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Absolute Value

Module by: Community College Online Textbook Project. E-mail the author

Summary: This document is a basic introduction to absolute value.

When we want to talk about how “large” a number is without regard as to whether it is positive or negative, we use the absolute value function.

Geometric Definition of Absolute Value

The absolute value of a number a, denoted ∣a∣∣a∣ size 12{ lline a rline } {}, is the distance from that number to the origin (zero) on the number line. Absolute value answers the question of “how far,” not “which way.” That distance is always given as a nonnegative number. In short:

  • If a number is positive (or zero), the absolute value function does nothing to it: ∣3∣=3∣3∣=3 size 12{ lline 3 rline =3} {}
  • If a number is negative, the absolute value function makes it positive: ∣−3∣=3∣−3∣=3 size 12{ lline -3 rline =3} {}

WARNING: If there is arithmetic to do inside the absolute value function, you must do it before taking the absolute value—the absolute value function acts on the result of whatever is inside it. For example, a common error is

∣5+−2∣=5+2=7∣5+−2∣=5+2=7 size 12{ lline 5+ left (-2 right ) rline =5+2=7} {} (Wrong!)

The mistake here is in assuming that the absolute value makes everything inside it positive. This is not true. It only makes the result positive. The correct result is

∣5+−2∣=∣3∣=3∣5+−2∣=∣3∣=3 size 12{ lline 5+ left (-2 right ) rline = lline 3 rline =3} {} (Correct)

Examples

Determine each value.

∣4∣=∣4∣= size 12{ lline 4 rline ={}} {}__

  • The answer is 4:
    Figure 1
    Figure 1 (graphics1.png)

∣−4∣=∣−4∣= size 12{ lline -4 rline ={}} {}__

  • The answer is 4:
    Figure 2
    Figure 2 (graphics2.png)

∣0∣=∣0∣= size 12{ lline 0 rline ={}} {}__

  • The answer is 0.

−∣5∣=−∣5∣= size 12{- lline 5 rline ={}} {}__

  • The quantity on the left side of the equal sign is read as “negative the absolute value of 5.” The absolute value of 5 is 5. Hence, negative the absolute value of 5 is −5.

−∣−3∣=−∣−3∣= size 12{- lline -3 rline ={}} {}__

  • The quantity on the left side of the equal sign is read as “negative the absolute value of −3.” The absolute value of −3 is 3. Hence, negative the absolute value of −3 is −(3) = −3.

Geometric Absolute Value Exercises

By reasoning geometrically, determine each absolute value.

Exercise 1

∣7∣=∣7∣= size 12{ lline 7 rline ={}} {}__

Solution

7

Exercise 2

∣−3∣=∣−3∣= size 12{ lline -3 rline ={}} {}__

Solution

3

Exercise 3

∣12∣=∣12∣= size 12{ lline "12" rline ={}} {}__

Solution

12

Exercise 4

∣0∣=∣0∣= size 12{ lline 0 rline ={}} {}__

Solution

0

Exercise 5

−∣9∣=−∣9∣= size 12{- lline 9 rline ={}} {}__

Solution

−9

Exercise 6

−∣−6∣=−∣−6∣= size 12{- lline -6 rline ={}} {}__

Solution

−6

Algebraic Definition of Absolute Value

From the preceding exercises, we can suggest the following algebraic definition of absolute value. Note that the definition has two parts.

The absolute value of a number a is

Figure 3
Figure 3 (graphics3.png)

The algebraic definition takes into account the fact that the number a could be positive or zero (a ≥ 0) or negative (a < 0).

  1. If the number a is positive or zero (a ≥ 0), the first part of the definition applies. The first part of the definition tells us that, if the number enclosed in the absolute value bars is a nonnegative number, the absolute value of the number is the number itself.
  2. The second part of the definition tells us that, if the number enclosed within the absolute value bars is a negative number, the absolute value of the number is the opposite of the number. The opposite of a negative number is a positive number.

Note:

The definition says that the vertical absolute value lines may be eliminated only if we know whether the number inside is positive or negative.

Examples

Use the algebraic definition of absolute value to find the following values.

∣8∣∣8∣ size 12{ lline 8 rline } {} = __

  • The number enclosed within the absolute value bars is a nonnegative number, so the first part of the definition applies. This part says that the absolute value of 8 is 8 itself. ∣8∣∣8∣ size 12{ lline 8 rline } {} = 8

∣−3∣∣−3∣ size 12{ lline -3 rline } {} = __

  • The number enclosed within absolute value bars is a negative number, so the second part of the definition applies. This part says that the absolute value of −3 is the opposite of −3, which is −(−3). By the definition of absolute value and the double-negative property, ∣−3∣∣−3∣ size 12{ lline -3 rline } {} = −(−3) = 3

Algebraic Absolute Value Exercises

Use the algebraic definition of absolute value to find the following values.

