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

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 how such a proof could work:

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 fraction of two integers.

2. **Assume the Contrary**:
- Let \( r \) be a rational number (e.g., \( r = \frac{a}{b} \)) and \( x \) be an irrational number.
- Assume that the product \( y = r \cdot x \) is rational.

3. **Manipulate the Assumption**:
- If \( y \) is rational, we can express \( y \) as \( \frac{m}{n} \) for some integers \( m \) and \( n \).
- Thus, we have \( y = \frac{a}{b} \cdot x \).

4. **Rearranging the Equation**:
- Rearranging gives:
\[
x = \frac{y \cdot b}{a}
\]
- Since \( y \) is assumed to be rational and \( a \) and \( b \) are integers, \(\frac{y \cdot b}{a}\) is also a rational number (as long as \( a \neq 0 \)).

5. **Contradiction**:
- This implies that \( x \), which we assumed to be irrational, is actually rational, leading to a contradiction.

Thus, we conclude that the assumption that \( y \) (the product of a rational and an irrational number) is rational must be false. Hence, \( y \) must be irrational.

This direct proof demonstrates that multiplying a rational number by an irrational number results in an irrational product.