prove that the product of two consecutive positive integer is divisible by 2
Answers
Answer :-
We have to prove that the product of two consecutive integers is divisible by 2 .
Let the integer be m
Then another integer = m +1
Now by multiplying
= m(m + 1)
= m² + m
Now if m is even it is in the form of 2n
= (2n)² + 2n
= 4n² + 2n
= 2n( 2n + 1)
Which is always divisible by 2
Now if m is odd it is in the form of 2n + 1
= (2n + 1)² + (2n + 1)
= (4n² + 2n + 1) + (2n + 1)
= 4n² + 4n + 2
= 2( 2n² + 2n + 1)
Which is always divisible by 2
Hence Proved
Answer:
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Let n and n + 1 are two consecutive positive integer
We know that n is of the form n = 2q and n + 1 = 2q + 1
n (n + 1) = 2q (2q + 1) = 2 (2q2 + q)
Which is divisible by 2
If n = 2q + 1, then
n (n + 1) = (2q + 1) (2q + 2)
= (2q + 1) x 2(q + 1)
= 2(2q + 1)(q + 1)
Which is also divisible by 2
Hence the product of two consecutive positive integers is divisible by 2