Bioperational Multisets in Various Semi-rings

TL;DR

This paper explores bioperational multisets, which are sets with equal sum and product, across various algebraic domains including semi-rings, extending beyond the traditional whole numbers.

Contribution

It systematically enumerates all sum-product pairs of bioperational multisets over whole numbers and six other algebraic domains, a novel extension in the study of such sets.

Findings

Enumerated all sum-product pairs over whole numbers.

Extended enumeration to six additional algebraic domains.

Provided foundational data for further algebraic exploration.

Abstract

One can find lists of whole numbers having equal sum and product. We call such a creature a bioperational multiset. No one seems to have seriously studied them in areas outside whole numbers such as the rationals, Gaussian integers, or semi-rings. We enumerate all possible sum-products for a bioperational multiset over whole numbers and six additional domains.

Equations

$$\log (1+2+3)=\log 1+\log 2+\log 3,$$

$$\log (ab)=\log a+\log b.$$

$$\sum _{i=1}^n{a}_i=\prod _{i=1}^n{a}_i.$$

$$\{2,7,2,2,3\}+\{1,2,7,7\}=\{1,2,2,2,2,3,7,7,7\}.$$

$$\{2,7,2,2,3\}-\{2,2,3\}=\{2,7\}.$$

$$3\{2,5,5\}=\{2,2,2,5,5,5,5,5,5\}.$$

$$2+2=2\cdot 2=4,$$

$$1+2+3=1\cdot 2\cdot 3=6,$$

$$\text{and}1+1+2+4=1\cdot 1\cdot 2\cdot 4=8.$$

$$3a\ge a+b+c=abc\ge 4a$$

$$4a\ge a+b+c=abc\ge 8a.$$

$$3a+1\ge a+b+c+1=abc\ge 4a$$

$$1+1+2+2+2=1\cdot 1\cdot 2\cdot 2\cdot 2=8,$$

$$1+1+1+3+3=1\cdot 1\cdot 1\cdot 3\cdot 3=9,$$

$$\text{and}1+1+1+2+5=1\cdot 1\cdot 1\cdot 2\cdot 5=10.$$

$$\prod _{i=1}^n{a}_i\ge \sum _{i=1}^n{a}_i.$$

$$a_{n+1}-1\ge \pi (S{)}^k\text{and}{a}_{n+1}<\pi (S{)}^{k+1},$$

$$\pi (S^{\prime })=a_{n+1}\pi (S)=(a_{n+1}-1)\pi (S)+\pi (S)\ge \pi (S{)}^k\pi (S)+\sigma (S)$$

$$=\pi (S{)}^{k+1}+\sigma (S)>a_{n+1}+\sigma (S)=\sigma (S^{\prime })$$

$$S^{\prime }=\{a_1,a_2,...,a_k,\stackrel{d\text{ times}}{\overbrace{1,...,1}}\}$$

$$\sigma (S^{\prime })=a_1+...+a_k+\stackrel{d\text{ times}}{\overbrace{1+...+1}}=\sigma (S)+(\pi (S)-\sigma (S))=\pi (S)=\pi (S^{\prime })=m.$$

$$1,1,1,3,1,2,2,2,2,3,2,4,2,...$$

$$n=2,5,13,25,37,41,61,85,113,181,361,421,433,...$$

$$S=\{1,2,3,-1,-1,1,1\}$$

$$S^{\prime }=S+13T$$

$$S^{\prime \prime }=S+13T-6\{-1,-1,1,1,1\}=\{3,-5,\stackrel{14\text{ times}}{\overbrace{-1,...,-1}},1\}$$

$$a_{n+1}=\frac{\sigma (S)}{\pi (S)-1}.$$

$$a_1^n=n{a}_1.$$

$$D(ab)=D(a)+D(b)-1\text{and}D(a+b)=\text{max}\{D(a),D(b)\}.$$

$$D(\sigma (S))=\text{max}\{D(a_i){\}}_{i=1}^n=D(a_1)$$

Full text

Bioperational Multisets in Various Semi-rings

O.M. Cain

onnomawcain@gmail.com

(July 2019)

Abstract. One can find lists of whole numbers having equal sum and product. We call such a creature a bioperational multiset. No one seems to have seriously studied them in areas outside whole numbers such as the rationals, Gaussian integers, or semi-rings. We enumerate all possible sum-products for a bioperational multiset over whole numbers and six additional domains.

