We have

The relation “≥” on the set R of all real numbers.

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.

So, let the relation having “≥” be P.

We can write

P = {(a, b): a ≥ b, a, b ∈ R}

Check for Reflexivity:

∀ a ∈ R

If (a, a) ∈ P

⇒ a ≥ a, which is true.

Since, every real number is equal to itself.

So, ∀ a ∈ R, then (a, a) ∈ P.

⇒ P is reflexive.

∴ P is reflexive.

Check for Symmetry:

∀ a, b ∈ R

If (a, b) ∈ P

⇒ a ≥ b

Now, replace a by b and b by a. We get

b ≥ a, might or might not be true.

Let us check:

Take a = 7 and b = 5.

a ≥ b

⇒ 7 ≥ 5, holds.

b ≥ a

⇒ 5 ≥ 7, is not true as 5 < 7.

⇒ b ≥ a, is not true.

⇒ (b, a) ∉ P

So, if (a, b) ∈ P, then (b, a) ∉ P

∀ a, b ∈ R

⇒ P is not symmetric.

∴ P is not symmetric.

Check for Transitivity:

∀ a, b, c ∈ R

If (a, b) ∈ P and (b, c) ∈ P

⇒ a ≥ b and b ≥ c

⇒ a ≥ b ≥ c

⇒ a ≥ c

⇒ (a, c) ∈ P

So, if (a, b) ∈ P and (b, c) ∈ P, then (a, c) ∈ P

∀ a, b, c ∈ R

⇒ P is transitive.

∴ P is transitive.

Thus, shown that the relation “≥” on the set R of all the real numbers are reflexive and transitive but not symmetric.