Some irrational numbers are transcendental, which makes them in some sense "more irrational" than algebraic numbers. There are also numbers, such as the golden ratio $\varphi$, which are poorly approximable by rationals. But I wonder if there is another sense in which one number is more irrational than another.

Consider the following well known irrationals: $\sqrt{2}$, $\varphi$, $\log_2{3}$, $e$, $\pi$, $\zeta(3)$.

The proofs of irrationality of these numbers increase in difficulty from grade-school arguments, to calculus, to advanced methods. Other probable irrationals such as $\gamma$ most likely have very difficult proofs.

Can this notion be made precise? Is there a well defined way in which, for example, $\pi$ is more irrational than $e?$

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    $\begingroup$ If irrationality of $\pi$ is harder to prove, I would say that $\pi$ is less irrational, not more. $\endgroup$ Commented Jan 29, 2011 at 23:30
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    $\begingroup$ Doesn't being poorly approximable by rationals make a number less irrational than others? $e$ and $\pi$ have been proved irrational by showing that they are relatively well-approximable by rationals. And rationals themselves are not so well approximated by rationals. $\endgroup$ Commented Jan 30, 2011 at 0:24
  • $\begingroup$ If your criterion is ease of proof of irrationality, normal numbers are particularly easy to see irrationality of. $\endgroup$
    – ziggurism
    Commented May 8, 2014 at 5:03

7 Answers 7


Yes, there is such a thing as the irrationality measure of a real number (I'm not sure if it can be / has already been extended to complex numbers). It is based on the idea that all algebraic numbers (including the golden ratio) are hard to approximate well by rationals, relative to the size of the denominator of the rational used, while it is sometimes possible for a transcendental number to be approximated better. In particular, if a number $\alpha\in\mathbb{R}\setminus\mathbb{Q}$ has the property that there are infinitely many rational approximations $\frac pq\in\mathbb{Q}$ with $|\,\alpha-\frac pq| < q^{-t}$, then $t$ is a lower bound for the irrationality measure of $\alpha$; the larger $t$ is, i.e. the better your approximations are relative to the denominator, the "more irrational" you are, at least from a Diophantine approximation point of view.

From Wikipedia: The irrationality measure of a rational number is 1; the very deep theorem of Thue, Siegel, and Roth shows that any algebraic number that isn't rational has irrationality measure 2; and transcendental numbers will have an irrationality measure $\geq2$. However, as Douglas Zare has pointed out in the comments, the set of transcendental numbers of irrationality measure $>2$ has measure 0, so that in most cases it's unfortunately not useful as a comparison.

It appears that the irrationality measure of $\pi$ is not currently known, but that there are upper bounds; the most recent one I could find is this, which would appear to show that $\mu(\pi)\leq7.6063$. The Wikipedia article claims that $\mu(e)=2$, so whether or not $\pi$ is "more irrational" than $e$ looks like an open question.

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    $\begingroup$ It is very misleading to say that transcendental numbers can be approximated better than algebraic numbers. Almost all transcendental numbers can't be approximated that well. Liouville numbers have measure $0$, and so do the reals with irrationality measure greater than $2$. $\endgroup$ Commented Jan 29, 2011 at 16:38
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    $\begingroup$ Interesting! I was unaware of that fact; thank you for the correction. So perhaps irrationality measure is not quite so helpful as a comparison after all. $\endgroup$ Commented Jan 29, 2011 at 16:43
  • $\begingroup$ By this type of measure, $e$ is easier to approximate well by rationals than most irrationals are since the coefficients of the simple continued fraction of $e = [2; 1,2,1,1,4,1,1,6,...,1,1,2n,...]$ are unusually large, on average, which might lead one to say that $e$ is unusually irrational, more irrational than $\pi$ is expected to be. $\endgroup$ Commented Jan 29, 2011 at 16:45
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    $\begingroup$ Mathworld lists upper bounds for a few other constants, though not necessarily the most recent ones: mathworld.wolfram.com/IrrationalityMeasure.html $\endgroup$ Commented Jan 29, 2011 at 21:38
  • $\begingroup$ This "irrationality measure" plays a role in the MO question on the reciprocal Fibonacci constant (mathoverflow.net/questions/51426/…), which (in some sense) is known to be quite irrational, but not (yet) known to be transcendental. $\endgroup$ Commented Jan 30, 2011 at 1:52

The other answers and comments are fascinating, particularly about the irrationality measure, but allow me to give a little more information along the lines of Mark Sapir's answer by mentioning that there are several very large, intensely studied hierarchies of complexity for reals numbers. After the initial familiar notions come several others...

