So how can I prove that there exists a bijection between R (real numbers) and C (complex numbers) I think that I should prove that there exists a bijection between RxR and C which is not too hard and then find a bijection between R and RxR, but isn't there a simple solution to find a direct bijection between C and R? P.s: I'm in terminal highschool so I'm fairly ignorant about set theory.
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I want to find a direct bijection not one that goes from R -> RxR and then RxR-> C, so I think my question is a bit different – AymaneMaaitat Sep 27 '18 at 12:22
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If $f\colon X\to Y$ is a bijection, $f^{-1}\colon Y\to X$ is a bijection too. – Asaf Karagila Sep 27 '18 at 12:25
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@Lazarus The composition of these functions is a direct bijection. – M. Winter Sep 27 '18 at 12:28