Chapter 1.5, Problem 61E

To determine

The values of $a$ and $b$ .

Answer to Problem 61E

The value of $a$ and $b$ can be any integer until they have opposite signs.

Explanation of Solution

Given information: The following equation:

$\left|a+b\right|=\left|a\right|+\left|b\right|$

To find the values of $a$ and $b$ , four cases are considered.

Case 1: Both $a$ and $b$ are positive.

  $\begin{array}{l}\left|a+b\right|=\left|a\right|+\left|b\right|\\ a+b=a+b\end{array}$

Case 2: Both $a$ and $b$ are negative

$\begin{array}{l}\left|(-a)+(-b)\right|=\left|-a\right|+\left|-b\right|\\ \left|-a-b\right|=\left|-a\right|+\left|-b\right|\\ \left|-(a+b)\right|=\left|-a\right|+\left|-b\right|\\ a+b=a+b\end{array}$

Case 3: $a$ is positive, $b$ is negative.

$\begin{array}{l}\left|a+(-b)\right|=\left|a\right|+\left|-b\right|\\ \left|a-b\right|=\left|a\right|+\left|-b\right|\\ a-b<a+b\end{array}$

Case 4: $a$ is negative and $b$ is positive.

$\begin{array}{l}\left|(-a)+b)\right|=\left|-a\right|+\left|-b\right|\\ \left|-a+b\right|=\left|-a\right|+\left|b\right|\\ b-a<b+a\end{array}$

Conclusion: Since the equation is satisfied in cases 3 and 4, it can be concluded that the values of $a$ and $b$ will have different signs.