A function R on the set N of natural numbers is defined as `R={[2n,2n+1]:ninN}`
The domain of R …

Answer

To find the domain of the function \( R \) defined as \( R = \{[2n, 2n+1] : n \in \mathbb{N}\} \), we will follow these steps:

### Step 1: Understand the Definition of the Relation
The relation \( R \) consists of ordered pairs where the first element is \( 2n \) and the second element is \( 2n + 1 \). Here, \( n \) is a natural number.

### Step 2: Identify the Values of \( n \)
The natural numbers \( \mathbb{N} \) are defined as \( \{1, 2, 3, 4, \ldots\} \). We will substitute these values into the expression \( 2n \).

### Step 3: Calculate the First Element of the Ordered Pairs
- For \( n = 1 \):
\[ 2n = 2 \times 1 = 2 \]
- For \( n = 2 \):
\[ 2n = 2 \times 2 = 4 \]
- For \( n = 3 \):
\[ 2n = 2 \times 3 = 6 \]
- For \( n = 4 \):
\[ 2n = 2 \times 4 = 8 \]
- Continuing this pattern, we see that as \( n \) increases, \( 2n \) will yield all even natural numbers.

### Step 4: Generalize the Result
From the calculations, we can see that the first element of each ordered pair in the relation \( R \) is always an even natural number. Therefore, the domain of \( R \) is the set of all even natural numbers.

### Conclusion
The domain of the relation \( R \) is:
\[ \{2, 4, 6, 8, 10, \ldots\} \text{ or } \{2n : n \in \mathbb{N}\} \]