Chapter 4.7, Problem 61STP

To determine

To determine an irrational number.

Answer to Problem 61STP

  $\sqrt{10}\text{ is an irrational number}\text{.}$

Explanation of Solution

Given information:

  $\begin{array}{l}\text{A}=-8\text{ C}=\sqrt{10}\\ \text{B}=\frac{3}{4}\text{ D}=\sqrt{16}\end{array}$

Concept used: Irrational number.

  $\begin{array}{l}\text{An irrational number is a number that cannot be expressed as }\frac{\text{a}}{\text{b}},\\ \text{where a and b are integers and b}\ne \text{0}\text{.}\\ \pi =\mathrm{3.14159........}\end{array}$

Calculation : Irrational number can be obtained as,

  $\begin{array}{l}\text{A}=-8\text{ This number is an integer and a rational number}\text{.}\\ \text{B}=\frac{3}{4}\text{ Since }\frac{3}{4}=0.75,\text{this number is a rational number}\text{.}\\ \text{C}=\sqrt{10}\text{ Since }\sqrt{10}=3.16227766\text{ ,it is not the square root of a perfect }\\ \text{ square so it is irrational number}\text{.}\\ \text{D}=\sqrt{16}\text{ Since }\sqrt{16}=4,\text{ this number is a natural number, a whole number, }\\ \text{ an integer, and a rational number}\text{.}\\ \text{Hence }\sqrt{10}\text{ is an irrational number}\text{.}\end{array}$