A question which is to be proved assuming Bertrand Postulate I am trying  exercise problem of David Burton Elementary Number Theory and I am unable to think about this problem. 


  Problem is->Prove that for every n$\geq$ 2 there exists a prime p with p$\leq$n&lt; 2p . 


Can someone please help how to think about this problem. 
I shall be really thankful.
According to Bertrand&rsquo;s postulate, there is a prime $p$ with $k\lt p\lt2k$ for all integers $k\gt1$. Apply this for $k=\left\lceil\frac n2\right\rceil$ to obtain a prime $p$ with $\left\lceil\frac n2\right\rceil\lt p\lt2\left\lceil\frac n2\right\rceil$ and thus $\frac n2\lt p\le n$. Now multiply the left inequality by $2$ and move it to the right.

(This proof doesn&rsquo;t apply for $n=2$, since in that case $\left\lceil\frac n2\right\rceil=1$, so we have to check separately that there is a prime $p=2$ such that $p\le2\lt2p$.)
Assuming $n&gt;2$ is non prime, now take the nearest prime (smaller than) $p$ of $n$, we can't have $p&lt;2p\leq n$ because that would imply there is a prime $q$ such that $p&lt;q&lt;2p\leq n$ .(bertrand) contradicting the assumption that $p$ is the nearest prime. so the exact inequality is : $p&lt;n&lt;2p$.(why $n\neq 2p ?$) for $n$ is prime, thats an easy task.