Q. 243 - Intermediate Algebra

Question

(a) Explain why ${(-\sqrt{n})}^2$ is always non-negative, for $n\ge 0$

(b) Explain why $-{\left(\sqrt{n}\right)}^2$ is always non-positive, for $n\ge 0$

Step-by-Step Solution

Answer

(a) ${(-\sqrt{n})}^2$ is always non-negative for $n\ge 0$ since the square is applicable to both negative sign and radical, so negative sign becomes positive and the resultant is non-negative number.

(b) $-{\left(\sqrt{n}\right)}^2$ is always non-positive for $n\ge 0$, since the square is only applicable to radical and not negative sign. So the resultant is non-positive number.

Part (a) Step 1. To explain

To explain why ${(-\sqrt{n})}^2$ is non-negative for $n\ge 0$

Part (a) Step 2. Consider a number

Consider a number $n\ge 0$

That is $n=5$

${(-\sqrt{n})}^2={(-\sqrt{5})}^2$

The square is applicable to both negative sign and the radical. So,

               $=(-\sqrt{5})(-\sqrt{5})$

               $=5$ which is a non-negative number.

So, for any $n\ge 0$, ${(-\sqrt{n})}^2$ is non-negative..

Part (b) Step 1. To explain

To explain why $-{\left(\sqrt{n}\right)}^2$ is non-positive.

Part (b) Step 4. Consider a number.

Consider a number $n\ge 0$

That is $5$

$-{\left(\sqrt{n}\right)}^2=-{\left(\sqrt{5}\right)}^2$

The root is applicable only to radical.

              $=-\left(\sqrt{5}\right)\left(\sqrt{5}\right)$

              $=-\left(5\right)$ which is a negative number.

So, for any $n\ge 0,-{\left(\sqrt{n}\right)}^2$ is non-positive.