Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

share|cite|improve this question
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
up vote 9 down vote accepted

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.

share|cite|improve this answer
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 $x^2 \lt x^3$.

share|cite|improve this answer

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

share|cite|improve this answer

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 $x-1>0$.

share|cite|improve this answer

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 $x$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.