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I had a discussion with a friend that say that

transcendental numbers can be better approximated by rationals than algebraic (irrational) numbers, so, is some sense, they are more ''near'' to rationals that the algebraic numbers.

I don't understand in what sense this statement can be true (and I suspect that it is false).

I know that the way to estimate how a irrational number is approximated by rational is the irrationality measure, and that this measure is $1$ for rational numbers, $2$ for algebraic numbers, $\ge 2$ for transcendental numbers and $\infty$ for Liouville numbers.

I interpret this as the Liouville numbers are the numbers better approximated by rationals (correct?) and we know that these numbers are transcendental. But I also know that the set of Liouville numbers has measure $0$, so almost all transcendental numbers have irrationality measure $\ge 2$ but $< \infty$. But we cannot prove that almost all have irrationality measure $>2$, so we cannot say that thay are generally better approximated by rationals. Or there is some proof?

Anyway, i think that the idea of ''near to'' cannot be applied in this situation since it is obviously false in standard topology of reals and I don't see a topology in which it can be true. Or there is someone?

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A proof is given in Bugeaud, Yann (2012). Distribution modulo one and Diophantine approximation. Cambridge Tracts in Mathematics 193. With theorem statement on page 246 and proof reaching page 247. Naturally, page 246 is not part of the online preview. The really good pages never are. It seems it is in Appendix E.

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  • $\begingroup$ I found this answer quite misleading and would recommend editing it to state exactly what is being proved. The OP said "we cannot prove that almost all [transcendental numbers] have irrationality measure > 2 ... Or there is some proof?" This book except appears to prove the opposite claim: that almost all transcendental numbers have irrationality measure equal to 2. $\endgroup$ – tparker Jan 13 '18 at 19:05
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A quadratic irrational $x$ has an eventually periodic continued fraction, and therefore a continued fraction with bounded terms. That in turn implies that $x$ can't have very good rational approximations: there is $c > 0$ such that $|x - p/q| > c/q^2$ for every rational $p/q$. On the other hand, almost all $x$ have unbounded terms, so there are $p/q$ with $q^2 |x - p/q| $ arbitrarily small.

Unfortunately, we know very little about the continued fractions for algebraic numbers of degree $>2$; AFAIK the betting is that they have unbounded terms.

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