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\vspace{-2em}

Suppose $x > 4$, and $2^k \le x < 2^{k + 1}$, for
some $k \ge 2$. Then
\[ \pc(x) \le \pc(2^{k + 1}) \le \frac{3 \times
2^{k + 1}}{k + 1} \le \frac{6 \times 2^k}{k} = 6
\log 2 \cdot \frac{2^k}{k \log 2} = 6 \log 2 \cdot
f(2^k) \]
where $f(x) = \frac{x}{\log x}$. Note that
\[ f'(x) = \frac{\log x - 1}{(\log x)^2} \]
and $f'(x) > 0$ when $x > e$. Hence $f(x)$ is
increasing on $(4, \infty)$. Hence
\[ \pc(x) \le 6 \log 2 \cdot f(x) = 6 \log 2
\frac{x}{\log x} .\]

We now find a lower bound for $\pi(x)$. Let $n \in
\NN$, $N = {2n \choose n}$. We know if $p$ is a
\gls{prime} and $p \divides N$, then $p \le 2n$. So
\[ N
= \prod_p p^{\padic_p(N)}
= \prod_{p \le 2n} p^{\padic_p(N)}
\le \prod_{p \le 2n} (2n)
= (2n)^{\pc(2n)} .\]
We also know that $N \ge \frac{2^{2n}}{2n}$, hence
\[ \frac{2^{2n}}{2n} \le N \le (2n)^{\pc(2n)} .\]
Hence
\begin{align*}
  \implies 2^{2n}
  &\le (2n)^{\pc(2n) + 1} \\
  \implies 2n \log 2
  &\le (\pc(2n) + 1) \log 2n \\
  \implies \pc(2n)
  &\ge \frac{2n}{\log 2n} \cdot \log 2 - 1
\end{align*}
Now suppose $X > 4$, and choose $n \in \NN$ so
that $2n \le x \le 2n + 2$. Then
\[ \pc(x) \ge \pc(2n)
\ge \frac{2n}{\log 2n} \cdot \log 2 - 1
\ge \frac{x - 2}{\log x} \cdot \log 2 - 1 .\]
\textbf{Claim:} If $x \ge 16$, then
\[ \frac{x - 2}{\log x} \log 2 - 1
\ge \frac{x}{\log x} \frac{\log 2}{2} .\]
Proof of the claim: Equivalent to
\[ \label{lec17_l48_eq} \frac{\log 2}{2}
\frac{x}{\log x} - \frac{2 \log 2}{\log x} - 1 \ge
0 \tag{$*$} \]
Plugging in $x = 16$, we get
\[ \frac{\log 2}{2} \cdot \frac{16}{4 \log 2} -
\frac{2 \log 2}{4 \log 2} - 1 = 2 - \half - 1 =
\half \ge 0 .\]
Note the $\RHS$ of \eqref{lec17_l48_eq} is
increasing when $x \ge 16$.

This claim now implies
\[ \pc(x) \ge \frac{\log 2}{2} \frac{x}{\log x} \]
when $x \ge 16$. Remains to consider $4 < x \le
16$. $\frac{x}{\log x}$ is increasing implies the
largest value of $\frac{\log 2}{2} \frac{x}{\log
x}$ in thsi range is
\[ \frac{\log 2}{2} \cdot \frac{16}{4 \log 2} = 2
.\]
Certainly $\pc(x) \ge 2$ when $4 < x \le 16$.
\fakeqedhere

\begin{theorem}[Bertrand's Postulate]
  \label{bertrands_postulate}
  If $n \in \NN$, $n > 1$, then there exists a
  \gls{prime} $p$ such that $n \le p < 2n$.
\end{theorem}
\vspace{-1em}

