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Q&A

Is this a proper measure for irrationality?

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It is often said that the golden mean $\phi$ is the "most irrational" number. However what I've never seen is an actual measure of irrationality which would make such a statement meaningful in a quantitative way.

Now I thought about what such a measure would look like. Obviously it would need to have the following properties:

  • The irrationality of rational numbers is always $0$.

  • The irrationality of irrational numbers is always strictly positive.

  • The irrationality of $\phi$ is the maximal attainable value.

Clearly if the third point doesn't hold, it could still be a proper irrationality measure, it would just measure another aspect if irrationality that the one maximised by $\phi$.

Now I thought about it and came up with the following candidate:

$$i(x) = \inf_{m,n\in\mathbb Z, n\ne 0} \left|nx - m\right|$$

Obviously this fulfils the first objective: If $x=p/q$, then for $m=p, n=q$ the expression $nx-m$ gets $0$, and thus $i(x)=0$.

However I have no idea how to check the second objective, let alone the third. Nor have I any idea how one might actually calculate $i(x)$.

Now obviously each single absolute value would be larger than $0$, but then, an infimum of positive values may still be $0$, e.g. $\inf_{n\in\mathbb Z_{>0}} 1/n = 0$.

Indeed, I can't even tell if $i(x)$ will be nonzero for any irrational number.

Therefore my questions:

  1. Is $i(x)$ a valid irrationality measure (i.e. does it give a strictly positive result for all irrational numbers)?

  2. Does it give $\phi$ as maximally irrational, i.e. is $i(x)\le i(\phi)$ for all $x\in\mathbb R$?

  3. Is there a way to actually calculate the value for at least some irrational numbers (in particular, $\phi$?)

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Look at the continued fraction for that number (1 comment)
Note: I'm not really happy with the tag but that's the best I could find. If you find a better one, f... (1 comment)

1 answer

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  1. Is \(i(x)\) a valid irrationality measure (i.e. does it give a strictly positive result for all irrational numbers)?

No, by Hurwitz's theorem, which states that for every irrational number \(x\), there are infinitely many relatively prime \(m\) and \(n\) such that

\[ \left|x - \frac{m}{n}\right| < \frac{1}{\sqrt{5}n^2} \mathrm{.} \]

Multiply both sides of the inequality by \(n\), and you can see that \(i(x)\) is always 0; the difference can be made arbitrarily small by choosing a sufficiently large \(n\), and since there are an infinite number of \((m, n)\) pairs from which to choose, there can't be an upper bound on \(n\).

Here are some articles that might point you in a more productive direction: Irrationality measure, Markov constant.

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