Answer :

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

**Check for Reflexivity:**

For a ∈ ℝ

If aRa,

⇒ |a| ≤ a, is true

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.

|a| = positive.

& |a| ≤ a

Hence, |a| ≤ a.

⇒ aRa is true.

So, ∀ a ∈ ℝ, then aRa is true.

⇒ R is reflexive.

**∴** **R is reflexive.**

**Check for Symmetry:**

∀ a, b ∈ ℝ

If aRb,

⇒ |a| ≤ b

Replace a by b and b by a, we get

⇒ |b| ≤ a, which might be true or not.

Let a = 2 and b = 3.

|a| ≤ b

⇒ |2| ≤ 3, is true

|b| ≤ a

⇒ |3| ≤ 2, is not true

⇒ 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 and |b| ≤ c

⇒ |a| ≤ c or not.

Let us check.

If |a| ≤ b and |b| ≤ c

b ≠ |b|

Say, if b = –2

⇒ –2 ≠ |–2|

⇒ –2 ≠ 2

But, from |a| ≤ b

b ≥ 0 in every case otherwise the statement would not hold true.

⇒ b can only accept positive values including 0.

⇒ b is a whole number.

∴ if |a| ≤ b and |b| ≤ c

⇒ |a| ≤ b, b ≤ c

⇒ |a| ≤ b ≤ c

⇒ |a| ≤ c

⇒ 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.**

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