To prove that the sum of a rational and irrational number is irrational, we rely on the definitions and properties of rational and irrational numbers. The correct response is:
The sum or difference of a rational and irrational number is a rational number. (This statement is incorrect; the correct statement is the opposite.)
However, let's clarify the concept:
- A rational number is a number that can be expressed as the quotient \( \frac{a}{b} \), where \( a \) and \( b \) are integers, and \( b \neq 0 \).
- An irrational number cannot be expressed in such a form.
Now, let's assume the contrary, that the sum \( r + x \) (where \( r \) is a rational number and \( x \) is an irrational number) is rational. If this were true, we could rearrange it to get \( x = (r + x) - r \). Here, the right-hand side is a difference of two rational numbers (since \( r \) is rational and \( r + x \) would also be rational by our assumption), which would mean that \( x \) must be rational as well. This contradicts the fact that \( x \) is irrational.
Therefore, the sum of a rational and an irrational number is indeed irrational.
The best choice from the provided options indicating the closure property is actually the negation of what was stated:
The sum or difference of a rational and irrational number is a rational number (this statement is incorrect; it's the opposite). Thus, not any of the provided responses correctly supports the claim.