# 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 **x ^{2}**, 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|**

### (IV) **|a/b| = |a| / |b|, for all** **a, b ****∈**** R, b ≠ 0**

**(V) -|a| ≤ a ≤ |a|**** for all** **a ****∈**** R**

**(VI) |a|**^{2} = a^{2}, **for all** **a ****∈**** R**

^{2}= a

^{2},

**(VII) |x – a| ≤ δ ****⇒ ****– δ ≤ x – a ≤ δ**

**(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: x ^{2} – 2|x| – 3 = 0.**

**Solution:**

Let |x| = y, since |x|^{2} = x^{2}, the given equation reduced to

y^{2} – 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} = a^{2} + b^{2} + 2ab

< a^{2} + b^{2} + 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.