Abstract
A ball can be cut into finitely many pieces and, by rotating and rearranging them alone, made into two balls of the same size. This paper computes every computable part of that claim and shows in numbers what it is that permits it. NO NEW MATHEMATICAL THEOREM AND NO NEW LAW IS CLAIMED.
First, two rotations generate a free group. For two rotations through $\arccos(1/3)=1.230959417340775$, checking $354{,}292$ words in exact integer arithmetic found not one of them equal to the identity (Section 3).
Second, the reason is a single prime. Keeping a point as $(a,\,b\sqrt2,\,c)/3^n$, $b$ is never once divisible by $3$ — over $118{,}098$ words, $b \bmod 3$ took only the values $1$ and $2$ (Section 4).
Third, four pieces rebuild the whole thing twice. The $39365$ words of length at most $9$ split exactly into the identity and four sets of $9841$, and moving two of them once each restores the whole — with $\mathbf{0}$ missing (Section 5).
Fourth, this is the core. The share of a set displaced by one move falls to $0.0002$ on a line and $0.01$ in a plane, and in the free group it does not fall at all but rises towards $1$ (Section 6).
Fifth, the pieces therefore cannot have volumes. Were they to have any, $V=2V$ and so $V=0$, contradicting $4\pi/3=4.1887902048$ (Section 7).
Scope (what is not claimed)
NO NEW MATHEMATICAL THEOREM AND NO NEW LAW IS CLAIMED — Hausdorff, Banach–Tarski, Banach's impossibility in one and two dimensions, and von Neumann's amenability are all standard. No theorem is proved — what is done here is to compute the computable parts. The decomposition itself is not constructed — constructing it requires the axiom of choice, and no decomposition can be written down. It is not shown that no Følner sequence exists — what is shown is that the most natural candidate fails, the non-existence being a theorem that is cited. The merits of the axiom of choice are not argued. Matter and physical balls are not treated — the subject is a set of points. Relation to earlier papers: Paper 209 was equal volume is not enough to cut and reassemble — this paper is its reverse: unequal volume is no obstacle, if the pieces have no volume. The argument modulo $3$ that settled 209 works here unchanged. In Paper 224 the bill kept being presented however the shape was altered — here there is no bill to present. Paper 179 separated the senses of cannot be constructed — the decomposition here belongs to the sense of existing but not writable. What is added is checking $354{,}292$ words in exact arithmetic, tabulating $b \bmod 3$ out to length $11$, verifying the four-way split and the rebuilding by actual set operations and finding nothing missing, and setting the displaced share side by side for three groups.
1. Notation and Position
Let $\sigma$ be a rotation about the $z$ axis and $\tau$ one about the $x$ axis, both through the angle $\arccos(1/3)$. A reduced word is a string of $\sigma^{\pm1}$ and $\tau^{\pm1}$ in which no two neighbours cancel.
Position: Paper 209 was equal volume is not enough to cut and reassemble — this paper is its reverse, the case where unequal volume is no obstacle at all. In Paper 224 the bill kept being presented — here there is no bill.
2. Two Rotations
Check 2.1:
| Rotation | Determinant | Departure from orthogonality | Angle |
|---|---|---|---|
| About the $z$ axis | $1.000000000000000$ | $2.220\times10^{-16}$ | $1.230959417340775$ |
| About the $x$ axis | $1.000000000000000$ | $2.220\times10^{-16}$ | $1.230959417340775$ |
($\arccos(1/3)=1.230959417340775$.)
Observation 2.2 Both are perfectly ordinary rotations. The only thing special about them is that the angle is incommensurable with $\pi$, and the whole matter hangs on the single fact that the cosine is $1/3$.
3. Three Hundred Thousand Words, and None the Identity
Check 3.1: keep each matrix entry as $(p+q\sqrt2)/3^n$ with $p$ and $q$ whole numbers, and multiply without rounding anything.
| Length | Reduced words | Any equal to $3^n$ times the identity? |
|---|---|---|
| $1$ | $4$ | none |
| $4$ | $108$ | none |
| $7$ | $2916$ | none |
| $10$ | $78732$ | none |
| $11$ | $236196$ | none |
| Total | $\mathbf{354292}$ | not one |
Observation 3.2 The two rotations generate a free group — different reduced words are always different rotations. So far this is a finite check and not a proof, and the next section gives the reason.
4. The Reason Is a Single Prime
Keep the point $(1,0,0)$ as $(a,\,b\sqrt2,\,c)/3^n$. That point lies on the axis of $\tau$, so words beginning with $\tau^{\pm1}$ leave it where it is, and the words considered are therefore those beginning with $\sigma^{\pm1}$.
Check 4.1:
| Length | Words of that kind | Values taken by $b \bmod 3$ |
|---|---|---|
| $1$ | $2$ | $\{1,2\}$ |
| $5$ | $162$ | $\{1,2\}$ |
| $9$ | $13122$ | $\{1,2\}$ |
| $11$ | $\mathbf{118098}$ | $\mathbf{\{1,2\}}$ |
Observation 4.2 $b$ is never once divisible by $3$. So the point never returns and the word is never the identity — the whole business of turning one ball into two rests on a single prime. What settled the impossibility in Paper 209 was also an argument modulo $3$ — the same tool, reaching the opposite conclusion.
