Solution:
Given: R = {(x, y): 3x – y = 0, where x , y ∈ A} and A = {1, 2, 3, …., 14}
⇒ R = {(1,3), (2, 6), (3, 9), (4, 12)}
Here, we can see that,
R is not reflexive as (x, x) ∉ R, ∀ x ∈ A.
Similarly, R is not symmetric because there exist at least one (x, y) ∈ R but (y, x) ∉ R.
Hence, R is not symmetric.
The given R is not transitive because (1, 3), (3, 9) ∈ R but (1, 9) ∉ R.
Hence, R is not transitive.
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