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The given function \( f: \mathbb{N} \to \mathbb{N} \), defined by \( f(n) = n^2 \), maps natural numbers to natural numbers. The question asks if this function is injective.

### Definition of Injectivity
A function is injective (or one-to-one) if different inputs result in different outputs. That is, for the function \( f \), if \( f(n_1) = f(n_2) \), then \( n_1 = n_2 \). 

### Analyzing the Function \( f(n) = n^2 \)
For natural numbers, the function squares each input. Let’s check the injectivity:
- Suppose \( f(n_1) = f(n_2) \), which means \( n_1^2 = n_2^2 \).
- In the natural numbers \( \mathbb{N} \), the only way this equality holds is if \( n_1 = n_2 \), because \( n_1 \) and \( n_2 \) are both positive or zero. Negative numbers are not included in \( \mathbb{N} \).

### Conclusion
Since \( f(n_1) = f(n_2) \) implies \( n_1 = n_2 \) for natural numbers, the function is injective.

Thus, the correct answer is **True**.

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Would you like further details or have any questions?

Here are 5 related questions you may find helpful:
1. What is the difference between injective, surjective, and bijective functions?
2. How can you prove a function is injective using a formal approach?
3. What would change if the function were defined over the integers \( \mathbb{Z} \) instead of natural numbers \( \mathbb{N} \)?
4. Can you provide an example of a non-injective function?
5. How does injectivity affect the invertibility of a function?

**Tip:** If you’re unsure whether a function is injective, check if distinct inputs always lead to distinct outputs by solving \( f(n_1) = f(n_2) \) and seeing if this forces \( n_1 = n_2 \).

The function f: N → N defined by f(n) = n^2 is injective.

Explore whether the function f(n) = n^2, mapping natural numbers to natural numbers, is injective. This problem discusses the concept of injectivity and how to prove it using basic properties of natural numbers.

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