Question

The value of x2 is always greater than the value of x

Answer 1

The statement is true.

Given: The value of x2 is always greater than the value of x.

To find: We have to prove the statement.

Solution:

x is an integer and $x^2$ is a multiple of x to x.

For any value of x, $x^2$ gives a greater value than x.

Let us take x us equal to 2.

So, $x^2$ will be-

${x}^{2} = {2}^{2} \\ = 4$

So, 4 will always be greater than 2.

Again if we take x is equal to -1.

Then the value of $x^2$ is -

${x}^{2} = { - 1}^{2} \\ = 1$

So, 1 is always greater than -1.

So, we can say the value of $x^2$ is always greater than the value of x.

Answer 2

The value of x² is always greater than the value of x is False

x² is square of x

Relation between x²  and x can be broken in  

(-∞ ,  0) , (0 , 1) , (1 , ∞) and {0 , 1}

For (-∞ , -0)

x² > x

as x² is  non negative for all real x or

x² is positive for all non zero real x .

For (0 , 1)

x² <  x

For  example

(1/2)² < 1/2  ( ∵  1/4 <  1/2)

For (1 , ∞)

x² > x  

For example 3² > 3

For { 0 , 1}

x² = x

0² = 0  and 1² = 1

As for x ∈  (0 , 1)  x² < x   and for  x ∈ {0 , 1}  x² =  x

Hence for [0 , 1]  , x² greater than x is not True

Value of x² is always greater than the value of x  iff x ∈ (-∞ , -0) U  (1 , ∞)

or x ∈ R - [0 , 1]

Hence given statement is false

The value of x² is always greater than the value of x is False

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