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

xy is the square of an integer. x, y ∈ N.

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

The relation can also be defined as

R = {(x, y): xy = a², a = √(xy), a ∈ N} on N

Check for Reflexivity:

∀ x ∈ N

We should have, (x, x) ∈ R

⇒ xx = a², where a = √(xx)

⇒ x² = a², where a = √(x²)

which is true every time.

Take x = 1 and y = 4

xy = a²

⇒ 1 × 4 = (√(1 × 4))² [∵ a = √(xy)]

⇒ 4 = (√4)²

⇒ 4 = (2)²

⇒ 4 = 4

So, ∀ x ∈ N, then (x, x) ∈ R

⇒ R is reflexive.

Check for Symmetry:

∀ x, y ∈ N

If (x, y) ∈ R

⇒ xy = a², where a = √(xy)

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

yx = a², which is as same as xy = a²

where a = √(yx)

⇒ yx = a²

⇒ (y, x) ∈ R

So, if (x, y) ∈ R, and then (y, x) ∈ R ∀ x, y ∈ N

⇒ R is symmetric.

Check for Transitivity:

∀ x, y, z ∈ N

If (x, y) ∈ R and (y, z) ∈ R

⇒ xy = a² and yz = a²

⇒ xz = a², may or may not be true.

Let us take x = 8, y = 2 and z = 50

xy = a², where a = √(xy)

⇒ (8)(2) = (√(8 × 2))²

⇒ 16 = (4)²

⇒ 16 = 16, which is true.

yz = a²

⇒ (2)(50) = (√(2 × 50))²

⇒ 100 = (10)²

⇒ 100 = 100, which is true

xz = a²

⇒ (8)(50) = (√(8 × 50))²

⇒ 400 = (20)²

⇒ 400 = 400

We won’t be able to find a case to show a contradiction.

⇒ xz = a²

⇒ (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 symmetric and transitivity, but not reflexive.