How to disprove there exists a real number $x$ with $x^2 < x < x^3$ 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
 A: First let $x^3\gt x^2\implies x(x+1)(x-1)\gt 0$ which has solutions $(-1,0)\cup (1,\infty)$. Now, taking $x\gt x^2\implies x(1-x)\gt0$ which has solution set $(0,1)$. Th intersection of these two solution sets is $\emptyset$. Thus we can't have $x^3\gt x^2$ and $x\gt x^2$ at the same time.
A: $\rm{\bf Hint}\quad\rm x^{-2}(x^2 < x^3)\ \Rightarrow\ 1 < x\ \Rightarrow\ x < x^2\ \Rightarrow\Leftarrow\ x > x^2$
A: Another way to prove it is to subtract $x^2$ from each term. Then you get
$$0 < x-x^2 < x^3-x^2$$ that is, $$0<x(1-x)<x^2(x-1)$$
Now $x^2>0$, thus we must have $x-1>0$. But then $1-x<0$, thus to have $x(1-x)>0$ we also need $x<0$. But then $x-1<-1<0$ in contradiction to $x-1>0$.
A: 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 $x$.
A: If $x \le 0$, then we cannot have $x^2 \lt x$, since $x^2\ge 0$ for all $x$. 
So now suppose that $x \gt 0$. If $x \lt 1$, then $x^3 \lt x^2$, contradicting one of our inequalities. If $x \gt 1$, then $x^2\gt x$, contradicting the other inequality. And of course if $x=1$ then both inequalities fail.
Remark: We do not really need a cases analysis. If the inequality holds, then clearly $x \gt x^2 \ge 0$. But then from $x^2 \lt x$ we can conclude that $x^3 \lt x^2$, which contradicts the given fact that $x^2 \lt x^3$. 
