Answer
a. Natural Number is $\sqrt 64$ only
b. Whole numbers are 0 and $\sqrt 64$
c. Integers are 0, $\sqrt 64$ and -11.
d. Rational numbers are -$\frac{5}{6}$, 0.75, 0 , $\sqrt 64$ , -11
e. Irrational numbers are $\sqrt 5$ and $\pi$.
f. All the numbers in the given set { -11, -$\frac{5}{6}$, 0, 0.75, $\sqrt 5$, $\pi$, $\sqrt 64$ } are real numbers.
Work Step by Step
a. Natural Numbers are the numbers used for counting. The only number in the set { -11, -$\frac{5}{6}$, 0, 0.75, $\sqrt 5$, $\pi$, $\sqrt 64$ } is $\sqrt 64$ because $\sqrt 64$ is 8.
b. Whole numbers consists of 0 and Natural numbers. So, the whole numbers in the set { -11, -$\frac{5}{6}$, 0, 0.75, $\sqrt 5$, $\pi$, $\sqrt 64$ } are 0 and $\sqrt 64$.
c. The set of integers includes the negative of the natural numbers and the whole numbers. So the set { -11, -$\frac{5}{6}$, 0, 0.75, $\sqrt 5$, $\pi$, $\sqrt 64$ } contains the integers 0, $\sqrt 64$ and -11.
d. The set of rational numbers is the set of all numbers that can be expressed as a quotient of two integers, with the denominator not 0. Rational numbers can be expressed as terminating or repeating decimals. In the set { -11, -$\frac{5}{6}$, 0, 0.75, $\sqrt 5$, $\pi$, $\sqrt 64$ }, -$\frac{5}{6}$, 0.75, 0 (0=0/1), $\sqrt 64$ ($\sqrt 64$ =8= 8 /1), -11 (-11/1 = -11) are rational numbers.
e. The set of irrational numbers is the set of all numbers whose decimal representations are neither terminating nor repeating and also cannot be expressed as a quotient of integers. In the given set { -11, -$\frac{5}{6}$, 0, 0.75, $\sqrt 5$, $\pi$, $\sqrt 64$ }, In decimal form, $\sqrt 5$ and \pi neither terminate nor have blocks of repeating digits. So the only irrational number in the given set are $\sqrt 5$ and $\pi$.
f. Real Numbers are either rational or irrational. So all the numbers in the given set { -11, -$\frac{5}{6}$, 0, 0.75, $\sqrt 5$, $\pi$, $\sqrt 64$ } are real numbers.
