Prove that the product of two consecutive positive integers is divisib

1.	Prove that the product of two consecutive positive integers is divisible by 2.
Answer» Let n – 1 and n be two consecutive positive integers, then the product is n(n – 1) n(n – 1) = n²– n We know that any positive integer is of the form 2q or 2q + 1 for same integer q Case 1: When n = 2 q n² – n = (2q)² – 2q = 4q² – 2q = 2q (2q – 1) = 2 [q (2q – 1)] n² – n = 2 r r = q(2q – 1) Hence n – n. divisible by 2 for every positive integer. Case 2: When n = 2q + 1 n² – n = (2q + 1)² – (2q + 1) = (2q + 1) [2q + 1 – 1] = 2q (2q + 1) n² – n = 2r r = q (2q + 1) n² – n divisible by 2 for every positive integer.

Prove that the product of two consecutive positive integers is divisible by 2.

Answer»

Let n – 1 and n be two consecutive positive integers, then the product is n(n – 1)

n(n – 1) = n²– n

We know that any positive integer is of the form 2q or 2q + 1 for same integer q

Case 1:

When n = 2 q

n² – n = (2q)² – 2q

= 4q² – 2q

= 2q (2q – 1)

= 2 [q (2q – 1)]

n² – n = 2 r

r = q(2q – 1)

Hence n – n. divisible by 2 for every positive integer.

Case 2:

When n = 2q + 1

n² – n = (2q + 1)² – (2q + 1)

= (2q + 1) [2q + 1 – 1]

= 2q (2q + 1)

n² – n = 2r

r = q (2q + 1)

n² – n divisible by 2 for every positive integer.