To prove that a rational number multiplied by an irrational number yields an irrational product, we can adopt a direct proof approach.
Proof:
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Definition of Rational and Irrational Numbers:
- A number \( r \) is rational if it can be expressed as the fraction of two integers \( r = \frac{a}{b} \), where \( a \) and \( b \) are integers, and \( b \neq 0 \).
- A number \( x \) is irrational if it cannot be expressed as a fraction of two integers.
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Assumption:
- Let \( r \) be a rational number, expressed as \( \frac{a}{b} \).
- Let \( x \) be an irrational number.
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Multiplication:
- We want to consider the product \( r \times x \). This can be represented as: \[ r \times x = \frac{a}{b} \times x = \frac{ax}{b} \]
- Here, \( ax \) represents the product of the rational number \( a \) and the irrational number \( x \).
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Proof by Contradiction:
- Assume, for the sake of contradiction, that \( r \times x \) is rational. Hence, we can write: \[ r \times x = \frac{c}{d} \] for some integers \( c \) and \( d \) (with \( d \neq 0 \)).
- Rearranging gives: \[ ax = \frac{bc}{d} \]
- Since \( \frac{bc}{d} \) is rational (product of rational numbers \( b \) and \( c \), and \( d \neq 0 \)), we see that \( ax \) must also be rational.
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Considering \( a \) is non-zero:
- If \( a \neq 0 \), we can divide both sides of the equation \( ax = \frac{bc}{d} \) by \( a \): \[ x = \frac{bc}{ad} \]
- The right side is expressed as a fraction of integers (since \( b, c, a, d \) are integers), implying that \( x \) is rational.
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Conclusion:
- This contradicts our initial assumption that \( x \) is irrational. Therefore, our assumption that \( r \times x \) is rational must be false.
Thus, we can conclude that the product of a rational number and an irrational number is indeed irrational.
Answer:
The correct response is:
- by creating a proof that shows directly how multiplying a rational and irrational number leads to an irrational answer.
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