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What is the maximum number of times a piece of paper (of non-zero thickness) can be folded in half (mathematically)?

I totally forgot to mention the "non-zero thickness" part; I've now included it.

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Mathematically, if you're modelling the paper as just an infinitely thin plane, you can fold it infinitely often. If you're referring to actual paper, then this is entirely dependent on the physical properties of paper, and it's not really a math question. – Adrian Petrescu Jan 16 '11 at 19:29
Are you asking about "physical" or "mathematical" limitations? – user1736 Jan 16 '11 at 19:29
"Mathematical" limitations. I'll make the edit. – iamsid Jan 16 '11 at 19:31
Once - when it's in half, folding it again will make it smaller than a half. – user139781 Apr 2 '14 at 13:40
up vote 6 down vote accepted

The zero thickness model is a bad model. For nonzero thickness the problem was studied by Britney Gallivan, who derived an upper bound on the number of folds based on the thickness and dimensions of the paper. The current world record for actual paper was set by her; it's $12$. There are some additional details and references at her Wikipedia article.

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Let me mention that there are also physical constraints unrelated to the specific properties of the material you're using, e.g. the piece of "paper" has to fit in the universe and the final size of the "paper" after folding has to be at most the diameter of an atom. – Qiaochu Yuan Jan 16 '11 at 20:08
My bad! I forgot to mention "non-zero thickness"! – iamsid Jan 16 '11 at 20:19
Interesting - I had assumed it was just an exponential thing - each fold is twice as hard as the previous one. – Andrew Grimm Jan 17 '11 at 1:07
@Andrew: this is approximately true for the first few folds, but as the diagram in the first link makes clear, the situation is more complicated because the amount of space the "joints" in the fold take up necessarily increases. – Qiaochu Yuan Jan 17 '11 at 2:28

Theoretically you can fold the paper into half infinite number of times since $1 - (\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \cdots) = 0$. However, practically the number of times you can fold the paper into half will depend on the dimensions of the paper (length, breadth and thickness), the stiffness of the paper, and more importantly your patience and mechanical strength to keep folding it.

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+1: Also see Zeno's paradox (meant for some of the readers of this answer, I am pretty sure Sivaram knows about it :-)) – Aryabhata Jan 16 '11 at 19:34

A very simple constzraint: Folding a paper of thickness $d$ $n$ times produces a block of thickness $2^nd$, hence requires the existence of points of distance $2^nd$ in the original sheet. Typical paper thickness is $0.1\,\text{mm}$, so already by this constraint, folding twelve times requires a paper of length at least $40.96\,\text{cm}$ long (about A3). Of course this would require a "tower" only $0.1\,\text{mm}$ thick, which is impossible - as there are about $2^{n-1}$ layers from the bends vertically! This already hints at the essential $2^{2n}$ factor in the strip length formula.

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