Question

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

Answer 1

SOLUTION:

Let (n – 1) and n be two consecutive positive integers. Then their product is n (n - 1)

= n² - n

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

Case : 1

When n = 2q

n² - n

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 divisible by 2

Case : 2

when n = 2q + 1

n² - n = (2q + 1)² -  (2q + 1)

n² - n = (2q + 1) [ (2q + 1) – 1)]

n² - n = (2q + 1)   (2q)

n² - n = 2r

[Where , r = q (2q + 1)]

n² - n is divisible by 2

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

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Answer 2

let the first integer be x .then second integer must be (x+1)
Now product = x (x+1) = x^2+x
i) if x is even
x=2k
Then ,
x^2+x= (2k)^2+2k
= 4k^2+2k
= 2 (2k^2+k)
hence divisible by 2
ii) if x is odd
x=2k+1
Then
x^2+x = (2k+1)^2+2k+1
= 4k^2+2k+1+2k+1
= 4k^2+4k+2
= 2 (2k^2+2k+1)
hence divisible by 2
Since both conditions are satisfied .
Hence product of two consecutive positive integers is divisible by 2 .Proved