Absolute Values

In this lecture we shall discuss:

  • Absolute values
  • Inequalities involving absolute values
  • Theorem 1.2.2 (√a2=|a|)
  • Theorem inequality


1.2.1 Definition

The absolute value or magnitude of a real number a is denoted by |a| and is defined by
absolute value definition


      Example
|5|=5     Since 5>0
|-4/7|= -(-4/7) = 4/7   Since -4/7<0
|0|=0     Since 0≥0


      Remark
|a| is a non-negative number for all values of a and
-|a|≤ a ≤ |a|
If a itself is negative, then -a is positive and +a is negative!!!


      Example
Solve       |x-3|=4
Solution

x-3= 4
    x= 7
    -(x-3)= 4
    x-3= -4
       x= -1


      Example
Solve |3x-2|=|5x+4|

3x-2   = 5x+4
3x-5x = 4+2
    -2x = 6
       x = -3
    3x-2 = -(5x+4)
    .
    .
    .
       x = -1/4


      SQUARE ROOTS AND ABSOLUTE VALUES
            b2 = a

          (3)2 = 9
          so b = 3
but!!!
  (-3)2 = 9 so b = -3


The positive square root of the square of a number is equal to that number.


      THEOREM 1.2.2
For any real number a
            √a2 = |a|
e.g.
      √(-4)2 = √16 = 4 = |-4|


      1.2.3 THEOREM
If a and b are real numbers then,

  1. |-a| = |a|    a number and its negative have the same absolute values.
  2. |ab| = |a||b|    The absolute value of a product is the product of the absolute values.
  3. |a/b| = |a|/|b|    The absolute value of a ratio is the ratio of the absolute values.


      Proof
From theorem 1.2.2

(a)  |-a| = √(-a)2 = √a2 = |a|

(b)  |ab| = √(ab)2 = √a2b2 = √a2b2 = |a||b|


      Examples

(a)  |-4| = |4|

(b)  |2.-3| = |-6| = 6 = |2|.|3| = 6

(c)  |5/4| = 5/4 = |5|/|4| = 5/4


     The result (b) of above theorem can be extended to three or more factors.
For n-many real numbers
a1, a2, a3,...an

(a) |a1 a2 ...an| = |a1| |a2| ...|an|
(b) |an| = |a|n


      Geometric interpretation of Absolute Value

Where A and B are points with coordinate a and b. The distance between A and B is


      Theorem 1.2.4 (Distance formula)
    If A and B are points on a coordinate line with coordinates a and b respectively, then the distance d between A and B
        d = |b - a|


      TABLE 1.2.2 (a)
                    |x-a| < k (k>0)

          Alternative Form     -k < x-a < k
          Solution Set           (a-k, a+k)


      Example
The inequality
  |x-3| < 4
rewritten as
  -4 < x-3 < 4
adding 3 throughout gives
  -1 < x < 7
solution set (-1,7)

                        On real line


      Example
Solve |x+4| ≥ 2
x+4 ≤ -2
x ≤ -6
    x+4 ≥ 2
x≥ -2
On combining these two sets
                (-∞ , -6] ∪ [-2 , +∞ )

                          On real line


      THE TRIANGLE INEQUALITY

It is not generally true that
|a+b| = |a| + |b|
e.g.
if a = 2 and b = -3, then a+b = -1 so that |a+b| = |-1| = 1
whereas
|a|+|b| = |2|+|-3| = 2+3 = 5 so |a+b| = |a|+|b|


      1.2.5 THEOREM - (Triangle Inequality)
If   a  b  then |a+b| ≤ |a|+|b|
      Proof
Since for any real number a and b, we know that
-|a| ≤ a ≤ |a|   and   -|b| ≤ b ≤ |b|
          -|a| ≤ a ≤ |a|
                   +
          -|b| ≤ b ≤ |b|
      ______________
= -|a| + -|b| ≤ a+b ≤ |a|+|b|
______________________________________________
Now we have two cases:

Case 1 where a+b ≥ 0
certainly a+b=|a+b|
Hence
        |a+b| ≤ |a|+|b|

And

Case 2 where a+b < 0
        |a+b| = -(a+b)
                or
        (a+b) = -|a+b|

Comparing with the intial inequality
-(|a|+|b|) ≤ -|a+b|
  The result follows
_______________________________

Contact email:

 
Copyright © 2005-2017.