The correct response is:
by creating a proof that shows directly how multiplying a rational and irrational number leads to an irrational answer.
To elaborate on this, we can prove that the product of a non-zero rational number and an irrational number is indeed irrational.
Let's assume:
- \( r \) is a non-zero rational number, which can be expressed as \( r = \frac{a}{b} \) where \( a \) and \( b \) are integers and \( b \neq 0 \).
- \( s \) is an irrational number.
We want to show that the product \( r \times s = \frac{a}{b} \times s \) is irrational.
Proof by Contradiction: Suppose \( r \times s \) is rational. This means we can express it as a fraction \( \frac{m}{n} \) where \( m \) and \( n \) are integers and \( n \neq 0 \).
Then we can rearrange:
\[ s = \frac{r \times s}{r} = \frac{\frac{m}{n}}{r} = \frac{m}{n} \times \frac{b}{a} = \frac{mb}{na} \]
Since \( \frac{mb}{na} \) is a ratio of integers (given that \( mb \) and \( na \) are integers), this means that \( s \) is rational.
However, this contradicts our original assumption that \( s \) is irrational. Therefore, our assumption that \( r \times s \) is rational must be false, meaning that \( r \times s \) must be irrational.
Hence, we have shown mathematically that a non-zero rational number multiplied by an irrational number results in an irrational number. This proves the claim.