# Countability of irrationals

Since the reals are uncountable, rationals countable and rationals + irrationals = reals, it follows that the irrational numbers are uncountable. I think I've found a "disproof" of this fact, and I can't see the error.

Since $Q$ is countable, list $q_1, q_2, \dots$ such that $q_i < q_{i+1}$. I want to show that between $q_i$ and $q_{i+1}$ there is exactly one irrational number; this will give us a bijection and prove the irrationals countable.

Since the irrationals are dense, it follows that there is at least one irrational number in $\left(q_i,q_{i+1}\right)$. Suppose there was more than one in this range, e.g. $x$ and $y$. Since $(x,y)$ is an open subset of $R$ and the rationals are dense, there must be some rational $q_c$ in this interval. But that means $q_i<q_c<q_{i+1}$, which contradicts our ordering. So there must be exactly one irrational in this range. QED.

Where is the problem? The only thing I can think of is that the rationals can be put into a sequence, but cannot be put into an increasing sequence, which seems odd.

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If the rationals could be put into an increasing sequence, with the usual ordering, would they be dense? – David Mitra Aug 20 '12 at 15:50
Uhmm, right at the beginning is a problem. How do introduce a total order on the rationals (which respects the usual order)? – user20266 Aug 20 '12 at 15:51
Note that if such ordering existed, there wouldn't be even one irrational number between $q_i$ and $q_{i+1}$. One can just repeat your argument. In fact, assume there is an irrational number $x$ such that $q_i\lt x\lt q_{i+1}$. Since $(q_i,x)$ is open and rationals are dense in $\mathbb{R}$, there must be some rational $q_c$ in this interval, that is $q_i\lt q_c\lt q_{i+1}$. Again, it contradicts our ordering. – Kuba Helsztyński Aug 20 '12 at 18:03
Here's a similarly odd property of the natural numbers: You can list them in increasing order, but you can't list them in decreasing order. – MJD Aug 20 '12 at 18:50
Something else to consider that may be enlightening: For each positive integer, there is a unique Roman numeral. For example, $100\mapsto C, 49\mapsto XLIX, 9876\mapsto MMMMMMMMMDCCCLXXVI$. Consider the set of positive integers ordered by alphabetical order of their roman numerals. This ordering begins with 100, 200, 300, 301… ($C, CC, CCC, CCCI\ldots$) and ends with 38 ($XXXVIII$). But it is infinite in the middle. – MJD Aug 20 '12 at 18:53

@Xodarap you might want to look into ordinal numbers then. $\mathbb{Q}$ is countable, just like $\mathbb{N}$, but they have different order types. – Robert Mastragostino Aug 20 '12 at 17:00
The assumption that you can list $\mathbb Q$ such that $q_1 < q_2 < \ldots$ is not true. You can prove that it is impossible to have this ordering with the argument you made.