Distinguishing properties of $\mathbb{Q}$ and $\mathbb{R} \setminus \mathbb{Q}$ that lead to differing cardinalities? I have what many on here would consider an elementary question, but I would very much appreciate responses that use only elementary ideas, if possible, so that I can understand them.  I would also appreciate detailed rather than brief responses.
By construction, $\mathbb{Q} \subseteq \mathbb{R}$.  The rationals are countably infinite, while the irrationals are uncountably infinite.  This got me thinking about what properties the irrationals have that the rationals do not have that would cause such a huge difference in their cardinalities (although this question isn't specific to their cardinalities -- the rationals have 0 Lebesgue measure and the irrationals have infinite Lebesgue measure).
Both the rationals and the irrationals are dense in $\mathbb{R}$.  So density does not play into cardinality.  But why doesn't it?  There are uncountably many irrationals, and so there are uncountably many intervals $(a,b)$ with irrational endpoints. Since each of these intervals contains a rational number, shouldn't there be uncountably many rationals?  There aren't uncountably many rationals, which means there is at least one rational that is contained in uncountably many of these intervals.  But which one?  Are they all contained in uncountably many of these intervals?  What if we only look at the subcollection of all intervals with irrational endpoints that also have infinitesimally small length (if possible)?  I know I'm rambling now...
I guess my main question is: what properties do the irrationals have that the rationals don't have that lead to the irrationals being uncountable?  Although, I would also like to hear thoughts on my statement above about the density of the rationals.
 A: In a metric space one can speak of a sequence of approximations which grow arbitrarily precise in the limit. One can phrase this in terms that do not actually assume there is a limit - namely, through the use of Cauchy sequences. A metric space fails to be complete if one can provide such a sequence of approximations growing arbitrarily precise but fails to converge to something. In a sense, we are specifying something that is not actually there - it is sort of a ghost, it transcends the space. It exists outside, in the completion of that space. One of the ingeniously clever tricks of math was instead of seeking to fully describe something infinite or beyond description, describe how one gets there (even though one never truly gets there). By identifying the destination with the journey, one gains the ability to speak of things like real numbers, limits from analysis, and even more exotic limits that exist in algebra (like, in the category of topological rings, the $p$-adic numbers).
There are different ways of going about "specifying" a real number. In general, any Cauchy sequence (modulo null sequences - those converging to $0$) will do, but kinds of specifications that come with rules are also nice. One can use digital expansions with respect to a chosen base - this is convenient and practical, if artificial. One can use continued fraction expansions (search this on google for more information). One can use Dedekind cuts, as a theoretical tool.
In these sorts of schemes (which are all metrically based - if one wants to use algebra and minimal polynomials to describe elements, or speak of computable/definable reals etc. then things can go a different way), rational numbers require only a finite amount of data to specify, whilst an arbitrary real will in general require an infinite amount of data. (I consider the repetition of digits a finite amount of data.) Since it shows up in all these metrical "specification" schemes, and the reals are defined metrically by such specifications (the reals are the unique archimedean linearly ordered complete field - and completeness refers to these things), I think this answer gets at the heart of why $\Bbb R$ is uncountable while $\Bbb Q$ isn't. To illustrate with binary: choosing a finite sequence of flips of a coin essentially encodes a natural number in binary, of which there are countably many, but choosing an infinite sequence of flips has a sample space of $2^{\aleph_0}$ which is $>\aleph_0$ by Cantor (and this theorem is not just auxillary to the discussion: to have a fair appreciation for set theory, it needs to be absorbed into one's intuition of what it means to be uncountable).
I think it is better to think of $\Bbb Q$ vs. $\Bbb R$ (the first contained in the second) instead of $\Bbb Q$ vs. $\Bbb R\setminus\Bbb Q$; for purposes of discussing cardinality there is no reason to partition $\Bbb R$ into $\Bbb Q$ and $\Bbb R\setminus\Bbb Q$.
Getting to the points you made: the fact that rationals are dense essentially means that we can use them in our specification of real numbers, but it fails to touch on the fact that it can take an infinite number of them to do that job, involving an infinite number of choices (even after getting rid of redundancies via null sequences), and the number of choices is what ultimately controls the cardinality. Yes, every rational $x$ is contained in uncountably many intervals with irrational endpoints - namely $(x-\epsilon,x+\epsilon)$ for every irrational $\epsilon>0$ - but this doesn't mean there are uncountably many rational numbers. If you try to pinpoint why you might suspect this implication of holding in the first place - a potential correspondence between intervals and points that forces $\Bbb Q$ to be uncountable - you will fail to find any meaningful correspondence. Why we begin with that kind of intuition in the first place I'm not sure, but intuition must be honed in light of facts.
A: $\mathbb R\setminus \mathbb Q$ has the property of being the complement of a proper subgroup of an uncountable group.
If $G$ is an uncountable group and $H$ is a subgroup with $H\neq G$, then $G\setminus H$ is uncountable.  If $H$ is uncountable, let $a$ be any element of $G\setminus H$, and then $aH$ is an uncountable subset of $G\setminus H$.  If $H$ is countable, then because $G$ is the union of cosets of $H$ and a countable union of countable sets is countable*, $H$ has uncountably many distinct (and therefore disjoint) cosets contained in $G\setminus H$.
*[assuming choice]
