Chapter 6.1, Problem 31OE

To determine

If $\left(x\right)$ is a real number then $\sqrt[4]{x^4}=\left|x\right|$ and $\sqrt[5]{x^5}=x$ . Using a numerical example show that why absolute sign must be there in first case but not in second.

Explanation of Solution

Given information: It is given that $\left(x\right)$ is a real number such that,

  $\begin{array}{c}\sqrt[4]{x^4}=\left|x\right|\mathrm{......}\left(1\right)\\ \text{and }\sqrt[5]{x^5}=x\mathrm{......}\left(2\right)\end{array}$

Proof: By definition, we know whenever a real number come outside the root-function, then there are two cases, either the number will be even or odd. For even number we place absolute sign on the number and for odd number no absolute sign is placed.

It can be shown with the help of a numerical example according to question. Therefore from (1) and (2),

  $\begin{array}{c}\sqrt[4]{{\left(-1\right)}^4}=\left|-1\right|\mathrm{......}\left(\text{For }x=-1\right)\\ {\left(-1\right)}^{\frac{4}{4}}=1\\ -1=\mathrm{1......}\left(\begin{array}{l}\text{Contradiction, so }\\ \text{absolute sign is used}\end{array}\right)\mathrm{....}\left(3\right)\\ \text{and }\sqrt[5]{{\left(-1\right)}^5}=-\mathrm{1......}\left(\text{For }x=-1\right)\\ {\left(-1\right)}^{\frac{5}{5}}=-1\\ -1=-\mathrm{1......}\left(\text{which is true}\right)\mathrm{....}\left(4\right)\end{array}$

From (4) it can be concluded that whenever there is odd index no absolute sign is required whereas from (3) it can be concluded that whenever there is even index the absolute sign is used.

Hence proved.