Is abstract algebra (mostly?) restricted to $2$-ary operators?

This may be due to my own pure ignorance but it's my experience that all abstract algebra I've been introduced to, both in actual courses and in self-studies only exclusively deals with algebraic objects consisting of a set together with one or more binary operators defined on that set, perhaps over some other algebraic structure. Not counting categories here, just "low-level"-ish stuff.

I'm an undergraduate so my knowledge is of course very limited but I can't help to wonder why I never stumble upon algebraic structures with $n$-ary operators? Is there a good reason for this, something along the lines of $n$-ary operators behave "badly" when $n>2$ or is it just my ignorance because I'm a beginner in the field?

• If you count additive or multiplicative inverses as operators, then they are unary operators. And identities are nullary operators. – lhf Oct 18 '14 at 20:08
• Also, those binary operations frequently want to be associative, but then they induce unique $n$-ary operations in a natural way, for all $n$. – Berci Oct 18 '14 at 20:10
• It's not something one would "stumble upon", but Wikipedia has an article about median algebras. $\hspace{.51 in}$ – user57159 Oct 18 '14 at 20:14
• Well, in a mysterious way each finite algebraic structure can be partitioned where the local terms are expressible by one of a known type of algebraic structures (having unary and/or binary operations only). – Berci Oct 18 '14 at 20:21
• I think that there was a question about a similar topic before. – Asaf Karagila Oct 18 '14 at 20:31

Associativity allows you to "connect" binary operations to produce $n$-ary ones. In modern algebra one can work with $n$-ary operations called $\infty$-structures (like $A_\infty$, $L_\infty$, $E_\infty$ etc...). Associativity is no more required.

• This connection is similar to $2$ being prime and numbers being factored into primes. – Ali Caglayan Oct 18 '14 at 21:24
• Associativity does allow you to do this, but is that because associativity guarantees the possibility of recursive definitions? Can't you connect some n-ary subtraction to some binary subtraction via a recursive definition of n-ary subtraction? – Doug Spoonwood Dec 8 '16 at 21:13
• what do you mean with n-ary subtraction? – Avitus Dec 9 '16 at 20:14

In logic the operators are 2-ary because any function $\mathbb Z_2^n\rightarrow \mathbb Z_2$ can be expressed by 2-ary operators. In mathematics it's only because that abstract algebra is a generalization of the numbers and their common operators.

To study general n-aries in the same manner would require a lot of new experiences and heuristic superstructures.

However, Heap-theory do study ternary (3-ary) operations. And an other interesting example is planar ternary rings.

But obviously, humans like 2-aries more.

• "In logic the operators are 2-ary" - No, but they are "generated" by 2-ary operators, and you actually mean boolean operators. – Martin Brandenburg Oct 18 '14 at 20:38
• @Martin: yes, all functions can be expressed by boolean operators. – Lehs Oct 18 '14 at 20:41

• You might be interested in reading up on universal algebra http://en.wikipedia.org/wiki/Universal_algebra. (I know virtually nothing about this subject beyond what one learns about binary operations in an undergraduate abstract algebra course, so I can't attest to the accuracy of the information in that wiki article.)
• This is purely speculation, but my guess is that most of the algebraic operations you're likely to encounter can be understood in terms of binary operations, so most introductory abstract algebra courses and textbooks will be centred around them.
• Around the time I took my undergraduate abstract algebra course, I wondered more or less the same thing. I explored it a little on my own (without any references, since I didn't know about universal algebra then). One of the difficulties I had was how to generalize some ideas from the binary case to the $n$-ary case. For example, suppose you're trying to construct an analogue of a group, but with a ternary operation instead of binary. How do you define the identity element and inverse of an element?

The reason is simply that many interesting mathematics can be done (or modelled by) operations of arities $\leq 2$. The purpose of abstract algebra is not to study algebraic structures defined by random operations and random rules between them, but rather to give a general framework for algebraic structures which appear from other contexts (which are often more geometric).

On the other hand, there are algebraic structures which are studied and which have $\geq 3$-ary operations which are not reducible to $\leq 2$-ary operations.

See also the introduction of arXiv:1403.7099 for some background and references on ternary algebras.

• Do we really know that there are no VERY interesting n-aries, not discovered yet? – Lehs Oct 18 '14 at 21:12
• Have you read my answer? – Martin Brandenburg Oct 18 '14 at 21:20
• Yes of course, but I reacted on the sentence "The purpose of abstract algebra is not to study algebraic structures defined by random operations and random rules between them". – Lehs Oct 18 '14 at 21:24
• There is nothing interesting about $n$-ary structures per se. Even the group axioms are not interesting for themselves. But specific groups (which motivated the group axioms) are interesting. – Martin Brandenburg Oct 18 '14 at 21:26
• Yes of course, but we don't know what is interesting or not among the unexplored. – Lehs Oct 18 '14 at 21:37