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Show $x^2$ in the interval $(0,1/3]$ has no fixed points.

I understand that the range of that domain is always lower than $y=x$, but what is a proper way of showing this? $$\left(0,\frac13\right] \to \left(0,\frac19\right]$$

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Funny looking intervals! – lhf Apr 12 '12 at 2:24
@lhf: I now have a pain in the neck, straining to tilt my head to the right... – Aryabhata Apr 12 '12 at 2:25

Let $f(x)=x^2$. You’re being asked to show that there is no $x\in(0,1/3]$ such that $f(x)=x$. Set up the equation $f(x)=x$, solve it, and discover that its only solutions are outside of $(0,1/3]$.

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@lhf: Perfectly true, but that wasn’t the question. – Brian M. Scott Apr 12 '12 at 1:09
Don't you mean that x > x^2 for (0,1) ? – Mary Apr 12 '12 at 1:19

The quadratic equation $x^2=x$ can be written as $x^2-x=0$, and then as $x(x-1)=0$. That has two solutions. Neither of them is in the interval $(0,1/3]$.

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