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Now consider the product of a nonzero rational number and an irrational number. Assume [tex]\( x = \frac{a}{b} \)[/tex], where [tex]\( a \)[/tex] and [tex]\( b \)[/tex] are integers and [tex]\( b \neq 0 \)[/tex]. Let [tex]\( y \)[/tex] be an irrational number. If we assume the product [tex]\( x \cdot y \)[/tex] is rational, we can set the product equal to [tex]\( \frac{m}{n} \)[/tex], where [tex]\( m \)[/tex] and [tex]\( n \)[/tex] are integers and [tex]\( n \neq 0 \)[/tex]. The steps for solving this equation are shown below.

\begin{tabular}{|c|c|}
\hline
Statement & Reason \\
\hline
[tex]\( x \cdot y = \frac{m}{n} \)[/tex] & \\
\hline
[tex]\( \frac{a}{b} \cdot y = \frac{m}{n} \)[/tex] & Substitution \\
\hline
[tex]\( \frac{b}{a} \cdot \frac{a}{b} \cdot y = \frac{m}{n} \cdot \frac{b}{a} \)[/tex] & Multiplication property of equality \\
\hline
[tex]\( y = \frac{m b}{n a} \)[/tex] & Simplify \\
\hline
\end{tabular}

Based on what we established about the classification of [tex]\( y \)[/tex] and using the closure of integers, what does the equation tell you about the type of number [tex]\( y \)[/tex] must be for the product to be rational? What conclusion can you make about the result of multiplying a rational and an irrational number?

Sagot :

GinnyAnsweridn
To address the question, let us carefully analyze each step and derive the final conclusion about the product of a nonzero rational number and an irrational number.

1. Initial Statement:
[tex]\[ x \cdot y = \frac{m}{n} \][/tex]
Here, [tex]\(x\)[/tex] is a nonzero rational number, and [tex]\(y\)[/tex] is an irrational number. The product [tex]\(x \cdot y\)[/tex] is assumed to be a rational number represented as [tex]\(\frac{m}{n}\)[/tex], where [tex]\(m\)[/tex] and [tex]\(n\)[/tex] are integers and [tex]\(n \neq 0\)[/tex].

2. Substitution:
[tex]\[ \frac{a}{b} \cdot y = \frac{m}{n} \][/tex]
Substituting [tex]\(x\)[/tex] with its definition [tex]\(\frac{a}{b}\)[/tex], where [tex]\(a\)[/tex] and [tex]\(b\)[/tex] are integers and [tex]\(b \neq 0\)[/tex].

3. Multiplication Property of Equality:
Multiply both sides by [tex]\(\frac{b}{a}\)[/tex] to isolate [tex]\(y\)[/tex]:
[tex]\[ \frac{b}{a} \cdot \frac{a}{b} \cdot y = \frac{m}{n} \cdot \frac{b}{a} \][/tex]
Simplifying the left-hand side:
[tex]\[ y = \frac{m \cdot b}{n \cdot a} \][/tex]

4. Simplification:
[tex]\[ y = \frac{m \cdot b}{n \cdot a} \][/tex]
We know that [tex]\(m\)[/tex], [tex]\(n\)[/tex], [tex]\(a\)[/tex], and [tex]\(b\)[/tex] are all integers, so their product and quotient will also result in a rational number. However, we initially assumed [tex]\(y\)[/tex] to be an irrational number. According to this equation, [tex]\(y\)[/tex] is expressed as the ratio of two integers, which contradicts our initial assumption that [tex]\(y\)[/tex] is irrational.

### Conclusion
By deriving the equation [tex]\(y = \frac{m \cdot b}{n \cdot a}\)[/tex], we see that the irrational number [tex]\(y\)[/tex] would have to be rational for the product [tex]\(x \cdot y\)[/tex] to be rational. Since this is a contradiction (as we assumed [tex]\(y\)[/tex] to be irrational):

[tex]\[ \text{The assumption that the product of a nonzero rational number and an irrational number can be rational is false.} \][/tex]

Therefore, we can conclude that:
[tex]\[ \boxed{\text{The product of a nonzero rational number and an irrational number is always irrational.}} \][/tex]
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