When we pass from the real numbers to the complex numbers, we lose total ordering. But what do we lose when we move from the rational numbers to the real numbers?

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    $\begingroup$ In this question I claimed that we lose "finite representability", but I'm not sure what that actually means or how to state it rigorously. $\endgroup$
    – Joe Z.
    Commented Jan 31, 2014 at 3:12
  • $\begingroup$ Would this be precision or exactitude? $\endgroup$
    – CAGT
    Commented Jan 31, 2014 at 3:15
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    $\begingroup$ We lose total disconnectedness. $\endgroup$ Commented Feb 1, 2014 at 7:42
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    $\begingroup$ That seems like a relatively weird property to lose, though, compared to the loss of total ordering in complex numbers and commutativity in quaternions. $\endgroup$
    – Joe Z.
    Commented Feb 1, 2014 at 21:53
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    $\begingroup$ Rationals have nice property: every additive function is linear. Reals doesnt. $\endgroup$
    – user135508
    Commented Apr 29, 2014 at 17:27

2 Answers 2


One of the biggest blows is the loss of decidable equality. Of course this is a constructive notion. By excluded middle two real numbers are either equal or they are not. The axiom does not compute though. Computers can't decide (in every case) whether two real numbers are the same.

Basically the real numbers become very cumbersome to compute with.

You may be aware that computer programs mostly work with bits and bytes. The natural numbers modulo 2^32 - 1 and so on. This is kind of true but really big integer and big natural number libraries are very common.

In a theorem prover like Coq defining the natural numbers inductively is very easy.

Inductive nat := O | S (n: nat).

This is not so efficient though.

While one would maybe like to define the integers more organically as the paths of the circle type there exist decently ok encodings of the integers and the rationals.

But as a completion the real numbers are fundamentally hard to encode and work with.

A sequence of rational approximations is a function of the natural numbers or a coinductive stream. Either gives up decidable equality. They are both pains to work with. You need extensionality to prove equality. But extensionality does not compute.

A pair of subsets of rational numbers is a pair of undecidable logical predicates. You do not have decidable equality here. Even proving equality is a headache because you need extensionality and proof irrelevance.


The rational numbers are all algebraic, whereas almost all of the real numbers are not.

What you mean by "finite representability" could be this. Or, you might mean that for any rational number $q$, there are (finite) integers $a$, $b$ where $a/b=q$, whereas if we define the reals as the limit points of Cauchy sequences of rational numbers, then for any real number $r$, we can only be sure that there are (infinite) integer sequences $(a_n)$, $(b_n)$ so that $a_n/b_n\to r$.

Or, you might mean that every rational number has a decimal expansion that either terminates or repeats, while the decimal expansion of irrationals continue infinitely without repeating.

Unrelated to the idea of "finite representability," there is also the property that there is no proper subset of the rational numbers that forms a field with the usual operations of addition and multiplication.

  • $\begingroup$ Finite representability could mean that there is a way of encoding rationals such that any given rational takes up only finite information. The reals lack such a property. $\endgroup$ Commented Aug 19, 2015 at 2:51

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