Question

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

Answer 1

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Let two consecutive positive integers be x and (x +1)

Product of two consecutive positive integers = x(x +1)

x²+x

In First Case

let x is even number

x = 2k

↪ x² + x = (2k)² + 2k

↪ 4k² + 2k

↪2k (2k +1)

Hence the product is divisible by 2

In second case

x is odd number

↪ x²+ x =(2k + 1)² + (2k+1)

↪ 4k²+ 4k + 1 + 2k + 1

↪4k² + 6k + 2

↪ 2(2k²+3k+1)

Clearly the product is divisible by 2

From the both the cases we can conclude that the product of two consecutive positive integers is divisible by 2.

Answer 2

Step-by-step explanation:

Let two consecutive positive integers be x and (x +1)

Product of two consecutive positive integers = x(x +1)

x²+x

In First Case

let x is even number

x = 2k

↪ x² + x = (2k)² + 2k

↪ 4k² + 2k

↪2k (2k +1)

Hence the product is divisible by 2

In second case

x is odd number

↪ x²+ x =(2k + 1)² + (2k+1)

↪ 4k²+ 4k + 1 + 2k + 1

↪4k² + 6k + 2

↪ 2(2k²+3k+1)

Clearly the product is divisible by 2

From the both the cases we can conclude that the product of two consecutive positive integers is divisible by 2.