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The irrationality measure $\mu(x)$ of a real number $x$ is defined to be the supremum of the set of real numbers $\mu$ such that the inequalities $$0 < \left| x - \frac{p}{q} \right| < \frac{1}{q^\mu} \qquad (1)$$ hold for an infinite number of integer pairs $(p, q)$ with $q > 0$. Wikipedia says that $\mu(x)$ measures "how 'closely' $x$ can be approximated by rationals," but I'm very unclear about exactly how it does it, because the "approximability" of a real number seems to depend non-monotonically on $\mu$, with real numbers with low and high values of $\mu(x)$ easily approximable by rationals, and real numbers with intermediate values of $\mu(x)$ difficult to approximate by rationals.

Specifically, we have $\mu(x) \geq 1$, with

  • the preimage of $\mu(x) = 1$ is exactly the rationals $\mathbb{Q}$.

  • the preimage of $\mu(x) = 2$ contains all of the irrational algebraic numbers $\bar{Q} \setminus Q$ (by Roth's theorem), as well as almost all of the transcendental numbers (in the Lebesgue-measure sense), including $e$ and $\varphi$.

  • the preimage of $\mu(x) \in (2, \infty)$ is a measure-zero subset of the transcendental numbers

  • the preimage of $\mu(x) = \infty$ is the set of Liouville numbers (this set is "large" in the sense of having the cardinality of the continuum and being dense in the reals, but "small" in the sense of having Lebesgue measure zero).

The name "irrationality measure" seems to imply that if $\mu(x) > \mu(y)$, then $x$ is "more irrational" than $y$, i.e. is harder to approximate by a sequence of rational numbers. But in fact the opposite is true; the Louiville numbers, which have $\mu(x) = \infty$, are unusually easy to approximate by a sequence of rationals, although of course not as easy as the rationals themselves, which have $\mu(x) = 1$. How do I understand this strange non-monotonicity?

As I understand it, the problem stems entirely from the first inequality in (1), which seems extremely arbitrary and conceptually unnatural. If we remove that inequality, then the second inequality has a very nice interpretation: the error in the Diophantine approximation sequence decreases with $q$ as a power law with exponent $\mu$, and higher $\mu$ means that the error decays faster. So under this proposed modification, $\mu$ would be interpreted as a rationality measure: almost all irrational numbers would have the minimal value $\mu(x) = 2$, but a few numbers would be unusally easy to approximate and have $\mu(x) > 2$. For a rational number we would trivially have $\mu = \infty$ (under this modified definition), because the errors would vanish identically after some finite $q$. Liouville numbers would be unusual in that their Diophantine approximations would vanish with $q$ faster than any power law, although never hitting zero, so they would also have $\mu(x) = \infty$ just like the rationals.

Is there some motivation for the first inequality that I'm missing? It seems to enormously decrease the conceptual clarity of $\mu$ by making it a non-monotonic measure of Diophantine approximability.

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  • $\begingroup$ I’m not sure what you mean. You say if $\mu(x)$ is big then $x$ is “very irrational” and your counterpoint is that Louiville numbers have $\mu(x)$ big? Do Louiville numbers not count as “very irrational?” And rational numbers which minimise $\mu(x)$ surely count as “not very irrational?” Perhaps one should say that to approximate something by a rational, the approximation must not be exact—apart from that one, all the approximations to a rational are bad! One could say this removes the case of an exact approximation as not being in the spirit of “infinitely many pairs $(p,q)$.” $\endgroup$ – Dan Robertson Jan 12 '18 at 23:55
  • $\begingroup$ @DanRobertson No, I'm saying the exact opposite! I'm saying that if $\mu(x)$ is big then $x$ is very rational (or better, very close to rational). Liouville numbers are not very irrational, because they can be approximated unusually well by rational sequences. $\endgroup$ – tparker Jan 13 '18 at 0:11
  • $\begingroup$ The intuition I’m used to is: rationals are very rational; algebraic numbers are slightly rational (there’s still countably many of them); computable numbers (which don’t fit nicely into most frameworks) are less rational; and those transcendental uncomputable numbers are the most irrational. I suppose the problem is that “sort-of rational” doesn’t mean “close on the number-line to rational numbers” or “well approximated by rationals” but rather “morally close to rationals” or “close to rationals in spirit”. $\endgroup$ – Dan Robertson Jan 13 '18 at 0:46
  • $\begingroup$ You do not have $\mu=1$ for rational numbers. That's part of the point. Rationals cannot really be well approximated by other rationals. $\endgroup$ – Andrés E. Caicedo Jan 13 '18 at 1:00
  • $\begingroup$ (You are getting $\mu=1$ by looking at pairs of integers rather than the rationals they represent. The "right" definition looks at the latter or, equivalently, at pairs of relatively prime integers.) $\endgroup$ – Andrés E. Caicedo Jan 13 '18 at 1:17
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I will attempt to record/interpret the comments as an answer for permanence.

It seems that the point is that the motivation behind excluding the zero-error case comes from taking a strict definition of the word "approximation" that excludes exact equality, i.e. $x = y \implies x \not\approx y$. Under this strict sense, the rationals are unusually difficult to approximate by other rationals, as the commentators said. So the "irrationality measure" really measures "strict-approximability by rationals". I'm still not clear either (a) why this notion of "strict approximation" is any more conceptually useful than the ordinary notion of "approximation", or (b) why this measure is called the "irrationality measure" when it seems to me that in almost all cases (i.e. for the irrationals), a higher "irrationality measure" seems to indicate that a number is closer to being rational. But my original question has been answered.

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  • $\begingroup$ Arnold Ross used to ask, rhetorically, "What is an approximation to 5?" and his answer was, "Any number other than 5". $\endgroup$ – Gerry Myerson Feb 13 at 23:41

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