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

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

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

Answer 1

The best response to prove that the product of a rational number and an irrational number is irrational 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 that proof:

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

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

3. **Product**: Consider the product \( r \cdot x = \frac{a}{b} \cdot x = \frac{a \cdot x}{b} \).

4. **Assuming the Product is Rational**: Suppose, for contradiction, that \( r \cdot x \) is rational. Then \( \frac{a \cdot x}{b} \) must also be rational.

5. **Isolating x**: Rearranging gives \( a \cdot x = b \cdot (r \cdot x) \). Since both \( b \) and \( (r \cdot x) \) are rational, this means \( a \cdot x \) is rational.

6. **Conclusion**: For \( a \) (an integer) multiplied by \( x \) (assumed to be irrational) to yield a rational product (\( a \cdot x \)), \( x \) must be rational, contradicting the assumption that \( x \) is irrational.

Thus, we have shown that the assumption that \( r \cdot x \) is rational leads to a contradiction, proving that the product of a rational and an irrational number must be irrational.