If function `f:ZtoZ` is defined by `f(x)={{:((x)/(2), "if x is even"),(0," if x is odd"):}`, then f is

Options

A. one-one but not onto
B. onto but not one-one
C. neither one-one nor onto
D. a bijection

Answer

To determine the nature of the function \( f: \mathbb{Z} \to \mathbb{Z} \) defined by:

\[ f(x) = \begin{cases} \frac{x}{2} & \text{if } x \text{ is even} \\ 0 & \text{if } x \text{ is odd} \end{cases} \]

we need to analyze whether this function is one-one (injective) and onto (surjective).

### Step 1: Check if the function is one-one (injective)

A function is one-one if different inputs map to different outputs.

1. **Consider even integers**:
- If \( x = 2 \), then \( f(2) = \frac{2}{2} = 1 \).
- If \( x = 4 \), then \( f(4) = \frac{4}{2} = 2 \).
- If \( x = 6 \), then \( f(6) = \frac{6}{2} = 3 \).
- Thus, different even integers yield different outputs.

2. **Consider odd integers**:
- If \( x = 1 \), then \( f(1) = 0 \).
- If \( x = 3 \), then \( f(3) = 0 \).
- If \( x = 5 \), then \( f(5) = 0 \).
- Here, multiple odd integers map to the same output (0).

Since the odd integers do not yield unique outputs, the function is **not one-one**.

### Step 2: Check if the function is onto (surjective)

A function is onto if every element in the codomain (in this case, all integers) has a pre-image in the domain.

1. **Outputs from even integers**:
- For any even integer \( x \), \( f(x) = \frac{x}{2} \) can produce all integers (both positive and negative) as outputs.
- For example, \( f(0) = 0 \), \( f(2) = 1 \), \( f(-2) = -1 \), \( f(4) = 2 \), etc.

2. **Outputs from odd integers**:
- All odd integers map to 0, which means they do not contribute any new outputs.

Since every integer can be obtained from an even integer, and all odd integers map to 0, every integer in the codomain has a pre-image in the domain. Therefore, the function is **onto**.

### Conclusion

The function \( f \) is **not one-one** but is **onto**.

### Final Answer
The function \( f \) is **onto but not one-one**.

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