To prove that the product of a rational number and an irrational number is always irrational, we can use a proof by contradiction. Here’s how it can be structured:
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Definitions:
- A rational number can be expressed as the fraction of two integers (i.e., \( r = \frac{p}{q} \), where \( p \) and \( q \) are integers and \( q \neq 0 \)).
- An irrational number cannot be expressed as a fraction of two integers.
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Let’s assume the contrary: that multiplying a rational number \( r \) by an irrational number \( x \) results in a rational number \( y \). So, we have: \[ y = r \cdot x \] where \( y \) is rational.
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Rearranging: Since \( r \) is rational (i.e., \( r = \frac{p}{q} \)), we can rewrite the equation: \[ x = \frac{y}{r} \]
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Analyzing \( \frac{y}{r} \): Since both \( y \) (assumed to be rational) and \( r \) (which is also rational) are rational numbers, their quotient \( \frac{y}{r} \) will also be rational (as long as \( r \neq 0 \)).
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Contradiction: This leads us to conclude that \( x \) is rational (because it equals \( \frac{y}{r} \)), which contradicts the assumption that \( x \) is irrational.
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Conclusion: Therefore, our assumption must be incorrect. Thus, it follows that the product of a rational number and an irrational number must be irrational.
The correct response to the initial question is: by creating a proof that shows directly how multiplying a rational and irrational number leads to an irrational answer.