Ordered numbers
3 Answers
This is not true. For example, take $a=\frac{1}{4}$ and $b=\frac{1}{3}$. Then, $x. So $x, 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
Hint: No, you can't necessarily if $a,b$ are given. What happens if $b=\frac 12$?
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0@Hurkyl: it is fine if $s=\frac 1{16}$, but not if $a=\frac 13$ – 2012-06-10
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.