Question 32 of 33intermediate🔍 AnalyzeShort Answer3 marks

Let RR be a relation from NN to NN defined by R={(a,b):a,b∈N and a=b2}R = \{(a, b) : a, b \in N \text{ and } a = b^2\}. Are the following true?

(i)

(a,a)∈R(a, a) \in R, for all a∈Na \in N

Answer

False. (a, a) ∈ R requires a = a², which gives a(a-1) = 0. This is satisfied only for a = 1 (since a ∈ N). For example, (2, 2) ∉ R because 2 ≠\neq 2² = 4. Hence, not all a ∈ N satisfy this condition.

(ii)

(a,b)∈R(a, b) \in R, implies (b,a)∈R(b, a) \in R

Answer

False. If (a, b) ∈ R, then a = b². For (b, a) ∈ R, we need b = a². Consider (4, 2) ∈ R since 4 = 2². But (2, 4) ∉ R because 2 ≠\neq 4² = 16. Thus, the converse does not hold.

(iii)

(a,b)∈R,(b,c)∈R(a, b) \in R, (b, c) \in R implies (a,c)∈R(a, c) \in R. Justify your answer in each case.

Answer

False. If (a, b) ∈ R and (b, c) ∈ R, then a = b² and b = c². This gives a = c⁴. For (a, c) ∈ R, we need a = c². Consider (16, 4) ∈ R (since 16 = 4²) and (4, 2) ∈ R (since 4 = 2²). But (16, 2) ∉ R because 16 ≠\neq 2² = 4. Hence, transitivity fails.

Explanation

This question tests understanding of relation properties: reflexivity, symmetry, and transitivity. The relation R = {(a, b) : a = b²} on natural numbers fails all three properties. For reflexivity, only a = 1 satisfies a = a². For symmetry, the square relationship is not reversible (4 = 2² but 2 ≠\neq 4²). For transitivity, composing two square relations gives a fourth power, not a square.

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