Question

please answer question 2

Answer 1

We will do this by considering various cases

Case 1:
Let two consecutive positive integers be 1 and 2
So 1*2=2 which is divisible by 2

Case 2:
Let two consecutive positive integers be 2 and 3
2*3=6 which is divisible by 2

Case 3:
Let two consecutive positive integers be 3 and 4
3*4=12 which is divisible by 2

Therefore after taking into consideration various cases we conclude that the product of two consecutive positive integers is always divisible by two

Or the alternative method

When we take any two positive consecutive integers it is obvious that one of them will be even and other will be odd and we know that the product of even and odd no. is even that is divisible by two

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Answer 2

Let the two consecutive positive numbers be 'x' and '(x+1)'
Product of two consecutive positive integers
=x(x+1)
$= x {}^{2} + 1$
Case 1...
x is even number
Let x =2k
$= > x {}^{2} + x = (2k) {}^{2} + 2k$
$= 4k {}^{2} + 2k$
$= 2k(2k + 1)$
Case 2....
x is odd number
Let x =2k+1
$= > x {}^{2} + x = (2k + 1) {}^{2} + (2k + 1 )$
$= 4k {}^{2} + 4k + 1 + 2k + 1$
$= 4k {}^{2} + 6k + 2$
$= 2(2k {}^{2} + 3k + 1)$
Clearly the product is divisible by 2.
From both the cases we can conclude that the product of two consecutive positive integers is divisible by 2.

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