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Re:How many real numbers exist? New proof moves closer to an... by drdeca on Thursday July 15, 2021 @02:55PM (#27848612)
1/(2^{\aleph_0}) isn’t something that has a clear meaning.

2^{\aleph_0} is a cardinal number, which isn’t really a number in the sense of “an element of a field” or something like that. Dividing by it isn’t a well defined thing.

And, you certainly can’t just multiply any real number (or, any real number between 0 and 1) by 2^{\aleph_0} and get a different integer as a result.

(Now, if you work in the surreal numbers, you can define things like n/(2^{\aleph_0}) (identifying cardinals with the first ordinal of that cardinality), but these would not be real numbers. They would all be infinitesimal , smaller than 1/k for all positive integers k, and yet bigger than 0. Similarly in the surreal numbers, you could multiply real numbers between 0 and 1 by 2^{\aleph_0}, but you would get surreal numbers which are larger than every integer (in fact, larger than any countable ordinal))

Summary: What you wrote doesn’t define a mapping from the integers to the real numbers . (It can be interpreted as defining a map from integers to something else though.)

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