Question

Challenge: Find values of a and b for which the statement is true.

|a + b| < |a| + |b|

Short answer

It can be concluded that the values of a and b will have different signs for which the statement is true.

Step-by-step solution

– What is absolute value and how to determine the value of variable within the absolute value. 

Absolute Value

Absolute value signs make the final result of the operations between them positive. When deciding on the order of operations, treat absolute value bars like parentheses.

|– 3| = 3: The absolute value bars make negative 3 positive.

| 3 | = 3: The absolute value bars leave positive 3 unchanged

– |– 3| = – 3: The absolute value bars make negative 3 positive. Then the negative sign takes effect, and the answer is a negative 3.

– | 3 | = – 3: The absolute value bars leave positive 3 unchanged. Then the negative sign takes effect, and the answer is a negative 3.

Step 2:  Name of the property of absolute value 

The simplest case being when aand bwill have different signs; in this case called

Subadditivity property: $|a+b|\le \left|a\right|+\left|b\right|$

Find values of a and b for which the statement is true.                            |a + b| < |a| + |b|

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

Case 1: Both a and b are positive.$\begin{array}{l}|a+b|<\left|a\right|+\left|b\right|\\ a+b<a+b\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}False}\end{array}$

Example: a = 2 and b = 3. $\begin{array}{l}|a+b|<\left|a\right|+\left|b\right|\\ |2+3|<\left|2\right|+\left|3\right|\\ \left|5\right|<2+3\\ 5<5\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}False}\end{array}$

Case 2: Both a and b are negative$\begin{array}{l}\left|(-a)+(-b)\right|<|-a|+|-b|\\ |-a-b|<|-a|+|-b|\\ \left|-(a+b)\right|<|-a|+|-b|\\ a+b<a+b\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}False}\end{array}$

Example: a = – 2 and b = – 3$\begin{array}{l}|a+b|<\left|a\right|+\left|b\right|\\ \left|(-2)+(-3)\right|<\left|(-2)\right|+\left|(-3)\right|\\ |-5|<|-2|+|-3|\\ 5<2+3\\ 5<5\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}False}\end{array}$

Case 3: a is positive,b is negative.$\begin{array}{l}\left|a+(-b)\right|<\left|a\right|+|-b|\\ |a-b|<\left|a\right|+|-b|\\ a-b<a+b\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}True}\end{array}$

Example: a = 2 and b = – 3$\begin{array}{l}|a+b|<\left|a\right|+\left|b\right|\\ \left|2+(-3)\right|<\left|2\right|+\left|(-3)\right|\\ |-1|<\left|2\right|+|-3|\\ -1<2+3\\ -1<5\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}True}\end{array}$

Case 4: a is negative and b is positive.$\begin{array}{l}\left|(-a)+b)\right|=|-a|+|-b|\\ |-a+b|=|-a|+\left|b\right|\\ b-a<b+a\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}True}\end{array}$

Example: a = – 2 and b = 3$\begin{array}{l}|a+b|<\left|a\right|+\left|b\right|\\ |-2+3|<|-2|+\left|3\right|\\ \left|1\right|<|-2|+\left|3\right|\\ 1<2+3\\ 1<5\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}True}\end{array}$

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.