why does $\mathbf{Q}$ lie dense in $\mathbf{R}$ and $\overline{\mathbf{Q}}$ lie dense in $\mathbf{C}$ Why is $\mathbf{Q}$ dense in $\mathbf{R}$?
Does this imply that (Edit: the field of algebraic numbers) $\overline{\mathbf{Q}}$ is dense in $\mathbf{C}$?
For the first question, I guess one has to show that for every $x$ in $\mathbf{R}$ and every $\epsilon >0$, there exists a rational number $q$ such that $\vert x- q \vert < \epsilon$. Does the proof of this rely upon methods from Diophantine Approximation or is it easier than this?
 A: The density of the rational numbers inside the real numbers follows from the very definition of real numbers as (classes of equivalence of) Cauchy sequences of rational numbers. By definition, every real number is the limit of some sequence of rational numbers.
As for your second question: which is the meaning of your $\overline{\mathbf{Q}}$?
Added:  If $\overline{\mathbb Q}\supset\mathbb Q+i\mathbb Q$, you can just prove density separately in the real and imaginary axes
A: As I said in my comment, the algebraic closure $\overline{\mathbb Q}$ of $\mathbb Q$ in $\mathbb C$ is dense because it contains $\mathbb Q+i\mathbb Q$, which, as others explained, is dense. 
A: For the first, just take a rational with large enough denominator ($2/\epsilon$ will work)
For the second, what do you mean by $\overline{\mathbf{Q}}$?  If it is the closure of $\mathbf{Q}$, that is $\mathbf{R}$, but you don't get very close to $i$ with any real numbers.  Maybe you want $\mathbf{Q}\times\mathbf{Q}$?
Added:  If $\overline{\mathbb Q}\supset\mathbb Q+i\mathbb Q$, you can just prove density separately in the real and imaginary axes
A: If q is a rational number, then the sequence {q,q,..,q,...,q} converges to q.
If $q$ is irrational, then, by the LUB property, it has a (non-periodic) decimal 
expansion $a.q_oq_1...q_n.....$ for $a, q_0,q_1,...,q_n,... $integers  . Then the sequence : {a,$a+\frac {q_0}{10}, a+\frac{q_0}{10}+\frac{q_1}{10^2},......., a+\frac{q_0}{10}+.......+\frac{q_n}{10^{n+1}},....$ converges to q.
A: Well, if $\mathbb R$ is defined as some complete ordered field, then first prove the archimedean property, so that there is no positive number $< 1/n$ for all natural numbers $n$, and then the "rational with large enough denominator" trick will work.
