Which of the following is an example of why irrational numbers are not closed under addition?

A. [tex]\sqrt{4}+\sqrt{4}=2+2=4[/tex], and 4 is not irrational

B. [tex]\frac{1}{2}+\frac{1}{2}=1[/tex], and 1 is not irrational

C. [tex]\sqrt{10}+(-\sqrt{10})=0[/tex], and 0 is not irrational

D. [tex]-3+3=0[/tex], and 0 is not irrational



Answer :

Let's go through each example step-by-step to determine which one demonstrates that irrational numbers are not closed under addition.

Example 1: [tex]\(\sqrt{4} + \sqrt{4} = 2 + 2 = 4\)[/tex]

- Here, [tex]\(\sqrt{4}\)[/tex] is equal to 2, which is not an irrational number. Thus, the sum [tex]\(2 + 2 = 4\)[/tex] does not involve irrational numbers. This example does not show anything about the closure property of irrational numbers under addition.

Example 2: [tex]\(\frac{1}{2} + \frac{1}{2} = 1\)[/tex]

- Both [tex]\(\frac{1}{2}\)[/tex] and [tex]\(\frac{1}{2}\)[/tex] are rational numbers. Their sum, [tex]\(1\)[/tex], is also a rational number. This example does not involve irrational numbers and therefore does not demonstrate anything about the closure of irrational numbers under addition.

Example 3: [tex]\(\sqrt{10} + (-\sqrt{10}) = 0\)[/tex]

- Here, [tex]\(\sqrt{10}\)[/tex] is an irrational number. When you add [tex]\(\sqrt{10}\)[/tex] and [tex]\(-\sqrt{10}\)[/tex], you get [tex]\(0\)[/tex]. Since [tex]\(0\)[/tex] is a rational number and not an irrational number, this example shows that adding two irrational numbers can result in a rational number.

Example 4: [tex]\(-3 + 3 = 0\)[/tex]

- Both [tex]\(-3\)[/tex] and [tex]\(3\)[/tex] are rational numbers. Their sum, [tex]\(0\)[/tex], is also a rational number. This example does not involve irrational numbers and thus does not demonstrate anything about the closure of irrational numbers under addition.

Therefore, the correct example that illustrates why irrational numbers are not closed under addition is:
- [tex]\(\sqrt{10} + (-\sqrt{10}) = 0\)[/tex], and [tex]\(0\)[/tex] is not irrational.

This is Example 3.