Let I be the set of positive integers. R is a | vedclass

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(B) The relation is defined as $a R b \iff \log_2(a/b) = k$,where $k \in \{0, 1, 2, \dots\}$. This implies $a/b = 2^k$,or $a = b \cdot 2^k$ for some n

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reflexive,transitive but not symmetric.

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(B) The relation is defined as $a R b \iff \log_2(a/b) = k$,where $k \in \{0, 1, 2, \dots\}$. This implies $a/b = 2^k$,or $a = b \cdot 2^k$ for some non-negative integer $k$.$1$. Reflexive: For any $a \in I$,$a/a = 1 = 2^0$. Since $0$ is a non-negative integer,$(a, a) \in R$. Thus,$R$ is reflexive.$2$. Symmetric: Consider $(2, 1) \in R$ because $\log_2(2/1) = 1$,which is a non-negative integer. However,$(1, 2) 
otin R$ because $\log_2(1/2) = -1$,which is not a non-negative integer. Thus,$R$ is not symmetric.$3$. Transitive: Let $(a, b) \in R$ and $(b, c) \in R$. Then $a = b \cdot 2^{k_1}$ and $b = c \cdot 2^{k_2}$ for some non-negative integers $k_1, k_2$. Substituting $b$,we get $a = (c \cdot 2^{k_2}) \cdot 2^{k_1} = c \cdot 2^{k_1+k_2}$. Since $k_1+k_2$ is a non-negative integer,$(a, c) \in R$. Thus,$R$ is transitive.Therefore,$R$ is reflexive and transitive but not symmetric.

Let $I$ be the set of positive integers. $R$ is a relation on the set $I$ given by $R = \{(a, b) \in I \times I \mid \log_2(a/b) \text{ is a non-negative integer} \}$. Then $R$ is:

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