### Embarrasing mistake

I just realized that my last post was far off the mark. There is a super-obvious example that shows that f(n) >= n/2, and so in fact f(n) = n/2. The example is the subset {n/2 + 1, n/2 + 2, ... , n} which contains n/2 elements and clearly has the non-divisible property. The problem has very little to do with prime numbers. The previously-published result was sharp.

### A simple (?) number theory conjecture

Call a set S of positive integers non-divisible if, whenever a and b belong to S, it is not true that (a|b or b|a). For example, the set of prime numbers is non-divisible. Let S(n) be the set of the first n positive integers, and let f(n) be the cardinality of the largest non-divisible subset of S(n). Then clearly f(n) >= pi(n), where pi(n) denotes the number of prime numbers <= n. Furthermore, I know a pretty proof (published, but not by me) that shows f(n) <= n/2 (n even). That is, in any subset of n/2 + 1 positive integers all <= n, at least one integer divides another integer in the set. This proof involves the pigeonhole principle. So, pi(n) <= f(n) <= n/2. (n even)
My conjecture is that, in fact, f(n) = pi(n) for all n > 1. Does anyone have a counterexample or a proof? It seems like it should be very simple.

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