We first prove:
\begin{lemma}
  \label{primorial_upper_bound_lemma}
  \glssymboldefn{primorial}{$P$}{$P$}
  Let $x \ge 1$, $P(x) = \prod_{p \le x} p$. Then
  $P(x) \le 4^x$.
\end{lemma}

\begin{proof}
  It suffices to show $\primorial(x) \le 4^x$ when
  $x = n \in \NN$. We do this b induction on $n$.
  It holds for $n = 1, 2$. For the finduction
  step, consider for $k \in \NN$,
  \[ 2{2k + 1 \choose k + 1}
  = {2k + 1 \choose k + 1} + {2k + 1 \choose k}
  \le (1 + 1)^{2k + 1} = 2^{2k + 1} .\]
  If $p$ is a \gls{prime} and $k + 2 \le p \le 2k + 1$,
  then $p \divides {2k + 1 \choose k + 1}$. So
  \[ \primorial (2k + 2) = \primorial(2k + 1)
  = \prod_{p \le 2k + 1} p = \prod_{p \le k + 1} p
  \prod_{k + 2 \le p \le 2k + 1} p .\]
  By induction,
  \[ \primorial(2k + 1) \le 4^{k + 1} {2k + 1
  \choose k + 1} \le 4^{k + 1} 4^k = 4^{2k + 1} .\]
  Hence, $\primorial(2k + 1) \le 4^{2k + 1}$, and
  $\primorial(2k + 2) = \primorial(2k + 1) \le
  4^{2k + 1} \le 4^{2k + 2}$.
\end{proof}

\begin{proof}[Proof of \cref{bertrands_postulate}]
  Let $n \in \NN$, $n > 1$, and suppose for
  contradiction that there are no \glspl{prime}
  $p$ with $n \le p < 2n$. Consider $N = {2n
  \choose n}$. We proved in
  \cref{chebyshev_lemmas} that if $p
  \divides N$, then either $p > n$ or $p \le
  \frac{2n}{3}$. So in fact (since we're assuming
  there are no primes between $n$ and $2n$),
  \[ N = \prod_{p \le \frac{2n}{3}}
  p^{\padic_p(N)} .\]
  Write $N = N_1 N_2$, where
  \[ N_1 = \prod_{\substack{p \divides N \\
  \padic_p(N) = 1}}, \qquad N_2 = \prod_{p
  \divides N \\ \padic_p(N) \ge 2} p^{\padic_p(N)}
  .\]
  By \cref{primorial_upper_bound_lemma}, we have
  \[ N_1 \le \primorial \left( \frac{2n}{3}
  \right) \le 4^{\frac{2n}{3}} .\]
  If $p$ is \gls{prime} and $\padic_p(N) \ge 2$,
  then (by \cref{chebyshev_lemmas}),
  $p^{\padic_p(N)} \le 2n \implies p \le
  \sqrt{2n}$. So
  \[ \frac{2^{2n}}{2n} \le N = N_1 N_2 \le
  4^{\frac{2n}{3}} (2n)^{\sqrt{2n}} .\]
  (as product over \glspl{prime} $p \le
  \sqrt{2n}$). Rearrange:
  \begin{align*}
    2^{2n} - \frac{4n}{3}
    &\le (2n)^{1 + \sqrt{2n}} \\
    \implies \frac{2n}{3} \log 2
    &\le (1 + \sqrt{n}) \log 2n
  \end{align*}
  This is a contradiction when $n$ is large enough
  (as $\frac{(1 + \sqrt{2n})\log 2n}{2n} \to 0$ as
  $n \to \infty$). In fact, this gives a
  contradiction when $n \ge 500$, so the theorem
  holds in this case. To complete the proof for $1
  < n < 500$, can either check every case by hand,
  or note that it's enough to find a sequence $2 =
  p_1, p_2, \ldots, p_r$ of primes such that:
  \begin{itemize}
    \item $\forall i = 1, \ldots, r - 1$, $p_{i +
      1} \le 2p_i + 1$.
    \item $p_r < 500$.
  \end{itemize}
  (as then the intervals $\left(\frac{p}{2},
  p\right]$ cover $\NN \cap (1, 500)$).

  We can take $2, 5, 11, 23, 47, 89, 179, 359,
  719$.
\end{proof}