Question 33 of 35intermediate🔍 AnalyzeShort Answer3 marks

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

(i)

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

Answer

False. For (a,a) to be in R, we need a = a2a^{2}, which holds only for a=1. For a=2, (2,2) ∉ R, so not true for all a ∈ N.

(ii)

(a,b)∈R(a, b) \in \mathbb{R}, implies (b,a)∈R(b, a) \in \mathbb{R}

Answer

False. (a,b) ∈ R implies a = b2b^{2}. For (b,a) to be in R, we need b = a2a^{2}, which is not generally true. For example, (4,2) ∈ R but (2,4) ∉ R.

(iii)

(a,b)∈R,(b,c)∈R(a, b) \in \mathbb{R}, (b, c) \in \mathbb{R} implies (a,c)∈R(a, c) \in \mathbb{R}.

Answer

False. From (a,b) ∈ R and (b,c) ∈ R, we get a = b2b^{2} and b = c2c^{2}, so a = c4c^{4}. For (a,c) ∈ R, we need a = c2c^{2}, which fails for c=2: (16,4) and (4,2) are in R, but (16,2) ∉ R.

Explanation

The relation R is defined as {(a,b): a,b ∈ N and a = b2b^{2}}. We check each property using the definition. (i) Reflexive property would require (a,a) for all a, but a = a2a^{2} only for a=1, so false. (ii) Symmetry would require if (a,b) then (b,a), but a = b2b^{2} does not imply b = a2a^{2} except when a=b=1. (iii) Transitivity would require if (a,b) and (b,c) then (a,c), but a = b2b^{2} and b = c2c^{2} gives a = c4c^{4}, not necessarily c2c^{2}. Counterexamples demonstrate falsehood.

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