Chapter 6.1, Problem 27OE

To determine

Whether the statement $\sqrt{x+1}$ is always a real number, is true or not if $\left(x\right)$ belongs to real.

Answer to Problem 27OE

No, the given statement is false.

Explanation of Solution

Given information: The given statement for variable $\left(x\right)$ which belongs to real is,

  $\left(\sqrt{x+1}\text{ is always a real number}\right)\mathrm{......}\left(1\right)$

Calculation: From (1),

  $\left(\begin{array}{c}\sqrt{x+1}=\text{real}\mathrm{......}\left(\text{For }x\ge -1\right)\\ =\text{non-real}\mathrm{......}\left(\text{For }x<-1\right)\end{array}\right)\mathrm{......}\left(2\right)$

From (2) the statement $\sqrt{x+1}$ gives non-real values for $\left(x<-1\right)$ . Hence it can be said that $\sqrt{x+1}$ is not always a real number. So the given statement is false.