Imagine I draw a number line, and I took two points. What's the distribution of rational and irrational numbers between them? If I put it in a diagram where I color rational with a color and irrational with another color what pattern will I get?


For any continuous probability distribution on $\mathbb{R}$, the probability of picking a rational number is zero. In other words, if $X$ is a random variable which has a density, $$ P(\mathbb{Q}) = 0 \text{.} $$

The reason is that for continuous probability distributions, its probability measure $\mathbb{P}$ is absolutely continuous compared to the lebesgue measure $\lambda$, meaning that for all sets $X$ with $\lambda(X) = 0$ you also have $\mathbb{P}(X) = 0$. And you have $\lambda(\mathbb{Q}) = 0$, because $$ \lambda(\mathbb{Q}) = \lambda\left(\bigcup_{k\in\mathbb{N}} \{q_k\}\right) = \sum_{k=1}^\infty \lambda(\{q_k\}) = \sum_{k=1}^\infty 0 = 0 $$ for every enumeration $(q_k)_{k\in\mathbb{N}}$ of the rational numbers. The same works, for the same reason, for any countable subset of $\mathbb{R}$.

  • $\begingroup$ I'm just a layman, so yeah :l $\endgroup$ – user143270 Apr 15 '14 at 10:34
  • $\begingroup$ @user143270 I would love to give you a simple reason for this, but I don't think that's easily done. The main problem here is formalizing the idea of "picking an arbitrary point" - that's not quite as easy as it looks. The result follows from how that formalization works out in detail, so without going into these details, it's a bit hard to give a satisfactory answer... $\endgroup$ – fgp Apr 15 '14 at 10:51

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