How does the Closure Property prove that the sum of a rational and irrational number is irrational ? ( 1 point) numbers can be added or subtracted The sum or difference of a rational and irrational number is a rational number The sum of difference of two rational numbers is a rational number The sum of an irrational and rational number can't be predicted

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:

1. **Rational Numbers**: A number is rational if it can be expressed as a fraction \( \frac{a}{b} \) where \( a \) and \( b \) are integers and \( b \neq 0 \).

2. **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.