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Here's a simple function from positive integers to primes. Let f : Z+ -> Z+ Let f(0) = 2 Let f(n) = p, where p is not divisible by f(n') for all positive int
by karpierz 3y ago
Here's a simple function from positive integers to primes.
Let f : Z+ -> Z+
Let f(0) = 2
Let f(n) = p, where p is not divisible by f(n') for all positive integers n' such that n' < n, and p <= p' for all positive integers p' such that p' is not divisible by f(n'') for all positive integers n'' such that n'' < n.
- kristopolous 3y agoSure. You just used jargon and formality to define it. That's not an actual algebraic definition in the same way you can express a peano successor function for say, all natural numbers. My claim is that's not an equivalent class of functions. You can define the irrationals in an almost identical way to that which you proposed You could claim that one can come to an agreement on the Nth prime but we cannot come to an agreement on the Nth irrational. But I say sure you can. You can number both of them off however you please. For every prime, we take a square root and get an irrational. There's a really simple one. Of course there's lots of holes but the point is the behaviors are far closer then they are to the rationals. I bet the farm that when Riemann Zeta is solved it will show just this connection. I can only see it in a blurry fuzzy way. But I don't think it's an illusion
- karpierz 3y ago> That's not an actual algebraic definition in the same way you can express a peano successor function for say, all natural numbers. I guess it's not clear to me how you define an "actual algebraic definition" of a function. Can you provide a formal definition for this? Like some rules that I could use to determine whether a function is algebraically defined. I think it'll clear up whatever misunderstanding I seem to have.