Question

prove that n√2+m if divisible by 2 for any positive integer n​

Answer 1

Answer:

Suppose the positive integer is n.

• n = 2q or n = 2q + 1

CASE 1:

n = 2q

.. n² - n = (2q)² - 29

= 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.

Answer 2

Answer:

∴ n = 2q or n = 2q + 1 where q∈Z. Thus, in any case, n² - n is divisible by 2. Thus, n² - n is divisible by 2 for every positive integer n.