The best response 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 outline of such a proof:
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Definitions: Recall that a rational number can be expressed as a fraction of two integers, \( r = \frac{a}{b} \), 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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Assumption: Assume \( r \) is a rational number and \( x \) is an irrational number.
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Multiplication Result: Consider the product \( y = r \times x \).
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Proof by Contradiction:
- Suppose \( y \) is rational. This means it can be expressed as \( \frac{m}{n} \), where \( m \) and \( n \) are integers and \( n \neq 0 \).
- Then we can rearrange the equation: \[ x = \frac{y}{r} = \frac{\frac{m}{n}}{\frac{a}{b}} = \frac{mb}{na} \]
- Since \( mb \) and \( na \) are both integers (since products and sums of integers are integers), this means that \( x \) can be expressed as a fraction of integers, which contradicts the fact that \( x \) is irrational.
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Conclusion: Hence, our assumption that \( y \) is rational must be false. Therefore, \( r \times x \) must be irrational.
This proof shows that the product of a rational number and an irrational number is indeed irrational.