1 Introduction

The numbers $1,2,$ and $3$ have the property that their sum is also their product. That is, $1+2+3=1\cdot 2\cdot 3=6$. As Matt Parker [1] has pointed out, this gives us as a strange sort of byproduct:

[TABLE]

due to the identity

[TABLE]

We coin the term bioperational here to refer to any such multiset $\{a_{i}\}_{i=1}^{n}$ having an equal sum and product. That is to say,

[TABLE]

A multiset, as can be guessed, is a set in which a number can occur multiple times (instead of just once or not at all).

There are some open conjectures about the number of bioperational multisets over $\mathbb{N}$ of length $n$ as $n$ gets bigger and bigger [2]. There is also a smattering of analysis available on math.stackexchange including uniqueness of solutions and connections with trigonometry [3][4][5][6]. One can also find a surprisingly complicated solution algorithm [7]. But it seems no one has formally categorized bioperational multisets over $\mathbb{N}$ (unless we count Matt’s “pseudoproof” and the passing comments of others).

In addition to $\mathbb{N}$, it is interesting to explore bioperational multisets in other environments. They are well-defined anywhere addition and multiplication are well-defined (hence in any semi-ring). We would suspect therefore to find these creatures lurking in $\mathbb{Z},\mathbb{C},\mathbb{F}_{p},\text{GL}_{n}(\mathbb{R})$, and many other places (even in non-Abelian rings!). In this paper, we limit ourselves to analyzing bioperational multisets over

• •

non-negative integers ($\mathbb{N}$) in Section 3,

• •

integers ($\mathbb{Z}$) in Section 4,

• •

general fields ($\mathbb{Q},\mathbb{F}_{p},$ etc) in Section 5,

• •

lunar integers ($\mathbb{L}$) in Section 6,

• •

Gaussian integers ($\mathbb{Z}[i]$) in Section 7,

• •

Eisenstein integers ($\mathbb{Z}[\omega]$) in Section 8,

• •

and integers with $\sqrt{2}$ appended ($\mathbb{Z}[\sqrt{2}]$) in Section 9.

2 Some Definitions

Firstly, we have to blow some dead leaves out of the way to see clearly. To do so requires the leafblower of vocabulary. Suppose we have a bioperational multiset $S=\{a_{i}\}_{i=1}^{n}$. We say

• •

$S$ is trivial if it contains only one element,

• •

$S$ vanishes if the sum-product is zero, and

• •

$S$ is minimal if no proper subset of its terms forms a bioperational set of the same sum-product.

Note that we are using ‘trivial’ differently than in [2].

All the examples we are considering are 1) integral domains and 2) Abelian. That means 1) if any two numbers have a product of zero then one or both of them must also be zero ($ab=0$ implies $a=0$ or $b=0$) And 2) the order we multiply stuff doesn’t matter (so $ab=ba$). This has some consequences on our analysis.

1. In an integral domain, any vanishing bioperational multisets must contain zero – which is rather boring. An analysis of vanishing bioperational multisets over non-integral domains might be interesting. In fact, the first example $(1,2,3)$ vanishes if we place it in $\mathbb{Z}/6\mathbb{Z}$. But the adventure of non-integral domains in general will be neglected here.

2. Since the order of multiplication doesn’t matter in our examples, we call our subjects bioperational multisets. However, the author greatly hopes that bioperational multisets will be explored in non-Abelian rings (in which case they would be bioperational sequences). The author would have loved to explore these objects in the quaternions ($\mathbb{H}$) themselves but was too ignorant for the attempt.