Exercise 7

∣7∣∣7∣ size 12{ lline 7 rline } {} = __

Solution

7

Exercise 8

∣9∣∣9∣ size 12{ lline 9 rline } {} = __

Solution

9

Exercise 9

∣−12∣∣−12∣ size 12{ lline -"12" rline } {} = __

Solution

12

Exercise 10

∣−5∣∣−5∣ size 12{ lline -5 rline } {} = __

Solution

5

Exercise 11

−∣8∣−∣8∣ size 12{- lline 8 rline } {} = __

Solution

−8

Exercise 12

−∣1∣−∣1∣ size 12{- lline 1 rline } {} = __

Solution

−1

Exercise 13

−∣−52∣−∣−52∣ size 12{- lline -"52" rline } {} = __

Solution

−52

Exercise 14

−∣−31∣−∣−31∣ size 12{- lline -"31" rline } {} = __

Solution

−31

Module Review Exercises

For the following problems, determine each of the values. <<I figured out these answers; still have to QA>> <<also to verify: have squares & cubes been covered in earlier modules? Subtracting negative numbers? >>

Exercise 15

∣5∣∣5∣ size 12{ lline 5 rline } {} = __

Solution

5

Exercise 16

∣3∣∣3∣ size 12{ lline 3 rline } {} = __

Solution

3

Exercise 17

∣6∣∣6∣ size 12{ lline 6 rline } {} = __

Solution

6

Exercise 18

∣−9∣∣−9∣ size 12{ lline -9 rline } {} = __

Solution

9

Exercise 19

∣−1∣∣−1∣ size 12{ lline -1 rline } {} = __

Solution

1

Exercise 20

∣−4∣∣−4∣ size 12{ lline -4 rline } {} = __

Solution

4

Exercise 21

−∣3∣−∣3∣ size 12{- lline 3 rline } {} = __

Solution

-3

Exercise 22

−∣7∣−∣7∣ size 12{- lline 7 rline } {} = __

Solution

-7

Exercise 23

−∣−14∣−∣−14∣ size 12{- lline -"14" rline } {} = __

Solution

-14

Exercise 24

∣0∣∣0∣ size 12{ lline 0 rline } {} = __

Solution

0

Exercise 25

∣−26∣∣−26∣ size 12{ lline -"26" rline } {} = __

Solution

26

Exercise 26

−∣−26∣−∣−26∣ size 12{- lline -"26" rline } {} = __

Solution

-26

Exercise 27

−−∣4∣−−∣4∣ size 12{- left (- lline 4 rline right )} {} = __

Solution

4

Exercise 28

−−∣2∣−−∣2∣ size 12{- left (- lline 2 rline right )} {} = __

Solution

2

Exercise 29

−−∣−6∣−−∣−6∣ size 12{- left (- lline -6 rline right )} {} = __

Solution

6

Exercise 30

−−∣−42∣−−∣−42∣ size 12{- left (- lline -"42" rline right )} {} = __

Solution

42

Exercise 31

∣5∣−∣−2∣∣5∣−∣−2∣ size 12{ lline 5 rline - lline -2 rline } {} = __

Solution

3

Exercise 32

∣−2∣2∣−2∣2 size 12{ lline -2 rline rSup { size 8{2} } } {} = __

Solution

4

Exercise 33

∣−2⋅3∣∣−2⋅3∣ size 12{ lline - left (2 cdot 3 right ) rline } {} = __

Solution

-6

Exercise 34

∣−2∣−∣−9∣∣−2∣−∣−9∣ size 12{ lline -2 rline - lline -9 rline } {} = __

Solution

-7

Exercise 35

∣−6∣+∣4∣2∣−6∣+∣4∣2 size 12{ left ( lline -6 rline + lline 4 rline right ) rSup { size 8{2} } } {} = __

Solution

100

Exercise 36

∣−1∣−∣1∣3∣−1∣−∣1∣3 size 12{ left ( lline -1 rline - lline 1 rline right ) rSup { size 8{3} } } {} = __

Solution

0

Exercise 37

∣4∣+∣−6∣2−∣−2∣3∣4∣+∣−6∣2−∣−2∣3 size 12{ left ( lline 4 rline + lline -6 rline right ) rSup { size 8{2} } - left ( lline -2 rline right ) rSup { size 8{3} } } {} = __

Solution

92

Exercise 38

−∣−10∣−62−∣−10∣−62 size 12{- left [ lline -"10" rline -6 right ] rSup { size 8{2} } } {} = __

Solution

-16 (Note that this is different from the answer for −∣−10∣−62−∣−10∣−62 size 12{ left [- left ( lline -"10" rline -6 right ) right ] rSup { size 8{2} } } {}; pay attention to the order of operations.)

Exercise 39

−−−∣−4∣+∣−3∣32−−−∣−4∣+∣−3∣32 size 12{- left lbrace - left [- lline -4 rline + lline -3 rline right ] rSup { size 8{3} } right rbrace rSup { size 8{2} } } {} = __

Solution

-1

Exercise 40

A Mission Control Officer at Cape Canaveral makes the statement “lift-off, T minus 50 seconds.” Using absolute value notation, how long is it before lift-off?

Solution

∣−50∣∣−50∣ size 12{ lline -"50" rline } {} = 50 seconds

Exercise 41

Due to a slowdown in the industry, a Silicon Valley computer company finds itself in debt $2,400,000. Use absolute value notation to describe this company’s debt.

Solution

∣−2,400,00∣∣−2,400,00∣ size 12{ lline -2,"400","00" rline } {} Debt is a negative number, but the company is $2,400,000 away from a 0 balance.

Exercise 42

A particular machine is set correctly if, when it runs, its meter reads 0. One particular machine has a meter reading of -1.6 when it runs. How far is this machine off its correct setting?

Solution

∣−1.6∣∣−1.6∣ size 12{ lline -1 "." 6 rline } {} = 1.6

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