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It was proven algebraically in 19th century that it is impossible to construct a square with an area equal to the area of a given circle using only a compass and straight edge. However, I once came across a remark that this pertained to constructions involving a finite number of steps.

  1. I was wondering if this is an accurate statement, and 2) if a procedure exists using a compass and straight edge which could square a circle with an infinite number of steps.
Euclid Looked On Beauty Bare
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  • It is all about computing $\pi$ graphically. You can do so by evaluating Leibnitz' formula, for instance, which only involves rationals. –  Sep 10 '21 at 15:19
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    I only remember a method involving $10^{50}$ pieces or so, and I am not sure whether this is actually possible or just related to the Banach-Tarski-paradox. I did not understand why this should be possible considering that $\pi$ is not just irrational, but transcendental. With infinite many steps, you can of course circle the square because there is a sequence of rational numbers converging to $\sqrt{\pi}$ and every rational number is constructable. – Peter Sep 10 '21 at 15:19

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Rational numbers are constructible with compass and straightedge, and it's very easy to construct sequences of rational numbers that converge to $\pi$, for example: $3, 3.1, 3.14, 3.141, 3.1415, \ldots$. That solves the squaring a circle problem. The same argument proves that practically everything can be constructed with compass and straightedge if an infinite number of steps are allowed.

jjagmath
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    If you want a more regular pattern, so you don't have to look up every single rational number you need to construct, but can rather just look at the last one you did, and figure out from there what the next one is, you could go for something like $\frac41-\frac43+\frac45-\frac47+\cdots$. I think that's the simplest one. – Arthur Sep 10 '21 at 15:35
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    @Arthur Leibniz's formula was already mentioned in the comments. The point of my answer is to notice that any real number whose decimal expansion can be computed is also constructible with compass and straightedge with an infinite number of steps. – jjagmath Sep 10 '21 at 15:54
  • So it could be said the transcendental number pi is "infinitely constructible" whereas the algebraic numbers are "finitely constructible". – Euclid Looked On Beauty Bare Sep 10 '21 at 15:59
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    @EuclidLookedOnBeautyBare Not all algebraic numbers are "finitely constructible", for example $\sqrt[3] 2$. – jjagmath Sep 10 '21 at 16:00
  • sorry I just deleted my own comment. my brain is soup today. – Euclid Looked On Beauty Bare Sep 10 '21 at 16:20
  • If the circle is a unit circle, then the area of the circle is pi and the corresponding square would have sides equal to the squareroot of pi. Could you directly construct the square root of pi as part of an infinite procedure rather than doing it as the last step at the end of an infinite procedure to construct pi? – Euclid Looked On Beauty Bare Sep 10 '21 at 16:36
  • @EuclidLookedOnBeautyBare Yes, you can construct directly $\sqrt \pi$ by constructing step by step $1, 1.7, 1.77, 1.772, 1.7724, 1.77245, \ldots$ – jjagmath Sep 10 '21 at 18:17
  • On the other hand, it is possible to square some circles on the unit sphere, and some in the hyperbolic plane. One must construct both figures at the same time... http://zakuski.utsa.edu/~jagy/papers/Intelligencer_1995.pdf – Will Jagy Sep 10 '21 at 23:56