Each of the following defines a relation on N:

(i) Given, x is greater than y; x, y ∈ N

If (x, x) ∈ R, then x > x, which is not true for any x ∈ N.

Thus, R is not reflexive.

Let (x, y) ∈ R

⇒ xRy

⇒ x > y

So, y > x is not true for any x, y ∈ N

Hence, R is not symmetric.

Let xRy and yRz

⇒ x > y and y > z

⇒ x > z

⇒ xRz

Hence, R is transitive.

(ii) x + y = 10; x, y ∈ N

Thus,

R = {(x, y); x + y = 10, x, y ∈ N}

R = {(1, 9), (2, 8), (3, 7), (4, 6), (5, 5), (6, 4), (7, 3), (8, 2), (9, 1)}

It’s clear (1, 1) ∉ R

So, R is not reflexive.

(x, y) ∈ R ⇒ (y, x) ∈ R

Therefore, R is symmetric.

Now (1, 9) ∈ R, (9, 1) ∈ R, but (1, 1) ∉ R

Therefore, R is not transitive.

(iii) Given, xy is square of an integer x, y ∈ N

R = {(x, y) : xy is a square of an integer x, y ∈ N}

It’s clearly (x, x) ∈ R, ∀ x ∈ N

As x2 is square of an integer for any x ∈ N

Thus, R is reflexive.

If (x, y) ∈ R ⇒ (y, x) ∈ R

So, R is symmetric.

Now, if xy is square of an integer and yz is square of an integer.

Then, let xy = m2 and yz = n2 for some m, n ∈ Z

x = m2/y and z = x2/y

xz = m2n2/ y2, which is square of an integer.

Thus, R is transitive.

(iv) x + 4y = 10; x, y ∈ N

R = {(x, y): x + 4y = 10; x, y ∈ N}

R = {(2, 2), (6, 1)}

It’s clearly seen (1, 1) ∉ R

Hence, R is not symmetric.

(x, y) ∈ R ⇒ x + 4y = 10

And (y, z) ∈ R ⇒ y + 4z = 10

⇒ x – 16z = -30

⇒ (x, z) ∉ R

Therefore, R is not transitive.