Question

Use the following definition of absolute value to prove the given statements: If x is a real number, then the absolute value of x , | x | , is defined by: | x | = { x if x ≥ 0 − x if x < 0 For any real number x , | x | ≥ 0 . Moreover, | x | = 0 implies x = 0 . For any two real numbers x and y , | x | ⋅ | y | = | x y | . For any two real numbers x and y , | x + y | ≤ | x | + | y | .

Answer 1

Answer:

Proved all parts below.

Step-by-step explanation:

As given ,

|x| = $\left \{ {{x , x\geq 0} \atop {-x, x< 0}} \right.$

To prove- a) For any real number x , | x | ≥ 0 . Moreover, | x | = 0 ⇒ x = 0

                 b) For any two real numbers x and y , | x | ⋅ | y | = | x y | .

                 c) For any two real numbers x and y , | x + y | ≤ | x | + | y | .

Proof -

a)

As given x is a real number

Also , by definition of absolute value of x , we get

| x | ≥ 0

Now,

if |x| = 0

⇒ x = 0 and -x = 0

⇒ x = 0 and x = 0

⇒ x = 0

∴ we get

| x | = 0 ⇒  x = 0

Hence proved.

b)

To prove - | x | ⋅ | y | = | x y |

As we have,

|x| = $\left \{ {{x , x\geq 0} \atop {-x, x< 0}} \right.$

|y| = $\left \{ {{y , y\geq 0} \atop {-y, y< 0}} \right.$

|xy| = $\left \{ {{xy , x,y > 0 and x,y < 0} \atop {-xy, x > 0, y< 0 and x <0 , y > 0}} \right.$

We have 4 cases : i)  when x > 0 , y > 0

                              ii) when x > 0 , y < 0

                              iii) when x < 0, y > 0

                              iv) when x < 0, y < 0

For Case I - when x > 0 , y > 0

⇒ |x| = x, |y| = y

⇒|x|.|y| = xy

For Case Ii - when x > 0 , y < 0

⇒ |x| = x, |y| = -y

⇒|x|.|y| = -xy

For Case Iii - when x < 0 , y > 0

⇒ |x| = -x, |y| = y

⇒|x|.|y| = -xy

For Case IV - when x < 0 , y < 0

⇒ |x| = -x, |y| = -y

⇒|x|.|y| = (-x)(-y) = xy

∴ we get , from all 4 cases

| x | ⋅ | y | = | x y |

Hence Proved.

c)

To prove - | x + y | ≤ | x | + | y |

Let

|x| = |x + y - y|

    ≥ |x + y| - |y|     ( Triangle inequality)

⇒ |x| + |y| ≥ |x + y|

Hence Proved.