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Prove $Z^+\to Z$ given by f(n)=$\left(-1\right)^n\cdot n$ Is injective$$

To prove that the function $ f: \mathbb{Z}^+ \to \mathbb{Z} $ given by $ f(n) = (-1)^n \cdot n $ is injective, we need to show that if $ f(n_1) = f(n_2) $, then $ n_1 = n_2 $.

### Proof

Assume $ f(n_1) = f(n_2) $ for $ n_1, n_2 \in \mathbb{Z}^+ $.

So, we have:
\[ (-1)^{n_1} \cdot n_1 = (-1)^{n_2} \cdot n_2 \]

We need to consider the possible values of $ (-1)^n $:

1. If $ n $ is odd, $ (-1)^n = -1 $.
2. If $ n $ is even, $ (-1)^n = 1 $.

### Case 1: Both $ n_1 $ and $ n_2 $ are even
If $ n_1 $ and $ n_2 $ are both even, then $ (-1)^{n_1} = (-1)^{n_2} = 1 $. The equation becomes:
\[ n_1 = n_2 \]

### Case 2: Both $ n_1 $ and $ n_2 $ are odd
If $ n_1 $ and $ n_2 $ are both odd, then $ (-1)^{n_1} = (-1)^{n_2} = -1 $. The equation becomes:
\[ -n_1 = -n_2 \]
\[ n_1 = n_2 \]

### Case 3: One of $ n_1 $ or $ n_2 $ is even and the other is odd
If one of $ n_1 $ or $ n_2 $ is even and the other is odd, then $ (-1)^{n_1} \neq (-1)^{n_2} $. In this case:
- Suppose $ n_1 $ is even and $ n_2 $ is odd. Then the equation becomes:
  \[ n_1 = -n_2 \]
  Since $ n_1 $ and $ n_2 $ are both positive integers, $ n_1 \neq -n_2 $.

- Similarly, if $ n_1 $ is odd and $ n_2 $ is even, the equation becomes:
  \[ -n_1 = n_2 \]
  Again, since $ n_1 $ and $ n_2 $ are both positive integers, $ -n_1 \neq n_2 $.

Thus, it is impossible for $ f(n_1) = f(n_2) $ when one is odd and the other is even.

### Conclusion

Since $ f(n_1) = f(n_2) $ implies $ n_1 = n_2 $ in all cases, the function $ f(n) = (-1)^n \cdot n $ is injective.

Do you want details or have any questions? Here are some more questions you might find interesting:

1. How do you prove that a function is surjective?
2. What is the difference between injective, surjective, and bijective functions?
3. How can you determine if a function is bijective?
4. What are some examples of non-injective functions?
5. How do you prove the injectivity of functions involving trigonometric identities?

**Tip:** When proving injectivity, always consider the definition and try to show that different inputs must result in different outputs.

Learn how to prove the injectivity of the function f(n) = (-1)^n * n. Detailed step-by-step proof showing that if f(n1) = f(n2), then n1 = n2. Suitable for high school students studying functions and injectivity in mathematics.

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