How to Solve: Absolute Value (Basics)

How to Solve: Absolute Value (Basics)

Hi All,

I have posted a video on YouTube to discuss Absolute Value (Basics)



After this post please go through Absolute Value Problems and Absolute Value + Inequality post

Attached pdf of this Article as SPOILER at the top! Happy learning!

Following is Covered in the Video

Theory

What is Absolute Value / Modulus of a number
Absolute Value on Number Line
Properties of Absolute Values
Absolute Value on Number Line Examples

What is Absolute Value / Modulus of a number

• Absolute Value or modulus (|x|) of a real number x is the non-negative value of the number (x), without any consideration to its sign
Ex
|12| = 12
|-12| = 12 (we just the value after ignoring the sign)

• |x| = x for x >0
= -x for x <0
= 0 for x = 0

Q1. Find the value of |-3| + | 2*3 – 4*2| + |25|

Q2. Find the value of | x+y| where x + z = 20 and y – z = -25

Sol1: 3 + | 6-8 | + 25 = 3 + 2 + 25 = 30

Sol2: x + z = 20 and y – z = -25
Adding both of them we get x + y = -5
=> | x+y | = |-5| = 5

Absolute Value on Number Line

• Absolute value of a number x can also be imagined as the distance of that number x from 0 on a number line

Let's say we have two numbers x and y and x is positive and y is negative. What we are saying is
|x| = x = distance of x from origin
|y| = -y = distance of y from origin
As, shown in the image below:



Properties of Absolute Values

• PROP 1: Absolute value of a number is always Non-negative
|a| ≥ 0 for all values of a
Ex: |3| = 3 ≥ 0
|-7| = 7 ≥ 0

• PROP 2: Minimum value of |a| = 0, when a=0
Ex: If |x| =0 => x=0

• PROP 3: Square root of a number is always positive
\(\sqrt{a^2}\) = |?|
Ex:
If x = \(\sqrt{25}\) => x = +5
But if \(x^2\) = 25 => x = ± \(\sqrt{25}\) => x = ±5

• PROP 4: Absolute value of negative of a number is same as absolute value of the number
|-a| = |a|
A derivative of this is
| a-b | = | b-a | because | b-a | = | -(a-b) |

• PROP 5: Product of absolute value of two numbers is same as product of their absolute values
|ab| = |a|*|b|
Ex:
|7*3| = |7| * |3| = 21

• PROP 6: Division of absolute value of two numbers is same as division of their absolute values
\(|\frac{?}{?}| = \frac{{|?|}}{{|?|} }\)
Ex:
\(|\frac{4}{2}| = \frac{{|4|}}{{|2|} }\) = 2

• PROP 7: Sum of absolute value of two numbers is always ≥ absolute value of their sum
|a| + |b| ≥ |a+b|
Ex:
|7| + |3| ≥ |7+3| => 10 ≥ 10
|5| + |-8| ≥ |5 + (-8) | => 13 ≥ 3

• PROP 8: Difference of absolute value of two numbers is always ≤ absolute value of their difference
|a| - |b| ≤ |a-b|
Ex:
|7| - |3| ≤ |7-3| => 4 ≤ 4
|5| - |-8| ≤ |5 - (-8) | => -3 ≤ 13

• PROP 9: Taking absolute value multiple times or taking it once gives the same result
||a|| = |a|
Ex:
||-4|| = |-4| => |4| = |-4| = 4

• PROP 10: If absolute value of difference of two numbers is zero => both numbers are equal
|a-b|=0 => a=b
Ex:
| x-4 | =0 > x=4

Next two will be used a lot to solve absolute values problem!

• PROP 11: If |a| ≤ b => -b ≤ a ≤ b

• PROP 12: If |a| ≥ b => a ≤ -b or a ≥ b

• PROP 13: |\(a^n\)| = \(|a|^n\)
Ex:
|\({-2}^4\)| = \(|-2|^4\) = 16

• PROP 14: |a-b| ≥ ||a|-|b||
Ex:
|7-4| ≥ ||7|-|4|| => 3 ≥ 3
|8-(-2)| ≥ ||8|-|-2|| => 10 ≥ 6

Q1. If |a-3| ≤ 9 then find the range of values of a.

Q2. If |b+5| ≥ 10 then find the range of values of b.

Sol1: -6 ≤ a ≤ 12
Check Video For solution

Sol2: b ≤ -15 or b ≥ 5
Check Video For solution

Absolute Value on Number Line Example

If a and b are two variables given then:
|a-b| always means the distance between points a and b
|a+b| = |a| + |b| when a and b have the same sign and
|a+b| = |b| - |a| when a and b have different sign and |b| > |a|

Case 1: a and b are positive and a > b



|a-b | = |a| - |b|
|a+b| = |a| + |b|

Case 2: a is positive and b is negative



Given: |a| > |b|
|a-b| = |a| + |b|
|a+b| = |a| - |b|

Case 3: Both a and b are negative



Given: |a| > |b|
|a-b| = |a| - |b|
|a+b| = |a| + |b|

Q1. Given the information (below), Simplify |b-a| + |c-b|



Sol:
Method 1
|b-a| = Distance between a and b = AB
|c-b| = Distance between c and b = BC
=> |b-a| + |c-b| = AB = BC = AC = |c-a|

Method 2
|b-a| = |b| - |a|
|c-b| = |c| - |b|
=> |b-a| + |c-b| = |b| - |a| + |c| - |b| = |c| - |a| = |c-a|
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