To determine which answer choice demonstrates that the set of irrational numbers is not closed under addition, we need to evaluate the sums given in each of the answer choices.
1. [tex]\(\pi + (-\pi) = 0\)[/tex]:
- Here, [tex]\(\pi\)[/tex] is an irrational number.
- [tex]\(-\pi\)[/tex] is also irrational since the negative of an irrational number is still irrational.
- The sum of [tex]\(\pi\)[/tex] and [tex]\(-\pi\)[/tex] is [tex]\(0\)[/tex], which is a rational number.
2. [tex]\(\frac{1}{2} + \left(-\frac{1}{2}\right) = 0\)[/tex]:
- Both [tex]\(\frac{1}{2}\)[/tex] and [tex]\(-\frac{1}{2}\)[/tex] are rational numbers.
- The sum of these two rational numbers is [tex]\(0\)[/tex], which is also a rational number.
- This does not involve any irrational numbers, so it does not pertain to the closure of irrational numbers under addition.
3. [tex]\(\pi + \pi = 2\pi\)[/tex]:
- Here, [tex]\(\pi\)[/tex] is an irrational number.
- The sum of [tex]\(\pi\)[/tex] and [tex]\(\pi\)[/tex] is [tex]\(2\pi\)[/tex].
- Since [tex]\(\pi\)[/tex] is irrational, [tex]\(2\pi\)[/tex] is also irrational (because a rational multiple of an irrational number is still irrational).
4. [tex]\(\frac{1}{2} + \frac{1}{2} = 1\)[/tex]:
- Both [tex]\(\frac{1}{2}\)[/tex]s are rational numbers.
- The sum of [tex]\(\frac{1}{2}\)[/tex] and [tex]\(\frac{1}{2}\)[/tex] is [tex]\(1\)[/tex], which is a rational number.
- This also does not involve any irrational numbers.
From these evaluations, the only answer choice that shows the sum of two irrational numbers resulting in a rational number (thus proving that the set of irrational numbers is not closed under addition) is:
[tex]\[
\pi + (-\pi) = 0
\][/tex]
This demonstrates that irrational numbers are not closed under addition because the sum is a rational number. Therefore, the correct answer choice is:
[tex]\[
\pi + (-\pi) = 0
\][/tex]
