Question

prove that n^2-n is divisible by 2​

Answer 1

Step-by-step explanation:

Suppose the positive integer is n.

∴ n = 2q or n = 2q + 1         where q∈Z.

CASE 1:-  

n = 2q

∴ n² - n = (2q)² - 2q

             = 4q² - 2q

             = 2(2q² - q)

CASE 2:-

n = 2q + 1

∴n² - n = (2q + 1)² - (2q + 1)

           = 4q² + 4q + 1 - 2q - 1

           = 4q² + 2q

           = 2(2q² + q)

Thus, in any case, n² - n is divisible by 2.

Thus, n² - n is divisible by 2 for every positive integer n.

4.7

Answer 2

Step-by-step explanation:

Casei: Let n be an even positive integer. When n = 2q In this case , we have n2 - n = (2q)2 - 2q = 4q2 - 2q = 2q (2q - 1 ) n2 - n = 2r , where r = q (2q - 1) n2 - n is divisible by 2 . Case ii: Let n be an odd positive integer. When n = 2q + 1 In this case n2 -n = (2q + 1)2 - (2q + 1)= (2q +1) ( 2q+1 -1)= 2q (2q + 1) n2 - n = 2r , where r = q (2q + 1) n2 - n is divisible by 2. ∴ n 2 - n is divisible by 2 for every integer n

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