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 $a<x^{2}<x<b$?

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 $a<x^2<x<b$.
