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

We analyze the relation $$R = \{(a,b) : \gcd(a,b) = 1, 2a \neq b, a, b \in \mathbb{Z}\}$$.

Reflexive?

For reflexivity, we need $$(a, a) \in R$$ for all $$a \in \mathbb{Z}$$. This requires $$\gcd(a, a) = 1$$ and $$2a \neq a$$.

$$\gcd(a, a) = |a|$$, which equals 1 only when $$a = \pm 1$$. For $$a = 2$$: $$\gcd(2,2) = 2 \neq 1$$. So $$(2, 2) \notin R$$.

Therefore, R is NOT reflexive.

Symmetric?

For symmetry, if $$(a, b) \in R$$, we need $$(b, a) \in R$$. We need $$\gcd(b, a) = 1$$ (same as $$\gcd(a, b) = 1$$, so this holds) AND $$2b \neq a$$.

Consider $$(1, 2)$$: $$\gcd(1, 2) = 1$$ and $$2(1) = 2 = b$$, so $$(1, 2) \notin R$$ (fails $$2a \neq b$$).

Consider $$(2, 1)$$: $$\gcd(2, 1) = 1$$ and $$2(2) = 4 \neq 1$$, so $$(2, 1) \in R$$.

But we need the reverse: $$(1, 2)$$. We showed $$(1, 2) \notin R$$.

So $$(2, 1) \in R$$ but $$(1, 2) \notin R$$. Therefore, R is NOT symmetric.

Transitive?

For transitivity, if $$(a, b) \in R$$ and $$(b, c) \in R$$, we need $$(a, c) \in R$$.

Consider $$(3, 2)$$: $$\gcd(3, 2) = 1$$, $$2(3) = 6 \neq 2$$. So $$(3, 2) \in R$$.

Consider $$(2, 3)$$: $$\gcd(2, 3) = 1$$, $$2(2) = 4 \neq 3$$. So $$(2, 3) \in R$$.

Now check $$(3, 3)$$: $$\gcd(3, 3) = 3 \neq 1$$. So $$(3, 3) \notin R$$.

We have $$(3, 2) \in R$$ and $$(2, 3) \in R$$ but $$(3, 3) \notin R$$. Therefore, R is NOT transitive.

The correct answer is Option 4: neither symmetric nor transitive.