For two multisets $A$ and $B$ we let “$A+B$” denote their multiset sum which is best explained with an example:

[TABLE]

In technical terms, we are summing the multiplicities of all elements involved. Similarly “$A-B$” will denote subtracting multiplicites:

[TABLE]

We also use coefficients of multisets to denote scaling multiplicities. Or in other words

[TABLE]

Lastly, to keep our equations less messy, for a multiset $S$ we will write $\sigma(S)$ for the sum of its elements and $\pi(S)$ for the product. They look nicer than $\sum_{i=1}^{n}a_{i}$ and $\prod_{i=1}^{n}a_{i}.$

3 Non-negative integers ($\mathbb{N}$)

We begin with

Theorem 3.1**.**

There is exactly one non-vanishing bioperational multiset over $\mathbb{N}$ of length $n$ for $n=2,3,4$ with constructions

[TABLE]

[TABLE]

[TABLE]

This confirms Matt’s conjecture for $n=2$ and is stated without proof in [2].

Proof.

We first take $n=2$. Suppose we have $ab=a+b$. Rearrangement yields $ab-a-b+1=(a-1)(b-1)=1$. Since the only way to factor $1$ as two positive integers is $1=1\cdot 1$ it follows that $a-1=b-1=1$ or equivalently, that $a=b=2$.

A phenomenal proof of $n=3$ was given by Mark Bennet in [3]. We repeat it here. Suppose we have $a+b+c=abc$. At least one term must be $1$ since if otherwise $a\geq b\geq c\geq 2$ and we would have

[TABLE]

which is true only if $a\leq 0$. But that is a contractiction since we are assuming $a\geq 2$..

Okay, so let $c=1$. Then we have $a+b+1=ab$. Rearranging yields $ab-a-b+1=(a-1)(b-1)=2$. It follows that $\{a-1,b-1\}=\{1,2\}$ or equivalently, that $\{a,b\}=\{2,3\}$.

Similarly for $n=4$, suppose $a+b+c+d=abcd$. Similar to the case $n=3$, we know at least one term must be $1$ since $a\geq b\geq c\geq d\geq 2$ would imply

[TABLE]

So let $d=1$. But we can play the same trick again. If we have $a\geq b\geq c\geq 2$ then

[TABLE]

from which it follows that $a\leq 1$ which again contradicts our assumption $a\geq 2$. Let $c=1$ also.

We have $a+b+2=ab$. Rearranging yields $(a-1)(b-1)=3$ from which it follows $\{a,b\}=\{4,2\}$. ∎

One may be tempted to generalize this proof technique and keep tackling larger and larger $n$ (In fact, we wrote a program to do exactly this [7]. See [8] for another such solution algorithm). For example, $n=5$ yields

Theorem 3.2**.**

There are $3$ non-vanishing bioperational multisets over $\mathbb{N}$ of length $n=5$ with constructions

[TABLE]

[TABLE]

[TABLE]

Proof.

From computation. ∎

But there turns out to be an easier way to catalog all bioperational multisets. We first need a lower foothold (or “lemma” as they’re called).

Lemma 3.3**.**

The product of one or more real numbers, all greater than or equal to $2$, is greater than or equal to their sum. That is, if $a_{i}\geq 2$, $i=1,..,n$ then

[TABLE]

Proof.

Induction will be used on $n$. The base case, $n=1$, of a single number is clearly true since every number is equal to itself – and therefore is greater than or equal to itself ($a_{1}\geq a_{1}$).

Next suppose we have some multiset $S=\{a_{i}\}_{i=1}^{n}$ for which the theorem statement is true. So $\pi(S)\geq\sigma(S).$ We have to show the theorem true for a new multiset $S^{\prime}$ formed by appending a new element $a_{n+1}\geq 2$ since any multiset can be built up one element at a time.

Let $k$ be the largest integer such that $a_{n+1}>\pi(S)^{k}.$ From this we grab two crimps,

[TABLE]

with which the last bit of the proof can be shown easily.

[TABLE]

[TABLE]

Technically this is a bit overkill since we’ve shown $\pi(S^{\prime})>\sigma(S^{\prime})$ when all we needed was $\pi(S^{\prime})\geq\sigma(S^{\prime})$. Oh well. ∎

With that Lemma we can catalog all bioperational multisets over $\mathbb{N}$ by their sum-product.

