Answer
$a.$ $\{\sqrt{81}\}$
$b.$ $\{0,\sqrt{81}\}$
$c.$ $\{-17,0,\sqrt{81}\}$
$d.$ $\Big\{-17,0,\sqrt{81},-\dfrac{9}{13},0.75\Big\}$
$e.$ $\{\pi,\sqrt{2}\}$
$f.$ $\Big\{-17,-\dfrac{9}{13},0,0.75,\sqrt{2},\pi,\sqrt{81}\Big\}$
Work Step by Step
$\Big\{-17,-\dfrac{9}{13},0,0.75,\sqrt{2},\pi,\sqrt{81}\Big\}$
$a.$ $\textbf{Natural numbers}$
The natural numbers are the positive integers and are commonly used for counting. In the given set, only $\sqrt{81}$ is a natural number because $\sqrt{81}=9$ and $9$ is a natural number.
$b.$ $\textbf{Whole numbers}$
The whole numbers are all the positive integers including the number $0$. The whole numbers included in the given set are $0$ and $\sqrt{81}$
$c.$ $\textbf{Integers}$
An integer is a number that can be represented without a fractional component. In the given set, $-17$, $0$ and $\sqrt{81}$ are integers
$d.$ $\textbf{Rational numbers}$
Any number that can be expressed as the quotient of two integers is a rational number. In the given set, $-17$, $-\dfrac{9}{13}$, $0$, $0.75$ and $\sqrt{81}$ are rational numbers.
$e.$ $\textbf{Irrational numbers}$
If a real number is not rational, then it is irrational. In the given set, $\sqrt{2}$ and $\pi$ are irrational numbers.
$f.$ $\textbf{Real numbers}$
Since the real numbers include all the numbers that represent a quantity along a line, the rational, irrational, whole and natural numbers, and integers are real numbers. All numbers in the set are real numbers.
