Question

prove that the product of two consecutive positive integer is divisible by 2

Answer 1

Answer:

Step-by-step explanation:

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Answer 2

Step-by-step explanation:

Let the two consecutive positive integers be x and x+1

Let x=2a (even)     and x+1 =2a+1(odd)

According to the given question

Product of two integers (x)(x+1)

$=x^{2}+x$

Case 1: if x is an even number

By putting the values

We have $x^{2}+x=2 a^{2}+2 a$

=2a (a+1)

Check and divide the above expression by 2

= 2a (a+1)/2

We get a (a+1)

Hence it is clearly proved that the product of 2 integers are divisible by 2

Case 2: If x is an odd number

By putting the values x=2a+1

We have $x^{2}+x=(2 a+1)^{2}+(2 a+1)$

$\begin{array}{l}{=4 a^{2}+1+4 a+2 a+1} \\ {=4 a^{2}+6 a+2} \\ {=2\left(2 a^{2}+3 a+1\right)}\end{array}$

Check: Divide the above expression by 2

$=2\left(2 a^{2}+3 a+1\right) / 2$

We get $2 a^{2}+3 a+1$

Hence proved