# Can $1+1$ be $1$?

$1+1=2$ is often quoted by laypeople (in mathematics) as an epitome of the absolute truth. Those who know a bit of mathematics know that that is not the case. For example, there is a number system where $1+1=0$. It is denoted by $\mathbb{Z}_2$. $\mathbb{Z}_2$ has only two numbers $0$ and $1$ and it is the smallest field. Here, we don’t mean a field by a vector field in physics but a number system where two numbers can be added, a number can be subtracted from another, a number can be multiplied by another, or a number can be divided by another nonzero number. $\mathbb{Z}_2=\{0,1\}$ is also important in computer science as $0$ and $1$ can be identified with two possible values of a bit and to the boolean values true and false.

The mind of young Thomas Edison, the one who would become one of the greatest American inventors in the future, was full of curiosities and the traditional school education unfortunately could not satisfy the young kid’s endless desire to know. Often, what’s being taught in school didn’t make much sense to his creative mind. One day, his teacher was telling students that $1+1=2$ and this happened to be one of the things that did not make sense to him. Edison told his teacher that he did not understand why it is always true and gave a counter example: “If you put a tiger and a rabbit together in a closed room, soon there will be only tiger left, so $1+1$ can be $1$ also.” Such a creative kid in class can be a nightmare for a teacher but a pleasant nightmare, I might add.

We have seen that $1+1$ is not necessarily $2$, so can there be a number system where $1+1=1$ like young Thomas Edison contended? It turns out the answer is not that simple and it is even confusing as the short answer is yes and no. In order for something to be called a number system, it needs to be a field. Let me first state the formal definition of a field. A field $\mathbb{F}$ is a nonempty set of numbers with two binary operations $+$ and $\cdot$ such that

1. for $a,b\in\mathbb{F}$, $a+b=b+a$.
2. for $a,b,c\in\mathbb{F}$, $(a+b)+c=a+(b+c)$.
3. there exists $0\in\mathbb{F}$ such that $a+0=a$ for all $a\in\mathbb{F}$.
4. for each $a\in\mathbb{F}$ there exists $-a\in\mathbb{F}$ such that $a+(-a)=0$.
5. for $a,b\in\mathbb{F}$, $ab=ba$.
6. for $a,b,c\in\mathbb{F}$, $(ab)c=a(bc)$.
7. there exists $1\in\mathbb{F}$ such that $a1=a$ for all $a\in\mathbb{F}$.
8. for each $a\in\mathbb{F}\setminus\{0\}$, there exists $\frac{1}{a}\in\mathbb{F}$ such that $a\frac{1}{a}=1$.
9. for $a,b,c\in\mathbb{F}$, $a(b+c)=ab+ac$.

The properties 1-4 says $\mathbb{F}$ with $+$ is an abelian group. The properties 5-8 says $\mathbb{F}\setminus\{0\}$ with $\cdot$ is an abelian group. In universal algebra, the identities $0$ and $1$ and inverses $-$ and $/$ are also considered as operations, more specifically the identities are nullary operations and the inverses are unary operations. Since the inverse $/$ (division) is not defined for $0$, a field is not considered as an algebra or an algebraic structure, i.e. a set with operations in universal algebra. Examples of fields include $\mathbb{Q}$, the set of all rational numbers, $\mathbb{R}$, the set of all real numbers, and $\mathbb{C}$, the set of all complex numbers. There are also finite fields $\mathbb{Z}_p=\{0,1,2,\cdots,p-1\}$ where $p$ is a prime number. Fields are not necessarily sets of numbers. There are also fields called function fields, an example being the set of rational functions of one variable such as $\frac{x^3+2x+1}{x^2-1}$. The definition tells that a field must have at least two numbers $0$ and $1$. So, what happens if a field satisfies $1+1=1$? It implies that the field must have a single element and that $0=1$. This pretty much rules out the possibility of $1+1$ being $1$, right? But it is not. First, allowing $0=1$ does not violate the definition of a field, although in conventional mathematics, it is implicitly assumed that $0\not=1$. The definition says any nonzero element has a multiplicative inverse but it doesn’t say $0$ cannot have a multiplicative inverse (although again it is implicitly assumed that $0$ does not have a multiplicative inverse in conventional mathematics). For a moment, let us allow $0=1$. There is a dire consequence of allowing this: any vector space over this field with one element is $0$-dimensional as ${\mathbf v}=1{\mathbf v}=0{\mathbf v}={\mathbf 0}$! Perhaps, we shouldn’t be allowing $0=1$ which seems to be good-for-nothing and should end discussion on the possibility of the field with one element, right? Not so fast, this is not over yet. While we don’t even know whether it can exist, mathematicians are taking the field with one element pretty serious. They want it so badly that they are willing to bend the definition of a field. It is usually denoted by $\mathbb{F}_1$ or $\mathbb{F}_{\mathrm{un}}$. The subscript “un” is from the French word meaning “one”. The field with one element was first introduced by a Belgian-French mathematician Jacques Tits in 1957 in his paper titled “Sur les analogues algébriques des groupes semi-simples complexes.” One of the axioms of projective geometry states that a line must have at least three points. If one replaces this axiom, while keeping others, by “a line admits only two points”, we obtain a degenerate geometry. Tits conjectured that such degenerate geometry can have equal footing with projective geometry if one were to consider such geometry on a field of characteristic one, i.e. a field satisfying $1=0$. Since the late 1980s, $\mathbb{F}_{\mathrm{un}}$ has gained a lot of attention from major league mathematicians after they realized profound implications of $\mathbb{F}_{\mathrm{un}}$ in number theory (including the Riemann hypothesis), algebraic geometry, and noncommutative geometry. Despite many attempts and proposals, as far as I know, there is no widely accepted notion of $\mathbb{F}_{\mathrm{un}}$, yet. Here is my own take on it. I propose to replace the above conventional definition of a field by the following: A field $\mathbb{F}$ is a nonempty set of numbers with two binary operations $+$ and $\cdot$ such that

1. $\mathbb{F}$ with $+$ is a abelian group.
2. $\mathbb{F}$, possibly excluding an element, with $\cdot$ is an abelian group.
3. for $a,b,c\in\mathbb{F}$, $a(b+c)=ab+ac$.

In this definition, there is no specific mentioning of particular additive and multiplicative identities, only that they exist. This definition then may allow the existence of a field with one element $\{1\}$ without requiring that $1=0$. I have not examined this idea much further yet, so whether this proposal can be something useful remains to be seen.

Thus far, $\mathbb{F}_{\mathrm{un}}$ is a phantom object in mathematics and mathematicians may be, after all, chasing a phantom. But throughout the history of mathematics, we have had phantom objects which turned out to be extremely important entities not only in mathematics but also in physics. Examples include 0, negative integers, irrational numbers, the imaginary number $i$ (complex numbers), quaternions, non-Euclidean geometry, and the summability of divergent series such as $1+2+3+\cdots=-\frac{1}{12}$ (read here for details if you are curious). One day, we may eventually understand what $\mathbb{F}_{\mathrm{un}}$ is and if and when that happens, it will take up its rightful place in mathematics.

Returning to the original question, can $1+1$ be $1$? Je ne sais pas. (I don’t know.) But those major league mathematicians who are way smarter than I am desperately want that to be true. Perhaps time will tell.

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