# And one more similar one…

May 6, 2012 1 Comment

Take any number . It’s divisible by at least one prime. Since it has no common factors with , then the new number is divisible by at least two primes. Then is divisible by at least 3 primes by similar reasoning. In general, is divisible by at least distinct primes. Since can grow arbitrarily large, there must be arbitrarily large primes.

This proof has a similar vibe to the first two I gave. In fact, if you unravel the topological proof I gave yesterday, you’ll see even that has its similarities. I promise, we’re going to move on to genuinely different things now.

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