Answer:

let the first integer be x

then the second integer shall be x+1

then their product be x(x+1) = x²+x

(i) If x is even

then x = 2k  

∴ x²+x= (2k)²+2k

=4k²+2k

=2(2k²+k)

hence DIVISIBLE by two.

(ii)Let x be odd.

∴ x= 2k+1

∴ x²+x = (2k+1)²+2k+1

=(2k)²+8k+1+2k+1

=4k²+10k+2

=2(2k²+5k+1)

hence divisible by two/.

since bothe of our conditions satisfy the statement, we can say that the product of two consecutive integers is divisible b

Step-by-step explanation:

one more here

It is true that product of two consecutive +ve integers is divisible by 2 as because of the following reasons:-

Let U consider two consecutive +ve integer as N and ANOTHER as n-1.

Now, the product of both is n2-n

Case 1

When n=2q ( Since any +ve integer can be in the form of 2q or 2q+1)

n2-n= (2q)2-2q

= 4q2- 2q

It is divisible by 2 as it leaves a remainder 0 after division.

Case 2

When n= 2q+1

n2-n = (2q+1)2 - 2q+1

= 4q2+ 4q+2q+2

= 4q2+ 6q+2

It is divisible by 2

So, in both the case of the consecutive integers, it is divisible by 2