Answer
Both proofs confirm the statement: the negative of any irrational number is irrational.
Work Step by Step
We wish to prove the statement:
\[
\text{"For all real numbers } r, \text{ if } r \text{ is irrational then } -r \text{ is irrational."}
\]
### (a) Proof by Contraposition
**Contrapositive Statement:**
The contrapositive of “if \( r \) is irrational then \(-r\) is irrational” is:
\[
\text{"If } -r \text{ is rational then } r \text{ is rational."}
\]
**Proof:**
Suppose \(-r\) is rational. By the definition of rational numbers, there exist integers \( p \) and \( q \) (with \( q \neq 0 \)) such that:
\[
-r = \frac{p}{q}.
\]
Multiplying both sides of the equation by \(-1\) yields:
\[
r = -\frac{p}{q} = \frac{-p}{q}.
\]
Since \(-p\) is an integer (as the integers are closed under negation) and \( q \) is a nonzero integer, \( r \) is expressed as a quotient of two integers. Therefore, \( r \) is rational.
Thus, we have shown: if \(-r\) is rational, then \( r \) is rational. This completes the proof by contraposition, which establishes the original statement.
---
### (b) Proof by Contradiction
**Proof:**
Assume for the sake of contradiction that there exists a real number \( r \) such that \( r \) is irrational but \(-r\) is rational.
Since \(-r\) is rational, there exist integers \( p \) and \( q \) (with \( q \neq 0 \)) such that:
\[
-r = \frac{p}{q}.
\]
Multiplying both sides by \(-1\) gives:
\[
r = -\frac{p}{q} = \frac{-p}{q}.
\]
This shows that \( r \) can be written as the quotient of two integers, meaning that \( r \) is rational.
This conclusion contradicts our assumption that \( r \) is irrational. Therefore, our assumption must be false. Consequently, if \( r \) is irrational, then \(-r\) cannot be rational, i.e., \(-r\) must be irrational.
