To prove that multiplying a rational number by an irrational number results in an irrational product, we can use a proof by contradiction. Here’s a structured way to do that:
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Definitions:
- A rational number can be expressed as the ratio of two integers, i.e., \( r = \frac{a}{b} \) where \( a \) and \( b \) are integers and \( b \neq 0 \).
- An irrational number cannot be expressed as a ratio of two integers.
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Assumption for Contradiction:
- Let's assume that \( r \) is a rational number and \( x \) is an irrational number, and their product \( y = r \cdot x \) is a rational number.
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Expressing the Product:
- Since \( r \) is rational, we can express \( y \) as: \[ y = \frac{a}{b} \cdot x \]
- Rearranging gives: \[ x = \frac{y \cdot b}{a} \]
- Here, \( y \) is the product we assumed to be rational, and \( a \) and \( b \) are integers.
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Analysis:
- The right-hand side of the equation \( x = \frac{y \cdot b}{a} \) is the ratio of two integers (since \( y \) is assumed to be rational, and \( a \) and \( b \) are integers), which means \( x \) is rational.
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Contradiction:
- This leads to a contradiction because we started with \( x \) being irrational, but derived that it must also be rational.
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Conclusion:
- Therefore, our initial assumption that \( r \cdot x \) (the product of a rational and an irrational number) is rational must be incorrect. This means \( r \cdot x \) must be irrational.
The correct answer is: by creating a proof that shows directly how multiplying a rational and irrational number leads to an irrational answer.
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