The best response to Ryiaz's claim is:
by creating a proof that shows directly how multiplying a rational and irrational number leads to an irrational answer.
To provide a proof, consider the following:
Let \( r \) be a non-zero rational number, which can be expressed as \( r = \frac{a}{b} \), where \( a \) and \( b \) are integers and \( b \neq 0 \). Let \( x \) be an irrational number. We want to show that the product \( rx \) is also irrational.
Assume, for the sake of contradiction, that \( rx \) is rational. If \( rx \) is rational, it can also be expressed as \( \frac{m}{n} \), where \( m \) and \( n \) are integers and \( n \neq 0 \).
Now, we have:
\[ rx = \frac{m}{n} \]
Substituting for \( r \):
\[ \frac{a}{b} x = \frac{m}{n} \]
Multiplying both sides by \( b \) (which is non-zero), we get:
\[ a x = \frac{bm}{n} \]
Now, solving for \( x \):
\[ x = \frac{bm}{an} \]
Since both \( b \) and \( n \) are integers, and \( a \) is also an integer (assuming \( a \neq 0 \)), this means \( x \) can be expressed as a ratio of integers, which contradicts the assumption that \( x \) is irrational.
Thus, our initial assumption that \( rx \) is rational must be false, which means that \( rx \) must be irrational.
This proof shows clearly that the product of a non-zero rational number and an irrational number is indeed irrational.