A function f given as `f:{(2,7),(3,4),(7,9),(-1,6),(0,2),(5,3)}`. Is this function one-one onto?
Interchange the order of the elements in the ordered pairs and form the new relation. Is this relation a function? If it is a function, is it one-one onto.

Answer

To determine whether the function \( f \) is one-one and onto, and to analyze the new relation formed by interchanging the ordered pairs, we will follow these steps:

### Step 1: Identify the function and its properties

The function \( f \) is given as:
\[ f = \{(2, 7), (3, 4), (7, 9), (-1, 6), (0, 2), (5, 3)\} \]

#### Check if \( f \) is one-one (injective):

A function is one-one if no two different elements in the domain map to the same element in the codomain.

- The outputs (codomain values) from the pairs are: \( 7, 4, 9, 6, 2, 3 \).
- All outputs are unique.

Since there are no repeated \( y \) values for different \( x \) values, **\( f \) is one-one**.

#### Check if \( f \) is onto (surjective):

A function is onto if every element in the codomain has a pre-image in the domain.

- The domain (input values) from the pairs are: \( 2, 3, 7, -1, 0, 5 \).
- The codomain values are \( 7, 4, 9, 6, 2, 3 \).

Since every value in the codomain has a corresponding value in the domain, **\( f \) is onto**.

### Conclusion for \( f \):
- \( f \) is one-one and onto.

### Step 2: Interchange the order of the elements in the ordered pairs

Now we need to form a new relation by interchanging the elements in each ordered pair:
\[ R = \{(7, 2), (4, 3), (9, 7), (6, -1), (2, 0), (3, 5)\} \]

### Step 3: Check if \( R \) is a function

To determine if \( R \) is a function, we need to check if each input (first element) maps to exactly one output (second element).

- The inputs from \( R \) are: \( 7, 4, 9, 6, 2, 3 \).
- The outputs from \( R \) are: \( 2, 3, 7, -1, 0, 5 \).

Since each input has a unique output and there are no repeated inputs, **\( R \) is a function**.

### Step 4: Check if \( R \) is one-one (injective)

To check if \( R \) is one-one, we look for repeated outputs:

- The outputs are: \( 2, 3, 7, -1, 0, 5 \).
- All outputs are unique.

Thus, **\( R \) is one-one**.

### Step 5: Check if \( R \) is onto (surjective)

To check if \( R \) is onto, we need to ensure every element in the codomain has a pre-image in the domain.

- The codomain values from \( R \) are: \( 2, 3, 7, -1, 0, 5 \).
- The inputs (domain values) are: \( 7, 4, 9, 6, 2, 3 \).

Since every value in the codomain has a corresponding value in the domain, **\( R \) is onto**.

### Conclusion for \( R \):
- \( R \) is one-one and onto.

### Final Summary:
- The function \( f \) is one-one and onto.
- The new relation \( R \) is also a function, and it is one-one and onto.