Question

Let \[ f : \mathbb{N} \rightarrow \mathbb{R} \] be defined by \[ f(x) = x^2 \] What will be the range of the function?

Answer 1

Given:

Let \[ f : \mathbb{N} \rightarrow \mathbb{R}, \quad f(x) = x^2 \]

Step-by-Step Explanation:

• \( \mathbb{N} = \{1, 2, 3, \dots\} \)
• When we square any natural number, the result is a positive real number.

So, \[ f(1) = 1,\quad f(2) = 4,\quad f(3) = 9,\quad f(4) = 16,\quad \dots \]

This gives: \[ \text{Range of } f = \{1, 4, 9, 16, 25, \dots\} \]

✅ Final Answer:

\[ \boxed{\text{Range} = \{x^2 \mid x \in \mathbb{N}\} \subset \mathbb{R}} \]

Answer 2

Check Whether the Function is Injective

Given: Let \[ f : \mathbb{N} \rightarrow \mathbb{N}, \quad f(x) = x^2 \]

Step-by-Step Verification:

Let us take any two natural numbers \( x_1, x_2 \in \mathbb{N} \) and suppose:

\[ f(x_1) = f(x_2) \Rightarrow x_1^2 = x_2^2 \]

Since \( x_1, x_2 \in \mathbb{N} \) (i.e., positive integers), we conclude:

\[ x_1 = x_2 \]

This satisfies the definition of an injective function: If \( f(x_1) = f(x_2) \Rightarrow x_1 = x_2 \), then \( f \) is one-one.

✅ Final Answer:

\[ \boxed{f(x) = x^2 \text{ is injective on } \mathbb{N}} \]

Answer 3

Given:

Let \[ f : \{1, 2, 3, 4, \dots\} \rightarrow \{1, 4, 9, 16, \dots\} \] be defined by \[ f(x) = x^2 \]

Step 1: Injectivity

We have already shown that \( f \) is injective (one-one): If \( f(x_1) = f(x_2) \), then \[ x_1^2 = x_2^2 \Rightarrow x_1 = x_2 \quad \text{(since } x_1, x_2 \in \mathbb{N}) \]

Step 2: Surjectivity

To prove that \( f \) is surjective (onto), we take any element \( y \) in the codomain: \[ y \in \{1, 4, 9, 16, \dots\} \Rightarrow y = x^2 \text{ for some } x \in \{1, 2, 3, \dots\} \] This shows that every \( y \) in the codomain has a pre-image in the domain.

✅ Final Conclusion:

Since \( f \) is both one-one and onto, we conclude: \[ \boxed{f(x) = x^2 \text{ is bijective}} \]

Answer 4

Given:

Let \[ f : \mathbb{R} \rightarrow \mathbb{R}, \quad f(x) = x^2 \]

1. Injectivity (One-One):

To test injectivity, suppose: \[ x_1 = 2,\quad x_2 = -2 \Rightarrow f(2) = 4 = f(-2) \] But clearly, \( 2 \ne -2 \), so: \[ f(x_1) = f(x_2) \text{ but } x_1 \ne x_2 \Rightarrow f \text{ is not injective} \]

2. Surjectivity (Onto):

To be surjective, for every \( y \in \mathbb{R} \), there must exist \( x \in \mathbb{R} \) such that: \[ f(x) = x^2 = y \] However, if \( y = -1 \), then: \[ x^2 = -1 \text{ has no real solution, since } x^2 \geq 0 \text{ for all } x \in \mathbb{R} \] So, \( f \) is not onto.

✅ Final Conclusion:

\[ \boxed{f(x) = x^2 \text{ is neither injective nor surjective on } \mathbb{R} \to \mathbb{R}} \]

💡 Quick Tip:

Changing the codomai