Question

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

Answer 1

Given:

Two consecutive numbers.

Prove:

That the product of two consecutive integer is divisible by 2.

Proof:

Let n - 1 and n be two consecutive positive integer.

Then there product is (n - 1) = n² - n

We know that every positive integer is of the form 2q or 2q + 1 for some integer q.

So, Let n = 2q

So, n² - n = (2q)² - (2q)

=> n² - n = (2q)² - (2q)

=> n² - n = 4q² - 2q

=> n² - n = 2q(2q - 1)

=> n² - n = 2r [where r = q(2q - 1)]

=> n² - n is even and divisible by 2

Let n = 2q + 1

So, n² - n = (2q + 1)(2q + 1) - 1

=> n² - n = (2q + 1)(2q)

=> n² - n = 2r(r = q(2q + 1))

=> n² - n is even and divisible by 2

Hence, it is proved that the product of two consecutive integer is divisible by 2.

I hope it will help you.

Regards.

Answer 2

Let the numbers as x and x+1

$→$ x(x + 1)

${\large{\boxed{\sf{Even×Odd=Even}}}}$

$→$ If x is even, (even × (even+1))

= even × odd = even

$→$ If x is odd, (odd × (odd+1))

+1)) = odd × even = even

All even numbers are divisible by 2.

So, product of two consecutive numbers is divisible by 2...

★ Hence proved