5. Four Pieces, Rebuilt into Two
Check 5.1: form every reduced word of length at most $9$ and sort them by their first letter.
| Piece | Words |
|---|---|
| The identity alone | $1$ |
| Beginning with $\sigma$ | $9841$ |
| Beginning with $\sigma^{-1}$ | $9841$ |
| Beginning with $\tau$ | $9841$ |
| Beginning with $\tau^{-1}$ | $9841$ |
| Total | $\mathbf{39365}$ (and there are exactly $39365$) |
Check 5.2: take the piece beginning with $\sigma$ together with the piece beginning with $\sigma^{-1}$ moved once by $\sigma$.
| Rebuilding | Words covered | Words of length $\le 8$ left uncovered |
|---|---|---|
| Using $\sigma$ | $19682$ | $\mathbf{0}$ |
| Using $\tau$ | $19682$ | $\mathbf{0}$ |
Observation 5.3 Two of the four pieces, moved once each, give back the whole. The other two do the same again — one becomes two. Nothing was created: the number of words was infinite to begin with, which is why it can appear to grow.
6. The Core: It Falls on a Line and in a Plane, and Not in the Free Group
That a group can average means that finite sets can be made as large as one likes which barely move off themselves when shifted.
Check 6.1: shift a set by one step and measure the share displaced.
| Group | Size of the set | Share displaced | Tending to |
|---|---|---|---|
| The line | $10$ | $0.2000000000$ | |
| The line | $1000$ | $0.0020000000$ | |
| The line | $10000$ | $\mathbf{0.0002000000}$ | $\mathbf{0}$ |
| The plane | $25$ | $0.4000000000$ | |
| The plane | $3600$ | $0.0333333333$ | |
| The plane | $40000$ | $\mathbf{0.0100000000}$ | $\mathbf{0}$ |
| The free group | $17$ | $1.0588235294$ | |
| The free group | $13121$ | $1.0000762137$ | |
| The free group | $118097$ | $\mathbf{1.0000084676}$ | $\mathbf{1}$ |
Observation 6.2 On the line and in the plane, the larger the set the less it moves off itself. In the free group it moves off itself almost entirely, however large it is made — the share tends to $1$. This is what it really is for the thing to be impossible in one and two dimensions and possible in three — what is at work is not the number of dimensions but whether the group of rotations contains a free group.
Observation 6.3 What is shown here is only that the most natural candidate fails. That every finite set fails is a theorem that is cited and not a computation of this paper.
7. So the Pieces Cannot Have Volumes
Observation 7.1 suppose the five pieces had volumes. Rotations do not change a volume, so the total after rebuilding equals the total before.
| Quantity | Status |
|---|---|
| The volume of the ball | $4\pi/3=4.1887902048$ |
| The volume of one piece | undefined |
| The total of the pieces | not a sum of numbers |
Observation 7.2 Had they volumes, $V=2V$ and so $V=0$, contradicting $4\pi/3$. There is only one way out — that the pieces have no volume. The volume did not double: there is no such quantity in between.
8. The Separator and the Fence
Observation 8.1 (the separator) What separates the senses of volume is which sets it is defined on.
| Phrase | On which sets |
|---|---|
| The volume of the ball | the measurable ones |
| The volume of a piece | not defined |
| A finitely additive volume | on a line and a plane, on every set |
| The same thing | in space, on no such family |
Observation 8.2 (the fence) The fence around volume is conserved is that every set involved be measurable. Outside it the sentence is not even false but empty — the same shape as this sequence is random in Paper 223 and this surface is flat in Paper 224.
Observation 8.3 Paper 209 and this one reach opposite conclusions with the same tool.
| Paper 209 | This paper | |
|---|---|---|
| The question | Does equal volume suffice? | Does unequal volume forbid? |
| The answer | it does not | it does not |
| The tool | an argument modulo $3$ | an argument modulo $3$ |
| The pieces | measurable (polyhedra) | not measurable |
Observation 8.4 What separates them is whether the pieces can be measured. Allow only measurable pieces and volume is not enough (209) — allow unmeasurable ones and volume is not needed (here).
9. The Census and the Close
What this paper measured:
- Both are ordinary rotations (Check 2.1): the angle is $1.230959417340775$.
- $354292$ words, none the identity (Check 3.1): in exact arithmetic.
- $b$ is never divisible by $3$ (Check 4.1): only $\{1,2\}$ over $118098$ words.
- The split is exactly $9841\times4+1$ (Check 5.1).
- Two pieces give back the whole (Check 5.2): nothing missing.
- The displaced share (Check 6.1): $0.0002$, $0.01$, and $1.0000084676$.
- The pieces cannot have volumes (Observation 7.1): else $V=2V$.
References
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Reference
Yamagishi, Y. (2026). One Ball Becomes Two of the Same Size [Paper 225]. YOI: https://www.yamagishiyuuki.com/yoi/225
DOI: 10.5281/zenodo.22120991
ORCID: 0009-0006-8587-3450
PDF (English) · PDF (Japanese) · Date proof of the English PDF (OpenTimestamps .ots) · SHA256SUMS
The record page for this paper (in Japanese, with the PDFs in each language and the date proof): YOI 225. This paper is a preprint and has not been peer reviewed.