Showing two sets are equivalent

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    Equivalent Sets
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Homework Statement



Show that R \approx R+ , that is, the set of all real numbers is equivalent to the set of all positive real numbers


Homework Equations



The only relevant equation is finding one such that F:R\rightarrowR+ is a bijection.

The Attempt at a Solution



I've attempted to use the tangent function as a bijection, but that doesn't work. This isn't for homework but for a test review sheet, so all guidance is gladly appreciated.
 
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Equivalent? You could probably show that the sets are isomorphic by finding an isomorphism. The function:
F: \mathbb{R} \to \mathbb{R}^+
F:x \to x^2
would map the real numbers into positive real numbers. I should think F is an isomorphism over these sets.
 
But that isn't a bijection as its not one-to-one. For example, -5 maps to 25 and 5 maps to 25. If I can find a bijection from R to R+ ,then I've proven they are equivalent sets. Just can't find the darn bijection.
 
Hi hwill205:smile:

You should think of exponential functions...
 
Damn, y=ex works. Thanks a lot man. Can't believe I didn't think of that.
 
hwill205 said:
But that isn't a bijection as its not one-to-one. For example, -5 maps to 25 and 5 maps to 25. If I can find a bijection from R to R+ ,then I've proven they are equivalent sets. Just can't find the darn bijection.

Oh, of course. That + sign made me think of positive real numbers, and I messed up codomain and domain. :(
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

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