Answer
See explanation
Work Step by Step
### a. Proof by Contradiction
**What to Suppose:**
To prove the statement by contradiction, assume that the statement is false for some real number. That is, assume there exists a real number \(r\) such that:
- \(r^2\) is irrational, **and**
- \(r\) is rational.
**What You Need to Show:**
From this assumption, derive a contradiction. Typically, one would show that if \(r\) is rational (i.e., can be written as \(r = \frac{p}{q}\) for integers \(p\) and \(q \neq 0\)), then \(r^2 = \frac{p^2}{q^2}\) must also be rational. This contradicts the assumption that \(r^2\) is irrational. Hence, the assumption leads to a contradiction, proving that if \(r^2\) is irrational then \(r\) must be irrational.
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### b. Proof by Contraposition
**What to Suppose:**
The contrapositive of the original statement is logically equivalent to it. The contrapositive is:
“For all real numbers \(r\), if \(r\) is rational then \(r^2\) is rational.”
So, to prove by contraposition, suppose that \(r\) is rational.
**What You Need to Show:**
Under the assumption that \(r\) is rational (i.e., \(r = \frac{p}{q}\) for some integers \(p\) and \(q\) with \(q \neq 0\)), show that \(r^2 = \frac{p^2}{q^2}\) is also rational. This directly establishes the contrapositive, and thus the original statement follows.
