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The constructible numbers are those that can be achieved as lengths of line segments via compass and straightedge, starting with a segment of length $1$. The origami (constructible) numbers are those that can be achieved as lengths of line segments by folding paper, starting with a segment of length $1$.

We say a length has been achieved when it lies between two points occurring as intersections, of drawn lines and circles for the constructible numbers, and of intersecting folds for the origami numbers. I believe that we also allow points to be identified with origami by marking the image of an existing one when it's folded onto a new location (anyone more familiar with the axioms, please correct me if necessary).

It turns out $r\in\mathbb{R}^+$ is constructible if and only if it's contained in some chain of real quadratic field extensions of $\mathbb{Q}$ (Wentzel, 1837), and $r\in\mathbb{R}^+$ is origami-constructible it it's contained in some chain of real degree $2$ or $3$ extensions of $\mathbb{Q}$ (Haga, 1999). So the difference is that origami-constructible numbers are closed under taking cube roots.

I'm interested in seeing the simplest possible origami construction that realizes a cube root. By simplest, I mean minimal number of folds.

j0equ1nn
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  • You jump back and forth between "figure" and "number" (or "length"). Which do you want? They may not be the same thing. – Gerry Myerson May 10 '19 at 04:51
  • @GerryMyerson You're quite right. Upon looking more closely, I see there are various definitions of these. To discuss the algebra, it would make more sense to define them as points you can get to in $\mathbb{C}$, but I'd rather appeal to the classical notion, as lengths of line segments. I've made the question more precise now, I hope. – j0equ1nn May 13 '19 at 22:59
  • @BassamKarzeddin I won't flag your answer because I don't want you to feel persecuted, but what you're typing is kind of incoherent. I find it interesting to have suspicions about the motives for anti-constructivism, etc. Perhaps expressing it via comments on posts in run-on sentences isn't the best way. – j0equ1nn Mar 08 '21 at 04:58

1 Answers1

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Copying from http://mathandmultimedia.com/2011/08/02/paper-folding-cube-root/ (but the diagram seems to be missing).

"Get a rectangular piece of paper and fold it in the middle, horizontally and vertically, and let the creases represent the coordinate axes. Let $M$ denote $(0,1)$ and let $R$ denote $(-r,0)$. Make a single fold that places $M$ on $y = -1$ and $R$ on $x=r$. The $x$-intercept of the fold is $\sqrt[3]{r}$."

You can't do better than one fold, nor better than degree 3, so this would seem to be the winner.

EDIT: Since the mathandmultimedia page leaves something to be desired, I looked for other duscussions of folding cube roots. Maybe https://www.researchgate.net/figure/Belochs-origami-construction-of-the-cube-root-of-two_fig2_233592288 will be helpful, as well as the link there to https://www.researchgate.net/publication/233592288_Solving_Cubics_With_Creases_The_Work_of_Beloch_and_Lill

There's a simple step-by-step $\root3\of2$ at http://www.cutoutfoldup.com/409-double-a-cube.php

An older question here on math.stackexchange is Solving Cubic Equations (With Origami)

The Hull article in the Monthly is also available at http://sigmaa.maa.org/mcst/documents/ORIGAMI.pdf

One more source for $\root3\of2$: http://www.math.ubc.ca/~cass/courses/m308-05b/projects/ayuen/doublecube.html

Gerry Myerson
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  • "You can't do better than one fold" I see three. – Arthur May 10 '19 at 06:08
  • Yes, I forgot to count the two that make the axes. Three it is. Still, probably hard to beat. – Gerry Myerson May 10 '19 at 06:20
  • Thanks for that, but I find that description kind of vague. (Admittedly, my question was pretty vague too -- I've just edited it to try to clear that up.) For one, you have to make additional folds in order to identify the lines $y=-1$ and $x=r$. Then you have to get pretty violent with the paper to do that last step. Have you tried it? It's hard. But I believe it's the essence of the answer: to get a cube root, you must be able to send 2 distinct points to 2 distinct lines. – j0equ1nn May 13 '19 at 23:08
  • I'd also like to a nice way to see why that intersection point is $\sqrt[3]r$. I think I can do this for myself, but I'm interested in a figure I can do on a board where students will follow it. – j0equ1nn May 13 '19 at 23:15
  • Any thoughts on my edits, j0e? – Gerry Myerson May 15 '19 at 06:00
  • Looks like I should have just searched the web more before posting. – j0equ1nn May 18 '19 at 21:03
  • Maybe you'd like to write up and post an answer, based on what you've found at those links? – Gerry Myerson May 19 '19 at 00:14
  • That's a good idea. I'd like to do that once I have time. – j0equ1nn May 19 '19 at 19:18
  • Not had the time, j0equ1nn? – Gerry Myerson Mar 09 '20 at 10:34
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    @GerryMyerson Unfortunately, no. I did create a handwritten set of notes that worked for what I was doing (a couple of staff training talks). That was when I was still at MoMath. Now that I'm not, the museum is keeping my notes... What you posted definitely gives a curious person the ingredients to make their own notes though. – j0equ1nn Mar 09 '20 at 20:11