Question

PROVE that x²_x
x²_x IS divisible
by
2 for all positive integers x
ܬܐܝܘ
divisible
FOR
positive integens
x.​

Answer 1

Note :

★ Every positive even integer can be expressed in the form 2n , where n is any natural number .

★ Every positive odd integer can be expressed in the form of (2n - 1) , where n is any natural number .

★ Any positive integer is either even or odd .

Proof :

Here ,

It is given that , x is any positive integer.

Thus ,

x can be either even or odd .

Case1 : If x is even

Let x = 2n , where n is natural number .

Now ,

x² - x

= (2n)² - 2n

= 4n² - 2n

= 2(2n² - n) { which is divisibile by 2 }

Clearly ,

x² - x is divisibile by 2 if x is even .

Case2 : If x is odd

Let x = 2n - 1 , where n is natural number .

Now ,

x² - x

= (2n - 1)² - (2n - 1)

= 4n² - 4n + 1 - 2n + 1

= 4n² - 6n + 2

= 2(2n² - 3n + 1) { which is divisibile by 2 }

Clearly ,

x² - x is divisibile by 2 if x is odd .

Now ,

From case1 and case2 it is clear that x is either even or odd , x² - x is always divisible by 2 .

Hence ,

For every position integer x , x² - x is divisibile by 2 .