To prove that a rational number multiplied by an irrational number gives an irrational product, the best approach is:
by creating a proof that shows directly how multiplying a rational and irrational number leads to an irrational answer.
Here's a brief outline of how such a proof could work:
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Definition of Rational and Irrational Numbers:
- A rational number can be expressed as \( \frac{a}{b} \), where \( a \) and \( b \) are integers and \( b \neq 0 \).
- An irrational number cannot be expressed as a fraction of two integers.
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Assume the Contrary:
- Let \( r \) be a rational number (e.g., \( r = \frac{a}{b} \)) and \( x \) be an irrational number.
- Assume that the product \( y = r \cdot x \) is rational.
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Manipulate the Assumption:
- If \( y \) is rational, we can express \( y \) as \( \frac{m}{n} \) for some integers \( m \) and \( n \).
- Thus, we have \( y = \frac{a}{b} \cdot x \).
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Rearranging the Equation:
- Rearranging gives: \[ x = \frac{y \cdot b}{a} \]
- Since \( y \) is assumed to be rational and \( a \) and \( b \) are integers, \(\frac{y \cdot b}{a}\) is also a rational number (as long as \( a \neq 0 \)).
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Contradiction:
- This implies that \( x \), which we assumed to be irrational, is actually rational, leading to a contradiction.
Thus, we conclude that the assumption that \( y \) (the product of a rational and an irrational number) is rational must be false. Hence, \( y \) must be irrational.
This direct proof demonstrates that multiplying a rational number by an irrational number results in an irrational product.
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