To determine which answer choice demonstrates that the set of irrational numbers is not closed under addition, we first need to understand the concept. A set is closed under an operation (such as addition) if performing that operation on members of the set always results in a member of the same set.
Here, we're discussing the set of irrational numbers and their closure properties under addition. Irrational numbers are numbers that cannot be expressed as a fraction [tex]\( \frac{a}{b} \)[/tex] where [tex]\( a \)[/tex] and [tex]\( b \)[/tex] are integers and [tex]\( b \neq 0 \)[/tex].
Let's analyze each given choice:
1. [tex]\(\pi + (-\pi) = 0\)[/tex]
- [tex]\(\pi\)[/tex] (pi) is an irrational number.
- [tex]\(-\pi\)[/tex] is also irrational because it is simply the negative of an irrational number.
- Adding [tex]\(\pi + (-\pi)\)[/tex] results in 0, which is a rational number (since it can be expressed as [tex]\( \frac{0}{1} \)[/tex]).
This choice shows that the sum of two irrational numbers can result in a rational number, indicating that the set of irrational numbers is not closed under addition.
2. [tex]\(\frac{1}{2} + \left(-\frac{1}{2}\right) = 0\)[/tex]
- [tex]\(\frac{1}{2}\)[/tex] and [tex]\(-\frac{1}{2}\)[/tex] are both rational numbers.
- Adding [tex]\(\frac{1}{2} + \left(-\frac{1}{2}\right)\)[/tex] results in 0, which is a rational number.
This choice involves rational numbers only and doesn't address the closure of the set of irrational numbers.
3. [tex]\(\pi + \pi = 2\pi\)[/tex]
- [tex]\(\pi\)[/tex] is an irrational number.
- Adding [tex]\(\pi + \pi\)[/tex] results in [tex]\(2\pi\)[/tex], which is also irrational.
This choice does not illustrate the set of irrational numbers not being closed under addition, as the result is still an irrational number.
4. [tex]\(\frac{1}{2} + \frac{1}{2} = 1\)[/tex]
- [tex]\(\frac{1}{2}\)[/tex] is a rational number.
- Adding [tex]\(\frac{1}{2} + \frac{1}{2}\)[/tex] results in 1, which is also a rational number.
This choice involves rational numbers only and, therefore, is not relevant to demonstrating the closure properties of irrational numbers.
Thus, the answer choice that shows the set of irrational numbers is not closed under addition is:
[tex]\[\boxed{\pi + (-\pi) = 0}\][/tex]
