I'm working through Velleman's "How to Prove It", and I decided to try and do a proof of a lemma used in one of the exercises. The question in the book were for the reader to think about whether there are any more triplet primes except 3, 5, 7.
Of course there aren't, and the easiest proof of that is showing that since one out of every three consecutive integers needs to be divisible by three, it's impossible for all those three numbers to be prime.
Since that proof uses the (unwritten) lemma that for every n consecutive integers, at least one of them will be divisible by n, I'd try my hand at proving it, before actually diving into the proof methods of the book.
In short, rate my proof Veeky Forums:
Theorem:
For every [math]n[/math] consecutive integers, at least one of those integers will be divisible by [math]n[/math].
Proof:
The case where n is 1 is trivial, since all numbers divide by (at least) 1.
Let [math]n[/math] be an integer such that [math]n > 1[/math]. Let [math]x, x-1,x-2,...,x-(n-1)[/math] be the list of the [math]n[/math] consecutive integers.
We proceed by contradiction. Since none of the integers in the list is divisible by [math]n[/math], [math]x[/math] is not divisible by [math]n[/math]. Thus,
[eqn]x = qn + r[/eqn],
[eqn](1) x - r = qn[/eqn]
Where [math]r[/math] is the remainder.
Since [math]r[/math] is the remainder, [math]0 < r < n[/math], but [math]r[/math] must be greater than [math]n-1[/math], since all integers from [math]x-1[/math] to [math]x - (n-1)[/math] are found in the list. Thus we have a contradiction.
Q.E.D (hopefully)
Feel free to point out the flaws in my proof, there are certain to be some. Anything that's unclear, or unconventional in proofs.