To prove that a rational number multiplied by an irrational number gives an irrational product, the most effective 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 that proof:
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Definitions:
- A rational number can be expressed as \( r = \frac{a}{b} \) where \( a \) and \( b \) are integers and \( b \neq 0 \).
- An irrational number cannot be expressed as a fraction of integers.
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Proof by contradiction:
- Assume \( r \) is a rational number and \( x \) is an irrational number. Assume for contradiction that the product \( rx \) is rational.
- Then, we could express \( rx \) as \( \frac{m}{n} \) where \( m \) and \( n \) are integers and \( n \neq 0 \).
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Solving for x:
- Rearranging gives us \( x = \frac{rx}{r} \). Since \( r \neq 0 \), we have \( x = \frac{m}{n \cdot r} \).
- But this expression shows that \( x \) can be expressed as a ratio of integers (hence, rational), which contradicts our initial assumption that \( x \) is irrational.
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Conclusion:
- Therefore, our assumption that the product \( rx \) is rational must be false, meaning that \( rx \) is irrational.
This proof demonstrates directly that the product of a rational number and an irrational number is indeed irrational.
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