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.

|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,

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

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

|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 = a², for all a ∈ R

(VII) |x – a| ≤ δ ⇒ – δ ≤ x – a ≤ δ

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

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.

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,

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

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.

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