# Let R = {(a, b) : a = b2} for all a, b ∈ N.Show that R satisfies none of reflexivity, symmetry and transitivity.

We have, R = {(a, b) : a = b2} relation defined on N.

Now,

We observe that, any element a N cannot be equal to its square except 1.

(a,a) R a N

For e.g. (2,2) R 2 ≠ 22

R is not reflexive.

Let (a,b) R a, b N

a = b2

But b cannot be equal to square of a if a is equal to square of b.

(b,a) R

For e.g., we observe that (4,2) R i.e 4 = 22 but 2 ≠ 42 (2,4) R

R is not symmetric

Let (a,b) R and (b,c) R a, b,c N

a = b2 and b = c2

a ≠ c2

(a,c) R

For e.g., we observe that

(16,4) R 16 = 42 and (4,2) R 4 = 22

But 16 ≠ 22

(16,2) R

R is not transitive.

Thus, R is neither reflexive nor symmetric nor transitive.

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