basic algebra inequalities it's been a while since I worked with inequalities. Im trying to figure out the steps taken to go from $p > n^{1/3}$ to $p/n < n^{2/3}$
I am assuming the first step they did, is take the inverse of $p$, so
$p > n^{1/3} \Rightarrow p^{-1} < n^{1/3}$ then multiplied both sides by $n$.
So
$n/p < n^{2/3}$
Is this correct? I am just not sure if one could simply take an inverse of a number he pleases in an equation and reverse the inequalities.
Thank you
 A: Nevermind I figured it out. For anyone who is curious 
$p>n^{1/3}⇒p^{−1}<n^{−1/3}⇒np−1<nn^{−1/3}⇒n/p<n^{−1/3+1}=n^{2/3}$
A: You are correct but you made several typos that it appears everything you say is wrong.
" Im trying to figure out the steps taken to go from $p>n^{1/3}$ to $p/n<n^{2/3}$"
You meant (I assume) to type $\frac np < n^{\frac 23}$.  $\frac pn < n^{\frac 23}$ is not a correct result.
"I am assuming the first step they did, is take the inverse of p
, so $p>n^{1/3}⇒p^{−1}<n^{1/3}$"
I assume you meant to type $p^{-1} < n^{-\frac 13}$ and not $p^{-1} < n^{\frac 13}$.
By coincidence the latter is true if $p > 1$ and $n > 1$ but it isn't true if $n < 1$.  And the latter will not get you anywhere.
"then multiplied both sides by n. So $n/p<n^{2/3}$"
$n^{-\frac 13}*n = n^{-\frac 13 + 1} = n^{\frac 23}$ so this would be correct.
However as you had incorrectly typed $n^{\frac 13}$, $n^{\frac 13}*n = n^{\frac 13 + 1} \ne n^{\frac 23}$ and so wouldn't be correct.  But as $n^{\frac 13}$ wasn't correct in the first place, this is ultimately correct.
So, yes, your reasoning was correct but you typos were all wrong.
