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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?

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There are in a definite sense more irrationals than rationals. In fact, one could say (informally) that the ratio of irrationals to rationals is infinity. Google "irrationals uncountable". –  David Mitra Dec 21 '11 at 17:32
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As @DavidMitra points out, the set of rationals is countably infinite and the set of irrationals is uncountably infinite, which is a bigger 'kind' of infinity. So, it is in this sense, there are more irrationals than rationals.But about the ratio of their cardinalities, some one should clarify. –  user21436 Dec 21 '11 at 17:38

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up vote 8 down vote accepted

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 $m(\mathbb{Q})/m(\mathbb{R}) = 0/\infty = 0$.

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You could see this answer. There are more irrationals and (depending on your definition) one could say the ratio is zero.

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By "the ratio", is the ratio of cardinality of rational reals to irrational reals meant or something else? –  user21436 Dec 21 '11 at 17:53
    
@KannappanSampath: Yes, the cardinality of the irrationals is larger, as you can inject the rationals into the irrationals but not the other way. Since the irrationals are "infinitely larger" cardinality, that gives the ratio zero. –  Ross Millikan Dec 21 '11 at 18:21

the set of all rationals can be "listed" (they are a subset of all pairs of integers) while the real numbers say between 0 and 1 cannot be listed. say you had such a list of all real numbers between 0 and 1 written in decimal notation $$ 0.123456789... $$ $$ 0.267489022... $$ $$ 0.657748930... $$ $$ ... $$ we can construct a number not on the list in the following way. the first digit after the decimal choose to be something other that the first digit of the first number in the list (in the above, something other than 1). the second digit will be something other that the second digit of the second number on the list (something other that 6 above) and in general the $n$th digit will be something other than the $n$th digit of the $n$th number on the list. we can be sure that the number is not on the list by comparing our new number against the numbers in the list. our new number will be different from the $n$th number on the list in the $n$ decimal digit, hence not the same number.

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protected by Asaf Karagila Dec 21 '13 at 16:59

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