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

Let, (n – 1) and n be two consecutive positive integers
∴ Their product = n(n – 1)
= n ² −n
We know that any positive integer is of the form 2q or 2q + 1, for some integer q
When n =2q, we have
n ² − n = (2q)² − 2
= 4q² − 2q
2q(2q − 1)
Then n ² − n is divisible by 2.
When n = 2q + 1, we have
n​² − n = (2q + 1)² − (2q + 1)
= 4q² + 4q + 1 − 2q − 1
= 4q² + 2q
= 2q(2q + 1)
Then n​² − n is divisible by 2.
Hence the product of two consecutive positive integers is divisible by 2.

Let’s consider two consecutive positive integers as (n-1) and n. 

∴ Their product = (n-1) n 

= n² – n 

And then we know that any positive integer is of the form 2q or 2q+1. (From Euclid’s division lemma for b= 2)

So, when n= 2q 

We have, 

⇒ n² – n = (2q)² – 2q 

⇒ n² – n = 4q² -2q 

⇒ n² – n = 2(2q² -q) 

Thus, n² – n is divisible by 2. 

Now, when n= 2q+1 

We have, 

⇒ n² – n = (2q+1)² – (2q-1) 

⇒ n² – n = (4q²+4q+1 – 2q+1) 

⇒ n² – n = (4q²+2q+2) 

⇒ n² – n = 2(2q²+q+1) 

Thus, n² – n is divisible by 2 again. 

Hence, the product of two consecutive positive integers is divisible by 2.

Let two consecutive positive integers is n, and (n + 1)

Product of both integers = n(n + 1) = n² + n

We know that any positive integer is in the form 2q

and 2q + 1. where q is an integer.

Here two cases are possible

Case. I. when n = 2q then

⇒ n² + n = (2q)² + (2q)

⇒ n² + n = 4q² + 2q

⇒ n² + n = 2q(2q + 1) [Let r = q(2q + 1)]

⇒ n² + n = 2r

⇒ n² + n, 2 can be divided by 2

⇒ n(n + 1), also divided by 2

So, product of two consecutive positive integer is divided by 2

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.