Choosing rationally independent irrational numbers Can one choose $n$ irrational numbers (call the set $S$) such a way that the $n$ numbers along with the products of any combination numbers from $S$ are all irrational and rationally independent? Are there uncountably many such choices?
 A: Greg Martin has given a concrete example. However, there is also a general theorem here, applicable to all "small" subfields of $\mathbb{R}$.
Suppose $K$ is a subfield of $\mathbb{R}$ of cardinality $<2^{\aleph_0}$ (e.g. $\mathbb{Q}$, the algebraic reals, $\mathbb{Q}(\pi)$, etc.). Say that $A\subseteq\mathbb{R}$ is very $K$-independent iff the set of (finite) products of elements of $A$ is $K$-linearly independent. Then we have:

For every very $K$-independent set $A$ with $\vert A\vert<2^{\aleph_0}$ there is some $a\in \mathbb{R}\setminus A$ such that $A\cup\{a\}$ is very $K$-independent.

The proof is just a cardinality argument: there are $<2^{\aleph_0}$-many nonconstant single-variable polynomials with coefficients from $K\cup A$, each of which has in turn only finitely many zeroes, while $\mathbb{R}\setminus A$ has cardinality $2^{\aleph_0}$. This means that there is some $a\in\mathbb{R}\setminus A$ such that no nonconstant single-variable polynomial with coefficients from $K\cup A$ is zero at $a$, and this means exactly that $A\cup\{a\}$ is very $K$-independent.
As a corollary, Zorn's lemma gives us:

There is a $K$-independent set of size $2^{\aleph_0}$.


EDIT: The appeal to Zorn's lemma at the end raises a natural question: do we actually need the axiom of choice to get a very independent set of size continuum? Certainly we don't for $\mathbb{Q}$, but what about for an arbitrary "small" subfield of $\mathbb{R}$?
For countable fields, choice is not necessary although this takes a trick: by a sort of "tree construction" (the set of reals we want corresponding to the set of paths through the tree) we can in fact show that if $K$ is a countable field then there is a perfect set (= nonempty, closed, and no isolated points; note that all perfect sets in $\mathbb{R}$ have size continuum) which is very $K$-independent. I'm avoiding details here since it's rather tedious; if you're interested, let me know and I'll elaborate.
But even with choice there could be a wide gap between $\aleph_0$ and $2^{\aleph_0}$, and without choice things could get truly bizarre. So in fact I suspect that the very general statement "For every subfield $K$ of $\mathbb{R}$ of cardinality $<2^{\aleph_0}$ there is a very $K$-independent set of size $2^{\aleph_0}$" is independent of $\mathsf{ZF}$ (= set theory without choice). However, I don't immediately see how to prove this.
