91Ó°ÊÓ

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.

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

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

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

Case 1: Both a and b are positive.$\begin{array}{l}|a+b|<\left|a\right|+\left|b\right|\\ a+b<a+b\text{ â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰F²¹±ô²õ±ð}\end{array}$

Example: a = 2 and b = 3. $\begin{array}{l}|a+b|<\left|a\right|+\left|b\right|\\ |2+3|<\left|2\right|+\left|3\right|\\ \left|5\right|\text{ â¶Ä‰}<\text{ â¶Ä‰}2+3\\ 5<5\text{ â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰F²¹±ô²õ±ð}\end{array}$

Case 2: Both a and b are negative$\begin{array}{l}\left|(−a)+(−b)\right|<|−a|+|−b|\\ |−a−b|<|−a|+|−b|\\ \left|−(a+b)\right|<|−a|+|−b|\\ a+b<a+b\text{ â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰F²¹±ô²õ±ð}\end{array}$

Example: a = – 2 and b = – 3$\begin{array}{l}|a+b|<\left|a\right|+\left|b\right|\\ \left|(−\text{ }2)\text{ }+\text{ }(−3)\right|<\left|(−2)\right|+\left|(−3)\right|\\ |−\text{ â¶Ä‰â€‰}5|<|−2|+|−3|\\ 5<2+3\\ 5<5\text{ â¶Ä‰â€‰â¶Ä‰â€‰False}\end{array}$

Case 3: a is positive,b is negative.$\begin{array}{l}\left|a+(−b)\right|<\left|a\right|+|−b|\\ |a−b|<\left|a\right|+|−b|\\ a−b<a+b\text{ â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰â€‰True}\end{array}$

Example: a = 2 and b = – 3$\begin{array}{l}|a+b|<\left|a\right|+\left|b\right|\\ \left|2\text{ }+\text{ }(−3)\right|<\left|2\right|+\left|(−3)\right|\\ |−\text{ }1|<\left|2\right|+|−3|\\ −\text{ â¶Ä‰}1<2+3\\ −1<5\text{ â¶Ä‰â€‰â¶Ä‰â€‰True}\end{array}$

Case 4: a is negative and b is positive.$\begin{array}{l}\left|(−a)+b)\right|=|−a|+|−b|\\ |−a+b|=|−a|+\left|b\right|\\ b−a<b+a\text{ â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰True}\end{array}$

Example: a = – 2 and b = 3$\begin{array}{l}|a+b|<\left|a\right|+\left|b\right|\\ |−\text{ â¶Ä‰}2\text{ }+3|<|−\text{ â¶Ä‰}2|+\left|3\right|\\ \left|\text{ }1\right|<|−\text{ â¶Ä‰}2|+\left|3\right|\\ \text{ â¶Ä‰}1<2+3\\ 1<5\text{ â¶Ä‰â€‰â¶Ä‰â€‰True}\end{array}$

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.