Chapter 1.3, Problem 75S

To determine

To find: the additive inverse of the given number.

Answer to Problem 75S

The additive inverse of the given number is $4\frac{1}{5}$.

Explanation of Solution

Given information:

A real number is given as

  $-4\frac{1}{5}$

Concept used:

For any real numbers $a,b$ , and $c$ real numbers properties are tabulated below:

Property	Addition	Multiplication
Commutative	$a+b=b+a$	$a\cdot b=b\cdot a$
Associative	$\left(a+b\right)+c=a+\left(b+c\right)$	$\left(a\cdot b\right)\cdot c=a\cdot \left(b\cdot c\right)$
Identity	$a+0=0+a=a$ 0 is identity element for addition	$a\cdot 1=1\cdot a=a$ 1 is identity element for Multiplication
Inverse	$a+\left(-a\right)=\left(-a\right)+a=0$$\left(-a\right)$ is inverse element for addition	$a\cdot \frac{1}{a}=\frac{1}{a}\cdot a=1$$\frac{1}{a}$ is inverse element for Multiplication
Closure	$a+b$ is real number	$a\cdot b$ is real number
Distributive	$\begin{array}{l}a\left(b+c\right)=ab+bc\\ \left(b+c\right)a=ba+ca\end{array}$	$\begin{array}{l}a\left(b+c\right)=ab+bc\\ \left(b+c\right)a=ba+ca\end{array}$

Calculation:

Consider the given number.

  $-4\frac{1}{5}$

Let the additive inverse of the number be $a$.

  $\begin{array}{c}-4\frac{1}{5}+a=0\\ -\frac{21}{5}+a+\frac{21}{5}=0+\frac{21}{5}\\ a=\frac{21}{5}\\ a=4\frac{1}{5}\end{array}$

Hence, the additive inverse of the number is $4\frac{1}{5}$.