The relation R = {(a, b) : operatorname{gcd}( | vedclass

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(D) Reflexivity: For $R$ to be reflexive,$(a, a) \in R$ for all $a \in \mathbb{Z}$.This requires $\operatorname{gcd}(a, a) = |a| = 1$ and $2a 
eq a$.

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neither symmetric nor transitive

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(D) Reflexivity: For $R$ to be reflexive,$(a, a) \in R$ for all $a \in \mathbb{Z}$.This requires $\operatorname{gcd}(a, a) = |a| = 1$ and $2a 
eq a$. This is not true for all $a \in \mathbb{Z}$ (e.g.,$a=2$),so $R$ is not reflexive.Symmetry: For $R$ to be symmetric,if $(a, b) \in R$,then $(b, a) \in R$.Let $a=2, b=1$. $\operatorname{gcd}(2, 1) = 1$ and $2(2) = 4 
eq 1$,so $(2, 1) \in R$.However,for $(1, 2)$,$\operatorname{gcd}(1, 2) = 1$ but $2(1) = 2 = b$. Since the condition $2a 
eq b$ is violated,$(1, 2) 
otin R$.Thus,$R$ is not symmetric.Transitivity: For $R$ to be transitive,if $(a, b) \in R$ and $(b, c) \in R$,then $(a, c) \in R$.Let $a=14, b=19, c=21$.$\operatorname{gcd}(14, 19) = 1$ and $2(14) = 28 
eq 19$,so $(14, 19) \in R$.$\operatorname{gcd}(19, 21) = 1$ and $2(19) = 38 
eq 21$,so $(19, 21) \in R$.However,$\operatorname{gcd}(14, 21) = 7 
eq 1$,so $(14, 21) 
otin R$.Thus,$R$ is not transitive.Conclusion: $R$ is neither symmetric nor transitive.

The relation $R = \{(a, b) : \operatorname{gcd}(a, b) = 1, 2a \neq b, a, b \in \mathbb{Z}\}$ is:

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