We have the set of real numbers, 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

R = {(a, b): a ≤ b³}

Check for Reflexivity:

For a ∈ R

If (a, a) ∈ R,

⇒ a ≤ a³, which is not true.

Say, if a = – 2.

a ≤ a³

⇒ – 2 ≤ – 8

⇒ –2 ≤ –8, which is not true as – 2 > – 8.

Hence, (a, a) ∉ R

So, ∀ a ∈ R, then (a, a) ∉ R.

⇒ R is not reflexive.

∴ R is not reflexive.

Check for Symmetry:

∀ a, b ∈ R

If (a, b) ∈ R

⇒ a ≤ b³

Replace a by b and b by a, we get

⇒ b ≤ a³

[Take a = –2 and b = 3.

a ≤ b³

⇒ –2 ≤ 3³

⇒ –2 ≤ 27, which is a true statement.

Now, b ≤ a³

⇒ 3 ≤ (–2)³

⇒ 3 ≤ –8, which is not a true statement as 3 > –8]

⇒ (b, a) ∉ R

So, if (a, b) ∈ R, then (b, a) ∉ R

∀ a, b ∈ R

⇒ R is not symmetric.

∴ R is not symmetric.

Check for Transitivity:

∀ a, b, c ∈ R

If (a, b) ∈ R and (b, c) ∈ R

⇒ a ≤ b³ and b ≤ c³

⇒ a ≤ c³ or not.

Let us check.

Take a = 3, and .

a ≤ b³

⇒ 3 ≤ 3.37, which is true.

b ≤ c³

⇒ 1.5 ≤ 1.728

a ≤ c³

⇒ 3 ≤ 1.728, which is not true as 3 > 1.728.

Hence, (a, c) ∉ R.

So, if (a, b) ∈ R and (b, c) ∈ R, then (a, c) ∉ R.

∀ a, b, c ∈ ℝ

⇒ R is not transitive.

∴ R is not transitive.

Hence, R is neither reflexive, nor symmetric, nor transitive.