element of R, then (y, x) is also an element of R. This can be verified using an example-
Let x = 10 and y = 3
3x + y = 3(10) + 3 = 33
Hence, (10, 3) is an element of R.
Now, let x = 3 and y = 10
3x + y = 3(3) + 10 = 19
(3, 10) is not an element of R.
This example does not satisfy the condition, so R is not symmetric as well.
For R to be transitive, if (x, y) and (y, z) are elements of R then (x, z) should also be an element of R.
This can be verified using an example-
Let x = 10 and y = 3
3x + y = 3(10) + 3 = 33
Hence, (10, 3) is an element of R.
Let y = 3 and z = 24
3x + y = 3(3) + 24 = 24
Hence, (3, 24) is an element of R.
Now we will check if (x, z) satisfies the condition or not-
3x + y = 3(10) + 24 = 54
Hence, (x, z) is not an element of R. Therefore, the relation R is not transitive.

So, the correct answer is “Option D”.

Note: It is important to check carefully for each condition. It is also recommended to check and verify each condition using a suitable example. Even if one case is false, the condition is not verified. Also if it is not possible to prove that relation is symmetric, reflexive or transitive, then use a suitable example to show that it is not.