Theorem 3.4**.**

For every composite integer $m\in\mathbb{N}$ there exists a non-trivial bioperational multiset over $\mathbb{N}$ with a sum-product of $m$.

Proof.

Suppose a composite integer $m=a_{1}a_{2}...a_{k}$ with $k>1$ and $a_{i}\geq 2$ for $i=1,...,k$. Let $S=\{a_{i}\}_{i=1}^{k}$. By Lemma 3.3, we know $\pi(S)\geq\sigma(S)$. So let the non-negative integer $d=\pi(S)-\sigma(S)$ be their difference. The multiset

[TABLE]

is bioperational with sum-product $m$ since

[TABLE]

∎

From the proof of the previous theorem we can also make a statement about the lengths of bioperational multisets.

Corollary 3.4.1**.**

For every factorization of a composite integer $m=a_{1}a_{2}...a_{k}$ there exists a non-vanishing bioperational multiset over $\mathbb{N}$ of length $m+k-\sum a_{i}.$

Proof.

Let $S$ and $S^{\prime}$ denote the same multisets as in the proof of Theorem 3.4. $S^{\prime}$ is bioperational and contains $k+d=k+(\pi(S)-\sigma(S))=m+k-\sum a_{i}$ elements. ∎

Starting at $n=2$ the number of non-vanishing bioperational multisets over $\mathbb{N}$ of length $n$ is

[TABLE]

(sequence A033178 in OEIS [8]). The positions of record in this list occur at

[TABLE]

(sequence A309230 in OEIS). The terms all appear to have fewer prime factors than their neighbors.

4 Integers ($\mathbb{Z}$)

An interesting thing happens once negatives are on the playing field. A multiset can be extended without changing either sum or product. Consider

[TABLE]

which is bioperational with sum-product $6$. This is the first example of a non-minimal bioperational multiset. In $\mathbb{N}$ every non-vanishing bioperational multiset is also minimal. Not so in $\mathbb{Z}$!

Accordingly, we now use bioperation as a verb as well. We say a multiset has been bioperated if it has been made bioperational by means of changing its sum with appendages. For example, we may bioperate $S=\{3,-5\}$ which has a sum $\sigma(S)=-2$ and product $\pi(S)=-15$. Since appending $T=\{-1,-1,1\}$ decrements $\sigma(S)$ and fixes $\pi(S)$, bioperation is accomplished by just repeatedly appending $T$. In particular,

[TABLE]

is bioperational. Note however $S^{\prime}$ is not minimal. We can trim it down to minimality by shaving off groups of $\{-1,-1,1,1\}$ which have no effect on the sum-product obtaining

[TABLE]

which is, in fact, minimal.

There are three important appendages in $\mathbb{Z}$ which fix the product.

[TABLE]

We now give the parallel of Theorem 3.4 for $\mathbb{Z}$.

Theorem 4.1**.**

For every composite integer $m\in\mathbb{Z}$ there exists a non-trivial minimal bioperational multiset over $\mathbb{Z}$ with a sum-product of $m$.

Proof.

Choose a factorization $m=a_{1}...a_{n}$ with $n\geq 2$ where each $a_{i}$ may be positive or negative and $|a_{i}|\geq 2$ for $i=1,...,n$. The multiset $S=\{a_{i}\}_{i=1}^{n}$ has the desired product. Bioperate $S$ producing $S^{\prime}$ such that $\sigma(S^{\prime})=\pi(S^{\prime})=\pi(S)$. This is done by appending $T_{\pm 1}$ as needed. To be

Finally, if $S^{\prime}$ is not minimal we may take a minimal bioperational multiset, $S^{\prime\prime}$, from it. $S^{\prime\prime}$ must include the non-units $a_{1},...,a_{n}$ (a “unit” by the way is a fancy name for a number with an inverse in its same ring; in this case $1$ and $-1$). Since $n\geq 2$ we are assured that $S^{\prime\prime}$ is non-trivial. ∎

5 Fields

Bioperational multisets turn out disappointingly abundant in fields.

