The best way to respond to Ryiaz's claim is:
by creating a proof that shows directly how multiplying a rational and irrational number leads to an irrational answer.
Here’s a brief explanation of how this proof can be structured:
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Definition of Rational and Irrational Numbers:
- A rational number can be expressed as the fraction \( \frac{p}{q} \), where \( p \) and \( q \) are integers and \( q \neq 0 \).
- An irrational number cannot be expressed as a fraction of two integers.
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Consider a Rational and an Irrational Number:
- Let \( r = \frac{p}{q} \) be a rational number and let \( x \) be an irrational number.
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Multiply Them:
- We look at the product \( r \cdot x = \frac{p}{q} \cdot x \).
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Assumption for Contradiction:
- Assume for contradiction that \( r \cdot x \) is rational. This means we could express \( r \cdot x \) as some fraction \( \frac{m}{n} \) for integers \( m \) and \( n \).
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Isolate the Irrational Number:
- Since \( r \) is rational, we can rearrange this equation: \[ x = \frac{r \cdot x \cdot q}{p} \]
- Thus: \[ x = \frac{\frac{m}{n} \cdot q}{p} \]
- This shows that \( x \) can be expressed as a quotient of integers (because \( m \), \( n \), \( p \), and \( q \) are all integers), which contradicts our initial assumption that \( x \) is irrational.
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Conclusion:
- Therefore, the assumption that the product \( r \cdot x \) is rational leads to a contradiction. It must be the case that the product \( r \cdot x \) is irrational.
This proof demonstrates that multiplying a rational number by an irrational number results in an irrational product, effectively countering Ryiaz's claim.