Elementary Methods in Number Theory Exercise 1.5.14

Prove that
\begin{equation}
\pi(x)>\log_2\log_2x
\end{equation}for all $x>1$.

 

 

Proof:That is ,
\begin{equation}
2^{\pi(x)}>\log_2x
\end{equation}
That is
\begin{equation}
2^{2^{\pi(x)}}>x
\end{equation}
According to Elementary Methods in Number Theory Exercise 1.5.13 ,
\begin{equation}
2^{2^{\pi(x)}}\geq p_1p_2\cdots p_{\pi(x)}+1
\end{equation}

where $p_1<p_2<\cdots <p_{\pi (x)}$ are all the prime number not larger than $x$.It is easy to verify that

\begin{equation}
p_1p_2\cdots p_{\pi(x)}+1>x
\end{equation}(Why?hint:Factorize [x]).
So
\begin{equation}
2^{2^{\pi(x)}}>x
\end{equation}

转载于:https://www.cnblogs.com/yeluqing/archive/2012/12/02/3827595.html