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The rational approximations of $\sqrt 2$ given by its continuous fraction are: 1.5, 1.333, 1.4, 1.417, 1.412, etc. which is not strictly increasing. Similarly, this sequence for $\phi=(1+\sqrt5)\big/2$ is 2.0, 1.5, 1.667, 1.6, 1.625, etc.

Is there a real number so that the sequence of the best rational approximations (in the sense of continuous fractions) is strictly increasing?

  • The continued-fraction representation of $\sqrt 2$ is $[1;2,2,2,\ldots]$. The first few terms are $\frac32=1.5,\frac75=1.4,\frac{17}{12}\approx 1.41667,\frac{41}{29}\approx 1.41379,\frac{99}{70}\approx 1.41429,\frac{239}{169}\approx 1.414201$. Note that these values alternately over- and under-estimate $\sqrt 2$, as expected (unlike the sequence that you give). – TonyK Mar 13 '22 at 12:13
  • BTW they are called continued fractions not "continuous" fractions. – David Mar 13 '22 at 22:27
  • @David sorry there is a single word for that in my native language –  Mar 14 '22 at 22:43
  • @laravel that's interesting, what language and what is the word? – David Mar 15 '22 at 23:41

2 Answers2

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Look at the paragraph Some Useful Theorems in the Wikipedia article on continued fractions. There you will find, for instance, the statement $$\frac{h_n}{k_n}-\frac{h_{n-1}}{k_{n-1}}={(-1)^{n+1}}{k_nk_{n-1}}$$ where $\frac{h_n}{k_n}$ is the $n$th convergent of the continued fraction.

What this means is that the successive continued-fraction approximations to a positive irrational number will successively over- and under-estimate the limit. So not even three successive terms can be increasing.

TonyK
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Recall that the $n$th approximation for $\alpha$ is $\lfloor \alpha\rfloor$ plus the $(n-1)$st approximation for $\frac1{\alpha-\lfloor \alpha\rfloor}$. You can show by induction that it does not depend on the irrational $\alpha$, but only on the parity of $n$ whether the $n$th approximation is two small or too big.

Hagen von Eitzen
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  • So, if I get it correctly, every other approximation is underestimating the real? –  Mar 12 '22 at 18:32