Ryaz 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)

Question

• 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
• 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

Answer 1

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

• 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 a direct proof:

1. Let \( r \) be a rational number and \( x \) be an irrational number.
2. By definition, a rational number can be expressed as \( r = \frac{a}{b} \), where \( a \) and \( b \) are integers and \( b \neq 0 \).
3. Assume, for the sake of contradiction, that the product \( r \cdot x \) is rational. Then we could express it as \( r \cdot x = \frac{c}{d} \) for some integers \( c \) and \( d \) (where \( d \neq 0 \)).
4. Rearranging gives us \( x = \frac{c/d}{r} = \frac{c}{b} \) (because multiplying by the reciprocal of a rational number is also rational).
5. Since \( \frac{c}{b} \) is a ratio of integers, \( x \) would also be rational, which contradicts our assumption that \( x \) is irrational.
6. Therefore, \( r \cdot x \) must be irrational.

This method of using proof by contradiction is a standard way to demonstrate the assertion.