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)
• 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 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.

Answer 1

The best way to prove Ryiaz's claim wrong is:

• by creating a proof that shows directly how multiplying a rational and irrational number leads to an irrational answer.

### Proof:

Let \( r \) be a rational number and \( i \) be an irrational number. By definition, a rational number can be expressed as \( r = \frac{a}{b} \), where \( a \) and \( b \) are integers and \( b \neq 0 \).

Now, we will examine the product \( p = r \cdot i = \frac{a}{b} \cdot i \).

Assume for the purpose of contradiction that \( p \) is rational. Then, we can express \( p \) as \( p = \frac{m}{n} \), where \( m \) and \( n \) are integers, and \( n \neq 0 \).

Rearranging this gives:

\[
i = \frac{p \cdot b}{a}
\]

Since \( p \) is assumed to be rational and both \( b \) and \( a \) are integers, the right-hand side represents a ratio of integers, which means that \( i \) would also have to be rational.

However, this contradicts our initial assumption that \( i \) is irrational.

Since our assumption that \( p \) is rational leads to a contradiction, we conclude that the product \( p = r \cdot i \) must be irrational.

Thus, we have proven that the product of a rational number and an irrational number is indeed irrational.