Consider the following two binary relations on the set $$A = \{a, b, c\}$$: $$R_1 = \{(c, a), (b, b), (a, c), (c, c), (b, c), (a, a)\}$$ and $$R_2 = \{(a, b), (b, a), (c, c), (c, a), (a, a), (b, b), (a, c)\}$$. Then:

We recall the definitions first. A binary relation $$R$$ on a set $$A$$ is called

$$\text{symmetric} \iff \forall (x,y)\in R,\; (y,x)\in R$$

$$\text{transitive} \iff \forall (x,y)\in R \text{ and } (y,z)\in R,\; (x,z)\in R$$

Now we inspect the two relations given on the set $$A=\{a,b,c\}$$:

$$R_1=\{(c,a),(b,b),(a,c),(c,c),(b,c),(a,a)\}$$ $$R_2=\{(a,b),(b,a),(c,c),(c,a),(a,a),(b,b),(a,c)\}$$

Checking symmetry of $$R_1$$.

• The pair $$(c,a)$$ is present, and its reverse $$(a,c)$$ is in $$R_1$$.
• $$(b,b)$$ is self-symmetric.
• $$(a,c)$$ is present, and its reverse $$(c,a)$$ we already have.
• $$(c,c)$$ is self-symmetric.
• $$(b,c)$$ is present, but its reverse $$(c,b)$$ is not in $$R_1$$.

Because $$(c,b)\notin R_1$$, we conclude that $$R_1$$ is not symmetric.

Checking transitivity of $$R_1$$.

To apply the definition, we look at every ordered pair whose second component matches the first component of another pair and then verify the required third pair.

1. Take $$(c,a)$$ and combine with every pair beginning with $$a$$:

  • $$(c,a),(a,c) \;\Rightarrow\; (c,c)$$ and $$(c,c)\in R_1$$ ✔️
  • $$(c,a),(a,a) \;\Rightarrow\; (c,a)$$ and $$(c,a)\in R_1$$ ✔️

2. Take $$(b,b)$$ and combine with every pair beginning with $$b$$:

  • $$(b,b),(b,b) \;\Rightarrow\; (b,b)$$ present ✔️
  • $$(b,b),(b,c) \;\Rightarrow\; (b,c)$$ present ✔️

3. Take $$(a,c)$$ and combine with every pair beginning with $$c$$:

  • $$(a,c),(c,a) \;\Rightarrow\; (a,a)$$ present ✔️
  • $$(a,c),(c,c) \;\Rightarrow\; (a,c)$$ present ✔️

4. Take $$(c,c)$$ and combine with every pair beginning with $$c$$:

  • $$(c,c),(c,a) \;\Rightarrow\; (c,a)$$ present ✔️
  • $$(c,c),(c,c) \;\Rightarrow\; (c,c)$$ present ✔️

5. Take $$(b,c)$$ and combine with every pair beginning with $$c$$:

  • $$(b,c),(c,a) \;\Rightarrow\; (b,a)$$ but $$(b,a)\notin R_1$$ ❌

Because the pair $$(b,a)$$ that is required by transitivity is missing, $$R_1$$ is not transitive.

Checking symmetry of $$R_2$$.

We list each pair and its reverse:

• $$(a,b)$$ has reverse $$(b,a)$$ which is in $$R_2$$.
• $$(b,a)$$ has reverse $$(a,b)$$ in $$R_2$$.
• $$(c,a)$$ has reverse $$(a,c)$$ in $$R_2$$.
• $$(c,c),(a,a),(b,b)$$ are all self-symmetric.

All required reverse pairs are present, so $$R_2$$ is symmetric.

Checking transitivity of $$R_2$$.

We again search for a violating instance. Consider

$$(b,a)\in R_2 \quad\text{and}\quad (a,c)\in R_2$$

The definition of transitivity demands $$(b,c)$$ to be in $$R_2$$ because the second element of the first pair equals the first element of the second pair. But the ordered pair $$(b,c)$$ is absent from $$R_2$$.

Therefore, $$R_2$$ is not transitive.

Summarising the properties we have derived:

$$R_1: \text{ not symmetric, not transitive}$$
$$R_2: \text{ symmetric, not transitive}$$

Only the statement “$$R_2$$ is symmetric but it is not transitive” is true.

Hence, the correct answer is Option A.