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 1 + ab > 0

Check for Reflexivity:

For a ∈ ℝ

If aRa,

⇒ 1 + aa > 0

⇒ 1 + a² > 0

If a is a real number.

[Positive or negative, large or small, whole numbers or decimal numbers are all Real Numbers. Real numbers are so called because they are ‘Real’ not ‘imaginary’.]

This means, even if ‘a’ was negative.

a² = positive.

a² + 1 = positive

And any positive number is greater than 0.

Hence, 1 + a² > 0

⇒ aRa is true.

So, ∀ a ∈ ℝ, then aRa is true.

⇒ R is reflexive.

∴ R is reflexive.

Check for Symmetry:

∀ a, b ∈ ℝ

If aRb,

⇒ 1 + ab > 0

Replace a by b and b by a, we get

⇒ 1 + ba > 0

Whether we write ab or ba, it is equal.

ab = ba

So, 1 + ba > 0

⇒ bRa is true.

So, if aRb is true, then bRa is true.

∀ a, b ∈ ℝ

⇒ R is symmetric.

∴ R is symmetric.

Check for Transitivity:

∀ a, b, c ∈ ℝ

If aRb and bRc.

⇒ 1 + ab > 0 and 1 + bc > 0

⇒ 1 + ac > 0 or not.

Let us check.

1 + ab > 0 means ab > –1.

1 + bc > 0 means bc > –1.

1 + ac > 0 means ac > –1.

If ab > –1 and bc > –1.

⇒ ac > –1 should be true.

Take a = –1, b = 0.9 and c = 1

ab > –1

⇒ (–1)(0.9) > –1

⇒ –0.9 > –1, is true on the number line.

bc > –1

⇒ (0.9)(1) > –1

⇒ 0.9 > –1, is true on the number line.

ac > –1

⇒ (–1)(1) > –1

⇒ –1 > –1, is not true as –1 cannot be greater than itself.

⇒ ac > –1 is not true.

⇒ 1 + ac > 0 is not true.

⇒ aRc is not true.

So, if aRb is true and bRc is true, then aRc is not true.

∀ a, b, c ∈ ℝ

⇒ R is not transitive.

∴ R is not transitive.