Chapter 1.3, Problem 3P

To determine

To find: the subset of real number in which the number belong to the subset.

Answer to Problem 3P

Here $\left(a\right)$ is positive integer, $\left(b\right)$ decimal number, $\left(c\right)$ decimal number and $\left(d\right)$ irrational number.

Explanation of Solution

Given:

  $\left(a\right)\sqrt{9},\left(b\right)\frac{3}{10},\left(c\right)0.45\text{ and }\left(d\right)\sqrt{12}$ .

Concept used:

The number system is as follows:

Number system or real number $\left(R\right)$

consist of two types namely:

Rational numbers and irrational numbers.

Rational numbers are all terminating or recurring decimals and irrational numbers are all non -terminating and non- recurring decimals.

Rational numbers $\left(Q\right)$ is divided by two types namely integers and non-integers.

Integers are those numbers which are in the form composite number without the fraction and non-integers are numbers in fractions.

Integers are further divided into two types as:

Negative numbers and whole numbers and whole number is further divided into two type as natural number and zero.

From the above description it is clear that the rational and irrational numbers together form the set of real numbers.

Calculation:

  $\left(a\right):\sqrt{9}$ .

This can be simplified as:

  $\begin{array}{l}\sqrt{9}=\sqrt{3^2}.\\ \Rightarrow {\left(3^2\right)}^{\frac{1}{2}}=3.\end{array}$

This number is positive integer.

  $\left(b\right):\frac{3}{10}=0.3$ .

The number is decimal number since, it is terminating and recurring decimals.

  $\left(c\right)-\mathrm{0.45.}$

The number is terminating decimal in negative direction.

  $\left(d\right)\sqrt{12}=3.46410161151$ .

Since, the number can be makes into exact fraction so, the number is irrational number.

Hence, $\left(a\right)$ is positive integer, $\left(b\right)$ decimal number, $\left(c\right)$ decimal number and $\left(d\right)$ irrational number.