Chapter 4.9, Problem 10OE

To determine

To justify the statements

Explanation of Solution

Given information:

The given statement is that the solution set of $x^2+1>0$ is $\left\{\text{real numbers}\right\}$ .

Proof:

First write the inequality

  $\begin{array}{l}{x}^2+1>0\\ x^2>-1\left[\text{add}\left(-1\right)\text{ each sides}\right]\end{array}$

As $-1$ is a negative integer therefore for all real value of x, $x^2$ is greater than $-1$ suppose that $x=0$ then $x^2$ becomes 0 which is also greater than $-1$ because square of any real number is a non-negative real number.

Hence, the solution set of $x^2+1>0$ is $\left\{\text{real numbers}\right\}$ .