To prove that a rational number multiplied by an irrational number gives an irrational product, the most effective method is by creating a proof that shows directly how multiplying a rational and irrational number leads to an irrational answer.
Here's how the proof works:
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Definition of Rational and Irrational Numbers:
- A rational number can be expressed as the fraction \( \frac{a}{b} \) where \( a \) and \( b \) are integers, and \( b \neq 0 \).
- An irrational number cannot be expressed as such a fraction.
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Assume a Rational and an Irrational Number:
- Let \( r = \frac{a}{b} \) be a rational number and \( x \) be an irrational number.
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Multiply the Rational Number by the Irrational Number:
- Consider the product \( r \cdot x = \frac{a}{b} \cdot x = \frac{a \cdot x}{b} \).
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Contradiction:
- Assume for the sake of argument that \( r \cdot x \) is a rational number, say \( \frac{c}{d} \) where \( c \) and \( d \) are integers.
- Then we can rearrange: \[ a \cdot x = b \cdot \frac{c}{d} \] \[ x = \frac{b \cdot \frac{c}{d}}{a} \]
- Since both \( b \), \( c \), and \( a \) are integers, the right side of the equation shows that \( x \) can be expressed as a ratio of two integers, which contradicts the assumption that \( x \) is irrational.
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Conclusion:
- Therefore, our initial assumption that the product \( r \cdot x \) is rational must be false. This proves that multiplying a rational number \( r \) by an irrational number \( x \) yields an irrational product.
Thus, the correct approach to prove Ryiaz's claim wrong is through direct proof as mentioned.
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