Question

Question
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)
Responses

by creating a proof that shows directly how multiplying a rational and irrational number leads to an irrational answer
by creating a proof that shows directly how multiplying a rational and irrational number leads to an irrational answer

by using a proof by contradiction to show multiplication is a closed operation for rational numbers
by using a proof by contradiction to show multiplication is a closed operation for rational numbers

by choosing random numbers to plug into multiplication problems until he’s convinced
by choosing random numbers to plug into multiplication problems until he’s convinced

He’s correct; it is impossible to prove in any way.

Answer 1

To address Ryiaz's claim, we can use the first option:

**"by creating a proof that shows directly how multiplying a rational and irrational number leads to an irrational answer."**

Here's the proof:

1. **Definition of Rational and Irrational Numbers**:
- A rational number can be expressed as \(\frac{a}{b}\), where \(a\) and \(b\) are integers and \(b \neq 0\).
- An irrational number cannot be expressed as a simple fraction, meaning it cannot be written as \(\frac{c}{d}\) for any integers \(c\) and \(d\).

2. **Consider Rational and Irrational Numbers**:
- Let \(r\) be a rational number, so \(r = \frac{a}{b}\) where \(a\) and \(b\) are integers and \(b \neq 0\).
- Let \(x\) be an irrational number.

3. **Multiply \(r\) and \(x\)**:
- We want to show that the product \(rx\) is irrational.
- Assume \(rx\) is rational. This means \(rx = \frac{m}{n}\) for some integers \(m\) and \(n\) where \(n \neq 0\).

4. **Rearranging the Equation**:
- We can rewrite the equation as \(x = \frac{m}{n} \cdot \frac{b}{a}\).
- Since \(r = \frac{a}{b}\), we can multiply both sides by \(\frac{b}{a}\), giving us:
\[
x = \frac{mb}{an}
\]
- Here, \(\frac{mb}{an}\) is a fraction where both the numerator \(mb\) and the denominator \(an\) are integers (assuming \(a \neq 0\) and \(n \neq 0\)).

5. **Conclusion**:
- Since \(x\) can be expressed as a ratio of two integers, it follows that \(x\) would have to be rational, which contradicts our original assumption that \(x\) was irrational.
- Therefore, our assumption that \(rx\) is rational must be false, which means the product \(rx\) is irrational.

Thus, we have proved that the product of a rational number and an irrational number is indeed irrational, contradicting Ryiaz's claim.