To prove that the product of a non-zero rational number and an irrational number is irrational, you can use a proof by contradiction. Here’s how to do it:
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Assume the Opposite: Assume that \( r \) is a rational number and \( x \) is an irrational number, and that their product \( r \cdot x \) is rational.
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Express Rational number: Since \( r \) is a rational number, it can be expressed as \( r = \frac{a}{b} \), where \( a \) and \( b \) are integers and \( b \neq 0 \).
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Express the Assumption: If \( r \cdot x \) is rational, we can let \( r \cdot x = \frac{p}{q} \) for some integers \( p \) and \( q \) (where \( q \neq 0 \)).
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Solve for x: From the equation \( r \cdot x = \frac{p}{q} \), we can solve for \( x \): \[ x = \frac{p}{q} \cdot \frac{b}{a} = \frac{pb}{qa} \] Here, since \( p, q, a, b \) are all integers, \( \frac{pb}{qa} \) is a rational number.
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Contradiction: This implies that \( x \), which we assumed to be irrational, is actually rational, leading to a contradiction.
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Conclusion: Therefore, our initial assumption that the product \( r \cdot x \) is rational must be false. Hence, the product of a non-zero rational number and an irrational number is irrational.
So the correct 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.
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