Prove that the relation "less than" in the set of natural number is transitive but not reflexive and symmetric.

Explanation

To prove that the relation "less than" in the set of natural numbers is transitive but not reflexive and not symmetric, we will analyze each property step by step.

### Step 1: Define the Relation
Let \( R \) be the relation defined on the set of natural numbers \( \mathbb{N} \) such that for any \( x, y \in \mathbb{N} \), \( (x, y) \in R \) if and only if \( x < y \).

### Step 2: Prove that the Relation is Not Reflexive
A relation is reflexive if for every element \( x \) in the set, \( (x, x) \in R \).

- For any natural number \( x \), we have \( x < x \) is false. No natural number is less than itself.
- Therefore, \( (x, x) \notin R \) for all \( x \in \mathbb{N} \).

**Conclusion**: The relation "less than" is not reflexive.

### Step 3: Prove that the Relation is Not Symmetric
A relation is symmetric if for any \( x, y \in \mathbb{N} \), if \( (x, y) \in R \), then \( (y, x) \in R \).

- Suppose \( (x, y) \in R \). This means \( x < y \).
- If \( x < y \), then it cannot be true that \( y < x \) at the same time.
- Therefore, \( (y, x) \notin R \).

**Conclusion**: The relation "less than" is not symmetric.

### Step 4: Prove that the Relation is Transitive
A relation is transitive if for any \( x, y, z \in \mathbb{N} \), if \( (x, y) \in R \) and \( (y, z) \in R \), then \( (x, z) \in R \).

- Assume \( (x, y) \in R \) and \( (y, z) \in R \).
- This means \( x < y \) and \( y < z \).
- From the properties of inequalities, if \( x < y \) and \( y < z \), then it follows that \( x < z \).
- Thus, \( (x, z) \in R \).

**Conclusion**: The relation "less than" is transitive.

### Final Summary
- The relation "less than" in the set of natural numbers is **transitive** but **not reflexive** and **not symmetric**.

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