Question
If one man says he multiplied to rational numbers and got an irrational number, but the other man says it would've come out rational, who is correct?
Answers
GPT-4o mini
In the context of multiplying rational numbers, the statement made by the first man (that he multiplied two rational numbers and got an irrational number) is incorrect. The second man is correct: the product of two rational numbers is always a rational number.
To clarify, a rational number is defined as any number that can be expressed as the quotient of two integers, where the denominator is not zero. When you multiply two rational numbers (for example, \( \frac{a}{b} \) and \( \frac{c}{d} \) where \( a, b, c, d \) are integers and \( b, d \neq 0 \)), the result will also be a rational number:
\[
\frac{a}{b} \times \frac{c}{d} = \frac{a \times c}{b \times d}
\]
Since both \( a \times c \) and \( b \times d \) are integers and \( b \times d \) is not zero, the result is a rational number. Therefore, the statement made by the first man is incorrect, and the second man is correct.
To clarify, a rational number is defined as any number that can be expressed as the quotient of two integers, where the denominator is not zero. When you multiply two rational numbers (for example, \( \frac{a}{b} \) and \( \frac{c}{d} \) where \( a, b, c, d \) are integers and \( b, d \neq 0 \)), the result will also be a rational number:
\[
\frac{a}{b} \times \frac{c}{d} = \frac{a \times c}{b \times d}
\]
Since both \( a \times c \) and \( b \times d \) are integers and \( b \times d \) is not zero, the result is a rational number. Therefore, the statement made by the first man is incorrect, and the second man is correct.