To prove that a rational number multiplied by an irrational number results in an irrational product, we can use a direct proof. Here’s how it is structured:
Let \( r \) be a rational number, which can be expressed as \( r = \frac{a}{b} \), where \( a \) and \( b \) are integers and \( b \neq 0 \). Let \( x \) be an irrational number.
We will show that the product \( rx = \frac{a}{b} \cdot x \) is irrational. Assume, for the sake of contradiction, that the product \( rx \) is rational. Then we can express \( rx \) as some rational number \( \frac{c}{d} \), where \( c \) and \( d \) are integers and \( d \neq 0 \).
From this assumption, we can manipulate the equation as follows: \[ rx = \frac{a}{b} \cdot x = \frac{c}{d} \] Multiplying both sides by \( bd \) (to eliminate the denominators), we have: \[ a \cdot d \cdot x = b \cdot c \]
Now, we can isolate \( x \): \[ x = \frac{b \cdot c}{a \cdot d} \]
Since \( a \), \( b \), \( c \), and \( d \) are all integers and \( a \cdot d \neq 0 \) (because \( a \) and \( b \) are nonzero integers), \( \frac{b \cdot c}{a \cdot d} \) is a rational number.
This leads to a contradiction because we initially assumed \( x \) is an irrational number. Hence, our assumption that the product \( rx \) is rational must be false, proving that the product of a rational number and an irrational number is indeed irrational.
Thus, the correct response is: by creating a proof that shows directly how multiplying a rational and irrational number leads to an irrational answer.
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