Answer
a. Natural Numbers : $\sqrt 100 $
b. Whole Numbers : 0, $\sqrt 100 $
c. Integers : -9, 0, $\sqrt 100 $
d. Rational Numbers : -4/5, 0.25, 9.2, 0, $\sqrt 100 $,-9
e. Irrational Numbers : $\sqrt 3 $
f. Real Numbers : -4/5, 0.25, 9.2, 0, $\sqrt 100 $, $\sqrt 3 $,-9
Work Step by Step
a. Natural Numbers are the numbers used for counting. The only number in the set { -9, -4/5, 0, 0.25, $\sqrt 3 $, 9.2, $\sqrt 100 $ } is $\sqrt 100 $ because $\sqrt 100 $ is 10.
b. Whole numbers consists of 0 and Natural numbers. So, the whole numbers in the set { -9, -4/5, 0, 0.25, $\sqrt 3 $, 9.2, $\sqrt 100 $ } are 0, $\sqrt 100 $
c. The set of integers includes the negative of the natural numbers and the whole numbers. So the set { -9, -4/5, 0, 0.25, $\sqrt 3 $, 9.2, $\sqrt 100 $ } contains the integers 0, $\sqrt 100 $ and -9.
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. So, -4/5, 0.25, 9.2, 0 (0=0/1), $\sqrt 100 $ ($\sqrt 100 $ =10= 10 /1), -9 (-9/1 = -9) 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 { -9, -4/5, 0, 0.25, $\sqrt 3 $, 9.2, $\sqrt 100 $ }, $\sqrt 3 $ $\approx$ 1.732. (1.7320508075688..). In decimal form, $\sqrt 3 $ neither terminate nor have blocks of repeating digits. So the only irrational number in the given set is $\sqrt 3 $.
f. Real Numbers are either rational or irrational. So all the numbers in the given set { -9, -4/5, 0, 0.25, $\sqrt 3 $, 9.2, $\sqrt 100 $ } are real numbers.
