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- Thread starter rbzima
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VietDao29

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f: (0, 1) --> R

x |--> x

That means, your f now, is

Or, if you want some other nicer function, then square root is a function to look for:

In the reals, the function [tex]y = \frac{1}{\sqrt{-x(x - 1)}}[/tex] is only defined on the interval (0, 1), and undefined everywhere else.

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For example, between (-1,1) f(x) = x/(x^2-1) works, but when I tried converting that into a form that is simply between (0,1), the inverse function looks exceptionally nasty! Using the inverse, I'm basically proving the function is onto, which in turn is proving that the (0,1) ~

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HallsofIvy

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As for a function that is one-to-one and onto from (0, 1) to the set of all real numbers, I would think a variation on tan(x) would work nicely. tan(x) covers all real numbers for x between [itex]-pi/2[/itex] and [itex]\pi/2[/itex] so [itex]f(x)= tan(-\pi/2+ \pi x)[/itex] should work nicely: when x= 0, [itex]-\pi/2+ \pi (0)= -pi/2[/itex]. When x= 1, [itex]-pi/2+ \pi (1)= \pi/2[/itex]. Unless I have misunderstood what you are looking for, that should be exactly what you want.

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As for a function that is one-to-one and onto from (0, 1) to the set of all real numbers, I would think a variation on tan(x) would work nicely. tan(x) covers all real numbers for x between [itex]-pi/2[/itex] and [itex]\pi/2[/itex] so [itex]f(x)= tan(-\pi/2+ \pi x)[/itex] should work nicely: when x= 0, [itex]-\pi/2+ \pi (0)= -pi/2[/itex]. When x= 1, [itex]-pi/2+ \pi (1)= \pi/2[/itex]. Unless I have misunderstood what you are looking for, that should be exactly what you want.

You are pretty much the man! Thanks a whole heap. And yes, I was looking for a 1-1 correspondence between (0,1) and R. I might have neglected to mention that.

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