Chapter R, Problem 1RE

To determine

Which element from $A=\left\{\sqrt{6},0,-8,1.\overline{45},\sqrt{9},-\frac{2}{3},3\pi \right\}$ belongs to the given set.

Answer to Problem 1RE

Solution:

(a) $\sqrt{9}$

(b) $\sqrt{9},0$

(c) $\sqrt{9},0-8$

(d) $0,-8,\sqrt{9},1.\overline{45},-\frac{2}{3}$

(e) $\sqrt{6}\text{,}3\pi$

(f) $\sqrt{6},0,-8,1.\overline{45},\sqrt{9},-\frac{2}{3},3\pi$

Explanation of Solution

Given:

The set is $A=\left\{\sqrt{6},0,-8,1.\overline{45},\sqrt{9},-\frac{2}{3},3\pi \right\}$.

(a) N represents the set of natural numbers.

i.e., $\text{N}=\left\{1,2,\mathrm{3....}\right\}$

As $\sqrt{9}=3$.

Thus, $3\in \text{N}$.

(b) W represents the set of whole numbers.

i.e., $\text{W}=\left\{0,1,2,\mathrm{3....}\right\}$.

As $\sqrt{9}=3$.

Thus, $\left\{\sqrt{9},0\right\}\in \text{W}$.

(c) Z represents the set of integers that includes zero, negative numbers and positive numbers.

i.e., $\text{Z}=\left\{\mathrm{....}-2,-1,0,1,2,\mathrm{3....}\right\}$.

As $\sqrt{9}=3$.

Then, $\left\{\sqrt{9},0-8\right\}\in \text{Z}$.

(d) Q represents set of rational numbers that are in the form of fractions, $\frac{p}{q}$ where $p,q\in \text{Z}$ $q\ne 0$.

Then,$\left\{0,-8,\sqrt{9},1.\overline{45},-\frac{2}{3}\right\}\in \text{Q}$.

(e) H represents the set of irrational numbers that are exactly opposite to rational numbers.

Then $\left\{\sqrt{6}\text{,}3\pi \right\}\in \text{H}$.

(f) R represents the set of real numbers, i.e., all the numbers that are represented on the number line:

$\left\{\sqrt{6},0,-8,1.\overline{45},\sqrt{9},-\frac{2}{3},3\pi \right\}\in \text{ R}$