Consider the following:
1. If `R={(a,b)epsilonNxxN:a` divides b in `N}` then the relation `R` is reflexive and symmetric but not transitive.
2. If `A={1,2,3,4,5,5}` and `R={(S_(1),S_(2)):S_(1),S_(2)` are subsets of `A,S_(1)cancelsubS_(2)}`, then the relation `R` is not reflexive, not symmetric and not transitive.
Which of the statement is /are correct?

Options

A. 1 only
B. 2 only
C. Both 1 and 2
D. Neither 1 nor 2

Explanation

To determine the correctness of the two statements regarding the relations defined, we will analyze each statement step by step.

### Statement 1:
**Relation R = {(a, b) ∈ N × N : a divides b in N}**

1. **Reflexive Check**:
- A relation R is reflexive if for every element a in the set, (a, a) is in R.
- For any natural number a, a divides a (since a/a = 1, which is a natural number).
- Therefore, (a, a) is in R for all a ∈ N.
- **Conclusion**: R is reflexive.

2. **Symmetric Check**:
- A relation R is symmetric if for every (a, b) in R, (b, a) is also in R.
- Consider (2, 4) in R: 2 divides 4 (4/2 = 2).
- However, (4, 2) is not in R because 4 does not divide 2 (2/4 = 0.5, which is not a natural number).
- **Conclusion**: R is not symmetric.

3. **Transitive Check**:
- A relation R is transitive if whenever (a, b) in R and (b, c) in R, then (a, c) must also be in R.
- Consider (2, 4) and (4, 8): 2 divides 4 and 4 divides 8.
- However, (2, 8) is also in R since 2 divides 8 (8/2 = 4).
- This example shows that R can be transitive, but we need to find a counterexample.
- Consider (1, 2) and (2, 4): 1 divides 2 and 2 divides 4, but (1, 4) is also valid since 1 divides 4.
- After checking various combinations, we find that R can be transitive in some cases, but not universally.
- **Conclusion**: R is not universally transitive.

**Final Conclusion for Statement 1**: R is reflexive, not symmetric, and not universally transitive. Therefore, Statement 1 is **incorrect**.

### Statement 2:
**Set A = {1, 2, 3, 4, 5} and Relation R = {(S1, S2) : S1, S2 are subsets of A, and S1 is not a subset of S2}**

1. **Reflexive Check**:
- For R to be reflexive, for every subset S1 of A, (S1, S1) must be in R.
- However, S1 is always a subset of itself, which contradicts the condition that S1 is not a subset of S2.
- **Conclusion**: R is not reflexive.

2. **Symmetric Check**:
- For R to be symmetric, if (S1, S2) is in R, then (S2, S1) must also be in R.
- If S1 is not a subset of S2, it does not guarantee that S2 is not a subset of S1.
- For example, if S1 = {1, 2} and S2 = {1, 2, 3}, then S1 is not a subset of S2, but S2 is a subset of S1.
- **Conclusion**: R is not symmetric.

3. **Transitive Check**:
- For R to be transitive, if (S1, S2) and (S2, S3) are in R, then (S1, S3) must also be in R.
- If S1 is not a subset of S2 and S2 is not a subset of S3, it does not imply that S1 is not a subset of S3.
- For example, if S1 = {1}, S2 = {1, 2}, and S3 = {1, 2, 3}, then S1 is not a subset of S2, and S2 is not a subset of S3, but S1 is a subset of S3.
- **Conclusion**: R is not transitive.

**Final Conclusion for Statement 2**: R is not reflexive, not symmetric, and not transitive. Therefore, Statement 2 is **correct**.

### Final Answer:
- **Statement 1**: Incorrect
- **Statement 2**: Correct