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)}`
`R_(2) ={(a,a),(a,b),(c,c),(b,a),(b,b),(a,c)}` , Then:

Options

A. `R_2` is symmetric but it is not transitive
B. both `R_(1)` , and `R_2` are not symmetric
C. both `R_1` and `R_2` are transitive
D. `R_(1)` is not symmetric but it is transitive

Answer

To determine the properties of the binary relations \( R_1 \) and \( R_2 \) on the set \( A = \{a, b, c\} \), we will check if these relations are symmetric, transitive, or reflexive.

### Step 1: Check if \( R_1 \) is symmetric
A relation is symmetric if for every pair \( (x, y) \) in the relation, the pair \( (y, x) \) is also in the relation.

- \( R_1 = \{(c,a), (b,b), (a,c), (c,c), (b,c), (a,a)\} \)
- Check pairs:
- \( (c,a) \) is in \( R_1 \) but \( (a,c) \) is also in \( R_1 \) (symmetric).
- \( (b,b) \) is in \( R_1 \) (symmetric).
- \( (a,c) \) is in \( R_1 \) but \( (c,a) \) is also in \( R_1 \) (symmetric).
- \( (c,c) \) is in \( R_1 \) (symmetric).
- \( (b,c) \) is in \( R_1 \) but \( (c,b) \) is not in \( R_1 \) (not symmetric).

Thus, \( R_1 \) is **not symmetric**.

### Step 2: Check if \( R_1 \) is transitive
A relation is transitive if whenever \( (x, y) \) and \( (y, z) \) are in the relation, then \( (x, z) \) must also be in the relation.

- Check pairs:
- \( (c,a) \) and \( (a,c) \) are in \( R_1 \) but \( (c,c) \) is also in \( R_1 \) (transitive).
- \( (b,c) \) and \( (c,a) \) are in \( R_1 \) but \( (b,a) \) is not in \( R_1 \) (not transitive).

Thus, \( R_1 \) is **not transitive**.

### Step 3: Check if \( R_2 \) is symmetric
- \( R_2 = \{(a,a), (a,b), (c,c), (b,a), (b,b), (a,c)\} \)
- Check pairs:
- \( (a,b) \) is in \( R_2 \) but \( (b,a) \) is also in \( R_2 \) (symmetric).
- \( (b,b) \) is in \( R_2 \) (symmetric).
- \( (a,c) \) is in \( R_2 \) but \( (c,a) \) is not in \( R_2 \) (not symmetric).

Thus, \( R_2 \) is **not symmetric**.

### Step 4: Check if \( R_2 \) is transitive
- Check pairs:
- \( (a,b) \) and \( (b,a) \) are in \( R_2 \) but \( (a,a) \) is also in \( R_2 \) (transitive).
- \( (a,c) \) and \( (c,c) \) are in \( R_2 \) but \( (a,c) \) is also in \( R_2 \) (transitive).

Thus, \( R_2 \) is **transitive**.

### Summary of Results
- \( R_1 \): Not symmetric, Not transitive
- \( R_2 \): Not symmetric, Transitive

### Final Conclusion
- Both \( R_1 \) and \( R_2 \) are not symmetric.
- \( R_1 \) is not transitive, while \( R_2 \) is transitive.