Recall that for any binary relation R on set A. We have,

R is reflexive if for all x ∈ A, xRx.

R is symmetric if for all x, y ∈ A, if xRy, then yRx.

R is transitive if for all x, y, z ∈ A, if xRy and yRz, then xRz.

We have

x > y, x, y ∈ N

This relation is defined on N (set of Natural Numbers)

The relation can also be defined as

R = {(x, y): x > y} on N

Check for Reflexivity:

∀ x ∈ N

We should have, (x, x) ∈ R

⇒ x > x, which is not true.

1 can’t be greater than 1.

2 can’t be greater than 2.

16 can’t be greater than 16.

Similarly, x can’t be greater than x.

So, ∀ x ∈ N, then (x, x) ∉ R

⇒ R is not reflexive.

Check for Symmetry:

∀ x, y ∈ N

If (x, y) ∈ R

⇒ x > y

Now, replace x by y and y by x. We get

y > x, which may or not be true.

Let us take x = 5 and y = 2.

x > y

⇒ 5 > 2, which is true.

y > x

⇒ 2 > 5, which is not true.

⇒ y > x, is not true as x > y

⇒ (y, x) ∉ R

So, if (x, y) ∈ R, but (y, x) ∉ R ∀ x, y ∈ N

⇒ R is not symmetric.

Check for Transitivity:

∀ x, y, z ∈ N

If (x, y) ∈ R and (y, z) ∈ R

⇒ x > y and y > z

⇒ x > y > z

⇒ x > z

⇒ (x, z) ∈ R

So, if (x, y) ∈ R and (y, z) ∈ R, and then (x, z) ∈ R

∀ x, y, z ∈ N

⇒ R is transitive.

Hence, the relation is transitive but neither reflexive nor symmetric.