To prove that a rational number multiplied by an irrational number gives an irrational product, the best approach 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 this 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 (with \( b \neq 0 \)).
- An irrational number cannot be expressed as a fraction of two integers.
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Assume a Rational and an Irrational Number:
- Let \( r \) be a rational number ( \( r = \frac{a}{b} \) ), and let \( x \) be an irrational number.
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Suppose the Product is Rational:
- Assume for contradiction that the product \( rx \) is rational. Then there exists some rational number \( q \) such that \( rx = q \).
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Isolate the Irrational Number:
- Rearranging gives \( x = \frac{q}{r} \).
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Argument for Contradiction:
- Since \( q \) and \( r \) are both rational (with \( r \neq 0 \)), \( \frac{q}{r} \) must also be rational.
- This implies that \( x \) is rational, contradicting the original assumption that \( x \) is irrational.
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Conclusion:
- Therefore, our assumption that the product \( rx \) is rational must be incorrect, proving that the product of a rational number and an irrational number is indeed irrational.
This proof effectively shows that Ryiaz's claim is incorrect, and it is indeed possible to prove that the product of a rational number and an irrational number is irrational.