Lemma 5.1**.**

Given any multiset $S=\{a_{i}\}_{i=1}^{n}$ whose elements are in a field $F$ and such that $\pi(S)\not=1$, one can bioperate $S$ into a unique multiset $S^{\prime}$ by appending a single element,

[TABLE]

This was stated for $F=\mathbb{Q}$ and $n=4$ by Robert Israel in [6].

Proof.

Any element $a_{n+1}\in F$ which might bioperate $S$ must satisfy $\sigma(S)+a_{n+1}=\pi(S)a_{n+1}.$ Rearranging yields $a_{n+1}=\frac{\sigma(S)}{\pi(S)-1}$ which exists if $\pi(S)\not=1$. ∎

The lemma turns out to be an exhaustive description.

Theorem 5.2**.**

In any field, all non-trivial bioperational multisets can be produced with Lemma 5.1.

Proof.

Suppose we have some bioperational multiset $S=\{a_{i}\}_{i=1}^{n}$ which cannot be produced by the lemma. Let $S_{i}^{\prime}$ be the multiset formed by removing $a_{i}$. It follows from the lemma $\pi(S_{i}^{\prime})=\frac{\pi(S)}{a_{i}}=1$ for all $i=1,...,n$. This in turn implies $\pi(S)=a_{i}$ for $i=1,...,n$ and we see that all $a_{i}$ are equal. We therefore have a solution to

[TABLE]

But dividing out an $a_{1}$ from both sides gives us $n=a_{1}^{n-1}=\pi(S_{1}^{\prime})=1$ showing $S$ is trivial. ∎

Before leaving this territory, we note that there are solutions to $a^{n-1}=n$ leading to bioperational sets of a single value. Take for instance $\{2,2,2,2,2\}$ which is bioperational in $\mathbb{F}_{11}.$

6 Lunar Integers ($\mathbb{L}$)

The Lunar Integers are the only strictly semi-ring to be considered. Their arithmetic is well analyzed in [10] (there called “Dismal” Arithmetic) and Neil Sloane gives a wonderful introduction in a Numberphile interview [11].

We neglect to explain the arithmetic here ourselves. We need only note some properties of the number of digits. If we let $D(a)$ denote the number of digits of a lunar integer $a\in\mathbb{L}$, then

[TABLE]

These give us

Lemma 6.1**.**

In any Lunar Bioperational Set, there is at most one element with $2$ or more digits.

Proof.

Suppose $S=\{a_{i}\}_{i=1}^{n}\subset\mathbb{L}$ is bioperational and that $D(a_{1})\geq D(a_{i})$ for $i=2,...,n$. From the aforementioned identities

[TABLE]

and

[TABLE]

Since $D(\pi(S))=D(\sigma(S))$, it follows that $\sum_{i=2}^{n}(D(a_{i})-1)=0$ and hence that $D(a_{i})=1$ for $i=2,...,n.$ ∎

Apparently bioperational multisets can’t breathe well on the moon:

Theorem 6.2**.**

Every minimal bioperational multiset of Lunar integers is trivial.

Proof.

We prove the contrapositive. Suppose $S=\{a_{i}\}_{i=1}^{n}\subset\mathbb{L}$ bioperational and non-trivial. From Lemma 6.1 we may assume $D(a_{i})=1$ for $i=2,...,n.$ For $a\in\mathbb{L}$, let $F(a)\in\mathbb{L}$ denote be the last digit of $a$. From the definitions of addition and multiplication over $\mathbb{L}$

[TABLE]

But this implies $F(a_{1})=a_{2}=...=a_{n}$. In which case the multiset $S^{\prime}=\{a_{1}\}$ is trivially bioperational with the same sum-product as $S$ and hence $S$ is not minimal. ∎

So there are bioperational multisets in $\mathbb{L}$, like $\{17,7\}$ and $\{2,2,2\}$, but they aren’t very interesting.

