Let I be the set of integer and f : I `rarr` I be defined as f(x) = `x^(2), x in I`, the function is

Options

A. bijection
B. injection
C. surjection
D. None of these

Explanation

To determine the nature of the function \( f: I \to I \) defined by \( f(x) = x^2 \) for all \( x \in I \) (where \( I \) is the set of integers), we need to analyze whether the function is injective (one-to-one), surjective (onto), or bijective (both).

### Step 1: Check for Injectivity (One-to-One)
A function is injective if different inputs produce different outputs. In other words, if \( f(a) = f(b) \) implies \( a = b \).

- Let's consider \( f(1) \) and \( f(-1) \):
\[ f(1) = 1^2 = 1 \]
\[ f(-1) = (-1)^2 = 1 \]
- Here, \( f(1) = f(-1) = 1 \) but \( 1 \neq -1 \). This shows that the function is not injective.

**Conclusion**: The function is not injective.

### Step 2: Check for Surjectivity (Onto)
A function is surjective if every element in the codomain (output set) has a pre-image in the domain (input set). In other words, for every \( y \in I \), there exists an \( x \in I \) such that \( f(x) = y \).

- We set \( y = f(x) = x^2 \). Rearranging gives us:
\[ x^2 = y \implies x = \pm \sqrt{y} \]
- For \( y \) to be an integer, \( y \) must be a non-negative integer (since \( x^2 \) is always non-negative). However, if we take a negative integer, say \( y = -2 \):
\[ x^2 = -2 \implies x = \pm \sqrt{-2} \]
- The square root of a negative number is not an integer, which means there is no integer \( x \) such that \( f(x) = -2 \).

**Conclusion**: The function is not surjective.

### Step 3: Check for Bijectivity
A function is bijective if it is both injective and surjective. Since we have already established that the function is neither injective nor surjective, it cannot be bijective.

### Final Conclusion
The function \( f(x) = x^2 \) is neither injective nor surjective, and therefore it is not bijective.

### Summary
- **Injective**: No (as shown with \( f(1) = f(-1) \))
- **Surjective**: No (as shown with \( y = -2 \))
- **Bijective**: No

The correct answer is that the function is none of these.

---