Definition 1. Given $x \in \Bbb{R}$, the algebraic degree of $x$ is the degree of the minimal polynomial of $x$ over $\Bbb{Q}$. If $x$ is transcendental, we will define its algebraic degree to be $\infty$.

Definition 2. Given $x \in \Bbb{R}$, the rational approximation degree of $x$ is the following number: $$\mu(x) := \sup \left\{ \mu > 0 : \mbox{There are infinitely many solutions to } \left| x - \frac p q \right| \le \frac 1 {q^\mu} \mbox{, with } p, q \in \Bbb{Z} \right\}$$

Observe that all rational numbers have the smallest possible algebraic degree, $1$, while having the largest possible rational approximation degree, $\infty$.

Taking a look at irrational numbers, the first definition may be viewed as an algebraic complexity. The higher the algebraic degree is, the "harder" it is to describe the number using rational numbers.

Similarly, the second definition may be viewed as a (descending) Diophantine complexity. The lower the approximation degree is, the further away this number is from the rationals.

This intuition crashes when one learns the true nature of irrational numbers. The Liouville numbers - irrational numbers with infinite approximation degree - are not algebraically close to rationals at all; they are all transcendental! On the other end of the spectrum, the simplest irrational numbers - those of algebraic degree $2$ - are easily shown to have the worst possible rational approximation degree, $2$; in fact, by the Thue-Siegel-Roth theorem, the rational approximation degree is always $2$ for algebraic numbers of any degree.

Why is the algebraic complexity of an irrational number the opposite of its Diophantine complexity?

I know how to prove all of the above (well, except the Thue-Siegel-Roth theorem), but I don't understand the deep reason why it is true.

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    $\begingroup$ "Infinite solutions" is rather crude language, where "infinitely many solutions" is meant. "Infinite solutions" means "solutions, each one of which, by itself, is infinite". $\endgroup$ Commented Apr 11, 2013 at 14:08

1 Answer 1


I deny that the algebraic complexity of an irrational number is the opposite of its Diophantine complexity.

If the irrational is algebraic, then (as you note) its Diophantine complexity is $2$, regardless of its degree.

If the irrational is transcendental (thus, of infinite algebraic complexity), its rational approximation degree can be anything from 2 to infinity.

I don't see where one is the opposite of the other.

EDIT: Except for a set of measure zero, all reals have rational approximation degree $2$ (or we may call it Diophantine complexity $2$). Thus, Diophantine complexity does not serve to distinguish algebraics from transcendentals; rather, it distinguishes one very thin subset of the transcendentals from all the rest of the transcendentals (and the algebraics). So it seems to me that the questions one might want answered are, 1) why do all algebraic irrationals have Diophantine complexity $2$, and 2) why do a few transcendentals have Diophantine complexity exceeding $2$.

For the first question, I'm afraid I don't know any "deep reason", other than, the Thue-Siegel-Roth Theorem says so. I could mutter something about algebraic irrationals "repelling" rationals, but it wouldn't sound very convincing to anyone who isn't already convinced.

For the second question, it's easy to see that there are irrationals that are unusually well approximated by rationals --- just take some decimals with strings of zeros of rapidly increasing length. The hard part is seeing that they're all transcendental --- but then we're back to Thue-Siegel-Roth again.

As an aside, one generally asks that rational approximations be in lowest terms. With that convention, rationals have the lowest possible rational approximation degree, namely $1$. Rationals certainly do repel one another.

  • $\begingroup$ Because the Diophantine complexity is "descending": The smaller it is, the more complex it is. $\endgroup$ Commented Apr 11, 2013 at 16:12
  • $\begingroup$ But no matter what the Diophantine complexity of a transcendental, its algebraic complexity is infinite --- so how are the algebraic and Diophantine complexities opposite, when extreme variation in one is reflected in no variation in the other? $\endgroup$ Commented Apr 14, 2013 at 6:43
  • $\begingroup$ They are discrete opposites then :) Either "2, finite" (simple algebraic, complex Diophantine) or ">2, infinite" (simpler Diophantine, complex algebraic). $\endgroup$ Commented Apr 14, 2013 at 9:28
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    $\begingroup$ I think that this is what the OP was getting at: almost all irrationals are transcendental and have irrationality measure 2. A few irrationals (i.e. set of measure 0) - the algebraic numbers - are "unusually tractable" in the algebraic sense. A few - those with irrationality measure $> 2$ - are "unusually tractable" in the Diophantine sense. The surprise is that these two sets of "unusally tractable" irrationals are completely disjoint. $\endgroup$
    – tparker
    Commented Jan 13, 2018 at 20:38
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    $\begingroup$ Perhaps "strictly orthogonal" is a better analogy than "opposite", if you imagine plotting all the irrationals in the "algebraic tractability"-"Diophantine tractability" plane. You might or might not find the approximate orthogonality surprising, depending on your prior intuition, but I think the exact orthogonality is a bit counterintuitive. $\endgroup$
    – tparker
    Commented Jan 13, 2018 at 20:42

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