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I'm not familiar with approximation, I have tried to get some of two irrational number which is equal to $\pi$ , I have looked to $\sqrt{2}+\sqrt{3}$ as simple example , but since i'm not familiar with approximation field , i'm not sure if $\sqrt{2}+\sqrt{3}$ could be equal to $\pi$ however $(\sqrt{2}+\sqrt{3})-pi \to 0$ , then my question here is :

Question: ls $\sqrt{2}+\sqrt{3}$ the only sum of two irrational which close to $\pi$?

zeraoulia rafik
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    Your question is highly unclear. $\pi$ is a trascental number over $\mathbb{Q}$, hence no identity like $\pi=\sqrt{2}+\sqrt{3}$ has any chance to hold. Did you mean Is there some particular reason for $\pi$ to be extremely close to $\sqrt{2}+\sqrt{3}$? – Jack D'Aurizio Jan 18 '18 at 18:16
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    take a look here and here. – Masacroso Jan 18 '18 at 18:20
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    I meant (sqrt(2)+sqrt(3)) very close to pi , why we can't take it equal's pi – zeraoulia rafik Jan 18 '18 at 18:22
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    @zeraouliarafik: as already remarked, $\sqrt{2}+\sqrt{3}$ is algebraic over $\mathbb{Q}$ while $\pi$ is not. – Jack D'Aurizio Jan 18 '18 at 18:24
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    $\sqrt 2+\sqrt 3$ is even worse an approximation to $\pi$ than $\frac{22}7$ – Hagen von Eitzen Jan 18 '18 at 18:25
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    As a teacher of mine said, if no bound for error is given, $4$ is a good approxiamtion for any number. – ajotatxe Jan 18 '18 at 18:27
  • I am sure one can find the sum of two other square roots which is a much better approximation of pi, then the one given here. – imranfat Jan 18 '18 at 18:28
  • An exercise that maybe is curious is try to find a better approximation than previous as $\sqrt{2}^r+\sqrt{3}^s$, for rational numbers $r$ and $s$. –  Jan 18 '18 at 18:29
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    Regarding the new question (after the edit): $(\pi-\sqrt{2})+\sqrt{2}$ is a sum of two irrationals which is pretty close to $\pi$... – Hans Lundmark Jan 18 '18 at 18:35
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    $\pi = (\frac{\pi}2 + \frac{1}{1000000}) + (\frac{\pi}2 - \frac{1}{1000000})$. For that matter, $\pi = \frac{\pi}2 + \frac{\pi}2$. – MPW Jan 18 '18 at 18:36
  • Remark about the new version: if $\mathbb{Q}$ is dense in $\mathbb{R}$, the set of (constructible) algebraic numbers is dense in $\mathbb{R}$ as well. – Jack D'Aurizio Jan 18 '18 at 18:38
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    It is interesting to consider differences of the form $$\sqrt{a}-\sqrt{b}.$$ A few examples: $$\sqrt{317}-\sqrt{215} \approx \pi +2.286\times 10^{-5};$$ $$\sqrt{5241}-\sqrt{4796} \approx \pi -2.353\times 10^{-7};$$ $$\sqrt{32854}-\sqrt{31725} \approx \pi +5.404\times 10^{-8};$$ $$\sqrt{912155}-\sqrt{906164} \approx \pi -3.841\times 10^{-11};$$ $$\sqrt{10490742}-\sqrt{10470401} \approx \pi -8.950\times 10^{-13};$$ – Oleg567 Jan 22 '18 at 12:48
  • This is what i meant in the view of approximation , but i don't know why the question was closed – zeraoulia rafik Jan 22 '18 at 13:57
  • So try rephrase the question (in terms of linear combinations, differences of radicals etc), and maybe it will be reopened. And avoid sentences like "$\pi$ is equal to the sum of two/three radicals" of integer numbers. $\pi$ is just approximated by them. – Oleg567 Jan 22 '18 at 14:12

1 Answers1

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There are many algebraic approximations of $\pi$, but they are just approximations, since $\pi$ is trascendental over $\mathbb{Q}$ as proved by Lindemann. For instance, by approximating a circumference with the union of a large number of arcs of parabola, then exploiting the nice formula (due to Archimedes) for the area of the parabolic segment, we have that

$$ \pi\approx 8\sqrt{2-\sqrt{3}}-1, $$ $$ \pi\approx 2\left(8\sqrt{2-\sqrt{2+\sqrt{3}}}-\sqrt{2-\sqrt{3}}\right) $$ and so on. If you are confident with Italian, you may have a look at it here (page 19), too.

Another famous approximation (besides the Archimedean $\pi\approx\frac{22}{7}$ and the Chinese $\pi\approx\frac{355}{113}$, giving two convergents of the continued fraction of $\pi$) is $\pi\approx\sqrt[3]{31}$, related to the fact that $$ \sum_{n\geq 0}\frac{(-1)^n}{(2n+1)^3}=\frac{\pi^3}{32}$$ as shown here.

Jack D'Aurizio
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