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 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 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

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

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

Answer 1

To prove that a rational number multiplied by an irrational number gives an irrational product, the best approach is:

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

Here’s a brief outline of this proof:

1. **Definition of Rational and Irrational Numbers:**
- A rational number can be expressed as \( \frac{a}{b} \), where \( a \) and \( b \) are integers (with \( b \neq 0 \)).
- An irrational number cannot be expressed as a fraction of two integers.

2. **Assume a Rational and an Irrational Number:**
- Let \( r \) be a rational number ( \( r = \frac{a}{b} \) ), and let \( x \) be an irrational number.

3. **Suppose the Product is Rational:**
- Assume for contradiction that the product \( rx \) is rational. Then there exists some rational number \( q \) such that \( rx = q \).

4. **Isolate the Irrational Number:**
- Rearranging gives \( x = \frac{q}{r} \).

5. **Argument for Contradiction:**
- Since \( q \) and \( r \) are both rational (with \( r \neq 0 \)), \( \frac{q}{r} \) must also be rational.
- This implies that \( x \) is rational, contradicting the original assumption that \( x \) is irrational.

6. **Conclusion:**
- Therefore, our assumption that the product \( rx \) is rational must be incorrect, proving that the product of a rational number and an irrational number is indeed irrational.

This proof effectively shows that Ryiaz's claim is incorrect, and it is indeed possible to prove that the product of a rational number and an irrational number is irrational.