Answer
Both methods lead to the same conclusion: If \(n^2\) is odd, then \(n\) must be odd.
Work Step by Step
**a. Proof by Contradiction**
- **Suppose:**
To prove “For all integers \(n\), if \(n^2\) is odd then \(n\) is odd” by contradiction, assume that there is some integer \(n\) for which \(n^2\) is odd but \(n\) is not odd. (That is, assume \(n^2\) is odd and \(n\) is even.)
- **Need to Show:**
Under this assumption, derive a contradiction. In this case, show that if \(n\) is even then \(n^2\) must also be even, which contradicts the assumption that \(n^2\) is odd.
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**b. Proof by Contraposition**
- **Suppose:**
To prove the statement by contraposition, we instead prove its contrapositive: “For all integers \(n\), if \(n\) is not odd (i.e. \(n\) is even) then \(n^2\) is not odd (i.e. \(n^2\) is even).”
So, assume that \(n\) is even.
- **Need to Show:**
Show that under this assumption, \(n^2\) is even. Once this is proved, by contraposition the original statement holds.
