Let R be a relation on the set of all natural | vedclass

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(A) The relation is defined as $aRb \iff a \mid b^2$ for $a, b \in \mathbb{N}$.$I.$ Reflexivity: For any $a \in \mathbb{N}$,$a^2$ is always divisible

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$I$ only

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(A) The relation is defined as $aRb \iff a \mid b^2$ for $a, b \in \mathbb{N}$.$I.$ Reflexivity: For any $a \in \mathbb{N}$,$a^2$ is always divisible by $a$ (since $a^2/a = a \in \mathbb{N}$). Thus,$(a, a) \in R$ for all $a \in \mathbb{N}$. So,$R$ is reflexive.$II.$ Symmetry: For $R$ to be symmetric,$aRb \implies bRa$. Let $a=8$ and $b=4$. Here $8 \mid 4^2$ $(8 \mid 16)$ is true,so $(8, 4) \in R$. However,$4 \mid 8^2$ $(4 \mid 64)$ is true,but consider $a=2, b=4$. $2 \mid 4^2$ $(2 \mid 16)$ is true,but $4 \mid 2^2$ $(4 \mid 4)$ is true. Let us take $a=2, b=6$. $2 \mid 6^2$ $(2 \mid 36)$ is true,but $6 \mid 2^2$ $(6 \mid 4)$ is false. Thus,$R$ is not symmetric.$III.$ Transitivity: For $R$ to be transitive,$(a, b) \in R$ and $(b, c) \in R \implies (a, c) \in R$. Let $a=8, b=4, c=2$. $(8, 4) \in R$ because $8 \mid 4^2$ $(8 \mid 16)$.$(4, 2) \in R$ because $4 \mid 2^2$ $(4 \mid 4)$.Check $(8, 2)$: $8 \mid 2^2$ $(8 \mid 4)$ is false. Thus,$R$ is not transitive.Therefore,only $I$ is satisfied.

Let $R$ be a relation on the set of all natural numbers $\mathbb{N}$ defined by $aRb \iff a \text{ divides } b^2$. Which of the following properties does $R$ satisfy?
$I.$ Reflexivity
$II.$ Symmetry
$III.$ Transitivity

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Let $R$ be a relation on the set of all natural numbers $\mathbb{N}$ defined by $aRb \iff a \text{ divides } b^2$. Which of the following properties does $R$ satisfy?$I.$ Reflexivity$II.$ Symmetry$III.$ Transitivity