Ordered numbers Let $0<a<b<1$, can we find a point $x\in (a,b)$ such that $a<x^{2}<x<b$. I know that we can find $x$ such that $a<x<b$ and this $x$ will satisfies $0<x^{2}<x<b$, but I'm not sure how to choose such $x$  with $a<x^{2}<x<b$?
 A: This is not true.  For example, take $a=\frac{1}{4}$ and $b=\frac{1}{3}$.  Then, $x<b \implies x^2<b^2=\frac{1}{9}$.  So $x<b \implies x^2<a$, for all $x$.  This contradicts your statement.
As Serkan noted in the comments, however, this holds if and only if $b \gt b^2 \gt a$ (it cannot be equal).  If this condition holds, then $b^2>b^2-\frac{1}{n}>a$, for some natural number $n$.  Set $x=\sqrt{b^2-\frac{1}{n}}$.  Then $b^2>x^2>a$, so $b>x$.  Since $x<1$, $x^2<x$, and the inequality is complete, with $a<x^2<x<b$.
A: Hint:  No, you can't necessarily if $a,b$ are given.  What happens if $b=\frac 12$?
A: Suppose you could choose an $x$ with $0 < a < x^2 < x < b < 1$. Then any larger choice of $x$ would work too, so long as it's less than $b$, correct?
So, we might as well pick $x$ as close to $b$ as possible. And as $x$ gets arbitrarily close to $b$ (but slightly less), that means $x^2$ can be made arbitrarily close to $b^2$ (but slightly less). But that's the limit; we can't make $x^2 = b^2$ or larger.
So, that suggests what restrictions we have to put on $a$ in order for the problem to be solvable.
Once you understand this idea, it should be straightforward to do this calculation rigorously.
