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

Question

Responses

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

The sum of an irrational and rational number can’t be predicted.
The sum of an irrational and rational number can’t be predicted.

Irrational numbers can be added or subtracted.
Irrational numbers can be added or subtracted.

The sum or difference of two rational numbers is a rational number.

Answer 1

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:

1. A rational number is a number that can be expressed as the quotient $\frac{a}{b}$, where $a$ and $b$ are integers, and $b\ne 0$.
2. 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.