Steve cracked that elementary proof so easily, I thought I'd throw up another interesting one - somewhat related - which I remember from my number theory class. I haven't tried to reconstruct this one yet, but I remember it being somewhat more difficult than the infinitude of primes one.
Prove that for any natural number n, there exist n consecutive composite numbers.
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Oh no, wait, this one is easy too. Real easy.
Sadly, Steve's elegant little proof of the infinitude of primes was flawed. Even more sadly, I , being the greater fool, accepted it as valid (until he busted himself).
Which means this proof is actually easier.
Hint: construct a number X such that 2 divides X + 2, 3 divides X + 3, ..., X + 1 divides X + (n + 1).
That's about it on a platter. Multiply the first n integers starting with 2, forming product k. k is a multiple of 2 and therefore composite, and so then is k + 2. Likewise, k + 3 is a multiple of 3, and therefore composite, as is k + 4, k + 5, on up to k + n + 1, forming the desired series of n consecutive composites.
I think you need (n+1)! rather than n!, but otherwise this is right.
Oh, and OCD compels me to point out that you meant "...first n positive integers..."
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