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 solve the problem, we need to analyze the function \( f: I \rightarrow I \) defined by \( f(x) = x^2 \), where \( I \) is the set of integers. We will determine whether this function is injective, surjective, or both.

### Step 1: Define the function and its domain
Let \( I \) be the set of integers, which includes all positive and negative whole numbers, as well as zero. The function is defined as:
\[ f(x) = x^2 \]

### Step 2: Check if the function is injective
A function is injective (or one-to-one) if different inputs produce different outputs. In other words, if \( f(a) = f(b) \), then \( a \) must equal \( b \).

- Consider \( f(1) \) and \( f(-1) \):
\[ f(1) = 1^2 = 1 \]
\[ f(-1) = (-1)^2 = 1 \]
Since \( f(1) = f(-1) \) but \( 1 \neq -1 \), the function is not injective.

### Step 3: Check if the function is surjective
A function is surjective (or onto) if every element in the codomain (in this case, the set of integers) has a pre-image in the domain.

- The codomain is the set of integers \( I \). We need to check if every integer \( y \in I \) can be expressed as \( f(x) = x^2 \) for some integer \( x \).
- The outputs of \( f(x) = x^2 \) are non-negative integers (0, 1, 4, 9, ...). There are no negative integers that can be expressed as \( x^2 \) for any integer \( x \).

Since there are integers in the codomain (like -1, -2, etc.) that do not have a corresponding \( x \) such that \( f(x) = y \), the function is not surjective.

### Step 4: Conclusion
Since the function \( f(x) = x^2 \) is neither injective nor surjective, we conclude that the function is:
\[ \text{neither injective nor surjective} \]

Thus, the correct option is (d) none of the above.

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