[cokernel]

Slogan: "rational numbers exist but do not take up space."

This is in a measure-theoretic sense, so literally 100% (not just 99.5%). On the number line, the measure of an interval is the (absolute) difference between its endpoints (the "length" of the interval). For any measurable set, you can approximate its size from above by covering its points with intervals and shrinking them as much as possible.

But because there are only countably many rational numbers, you can cover all of the rationals with a set of intervals with a finite total sum of lengths.

To be a bit more specific, if your rationals are r_1, r_2, r_3, ..., then for any small number e > 0, you can form the set of intervals

(r_1 - e/2^2, r_1 + e/2^2), (r_2 - e/2^3, r_2 + e/2^3), (r_3 - e/2^4, r_3 + e/2^4), ...

which have a total sum of lengths e/2 + e/4 + e/8 + ... = e and contain all of the rational numbers.

So for any positive number e > 0, the measure of the rational numbers is less than e, meaning that the measure of the rational numbers is 0.

To turn this into actual probabilities requires a little bit more work, since the measure of the number line, and hence that of the irrational numbers, is infinite, not "100%". But you could look at the probability of selecting a rational number in the interval [0, 1] and use the same reasoning as above to get a probability of 100% for an irrational number and 0% for a rational number.

[enedil]

It's not less then even, it's exactly e.

[cokernel]

(Edited to try to answer the objection more clearly.)

The intervals overlap heavily. Each of the intervals R_i contains infinitely many rational numbers, and all but finitely many of those will have corresponding intervals which are completely contained within the interval R_i. As a result, the estimate obtained by the construction described is always strictly smaller than the chosen e.