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)

The correct response to prove that a rational number multiplied by an irrational number gives an irrational product 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 how this proof can be structured:

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

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

3. **Multiply the Rational Number by the Irrational Number:** Consider the product \( r \times x = \frac{a}{b} \times x = \frac{ax}{b} \).

4. **Prove by Contradiction:** Now, assume for the sake of contradiction that the product \( r \times x \) is rational. This would mean that \( ax \) must be rational since \( b \) is non-zero and rational.

5. If \( ax \) is rational and \( a \) (a non-zero integer) is rational, then \( x \) must also be rational (because the product of a non-zero rational number and an irrational number cannot be rational). This is a contradiction because \( x \) was assumed to be irrational.

6. **Conclusion:** Therefore, our initial assumption that \( r \times x \) is rational must be false, which means that the product \( r \times x \) is irrational.

This proof successfully demonstrates that multiplying a rational number by an irrational number leads to an irrational number.