To determine which answer choice demonstrates that the set of irrational numbers is not closed under addition, let's analyze each option and see if it satisfies the properties of irrational numbers.
1. \(\pi + (-\pi) = 0\)
Here, \(\pi\) is an irrational number. Adding \(-\pi\) (which is also irrational), results in \(0\).
- \(0\) is a rational number.
- This does not demonstrate the property because the sum of two irrationals in this case is rational.
2. \(\frac{1}{2} + \left( -\frac{1}{2} \right) = 0\)
Here, \(\frac{1}{2}\) is a rational number. Adding \(-\frac{1}{2}\), also rational, results in \(0\).
- This operation involves rational numbers and not irrational numbers.
- This cannot be used to demonstrate properties of irrational sets.
3. \(\pi + \pi = 2\pi\)
Here, \(\pi\) is an irrational number. Adding \(\pi\) to itself results in \(2\pi\).
- \(\pi\) being irrational implies \(2\pi\) is also irrational, because a non-zero rational number times an irrational number is still irrational.
- This example shows that the addition of two irrationals does not result in a rational number, but in another irrational number.
4. \(\frac{1}{2} + \frac{1}{2} = 1\)
Here, \(\frac{1}{2}\) is a rational number. Adding \(\frac{1}{2}\) to itself results in \(1\).
- This operation involves rational numbers and not irrational numbers.
- This cannot be used to demonstrate properties of irrational sets.
The correct answer is:
[tex]\[
\boxed{3}
\][/tex]
[tex]\(\pi + \pi = 2\pi\)[/tex] demonstrates that the sum of two irrational numbers (in this case [tex]\(\pi\)[/tex]) is an irrational number ([tex]\(2\pi\)[/tex]), showing that irrational numbers are not closed under addition.