7 Gaussian Integers ($\mathbb{Z}[i]$)

Gaussian integers are numbers of the form $a+bi$ where $a$ and $b$ are integers and $i^{2}=-1$ (so like $2+3i$ or $-1-19i$ for example). In addition to the appendages $T_{-1},T_{0},$ and $T_{1}$ given in Section 4, two more appear in $\mathbb{Z}[i]$,

[TABLE]

which perturb the sum by $\pm 2i$ and fix the product. So sometimes bioperate the imaginary part of a multiset sum.

Take, for instance, $S=\{1+2i,2+3i\}$. We have

[TABLE]

The difference is $\pi(S)-\sigma(S)=-7+2i$. We bioperate by 1) appending $T_{-1}$ seven times, 2) appending $T_{2i}$ once, and 3) shaving off $T_{0}$ until minimality is reached. The result is

[TABLE]

which is bioperational with $\pi(S^{\prime})=\sigma(S^{\prime})=-4+7i.$

We need a couple lemmas before the result analogous to Theorem 3.4.

Lemma 7.1**.**

A Gaussian integer $a+bi$ is a multiple of $1+i$ if and only if $a$ and $b$ have the same parity (that is, are both odd or both even).

Proof.

Firstly, suppose $a+bi=(1+i)(c+di)$ is a multiple of $1+i$. Then $a=c-d$ and $b=c+d$. $a$ and $b$ therefore have the same parity since $b=a+2d$.

Conversely, suppose $a$ and $b$ have the same parity. If both even, then we may write

[TABLE]

and are done. If both odd, then we may write

[TABLE]

∎

Lemma 7.2**.**

For any Gaussian integers $\alpha_{1},...,\alpha_{n}\in\mathbb{Z}[i]$ such that $1+i$ does not divide any $\alpha_{i}$,

[TABLE]

Proof.

Let $\varphi(a+bi)=\overline{b\ \%\ 2}\in\mathbb{F}_{2}.$ It follows that $\varphi(\alpha+\beta)=\varphi(\alpha)+\varphi(\beta)$. But more interestingly, it turns out that when neither of $\alpha$ nor $\beta$ are multiples of $1+i$ we have also $\varphi(\alpha\beta)=\varphi(\alpha)+\varphi(\beta)$. From lemma 7.1 it follows that the residues of $\alpha$ and $\beta$ in $\mathbb{Z}[i]/(2)\cong\mathbb{F}_{2}[i]$ are in $\{1,i\}$. It is enought to check that $\varphi$ has the desired property on $\{1,i\}:$

[TABLE]

[TABLE]

[TABLE]

The lemma follows noting

[TABLE]

∎

The enzymes of $\mathbb{Z}[i]$ have been assembled. We are ready to digest the theorem.

Theorem 7.3**.**

For every $\mu\in\mathbb{Z}[i]$ which factors into non-units (i.e. $\mu=\alpha\beta$ with $\alpha,\beta\not\in\{1,-1,i,-i\}$) there exists a non-trivial minimal bioperationl multiset over $\mathbb{Z}[i]$ with a sum-product of $\mu$.

Proof.

Pick some factorization $\mu=a_{1}...a_{n}$ and let $S=\{a_{i}\}_{i=1}^{n}$ with at least two $a_{i}$ non-units. We break into two cases.

Case 1) if $\text{Im}(\pi(S))$ and $\text{Im}(\sigma(S))$ have the same parity, we may bioperate $S$ by appending $T_{\pm 1}$ and $T_{\pm 2i}$ as needed. The result is $S^{\prime}$; bioperational with sum-product $\mu$. If $S^{\prime}$ is not minimal, we may take a minimal subset $S^{\prime\prime}$. And we are assured $S^{\prime\prime}$ is non-trivial since otherwise $S^{\prime\prime}=\{\mu\}$ which implies $a_{i}=\mu$ for some. And $a_{i}=\mu$ implies all other $\alpha_{j}$ for $j\not=i$ are units since $\alpha_{1}...\alpha_{i-1}\alpha_{i+1}...\alpha_{n}=1$ (note we are using in this last step the fact that $\mathbb{Z}[i]$ is an integral domain).

