It is proved that, for every positive integer n, there are n consecutive composite integers.
Given statement,
For every positive integer n, there are n consecutive composite integers.
Proof:
Assume that m is less than n and that both m and n are positive integers. When m is less than n,
n! = 1 ⋅ 2 ⋅ 3 ⋅ ⋅ ⋅ m ⋅ ⋅ ⋅ n,
which is to say that m is a factor of n!
So,
m + n! = m + ( 1 ⋅ 2 ⋅ 3 ⋅ ⋅ ⋅ m ⋅ ⋅ ⋅ n )
= m ((1 ⋅ 2 ⋅ 3 ⋅ ⋅ ⋅ m ⋅ ⋅ ⋅ n) / m + 1)
Since m is a multiple of n!, n!/m is still an integer. Since the integer m is bounded between 1 and n, it follows that any number you choose up to n can divide m + n!, making it composite until the nth integer.
However, since n! contains the nth integer, the nth integer is also composite, meaning you can choose any integer between 1 and n inclusively and it will be composite.
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