  • rational

  • algebraic

  • computable

The computable reals are those for which we can compute rational approximations to any desired accuracy, by Turing machine. (A concept used in computable analysis.) The computable subsets of $\mathbb{N}$ are those for which we can compute yes/no answers for membership in finite time. For example, all the numbers you mention in the question, such as $\pi$ and $e$, are computable.

  • computably enumerable

The c.e. subsets of $\mathbb{N}$ are those for which there is a computable enumeration procedure. Equivalently, you can compute the yes answers for membership in finite time. The concept of relative (oracle) computability leads to the hierarchy of Turing degrees, which measures the comparative computable complexity of a real.

  • arithmetic

A real $x$ is arithmetic if it's digits can be defined by a definition involving only quantification over the natural numbers and primitive operations. Equivalently, the arithmetic subsets of $\mathbb{N}$ arise from the computable subsets of $\mathbb{N}^k$ by projection and complement. The arithmetic hierarchy breaks naturally into levels, such as $\Sigma^0_n$ and $\Pi^0_n$, corresponding to the logical complexity of these definitions, and these levels are refined by the Turing degrees. For example, the set of Turing machine programs $p$ which compute total functions forms a complete $\Pi^0_2$ set. The relativized notion leads to the arithmetic degrees.

  • hyperarithmetic

A real is hyperarithmetic if it can be defined by two equivalent definitions, one involving just one universal quantifier over the reals and another having just one existential quantifier over the reals, and otherwise any level of arithmetic quantifiers. This is the same as $\Delta^1_1$. The hyperarithmetic hierarchy is stratified in a hierarchy of length $\omega_1^{CK}$, a lightface version of the Borel hierarchy, in which one uses uniformly computable countable unions and complements. The relativized notion leads to the hyperarithmetic degrees, a hyperarithmetic analogue of the Turing degrees.

  • projective

A real is projective if it can be defined by a description that quantifies only over the set of real numbers, plus natural number quantification and the primitive operations. The projective hierarchy is stratified by considering the logical complexity of these definitions, with levels $\Sigma^1_n$ and $\Pi^1_n$. For example, the lightface analytic sets are $\Sigma^1_1$ and co-analytic is $\Pi^1_1$, with hyperarithmetic being $\Delta^1_1=\Sigma^1_1\cap\Pi^1_1$.

  • constructible

A real is constructible if it exists in Gödel's constructible universe $L$. The concept of relative constructibility gives rise to the constructibility degrees, by which $x\sim y\leftrightarrow L[x]=L[y]$, forming a rich hierarchy.

  • ordinal-definable

A real (or set) is ordinal-definable if there is a definition of it in the language of set theory, using ordinal parameters. For example, the real whose $n^{th}$ binary digit is $1$ just in case $2^{\aleph_n}=\aleph_{n+1}$ is ordinal definable. The class HOD of all hereditarily ordinal definable sets satisfies ZFC, but can be strictly smaller than the universe of all sets.

  • generic

A real is generic over $L$ (or some other fixed universe $V$) if it exists in a forcing extension of $L$ (or $V$) by set forcing. Of course, it is relatively consistent with ZFC that every real is generic over $L$, since this is true in $L$ itself, but under some large cardinal axioms, there are reals, such as $0^\sharp$, that cannot be added by forcing over $L$.

The higher levels of these latter hierarchies are further developed and stratified by the enormous variety of models of set theory arising from large cardinals, various inner model constructions, forcing extensions and so on, so that the hierarchy loses its linear nature, becoming instead a dense jungle of various interacting concepts of set theory.


It is easy to invent criteria to compare the irrationality of different numbers, but I doubt that anyone understands irrationality well enough to give a serious criterion. We do not even know the continued fraction for the cube root of 2. Nor is the "difficulty" or the length of an irrationality proof a reasonable criterion. A theorem only has a difficult or long proof until one finds an easy or short proof.

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    $\begingroup$ There are only finite number of proofs of length $10^{10}$ so your last statement is wrong. Nevertheless computing the shortest length is impossible (same as the Kolmogorov complexity). Moreover, by Goedel's theorem adding axioms to a formal proof system can dramatically reduce the lengths of proofs. Hence the "length of proof" invariant depends on the proof system very much (unlike the Kolmogorov complexity). That is a more serious objection against this invariant $\endgroup$
    – user6976
    Commented Jan 31, 2011 at 11:32
  • $\begingroup$ On the other hand, Godel's theorem produces some ``artificial" statements with short proofs, so a proof of irrationality of some numbers may not be affected much by changing a proof system. $\endgroup$
    – user6976
    Commented Jan 31, 2011 at 13:59
  • $\begingroup$ @ Mel Nathanson: did you mean "... know the simple continued fraction for the cube root of 2." ? A generalized continued fraction of $2^{1/3}$ does exist: en.wikipedia.org/wiki/Generalized_continued_fraction . (See examples, example 1, cube root of two.) $\endgroup$
    – Max Muller
    Commented Feb 5, 2011 at 16:54

In principle, yes, you can measure irrationality of a number by the length of a shortest formal proof (in some formal proof system), something like the Kolmogorov complexity of a sequence. But it is difficult (if at all possible) to compute and the usefulness of it is not clear.

