# Rational approximation of transcendental and algebraic numbers

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?

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.