TITLE: How to disprove there exists a real number $x$ with $x^2 < x < x^3$
QUESTION [5 upvotes]: I realize that the only method is to show various cases:
I must test for $x > 1$, $x < -1$, $0 \leq x \leq 1$, and $-1\leq x \leq0$.
But even with this, I don't understand how to inject the properties of these four distinct possible $x$'s into the inequality (from the title) in order to show that none of these work.
Thanks for any help

REPLY [2 votes]: For the full inequality to be true, both halves must be true. Let's take $x^2 < x$ first. For any $x<0$, $x^2 > x$ because $x^2 = |x|^2$; all squares are nonnegative and so are greater than any negative root. For any $x>1$, $x^2>x$. That means that the only values for which the first inequality is true is where $0 < x < 1$.
So, for the entire inequality to be true, $x < x^3$ for some $0<x<1$. however, $x^3$ behaves much the same way $x^2$ does in this range, for the same reason; multiplying any $0<x<1$ by any other $0<y<1$ (including $x=y$) will result in a number $z$ such that $z < x$ and $z < y$, so any $x^n < x$ when $0<x<1$ and so the inequality can never hold for any $x$.