To prove that the product of a rational number and an irrational number is irrational, you can use a proof by contradiction. Here's how you can do it:
Proof by Contradiction:
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Assume the contrary: Suppose that \( r \) is a rational number and \( x \) is an irrational number, and let's assume that their product \( rx \) is rational.
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Definition of rational number: Since \( r \) is rational, we can express it as \( r = \frac{a}{b} \), where \( a \) and \( b \) are integers and \( b \neq 0 \).
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Assuming the product is rational: According to our assumption, let \( rx = q \), where \( q \) is a rational number.
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Rearranging the equation: Hence, we can write: \[ x = \frac{q}{r} = \frac{q}{\frac{a}{b}} = \frac{qb}{a} \] Since \( q \), \( a \), and \( b \) are all integers and \( b \) and \( a \) are not zero, \( \frac{qb}{a} \) is a ratio of integers, which is a rational number.
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Contradiction arises: This implies that \( x \) must also be rational, which contradicts our assumption that \( x \) is irrational.
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Conclusion: Therefore, our initial assumption that the product \( rx \) is rational must be false. Hence, the product of a rational number and an irrational number must be irrational.
This reasoning shows that it is indeed possible to prove that the product of a rational number and an irrational number is irrational. Thus, the correct response would be:
by creating a proof that shows directly how multiplying a rational and irrational number leads to an irrational answer.
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