Question

prove that n2-n is even for every positive integer n​

Answer 1

Answer:

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.

Step-by-step explanation:

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Answer 2

Step-by-step explanation:

Case i: 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.

∴ n2−n is divisible by 2 for every integer n

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