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Sagot :
To determine which answer choice shows that the set of irrational numbers is not closed under addition, let's examine each option in detail.
1. [tex]\(\pi + (-\pi) = 0\)[/tex]
- [tex]\(\pi\)[/tex] (pi) is an irrational number.
- [tex]\(-\pi\)[/tex] is also an irrational number because the negation of an irrational number is still irrational.
- However, when we add [tex]\(\pi\)[/tex] and [tex]\(-\pi\)[/tex] together, we get:
[tex]\[ \pi + (-\pi) = 0 \][/tex]
- The result, [tex]\(0\)[/tex], is a rational number (since it can be expressed as the fraction [tex]\(\frac{0}{1}\)[/tex]).
- This example demonstrates that when adding two irrational numbers, the result can be 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]
- Both [tex]\(\frac{1}{2}\)[/tex] and [tex]\(-\frac{1}{2}\)[/tex] are rational numbers.
- The result is also [tex]\(0\)[/tex], which is a rational number.
- This example involves only rational numbers and does not address the closure property of irrational numbers under addition.
3. [tex]\(\pi + \pi = 2\pi\)[/tex]
- [tex]\(\pi\)[/tex] is an irrational number.
- Adding [tex]\(\pi\)[/tex] to itself yields [tex]\(2\pi\)[/tex].
- [tex]\(2\pi\)[/tex] is still an irrational number.
- This example shows that adding two irrational numbers (in this case, [tex]\(\pi\)[/tex] + [tex]\(\pi\)[/tex]) results in another irrational number and does not demonstrate that the set of irrational numbers is not closed under addition.
4. [tex]\(\frac{1}{2} + \frac{1}{2} = 1\)[/tex]
- Both [tex]\(\frac{1}{2}\)[/tex] and [tex]\(\frac{1}{2}\)[/tex] are rational numbers.
- The result is [tex]\(1\)[/tex], which is also a rational number.
- Like the second option, this example involves only rational numbers and does not pertain to the closure property of irrational numbers under addition.
After examining all the choices, we can conclude that:
The first choice, [tex]\(\pi + (-\pi) = 0\)[/tex], is the correct example showing that the set of irrational numbers is not closed under addition. The resulting sum is a rational number, thereby highlighting that the addition of two irrational numbers does not always yield another irrational number.
1. [tex]\(\pi + (-\pi) = 0\)[/tex]
- [tex]\(\pi\)[/tex] (pi) is an irrational number.
- [tex]\(-\pi\)[/tex] is also an irrational number because the negation of an irrational number is still irrational.
- However, when we add [tex]\(\pi\)[/tex] and [tex]\(-\pi\)[/tex] together, we get:
[tex]\[ \pi + (-\pi) = 0 \][/tex]
- The result, [tex]\(0\)[/tex], is a rational number (since it can be expressed as the fraction [tex]\(\frac{0}{1}\)[/tex]).
- This example demonstrates that when adding two irrational numbers, the result can be 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]
- Both [tex]\(\frac{1}{2}\)[/tex] and [tex]\(-\frac{1}{2}\)[/tex] are rational numbers.
- The result is also [tex]\(0\)[/tex], which is a rational number.
- This example involves only rational numbers and does not address the closure property of irrational numbers under addition.
3. [tex]\(\pi + \pi = 2\pi\)[/tex]
- [tex]\(\pi\)[/tex] is an irrational number.
- Adding [tex]\(\pi\)[/tex] to itself yields [tex]\(2\pi\)[/tex].
- [tex]\(2\pi\)[/tex] is still an irrational number.
- This example shows that adding two irrational numbers (in this case, [tex]\(\pi\)[/tex] + [tex]\(\pi\)[/tex]) results in another irrational number and does not demonstrate that the set of irrational numbers is not closed under addition.
4. [tex]\(\frac{1}{2} + \frac{1}{2} = 1\)[/tex]
- Both [tex]\(\frac{1}{2}\)[/tex] and [tex]\(\frac{1}{2}\)[/tex] are rational numbers.
- The result is [tex]\(1\)[/tex], which is also a rational number.
- Like the second option, this example involves only rational numbers and does not pertain to the closure property of irrational numbers under addition.
After examining all the choices, we can conclude that:
The first choice, [tex]\(\pi + (-\pi) = 0\)[/tex], is the correct example showing that the set of irrational numbers is not closed under addition. The resulting sum is a rational number, thereby highlighting that the addition of two irrational numbers does not always yield another irrational number.
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