`R={(a,b):a, b in N, a !=b " and a divides b" }`. Is R reflexive ? Give reason

Answer

To determine if the relation \( R = \{(a, b) : a, b \in \mathbb{N}, a \neq b \text{ and } a \text{ divides } b\} \) is reflexive, we need to analyze the definition of reflexivity in the context of this relation.

### Step-by-Step Solution:

1. **Understanding Reflexivity**: A relation \( R \) on a set \( A \) is said to be reflexive if for every element \( a \) in \( A \), the pair \( (a, a) \) is in \( R \). This means that every element must relate to itself.

2. **Identifying the Set**: In this case, the set \( A \) is the set of natural numbers \( \mathbb{N} \). Therefore, we need to check if \( (a, a) \) belongs to \( R \) for all \( a \in \mathbb{N} \).

3. **Analyzing the Relation**: The relation \( R \) is defined such that:
- \( a \) and \( b \) must be natural numbers.
- \( a \) must not be equal to \( b \) (i.e., \( a \neq b \)).
- \( a \) must divide \( b \) (i.e., \( b = k \cdot a \) for some integer \( k \)).

4. **Checking Reflexivity Condition**: For reflexivity, we need to see if \( (a, a) \) can be in \( R \):
- If we take any natural number \( a \), then \( (a, a) \) implies \( a = a \), which does not satisfy the condition \( a \neq b \) because \( a \) is equal to \( a \).

5. **Conclusion**: Since \( (a, a) \) cannot be in \( R \) for any \( a \in \mathbb{N} \) (due to the requirement that \( a \) must not equal \( b \)), we conclude that the relation \( R \) is **not reflexive**.

### Final Answer:
No, the relation \( R \) is not reflexive because \( (a, a) \) does not belong to \( R \) for any \( a \in \mathbb{N} \).

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