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

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

Remark
1a1 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 VALUETS
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,

|-a| = |a| a number and its negative have the same absolute values.
|ab| = |a||b| The absolute value of a product is the product of the absolute values.
|a/b| = |a|/|b| The absolut 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 = √a²√b² = |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,...a_n
(a) |a₁ a₂ ...a_n| = |a₁| |a₂| ...|a_n|
(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
←^{_______________________________}→

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