I want to find the asymptotic number of natural numbers smaller than $n$ which are not prime. I know that the number of primes smaller than $n$ can be expressed by the prime counting function $\pi (n)$ which is asymptotically the same as $n/\log(n)$. So the total number of non-primes smaller than $n$ is asymptotically the same as $n - n/\log(n)$. As $n$ grows slightly faster than $n/\log(n)$, does this mean that the number of non primes smaller than $n$ is $\Theta (n)$ and therefore there exists a constant $c > 0$ s.t. the number of non-primes smaller than $n$ is greater than $c*n$?
Is this reasoning correct?
Yes, I think the asymptotic properties of the primes due to the prime number theorem, guarantees the existence of a global constant $c>0.$
If I edit your end statement slightly, to:
I expect the answer to be "Yes, and $c=1/3,$ is best." For example:
The number of non-primes in $\mathbb{N},\ \leq 1,$ is $\geq \frac{1}{3} \cdot 1;$
The number of non-primes in $\mathbb{N},\ \leq 2,$ is $\geq\frac{1}{3} \cdot 2;$
The number of non-primes in $\mathbb{N},\ \leq 3,$ is $\geq\frac{1}{3} \cdot 3;$
The number of non-primes in $\mathbb{N},\ \leq 4,$ is $\geq\frac{1}{3} \cdot 4;$
The number of non-primes in $\mathbb{N},\ \leq 5,$ is greater than $\frac{1}{3} \cdot 5;$
and so on. And we can't do better than $1/3,$ in the sense that, if we replace $1/3$ by a larger number $c> 1/3$, then the above fails, since then, the statement, "The number of non-primes in $\mathbb{N},\ \leq 3,$ is $\geq c \cdot 3$" would be false. Or another way to view it is that any $0\leq c \leq 1/3$ satisfies all the above statements.
However, I am not sure of a proof for this, although I would expect a proof for this to be known by some number theorists, perhaps with the help of known inequalities of the prime counting function.