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 such a proof:
- Assume a rational number \( r \) can be expressed as \( \frac{a}{b} \), where \( a \) and \( b \) are integers and \( b \neq 0 \).
- Let \( x \) be an irrational number.
- We want to show that \( r \cdot x \) is irrational.
Assuming for the sake of contradiction that \( r \cdot x \) is rational, it would mean that there exist integers \( c \) and \( d \) (with \( d \neq 0 \)) such that:
\[ r \cdot x = \frac{c}{d} \]
- Rearranging this, we have:
\[ x = \frac{c}{d} \cdot \frac{b}{a} = \frac{cb}{da} \]
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Since \( c, b, d, \) and \( a \) are all integers, \( \frac{cb}{da} \) would be a rational number.
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This leads to a contradiction because we assumed \( x \) is irrational.
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Thus, we conclude that \( r \cdot x \) must be irrational.
This proof effectively demonstrates that multiplying a rational number by an irrational number results in an irrational product, thus disproving Ryiaz's claim.