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    $\begingroup$ Could you clarify more precisely what you mean? e.g. proof of what? in which formal system? Kolmogorov complexity itself applies only to computable reals, a countable set. $\endgroup$ Commented Jan 30, 2011 at 2:40
  • $\begingroup$ Proof of irrationality. Formal system could be ZFC for example but not necessarily. The number must be effectively defined, for example by an algorithm which lists its decimal digits (or better decimal approximations), i.e. the number must be computable. So yes, we are talking about a countable set of numbers, or, more precisely, we say that the degree of irrationality of a non-computable number is infinity. $\endgroup$
    – user6976
    Commented Jan 30, 2011 at 3:02

For algebraic numbers one could take as a simple measure the degree of the number (lowest degree of non-zero rational polynomial having the number as a zero).

Of course, degree one are the rationals. And, degree two (quadratic irrationalities) have for example a nice characterization via being precisely those numbers with infinite yet periodic continued fraction expansion.

Though, others have already commented regarding classification by length/complexity proof and its problems, I wanted to add a (somewhat naive and subjective) remark:

I take the question as seeking (at least partially) some notion that matches (to a certain extent) an intuitive idea of what is more or less irrational. However, if this is so, then at least for my intuition (of course this being subjective), this would not work well at all, as there are quite different reasons why there is a simple/short proof of irrationality.

For example, comparing the perhaps two simplest arguments for irrationality: .) the irrationality of roots of integers (other then perfect powers) .) the irrationality of a number given by a description of its decimal expansion (provided it can be easily seen to be non-periodic)

The former yields only irrational numbers that I consider as very nice and not 'strange' at all.

Yet, using the latter one can get numbers that I consider as rather 'strange' (The obviously irrational number with decimal-expansion all 0 except at prime-place where the digit is 1, and at prime-power places where the digit is 7, has no intuitive meaning for me at all.)

So, a classification that puts those two types close together is not intuitive to me.


In response to a question about the comparative irrationality of real numbers, I wrote, "A theorem only has a difficult or long proof until one finds an easy or short proof." Mark Sapir replied, "There are only [a] finite number of proofs of length 10^10 so your last statement is wrong." I would argue, instead, that there are many contentious and, perhaps, wrong assumptions and philosophical misconceptions implicit in Sapir's sentence. For example, one could state and prove the theorem, "The sum of the first three odd numbers is nine." One could state and prove another theorem, "The sum of first five odd numbers is 25." One could state and prove another theorem, "The sum of first 100 odd numbers is 10,000." Continuing, one gets to statements of length greater than 10^10, and whose proofs would be even longer. Of course, someone might have the clever idea of proving inductively that, for all positive integers n, the sum of the first n odd numbers is n^2, and then one has a short proof of infinitely many theorems. It is exactly this kind of "exponential collapse," often called "progress in science," that I meant when I wrote a proof is long until one finds a short proof. I would be very interested to know if someone can really prove that there exist theorems that do not have short proofs. It is more likely that there are only finitely many theorems, and they all have short, simple, and elegant proofs. This, of course, is simply another description of Erdos' famous book.

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    $\begingroup$ At least in formalized mathematics, it can be proven that "there exist theorems that do not have short proofs." See Chapter 5 of Feasible Computations and Provable Complexity Properties (available via Google Books) by Juris Hartmanis: "Long Proofs of Trivial Theorems." $\endgroup$ Commented Feb 5, 2011 at 16:44
  • $\begingroup$ I think you mean: "In any particular formalization of mathematics, there are theorems that do not have short proofs." $\endgroup$ Commented Feb 6, 2011 at 1:05
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    $\begingroup$ Why is it "more likely that there are only finitely many theorems?" That seems profoundly unlikely to me. In particular, I'd argue that the fact that the set of theorems of PA (say) is incomputable, means that there is no real sense in which there are only finitely many theorems. $\endgroup$ Commented May 8, 2014 at 3:34

I agree with Mark Sapir's comments on shortest formal proofs. Also, Douglas Zare is correct to point out the difficulties of using irrationality measure. A refinement of irrationality measure was proposed by Mahler many years ago. It's rather technical, and instead of presenting the details here I suggest you type the words "Mahler" and "classification" into a search engine and peruse the many items your search will turn up.

Mahler's classification, like the irrationality measure, suffers from the difficulty of actually calculating which class any given number falls in.


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