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Compute R∘R⁻¹ for the "less than" Relation A to B | JEE

JEE Maths question with a full step-by-step solution.

Question

Let

R

be the relation "less than" from

A = \{1, 2, 3, 4\}

B = \{1, 3, 5\}

, i.e.

(a, b) \in R \Leftrightarrow a < b

. Then

R \circ R^{-1}

is:

\{(1, 3), (1, 5), (2, 3), (2, 5), (3, 5), (4, 5)\}

\{(3, 1), (5, 1), (3, 2), (5, 2), (5, 3), (5, 4)\}

\{(3, 3), (3, 5), (5, 3), (5, 5)\}

correct

\{(3, 3), (3, 4), (4, 5)\}

Solution

Step 1: Write

R

(pairs with

a < b

) and its inverse:

R = \{(1, 3), (1, 5), (2, 3), (2, 5), (3, 5), (4, 5)\}

R^{-1} = \{(3, 1), (5, 1), (3, 2), (5, 2), (5, 3), (5, 4)\}.

Step 2:

(x, z) \in R \circ R^{-1} \Leftrightarrow \exists\, y : (x, y) \in R^{-1} \text{ and } (y, z) \in R.

x = 3: \ y \in \{1, 2\} \Rightarrow z \in \{3, 5\} \Rightarrow (3, 3), (3, 5)

x = 5: \ y \in \{1, 2, 3, 4\} \Rightarrow z \in \{3, 5\} \Rightarrow (5, 3), (5, 5)

\therefore R \circ R^{-1} = \{(3, 3), (3, 5), (5, 3), (5, 5)\}.

Correct answer: (3)

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