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  • Let   be a relation defined on <img alt="\mathbb{N}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathbb%7BN%7

Let R  be a relation defined on \mathbb{N} as a R\, \, b if  2 a+3 b is a multiple of 5, a$, $b \in \mathbb{N}. Then R is

Option: 1

an equivalence relation


Option: 2

transitive but not symmetric


Option: 3

not reflexive


Option: 4

symmetric but not transitive


Answers (1)

best_answer

Reflexive

Let  \mathrm{a} \in \mathrm{N}

\begin{aligned} & \mathrm{a} \mathrm{R} \mathrm{a} \Rightarrow \quad 2 \mathrm{a}+3 \mathrm{a} \text { is a multiple of } 5 \\ & \quad \Rightarrow \quad 5 \mathrm{a} \text { which is a multiple of } 5 \\ & \quad \Rightarrow \quad \mathrm{R} \text { is reflexive } \\ & \text { Symmetric } \\ & \text { Let } \quad \mathrm{a}, \mathrm{b} \in \mathrm{N} \\ & \mathrm{a} \mathrm{R} \mathrm{b} \Rightarrow 2 \mathrm{a}+3 \mathrm{~b}=5 \lambda_1 \quad \lambda_1 \in \mathrm{N} \\ & \mathrm{b} \mathrm{R} \mathrm{a} \Rightarrow 2 \mathrm{~b}+3 \mathrm{a}=5 \lambda_2 \quad \lambda_2 \in \mathrm{N} \\ & \text { On Adding } \\ & (2 \mathrm{a}+3 \mathrm{~b})+(2 \mathrm{~b}+3 \mathrm{a})=5\left(\lambda_1+\lambda_2\right) \\ & 5 \mathrm{a}+5 \mathrm{~b}=5\left(\lambda_1+\lambda_2\right) \\ & \Rightarrow \mathrm{Both} \text { sides are multiple of } 5 \\ & \Rightarrow \mathrm{R} \text { is symmetric } \end{aligned}

Transitive\\ Let a, b, c \in \mathrm{N}\\\\ $$ \begin{aligned} & \mathrm{a} \mathrm{R} \mathrm{b} \Rightarrow 2 \mathrm{a}+3 \mathrm{~b}=5 \lambda_1 \ldots(1) \\ & \mathrm{b} \mathrm{R} \Rightarrow 2 \mathrm{a}+3 \mathrm{c}=5 \lambda_2 \ldots(2) \\ & 2 \mathrm{a}+5 \mathrm{~b}+3 \mathrm{c}=5\left(\lambda_1+\lambda_2\right) \\ & \Rightarrow(2 \mathrm{a}+3 \mathrm{c})=5\left(\lambda_1+\lambda_2-\mathrm{b}\right) \\ & 2 \mathrm{a}+3 \mathrm{c} \text { is divisible by } 5 \\ & \Rightarrow \mathrm{a} \mathrm{R} \text { is true } \\ & \Rightarrow \mathrm{R} \text { is transitive } \end{aligned}

R is Equivalence Relation

Posted by

Irshad Anwar

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