Ryiaz claims it is impossible to prove that a rational number multiplied by an irrational number gives an irrational product. How can you prove this?(1 point)

To prove that a rational number multiplied by an irrational number gives an irrational product, we can use a proof by contradiction. Here’s how the proof works:

1. **Definitions**:
- A rational number can be expressed as a fraction \( \frac{p}{q} \), where \( p \) and \( q \) are integers and \( q \neq 0 \).
- An irrational number cannot be expressed as a fraction of two integers.

2. **Assumption**:
- Suppose \( r \) is a rational number (let's say \( r = \frac{p}{q} \)) and \( x \) is an irrational number. We will assume that their product \( rx \) is a rational number.

3. **Expression**:
- If \( rx \) is rational, then we can express it as \( \frac{m}{n} \), where \( m \) and \( n \) are integers and \( n \neq 0 \).

4. **Rearranging**:
- From our assumption, we have \( rx = \frac{m}{n} \).
- Therefore, \( x = \frac{rx}{r} \).
- Since both \( r \) and \( rx \) are rational, the quotient \( \frac{rx}{r} \) would also be rational (because the set of rational numbers is closed under multiplication and division by non-zero rationals).

5. **Contradiction**:
- According to our assumption, \( x \) was irrational, but we have derived that \( x \) would be rational, which is a contradiction.

6. **Conclusion**:
- Therefore, our initial assumption that multiplying a rational number \( r \) by an irrational number \( x \) yields a rational product must be false. Thus, the product \( rx \) must be irrational.

This proof shows that multiplying a rational number by an irrational number indeed results in an irrational product. Therefore, the correct response is:
**by creating a proof that shows directly how multiplying a rational and irrational number leads to an irrational answer.**