To prove that for every integer $n$, if $n^3$ is odd, then $n$ is also odd, we will proceed with a direct proof.

Proof:

1. Assume that $n^3$ is odd.

   By the definition of odd numbers, if $n^3$ is odd, we know that it cannot be divisible by 2. That is, there does not exist an integer $k$ such that: n^3 = 2k \quad \text{(i.e., n^3 is not divisible by 2)}.

2. Goal: We need to prove that $n$ is also odd. To do so, we'll prove its contrapositive, which is logically equivalent.

   The contrapositive of the statement "If $n^3$ is odd, then $n$ is odd" is: \text{If n$is even, then$n^3 is even.} So, we will show that if $n$ is even, $n^3$ must be even.

3. Assume $n$ is even.

   If $n$ is even, then by definition, we can write $n$ as: $n = 2m \quad \text{for some integer } m.$

4. Cube $n$:

   Now, calculate $n^3$: $n^3 = (2m)^3 = 8m^3.$

   Since $8m^3$ is clearly divisible by 2, we conclude that $n^3$ is even.

5. Conclusion:

   We have shown that if $n$ is even, then $n^3$ is even. This proves the contrapositive statement, and by logical equivalence, it also proves the original statement that if $n^3$ is odd, then $n$ must be odd.

Thus, the proof is complete.

Summary:

If $n^3$ is odd, then $n$ must be odd.

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Would you like further explanation on any step, or have additional questions? Here are some related questions for further exploration:

1. What is the contrapositive of a logical statement, and why is it important in proofs?
2. How do you prove statements about even and odd numbers algebraically?
3. Can the proof be extended to other powers, such as $n^4$ or $n^5$?
4. What would the proof look like if we were to show that "if $n^3$ is even, then $n$ is even"?
5. How do you prove that if a number is divisible by 3, its cube is also divisible by 3?

Tip: The contrapositive of a statement always has the same truth value as the original statement, so it is a powerful technique in proofs.