Question

Prove that theproduct of two consecutive positive integers in divisible by 2

Answer 1

Let a be any +ve integer and b=2
Acc. to Euclid's Division Lemma ,
a=2q+r, where 0≤r<2 Hence , r=0,1
Possible values of a:
a=2q
a=2q+1
(2q)(2q+1) = 4q²+2q = 2(2q²+1) = 2m (for some integer 'm')
Hence , product of two consecutive integers is divisible by 2.

Answer 2

Hey there !!

Let the two consecutive integers be n , n + 1

Product = n(n+1) = n² + n

Any positive integer is of the form 2q , 2q + 1


When n = 2q ,
n² + n = (2q)² + 2q
= 4q² + 2q
= 2q (2q + 1 ) ------> (1)

When n = 2q + 1
n² + n = (2q + 1)² + 2q + 1
= 4q² + 4q + 1 + 2q + 1
=4q² + 4q + 2q + 2
= 2(2q² + 2q + q + 1) ------> (2)


Hence ,

From (1) and (2),
its clear that the product of two consecutive positive integers in divisible by 2