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The existence of transcendental numbers can be shown easily by considering the cardinality of the set of solutions to polynomials with integer cofficents and the cardinality of the real numbers.

It is written here that Liouville first proved the existence of transcendental numbers in 1844. I doubt that he proved this result by a cardinality argument, as I think that these ideas were first brought by Cantor and this was well after 1844. How did Liouville prove the existence of transcendental numbers.

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Cantor's argument would only appear 15-20 years later. Read about Liouville numbers. – Asaf Karagila Jan 3 '13 at 21:08
See Basically, the irrational algebraic numbers (non-transcendental numbers) "poorly approximate" rational numbers. – Andrew Salmon Jan 3 '13 at 21:09
Roughly speaking, irrational algebraic numbers are poorly approximable by rationals. Liouville showed that numbers whose decimal expansion involves only $0$'s and $1$'s, where the number of $0$'s between consecutive $1$'s grows rapidly, are well approximable by rationals. There is a hint of diagonalization in the construction. – André Nicolas Jan 3 '13 at 21:15

1 Answer 1

up vote 6 down vote accepted

He came up with the idea of what we now call a Liouville number, and then showed that all of them are transcendental.

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Thanks. Seeing this Liouville's argument makes me appreciate Cantor's argument more.(+1) – Amr Jan 3 '13 at 21:14
@Amr: Haha... But it is still a beautiful argument, though. – Haskell Curry Jan 3 '13 at 21:17
I still didn't read Liouville's argument. It just seems that Cantor's argument is more economic – Amr Jan 3 '13 at 21:19
@Amr economic maybe, but totally unconstructive. – Hagen von Eitzen Jan 3 '13 at 21:52
@HagenvonEitzen: Cantor's argument is, or can be made, totally constructive. One can explicitly enumerate the real algebraics, and using this enumeration one can explicitly construct the decimal expansion of a transcendental number. – André Nicolas Jan 3 '13 at 22:10

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