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]