Case 2) if $\text{Im}(\pi(S))$ and $\text{Im}(\sigma(S))$ have different parities, we may suppose from Lemma 7.2 that some $\alpha_{j}$ is divisible by $1+i$. We create a new multiset by removing $\alpha_{j}$ from $S$ and appending $\{i\alpha_{j},i,-1\}$. In notation,

[TABLE]

The product remains unchanged since

[TABLE]

More importantly, it is claimed that $\text{Im}(\sigma(S^{\prime}))$ and $\text{Im}(\sigma(S))$ have different parity. Their difference is

[TABLE]

Let $\alpha_{j}=a+bi$ for some integers $a$ and $b$. Substitution gives

[TABLE]

From Lemma 7.1 we may suppose that $a$ and $b$ have the same parity since $1+i|\alpha_{j}$. It follows that $a-b+1$ is odd, that $\text{Im}(\sigma(S^{\prime}))$ and $\text{Im}(\sigma(S))$ have different parity, and therefore that $\text{Im}(\sigma(S^{\prime}))$ and $\text{Im}(\pi(S))=\text{Im}(\pi(S^{\prime}))$ have the same parity. And so we return to the first case to bioperate $S^{\prime}$. ∎

8 Eisenstein Integers ($\mathbb{Z}[\omega]$)

Eisenstein integers are similar to the Gaussians in that they are all of the form $a+b\omega$ where $a$ and $b$ are integers and $\omega$ is a strictly complex number such that $\omega^{3}=1$. Right away this gives us our first appendage,

[TABLE]

One can show further show that $\omega^{2}=-1-\omega$ from which we get

[TABLE]

Thus we have $T_{\omega}=T_{3\omega}+T_{-2\omega}$ and $T_{-\omega}=T_{3\omega}+2T_{-2\omega}$ at our disposal. Suprisingly, we can therefore bioperate any multiset over $\mathbb{Z}[\omega]$. Our main theorem in this section will therefore run almost identically to its analog over $\mathbb{Z}$.

Theorem 8.1**.**

For $\mu\in\mathbb{Z}[\omega]$ which factors into non-units there exists a non-trivial minimal bioperational multiset over $\mathbb{Z}[\omega]$ with a sum-product of $\mu$.

Proof.

Choose a factorization $\mu=\alpha_{1}...\alpha_{n}$ with 2 non-units and let $S=\{\alpha_{i}\}_{i=1}^{n}$. Bioperate $S$ with $T_{\pm 1}$ and $T_{\pm\omega}$. The resulting bioperational multiset $S^{\prime}$ can be shaved down to minimality without becoming trivial since the non-units cannot be trimmed off. ∎

9 Integers $\sqrt{2}$ Appended ($\mathbb{Z}[\sqrt{2}]$)

Lastly, we consider the integer ring of a real quadratic number field. Numbers in $\mathbb{Z}[\sqrt{2}]$ are of the form $a+b\sqrt{2}$ where $a$ and $b$ are integers (again, very similar to $\mathbb{Z}[i]$ and $\mathbb{Z}[\omega]$). We use the fact that

[TABLE]

to create

[TABLE]

There’s good reason to think that this is the best we can do (I.e. that $T_{\pm\sqrt{2}}$ doesn’t exist over $\mathbb{Z}[\sqrt{2}]$). But the best proof the author could come up with for such a fact uses difficult results about quadratic number fields and complicated induction. Instead, we take a route similar to that taken through $\mathbb{Z}[i]$.

Lemma 9.1**.**

The number $a+b\sqrt{2}\in\mathbb{Z}[\sqrt{2}]$ is a multiple of $\sqrt{2}$ if and only if $a$ is even.

Proof.

Note $(c+d\sqrt{2})\sqrt{2}=2d+c\sqrt{2}.$ ∎

Lemma 9.2**.**

For numbers $\alpha_{1},...,\alpha_{n}\in\mathbb{Z}[\sqrt{2}]$ let

[TABLE]

If no $\alpha_{i}$ is divisible by $\sqrt{2}$ then $b\equiv d\mod 2$.

