# 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

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Instead of the inequality $x^2 < x < x^3$, investigate the inequalities $x^2 < x$ and $x < x^3$ separately. Then put your results together. – ShreevatsaR Jul 31 '12 at 19:10
You can also factor it as $x(x-1) < 0 < x(x-1)(x+1),$ and argue about the contradictions of signs. – user2468 Jul 31 '12 at 19:18
The test is that given a value of $x$, it must satisfy all of those inequalities! – Code-Guru Jul 31 '12 at 19:23

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.

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doh! I did it in my head and made a sign error ;-( – Code-Guru Jul 31 '12 at 19:26

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 $x^2 \lt x^3$.

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$\rm{\bf Hint}\quad\rm x^{-2}(x^2 < x^3)\ \Rightarrow\ 1 < x\ \Rightarrow\ x < x^2\ \Rightarrow\Leftarrow\ x > x^2$

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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 $x-1>0$.

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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 $x$.

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