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The central binomial coefficient satisfies $4^m/(2m+1)\le \left(\genfrac{}{}{0ex}{}{2m}{m}\right)\le 4^m$. Prime factors between $m$ and $2m$ give a dyadic upper bound for the Chebyshev theta function. The prime-power exponents in the factorial quotient are sums of zeros and ones, so the central coefficient gives a linear lower bound for the Second Chebyshev function. Higher prime powers contribute only $O(\sqrt{x}\log x)$; hence the prime-counting function has upper and lower bounds of order $x/\log x$.

A positive integer $n$ is decisive when every integer $a$ satisfying $1<a<n$ and $\gcd (a,n)=1$ is a prime number. There are only finitely many decisive numbers: every prime $p<\sqrt{n}$ must divide a decisive $n$, so the Chebyshev theta function gives $\log n\ge \vartheta (\sqrt{n})$, which eventually contradicts the linear lower bound for $\vartheta (\sqrt{n})$.

Use the Chebyshev estimate from central binomial coefficients. DefineThe first is the Chebyshev theta function; the second counts prime powers with the same logarithmic prime weight. For a positive integer $m$, every prime $m<p\le 2m$ divides $\left(\genfrac{}{}{0ex}{}{2m}{m}\right)$, henceSumming over dyadic intervals yields $\vartheta (2^j)\le 2^{j+1}\log 2$. By monotonicity and rounding $x$ upward to a power of two,

For the lower bound, the central binomial coefficient is the largest of the $2m+1$ coefficients whose sum is $4^m$, soThe exponent of a prime in the central coefficient isEach summand is zero or one. Therefore $\log \left(\genfrac{}{}{0ex}{}{2m}{m}\right)\le \psi (2m)$. Higher prime powers contribute onlyCombining the lower bound with $2m$ just below $x$ shows $\vartheta (x)\ge cx$ for sufficiently large $x$, with, for example, $c=(\log 2)/2$.

Let $\pi (x)$ denote the number of primes at most $x$. Since every prime weight is at most $\log x$,For the upper bound, separate the primes at $\sqrt{x}$. The small ones number at most $\sqrt{x}$, and every larger one has weight at least $(\log x)/2$, givingSince $\sqrt{x}=o(x/\log x)$, this provesfor all sufficiently large $n$, for instance with $A=(\log 2)/2$ and $B=8\log 2+1$. This elementary Chebyshev estimate does not assume the Prime number theorem.

Write the Chebyshev theta function asTaking natural logarithms in part a givesand henceBecause $\sqrt{n}\log n/n\to 0$, the right-hand side is at least $n/2$ for every sufficiently large integer $n$.

Now let $x$ be sufficiently large and set $n=\lfloor x/2\rfloor$ using the floor function. Then $2n\le x$ and $n\ge x/3$, so the monotonicity of $\vartheta$ gives