# Absolute Value or Modulus of a Real Number – An Idea of Number System

Number System / Monday, November 25th, 2019

# What is Absolute value or Modulus?

Let a be a real number, then the modulus or absolute value of a is denoted by |a| and is defined as follows:

|a| = a, if a > 0

= 0, if a = 0

= -a, if a < 0

From the definition, it is obvious that for all a R, |a| ≥ 0, |a| ≥ a, |a| ≥ -a

In fact |x| = max{-x, x}, where max{-x, x} denotes the greater of the two numbers –x and x. we may also say that |x| denotes the positive square root of x2, so that

$\left| x \right|=+\sqrt{\left( {{x}^{2}} \right)}$

From definition, it follows that |5| = 5, |-5| = 5, |0| = 0, |8 – 5| = 3, |5 – 8| = 3.

The modulus or absolute value of the difference between two real numbers as the measure of the distance between the corresponding points on the number line.

## Some results involving modulus or absolute value

Some simple and useful results involving the modulus or absolute value of real numbers are discussed below:

### (I) For a, b ∈ R, |a + b| ≤ |a| + |b|

i.e., the moduli of the sum of two real numbers is less than or equal to the sum of their moduli.

We have a ≤ |a|, b ≤ |b| a + b ≤ |a| + |b|

Also, -a ≤ |a|, -b ≤ |b| -(a + b ) ≤ |a| + |b|

Now, |a + b| = max {(a + b), -(a + b)}

Thus |a + b| ≤ |a| + |b|

Note: In fact, |a + b| = |a| + |b| if a, b have same sign and |a + b| < |a| + |b| if a, b have opposite sign.

 Example

|5 + 2| = 7, |5| + |2| = 7 so, |5 + 2| = |5| + |2|

|-6 – 2| = |-8| = 8, |-6| = 6, |-2| = 2 so, |-6 – 2| = |-6| + |-2|

But |7 – 3| = 4 and |7| = 7, |-3| = 3 so, |7| + |-3| = 10

Since, 4 < 10 we have |7 – 3| < |7| + |-3|

### (II) For a, b ∈ R, |ab| = |a||b|

i.e., the modulus of the product of two numbers is equal to the product of their moduli.

We have,

$\left| a \right|=+\sqrt{\left( {{a}^{2}} \right)},\left| b \right|=+\sqrt{\left( {{b}^{2}} \right)}$

$\Rightarrow \left| a \right|\left| b \right|=\sqrt{\left( {{a}^{2}} \right)}\sqrt{\left( {{b}^{2}} \right)}=\sqrt{\left( {{a}^{2}}{{b}^{2}} \right)}=\sqrt{{{\left( ab \right)}^{2}}}=\left| ab \right|$

 Example

|3 . 5| = |15| = 15 and |3| . |5| = 3 . 5 = 15

So |3 . 5| = |3| . |5|

Again, |(-4) . (-3)| = 12 = |-4| . |-3|

### (III) For a, b ∈ R, |a – b| = ||a| – |b||

|a| = |(a – b) + b| ≤ |a – b| + |b|

|a| – |b| ≤ | a – b| ………..(1)

Again, |b| = |(b – a) + a| ≤ |b – a| + |a|

|b| – |a| ≤ |b – a| = |a – b|………..(2)

Since ||a| – |b|| = max{|a| – |b|, -(|a| – |b|)},

We conclude from (1) and (2), that

||a| – |b|| ≤ |a – b|

### (VIII) 0 < |x – a| ≤ δ ⇒ a – δ ≤ x ≤ a + δ

 Example 01

Solve the equation |2x – 3| = 7

Solution:

|2x – 3| = 7

2x – 3 = 7 …………(1)

2x – 3 = -7 …………(2)

From (1), 2x = 10, i.e., x = 5

And from (2), 2x = -4, i.e., x = -2

Required solutions are: x = 5 and x = -2.

 Example 02

Solve the inequality |x – 5| < 9 and indicate the solution on the number line.

Solution:

|x – 5| < 9

⇒ -9 < x – 5 < 9

⇒ -9 + 5 < x < 9 + 5

⇒ -4 < x < 14

Solution is the open interval -4 < x < 14 or (-4, 14).

The solution is indicated below on the number line,

 Example 03

Solve: x2 – 2|x| – 3 = 0.

Solution:

Let |x| = y, since |x|2 = x2, the given equation reduced to

y2 – 2y – 3 = 0

⇒ (y – 3)(y + 1) = 0

i.e., y = 3, -1

Thus |x| = 3 i.e., x = +3, -3

|x| = -1 is not possible

Therefore the solutions are x = 3, -3

 Example 04

Show that |a + b| < |a| + |b| if and only if ab < 0.

Solution:

We know |a + b|2 = (a + b)2 = a2 + b2 + 2ab

< a2 + b2 + 2|ab|, if and only if ab < 0

= |a|2 + |b|2 + 2|ab| = (|a| + |b|)2

Hence |a + b| < |a| + |b| if and only if ab < 0.