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 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 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

Answer 1

To prove that the product of a rational number and an irrational number is always irrational, we can use a proof by contradiction. Here's how the proof works:

1. Definitions:

   • 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 two integers.

2. Assumption:

   • Let \( r \) be a rational number, so \( r = \frac{a}{b} \), where \( a \) and \( b \) are integers and \( b \neq 0 \).
   • Let \( x \) be an irrational number.
   • We want to show that the product \( r \cdot x \) is irrational.

3. Proving by Contradiction:

   • Assume, for contradiction, that the product \( r \cdot x \) is rational. Thus, \( r \cdot x = \frac{m}{n} \) for some integers \( m \) and \( n \neq 0 \).

4. Rearranging the equation:

   • From the equation \( r \cdot x = \frac{m}{n} \), we can express \( x \) as follows: \[ x = \frac{m}{n} \cdot \frac{b}{a} = \frac{mb}{na} \]
   • Here, \( \frac{mb}{na} \) is the product of integers divided by a non-zero integer, which means \( x \) would have to be rational (since it can be expressed as a fraction of integers).

5. Contradiction:

   • This contradicts our initial assumption that \( x \) is irrational. Therefore, our assumption that \( r \cdot x \) is rational must be false.

6. Conclusion:

   • Hence, the product of a rational number and an irrational number is always irrational.

Thus, the correct approach to prove Ryaz's claim incorrect is to create a proof that shows directly how multiplying a rational and an irrational number leads to an irrational answer.