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 rational numbers. The correct reasoning can be summarized as follows:

1. **Rational Number Definition**: A rational number can be expressed as a fraction \( \frac{a}{b} \) where \( a \) and \( b \) are integers and \( b \neq 0 \).
2. **Irrational Number Definition**: An irrational number cannot be expressed as a fraction of integers.
3. **Closure Property of Rational Numbers**: The sum or difference of two rational numbers is always a rational number.

Now, let's assume you have a rational number \( r \) and an irrational number \( i \). If \( r + i \) were rational, then we could express it as \( r + i = q \) where \( q \) is a rational number.

If we rearrange this equation, we get \( i = q - r \). Since \( q \) (being a rational number) and \( r \) (also rational) are both rational, their difference \( q - r \) must also be a rational number. This would mean that \( i \) is rational, which contradicts our initial assumption that \( i \) is irrational.

Therefore, we conclude that the sum of a rational number and an irrational number must be irrational.

The correct response highlighting this reasoning is not in the provided responses. However, the closest option that denotes the idea would be:

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

That response is technically incorrect, and the reasoning demonstrates the opposite.

Thus, the correct conclusion is that the sum of a rational number and an irrational number is indeed irrational, and the responses provided fail to capture this essence accurately.