The closure property helps us understand the behavior of sums within specific sets of numbers (like rational and irrational numbers). Here’s how it applies to the sum of a rational and an irrational number:
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Rational Numbers: A number is rational if it can be expressed as a fraction \( \frac{a}{b} \) where \( a \) and \( b \) are integers and \( b \neq 0 \).
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Irrational Numbers: A number is irrational if it cannot be expressed as a fraction of two integers; in other words, it cannot be exactly represented as \( \frac{a}{b} \) for any integers \( a \) and \( b \).
To show that the sum of a rational number \( r \) and an irrational number \( i \) is irrational, consider the contrary:
- Assume \( r + i \) is rational. Then we can write \( r + i = k \) for some rational number \( k \). Rearranging gives us \( i = k - r \), which is the difference of two rational numbers (since \( k \) and \( r \) are both rational).
According to the closure property of rational numbers, the difference of two rational numbers is also rational. Thus, \( i \) would have to be rational, which contradicts our original assumption that \( i \) is irrational.
Therefore, the assumption that the sum \( r + i \) is rational must be false. Hence, the sum of a rational and an irrational number is irrational.
Correct Answer: The sum of a rational and irrational number is irrational.