Question

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

Answer 1

Hint:In this problem, to prove the above statement, we will assume the numbers which are in order and are also multiple of 2. For the product to be divisible by 2, it should be multiple of 2. We will prove the statements in two different cases of even and odd numbers. If 2 is the common number in the product of both the cases, then the statement is verified.

Complete answer:
Assume two consecutive positive integers as $x$and$x + 1$.
 Now we will find the product of these numbers. The product of the numbers will be $ = x\left( {x + 1} \right)$. If the x is divisible by 2 then it must be the multiple of 2.

For even numbers, assume $x = 2k$.
Hence the number becomes 2k and $2k + 1$.
The product of these numbers can be expressed as $ = 2k\left( {2k + 1} \right)$.
Since, 2 is the common factor in the above expression. Therefore, it is clear from the above expression that the product is divisible by 2.

For odd numbers, assume $x = 2k + 1$.
Hence, the number becomes $2k + 1$ and $\left( {2k + 1} \right) + 1$.
 The product of the number now can be expressed as,
 $\begin{array}{l}
 = \left( {2k + 1} \right)\left( {\left( {2k + 1} \right) + 1} \right)\\
 = \left( {2k + 1} \right)\left( {2k + 2} \right)\\
 = 4{k^2} + 6k + 2\\
 = 2\left( {2{k^2} + 3k + 1} \right)
\end{array}$
Since, 2 is the common factor in the above expression. Therefore, it is clear from the above expression that the product is divisible by 2.
Hence it is proved that the product of two consecutive positive integers is divisible by 2.

Note:Consecutive numbers are the number which are in order. The consecutive number must be in order and hence for both the cases assume numbers which are in order. Try to make an assumption in such a way that the values are multiple of 2 and in the final answer 2 must be in common.