I am in the process of reading this brilliant little book Prove It: A Structured Approach--very brilliant, have I mention that already?
Anyways, here is the theorem: For every positive integer $n$,there is a sequence of $n$ consecutive positive integers containing no primes.
I understand the prove, for the most part; but there is a little bit giving me difficulty.
Here is the entire proof:
Suppose $n$ is a positive integer. Let $x=(n+1)!1+2$.We will show that none of the numbers $x,x+1,x+2,\ldots,x+(n - 1)$ is prime. Since this is a sequence of n consecutive positive integers, this will prove the theorem. To see that $x$ is not prime, note that $x=1\cdot2\cdot3\cdot4\cdots(n+1)+2\rightarrow x = 2\cdot(1\cdot3\cdot4\cdots(n+1)+1)$ Thus, $x$ can be written as a product of two smaller positive integers, so $x$ is not prime. Similarly, we have $x+1=1\cdot2\cdot3\cdot4\cdots(n+1)+3\rightarrow x=3\cdot(1\cdot2\cdot4\cdots(n+l) +1)$, so $x+1$ is also not prime. In general, consider any number $x+i$, where $0\leq i \leq n-1$. Then we have
$$x+i = 1\cdot2\cdot3\cdot4\cdots(n+1)+(i +2)$$ $$x+i= (i+2)\cdot(1\cdot2\cdot3\cdots(i+1)\cdot(i+3)\cdots(n+1)+1)$$
The last step is the one I can't quite apprehend. I see that $(i+2)$ is factored out, but I why are there remaining $i$'s in the second factor?