To prove that multiplying a rational number by an irrational number yields an irrational product, we can use a proof by contradiction. Here is how the proof works:
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Definition of Rational and Irrational:
- A rational number can be defined as any number that can be expressed as the quotient of two integers (where the denominator is not zero). For example, \( r = \frac{a}{b} \) where \( a \) and \( b \) are integers and \( b \neq 0 \).
- An irrational number cannot be expressed as the quotient of two integers. Examples include numbers like \( \sqrt{2} \) or \( \pi \).
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Assumption for Proof by Contradiction:
- Assume that multiplying a rational number \( r \) by an irrational number \( x \) results in a rational product. So, we have \( p = r \cdot x \), where \( p \) is assumed to be a rational number.
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Rearranging the Equation:
- Since \( r \) is rational and non-zero, we can express \( x \) as follows: \[ x = \frac{p}{r} \]
- The quotient \( \frac{p}{r} \) is the division of two rational numbers (since both \( p \) and \( r \) are rational), which means \( x \) must also be rational.
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Contradiction:
- This contradicts our original assumption that \( x \) is irrational. Thus, our assumption that multiplying a rational number by an irrational number results in a rational product must be false.
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Conclusion:
- Therefore, we conclude that the product of a rational number and an irrational number must be irrational.
So, the correct 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.
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