Question

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

Answer 1

Answer:

Yes it is true.

Ex: a. 1 x 2 = 2

    b. 5 x 4 = 20

    c. 6 x 7 = 42

    d. 8 x 9 = 72

All the products of the above questions are divisible by 2.

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Answer 2

$\large\underline{\sf{Solution-}}$

To prove that product of two consecutive positive integers is divisible by 2.

To prove this statement, we use The concept of Principal of Mathematical Induction.

Let n and n + 1 be two consecutive positive integers.

So, we have to prove that n (n + 1) is divisible by 2.

Let assume that

$\rm :\longmapsto\: \sf \: P(n) : \: n(n + 1) \: is \: divisible \: by \: 2$

Step :- 1 For n = 1

$\rm :\longmapsto\: \sf \: P(1) : \: 1 \times (1 + 1)$

$\rm :\longmapsto\: \sf \: P(1) : 2$

$\rm :\longmapsto\: \sf \: P(1) : 2 \: is \: divisible \: by \: 2$

$\bf\implies \:P(n) \: is \: true \: for \: n = 1$

Step :- 2 Let assume that for n = k, P(n) is true

$\sf :\longmapsto\:P(k) : k(k + 1) \: is \: divisible \: by \: 2$

$\sf\implies \:k(k + 1) = 2m$

$\sf\implies \: {k}^{2} + k= 2m$

$\sf\implies \: {k}^{2}= 2m - k - - - (1)$

Step :- 3 We have to prove that P(n) is true for n = k + 1

$\sf :\longmapsto\:P(k + 1) : (k + 1)(k + 1 + 1)$

$\sf \:  =  \: (k + 1)(k + 2)$

$\sf \:  =  \: {k}^{2} + 2k + k + 2$

$\sf \:  =  \: 2m - k + 3k + 2$

$\sf \:  =  \: 2m + 2k + 2$

$\sf \:  =  \: 2(m + k + 1)$

$\bf\implies \:P(n) \: is \: true \: for \: n = k + 1$

Hence, by the process of Principal of Mathematical Induction, product of two consecutive positive integers is divisible by 2.