Question

Name the set or sets of numbers to which each real number belongs.

$-\sqrt{64}$

Short answer

The real number $-\sqrt{64}$belongs to the sets of integers and rational numbers

Step-by-step solution

Step-1. Apply the concept of the real number system

Real numbers include all the numbers except complex numbers and have the following five subsets:

1. Natural numbers: Includes counting objects and start from 1.
2. Whole numbers: Includes the set of natural numbers along with 0.
3. Integers: Z = Includes numbers that are not fraction (positive and negative whole numbers)
4. Rational numbers: Includes the numbers which can be written in the form of$\frac{p}{q}$ where p and q are integers, $q\ne 0$.
5. Irrational numbers: Includes numbers that cannot be written in the form of $\frac{p}{q}$where p and q are integers, $q\ne 0$.

Step-2. Examples of the real number system.

1. Natural numbers:$\left\{1,2,3,4,...\right\}$

2. Whole numbers:$\left\{0,1,2,3,4,...\right\}$

3. Integers: Measurement of debts, temperatures, etc., fall under the set of integers $\left\{...,-3,-2,-1,0,1,2,3,...\right\}$

4. Rational numbers: If we cut a cake into equal pieces, then we may have a piece that represents a fraction like$\frac{5}{6},1.5\left(=\frac{3}{2}\right),\frac{6}{9},\frac{8}{3}$

5. Irrational numbers: The numbers that are square roots of positive rational numbers, cube roots of rational numbers, etc., such as$\sqrt{2},\sqrt[3]{5},-\sqrt{3}$

Step-3. Analyze the given real number.

Consider the given real number

We can rewrite it using the square root property as:

Now we will check each subset of real numbers:

1. Natural numbers: They are positive and start from 1. So, $-\sqrt{64}$being negative is not a natural number.
2. Whole numbers: They are natural numbers including 0. So, $-\sqrt{64}$being negative is not a whole number.
3. Integers: They are numbers that are not fraction. Since $-\sqrt{64}=-8$, it is an integer.
4. Rational numbers: They can be written in the form $\frac{p}{q}$where p and q are integers, $q\ne 0$. Since $-\sqrt{64}=-\frac{8}{1}$, it is a rational number.
5. Irrational numbers: They cannot be written in the form $\frac{p}{q}$where p and q are integers, $q\ne 0$. Since from the above point,$-\sqrt{64}$ can be written in the form $\frac{p}{q}$, it is not an irrational number.

Therefore, the real number $-\sqrt{64}$belongs to the sets of integers and rational numbers.