There exists an infinite amount of rational and irrational numbers. But is there more irrational numbers than rational? And if so can a ratio of one to the other be calculated?
Number Theory – Ratio of Rational to Irrational Real Numbers
irrational-numbersnumber theoryrational numbers
Best Answer
I just want to point out that, in a hard-nosed sense, the ratio of cardinalities $\# \mathbb{Q} / \# \mathbb{R}$ is not well-defined.
What should $\# \mathbb{Q} / \# \mathbb{R} = \kappa$ mean? It should mean -- shouldn't it? -- that (i) $\# \mathbb{R} \cdot \kappa = \# \mathbb{Q}$ and (ii) $\kappa$ is the only cardinal number which makes (i) hold.
But in fact there are no cardinals $\kappa$ satisfying (i), and this only uses the fact that $0 < \# \mathbb{Q} < \# \mathbb{R}$: we have $\# \mathbb{R} \times 0 = 0 < \# \mathbb{Q}$, and if $\kappa \geq 1$, then $\# \mathbb{R} \times \kappa \geq \# \mathbb{R} > \# \mathbb{Q}$.
(Now you might think: ah, well, you're equally well saying that there's no cardinal number $\kappa$ such that $5 \cdot \kappa = 2$, which is true but only because we don't call $\frac{2}{5}$ a cardinal number. However, extending the cardinal numbers by adding additive or multiplicative inverses doesn't work so well because the operations of infinite cardinal addition and multiplication are both $\kappa_1 + \kappa_2 = \kappa_1 * \kappa_2 = \max(\kappa_1,\kappa_2)$, which loses all information on the smaller cardinal. Thus group completions of the relevant Semigroups will be trivial.)
It seems that the claim that (size of $\mathbb{Q}$)/(size of $\mathbb{R}$) $= 0$ is better interpreted measure-theoretically: with respect to Lebesgue measure on $\mathbb{R}$ we have $m(\mathbb{Q})/m(\mathbb{R}) = 0/\infty = 0$.