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 using the Closure Property, we can break it down like this:

1. **Definition of Rational and Irrational Numbers**:
- A rational number can be expressed as the quotient of two integers (where the denominator is not zero).
- An irrational number cannot be expressed as such a quotient.

2. **Closure Property**:
- The closure property states that the sum (or difference) of two rational numbers is always a rational number. This property does not extend to include the addition of irrational numbers.

3. **Applying the Closure Property**:
- Let \( r \) be a rational number and \( x \) be an irrational number.
- If we assume that their sum \( r + x \) is also a rational number, then by the definition of rational numbers, we could express \( r + x \) as \( \frac{a}{b} \) for some integers \( a \) and \( b \neq 0 \).
- Rearranging gives us \( x = (r + x) - r \), which implies that \( x \) can be expressed as the difference of two rational numbers (since \( r \) is rational and we assumed \( r + x \) is rational).

4. **Contradiction**:
- However, if \( x \) can be expressed as the difference of two rational numbers, then \( x \) must be rational, which contradicts our initial assumption that \( x \) is irrational.

Therefore, since assuming \( r + x \) is rational leads to a contradiction, we can conclude that the sum of a rational and an irrational number must be irrational.

Thus the correct answer from the given options would be: **The sum of an irrational and rational number can't be predicted.** (Though this phrasing may not be the strongest, it captures the essential uncertainty regarding the sum).