To show that multiplying a rational number by an irrational number always results in an irrational product, we can prove it by contradiction.
Definitions:
- A rational number is a number that can be expressed as the quotient of two integers, \( \frac{a}{b} \), where \( a \) and \( b \) are integers and \( b \neq 0 \).
- An irrational number is a number that cannot be expressed as a quotient of two integers.
Proof by contradiction:
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Let's denote the rational number as \( r \) (where \( r = \frac{a}{b} \) for some integers \( a \) and \( b \neq 0 \)) and the irrational number as \( x \).
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Assume, for the sake of contradiction, that the product \( r \cdot x \) is a rational number. Let's denote this product as \( p \) (where \( p \) can also be expressed as \( \frac{c}{d} \) for integers \( c \) and \( d \neq 0 \)).
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Based on our assumption, we can write: \[ r \cdot x = p \]
If we solve for \( x \), we get: \[ x = \frac{p}{r} \]
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Since \( p \) is a rational number and \( r \) is a non-zero rational number, the division \( \frac{p}{r} \) results in a rational number. This follows from the fact that the quotient of two rational numbers is also rational (as long as the denominator is not zero).
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Therefore, we conclude that \( x \) is rational.
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However, this contradicts our original assumption that \( x \) is an irrational number.
Since our assumption that \( r \cdot x \) is rational leads to a contradiction, we conclude that the product of a non-zero rational number and an irrational number must be irrational.
This completes the proof, showing that multiplying a rational number by an irrational number always results in an irrational product.