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. Jan 31 '14 at 3:12
  • $\begingroup$ Would this be precision or exactitude? $\endgroup$ – CAGT Jan 31 '14 at 3:15
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    $\begingroup$ We lose total disconnectedness. $\endgroup$ – Matemáticos Chibchas Feb 1 '14 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. Feb 1 '14 at 21:53
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    $\begingroup$ Rationals have nice property: every additive function is linear. Reals doesnt. $\endgroup$ – user135508 Apr 29 '14 at 17:27

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. $\endgroup$ – PyRulez Aug 19 '15 at 2:51

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