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I'm having trouble proving that $\mathbb R$ is equinumerous to $(-π,π)$. I'm think about using a trigonometric function such as $\cos$ or $\sin$, but there are between the interval of $(0,1)$. Could someone help me define a bijective map from $(−π,π)→\mathbb R$?

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You could use a trigonometric function such as $\tan$, although you must first divide the number by $2$ to instead get $\tan$ applied to a number in the interval $(-\pi/2,\pi/2)$.

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  • $\begingroup$ That's very clever! Thank you. I'll accept your answer in a few :-) $\endgroup$ – Anna Jeanine Oct 25 '17 at 14:38
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For any open interval $(a,b)$, you can use:

$$f(x) = \frac{x -(a+b)/2}{(x-a)(x-b)}$$

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