## College Algebra (6th Edition)

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
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.