Q10. - Algebra 2

Question

Identify the additive inverse and multiplicative inverse for each number 

10. $-8$

Step-by-Step Solution

Answer

Additive inverse of $-8$ is 8 and multiplicative inverse of $-8$ is $-\frac{1}{8}$ 

Step 1 - Define additive inverse

Let $‘a’$ be a real number such that

$a+\left(-a\right)=0=\left(-a\right)+a$

Then $‘-a’$ is the additive inverse of $‘a’$

Step 2 - Additive inverse of - 8

Substitute $-8$ for $a$ 

 $a=-8$

Multiply both sides by -1

$\begin{array}{c}\left(-1\right)\times a=\left(-1\right)\times \left(-8\right)\\ -a=8\end{array}$

Thus additive inverse of $-8$ is 8

Step 3 - Define multiplicative inverse

Let ‘a’ be a real number with $a\cancel{=}0$ such that

$a·\frac{1}{a}=1=\frac{1}{a}·a$

Then '$\frac{1}{a}$' is the multiplicative inverse of ‘a’

Step 4 - Multiplicative inverse of - 8

Substitute $-8$ for a

$a=-8$

Reciprocal on both sides

 $\begin{array}{c}\frac{1}{a}=\frac{1}{-8}\\ \frac{1}{a}=-\frac{1}{8}\end{array}$

Thus multiplicative inverse of $-8$ is $-\frac{1}{8}$ .