Proof.

We again create a strange homomorphism. Let $\varphi(a+b\sqrt{2})=\overline{b\ \%\ 2}\in\mathbb{F}_{2}$. It follows that $\varphi(\alpha+\beta)=\varphi(\alpha)+\varphi(\beta)$. We claim if $\sqrt{2}$ divides neither $\alpha$ nor $\beta$ then $\varphi(\alpha\beta)=\varphi(\alpha)+\varphi(\beta)$ as well. From the previous lemma, we see that the residues of such $\alpha$ and $\beta$ with coefficients in $\mathbb{F}_{2}$ are in $\{1,1+\sqrt{2}\}$. We check $\varphi$ by hand:

[TABLE]

[TABLE]

[TABLE]

We end noting

[TABLE]

∎

It’s probable that if the author knew more about ring isomorphisms, the results of this section and those of Section 7 could have been demonstrated simultaneously.

Theorem 9.3**.**

For every $\mu\in\mathbb{Z}[\sqrt{2}]$ which factors into non-units there exists a non-trivial minimal bioperationl multiset over $\mathbb{Z}[\sqrt{2}]$ with a sum-product of $\mu$.

Proof.

Pick a factorization $\mu=\alpha_{1}...\alpha_{n}$ and let $S=\{\alpha_{i}\}_{i=1}^{n},\ a+b\sqrt{2}=\sigma(S),$ and $c+d\sqrt{2}=\pi(S)$. If $b$ and $d$ have the same parity, $S$ can be bioperated into the desired result. If not, we may pick some $\alpha_{j}$ a multiple of $\sqrt{2}$. Letting

[TABLE]

Letting $\alpha_{j}=x+y\sqrt{2}$, the change $\sigma(S)$ is

[TABLE]

But from Lemma 9.1, we may suppose that $x$ is even and that therefore $\sigma(S^{\prime})$ and $\sigma(S)$ have $\sqrt{2}$ coefficients of different parity. It follows that $\sigma(S^{\prime})$ and $\pi(S^{\prime})=\pi(S)$ have $\sqrt{2}$ coefficients of the same parity and that $S^{\prime}$ can therefore be bioperated into the desired result. ∎

10 Generalization and Open Problems

Let’s start this section by bundling up our main theorems into a single statement

Theorem 10.1**.**

If $R$ is one of $\mathbb{N},\mathbb{Z},\mathbb{Z}[i],\mathbb{Z}[\omega],$ or $\mathbb{Z}[\sqrt{2}]$ then for every $\mu\in R$ which factors into non-units, there exists a non-trivial minimal bioperational multiset over $R$ with a sum-product of $\mu$.

Proof.

Theorems 3.4, 4.1, 7.3, 8.1, 9.3. ∎

Some open problems of interest:

• •

Does Theorem 10.1 hold over the quaternions?

Order of multiplication now matters. We at least have $T_{\pm 2i},T_{\pm 2j},$ and $T_{\pm 2k}$ at our disposal since

[TABLE]

has a product of $1$ for $v\in\{i,j,k\}$.

• •

Does Theorem 10.1 hold for all integer rings of real quadratic number fields?

There are families of such rings that admit easy attack. Take for instance $\mathbb{Z}[\sqrt{t^{2}\pm 1]}$. From

[TABLE]

we can construct appendages $T_{\pm 2\sqrt{t^{2}\pm 1}}$ which gives us pretty good flexibility for bioperation. And in general, for $d=t(b^{2}t\pm 2)$ we can construct appendages $T_{\pm 2b\sqrt{d}}$. The first values not covered by these parametrizations are

[TABLE]

Perhaps $\mathbb{Z}[\sqrt{13}]$, which has a relatively large fundamental unit, is our first example for which Theorem 10.1 fails. One would think it easy to construct a counter-example ring to the theorem. However the handful of examples the author toyed with proved dead ends.