To address Ryiaz's claim, we can use the first option:
"by creating a proof that shows directly how multiplying a rational and irrational number leads to an irrational answer."
Here's the proof:
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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 simple fraction, meaning it cannot be written as \(\frac{c}{d}\) for any integers \(c\) and \(d\).
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Consider Rational and Irrational Numbers:
- Let \(r\) be a rational number, so \(r = \frac{a}{b}\) where \(a\) and \(b\) are integers and \(b \neq 0\).
- Let \(x\) be an irrational number.
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Multiply \(r\) and \(x\):
- We want to show that the product \(rx\) is irrational.
- Assume \(rx\) is rational. This means \(rx = \frac{m}{n}\) for some integers \(m\) and \(n\) where \(n \neq 0\).
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Rearranging the Equation:
- We can rewrite the equation as \(x = \frac{m}{n} \cdot \frac{b}{a}\).
- Since \(r = \frac{a}{b}\), we can multiply both sides by \(\frac{b}{a}\), giving us: \[ x = \frac{mb}{an} \]
- Here, \(\frac{mb}{an}\) is a fraction where both the numerator \(mb\) and the denominator \(an\) are integers (assuming \(a \neq 0\) and \(n \neq 0\)).
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Conclusion:
- Since \(x\) can be expressed as a ratio of two integers, it follows that \(x\) would have to be rational, which contradicts our original assumption that \(x\) was irrational.
- Therefore, our assumption that \(rx\) is rational must be false, which means the product \(rx\) is irrational.
Thus, we have proved that the product of a rational number and an irrational number is indeed irrational, contradicting Ryiaz's claim.