To prove that a rational number multiplied by an irrational number gives an irrational product, the best approach would be option 4: "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 can be structured:
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Definitions:
- A rational number can be expressed as \( r = \frac{a}{b} \), where \( a \) and \( b \) are integers, and \( b \neq 0 \).
- An irrational number cannot be expressed as a fraction of integers.
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Assumption:
- Let \( r \) be a rational number and \( x \) be an irrational number.
- We want to prove that the product \( rx \) is irrational.
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Proof by Contradiction:
- Assume for the sake of contradiction that the product \( rx \) is rational. Then, there exist integers \( m \) and \( n \) (with \( n \neq 0 \)) such that \( rx = \frac{m}{n} \).
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Rearranging:
- From the equation \( rx = \frac{m}{n} \), we can express \( x \) as: \[ x = \frac{m}{n} \cdot \frac{b}{a} = \frac{mb}{na} \]
- Since \( a \), \( b \), \( m \), and \( n \) are integers (and assuming \( r \neq 0 \)), the right side is a fraction of integers, which means \( x \) would be rational.
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Contradiction:
- This contradicts our original assumption that \( x \) is irrational. Therefore, our assumption that \( rx \) is rational must be false.
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Conclusion:
- We conclude that the product \( rx \) must be irrational whenever \( r \) is rational (and non-zero) and \( x \) is irrational.
Thus, the statement that a rational number multiplied by an irrational number gives an irrational product is indeed provable and demonstrated through the proof above.