Chapter A.10, Problem 117AYU

To determine

To find: The example to show that $\sqrt{{\left(a\right)}^2}$ is not equal to $a$ and to explain $\sqrt{{\left(a\right)}^2}=\left|a\right|$ .

Answer to Problem 117AYU

The below example and explanation shows $\sqrt{{\left(a\right)}^2}$ is not equal to $a$ and $\sqrt{{\left(a\right)}^2}=\left|a\right|$ .

Explanation of Solution

Concept used:

Modulus of any number $x$ , is given by:

  $\left|x\right|=x$ , $x\ge 0$ and

  $\left|x\right|=-x$ , $x\le 0$

Choose, $a=-2$

Then $\begin{array}{c}\sqrt{{\left(-2\right)}^2}=\sqrt{4}\\ =2\end{array}$

Therefore, by above example it is shown $\sqrt{{\left(a\right)}^2}$ is not equal to $a$ .

To explain $\sqrt{{\left(a\right)}^2}=\left|a\right|$ further simplified as:

Case1:

If $a\ge 0$ ,then

  $\sqrt{{\left(a\right)}^2}=a$ .

Case2:

If $a\le 0$ ,then

Suppose $a=-b$ where $b\ge 0$ then

  $\begin{array}{c}\sqrt{{\left(a\right)}^2}=\sqrt{{\left(-b\right)}^2}\\ =\sqrt{{\left(b\right)}^2}\end{array}$

As $a=-b$ & $b\ge 0$ using case1 further simplified as,

  $\begin{array}{c}\sqrt{{\left(a\right)}^2}=b\\ =-a\end{array}$

Therefore, by using case1 and case2 it is shown that $\sqrt{{\left(a\right)}^2}=\left|a\right|$ .