How does the Closure Property prove that the sum of a rational and irrational number is irrational?(1 point)

To prove that the sum of a rational and an irrational number is irrational, we can use the closure property of numbers.

The closure property states that when you perform a mathematical operation (like addition) on a set of numbers, the result is still within that set if the set is closed under that operation.

1. **Rational Numbers**: The sum of two rational numbers is always rational (closure property of rational numbers).
2. **Irrational Numbers**: The closure property does not apply to irrational numbers in the same manner.

Let's assume that the sum of a rational number (let's call it \( r \)) and an irrational number (let's call it \( i \)) is rational. If we denote the sum as \( s = r + i \), we can rearrange that to \( i = s - r \). Since \( s \) would be rational (as we assumed) and \( r \) is rational, \( s - r \) must also be rational (closure property of rational numbers). But that leads to a contradiction because we assumed \( i \) is irrational.

Thus, the assumption that the sum \( s \) is rational must be false, proving that the sum of a rational number and an irrational number is indeed irrational.

Given the answer choices provided, none of them accurately captures this reasoning. However, the correct reasoning can be summarized as:

- The sum or difference of a rational and irrational number is irrational.

So, the correct understanding would be that they could indeed rephrase this as:

- The sum or difference of a rational and an irrational number is not rational, thus it must be irrational.

None of the provided options correctly reflect this.