Let set of real numbers be ℝ.

So, 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

aRb if a – b > 0

Check for Reflexivity:

For a ∈ ℝ

If aRa,

⇒ a – a > 0

⇒ 0 > 0

But 0 > 0 is not possible.

Hence, aRa is not true.

So, ∀ a ∈ ℝ, then aRa is not true.

⇒ R is not reflexive.

∴ R is not reflexive.

Check for Symmetry:

∀ a, b ∈ ℝ

If aRb,

⇒ a – b > 0

Replace a by b and b by a, we get

⇒ b – a > 0

[Take a = 12 and b = 6.

a – b > 0

⇒ 12 – 6 > 0

⇒ 6 > 0, which is a true statement.

Now, b – a > 0

⇒ 6 – 12 > 0

⇒ –6 > 0, which is not a true statement as –6 is not greater than 0.]

⇒ bRa is not true.

So, if aRb is true, then bRa is not true.

∀ a, b ∈ ℝ

⇒ R is not symmetric.

∴ R is not symmetric.

Check for Transitivity:

∀ a, b, c ∈ ℝ

If aRb and bRc.

⇒ a – b > 0 and b – c > 0

⇒ a – c > 0 or not.

Let us check.

a – b > 0 means a > b.

b – c > 0 means b > c.

a – c > 0 means a > c.

If a > b and b > c,

⇒ a > b, b > c

⇒ a > b > c

⇒ a > c

Hence, aRc is true.

So, if aRb is true and bRc is true, then aRc is true.

∀ a, b, c ∈ ℝ

⇒ R is transitive.

∴ R is transitive.