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Le 05/04/2024 à 12:57, FromTheRafters a écrit :No, not "No Bijection", but that mapping isn't a bijection.WM explained on 4/4/2024 :Of course. But if someone doubts it, I could directly map the naturals n/1 to the fractions with the result that there is no bijection.Explain why first bijecting n and n/1 should destroy an existing bijection!>
You still seem to think that sets change. If you mean 'n' is an element of the naturals then of course N bijects with the naturals as embedded in Q.
Which only shows that this one mapping doesn't work.Also, the complement of the naturals over one in Q is the same size as the proper subset you created.No, that is disproved by the remaining Os.
Regards, WM
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