# Ternary relations that are not binary functions

By far the most prominent elementary relations that are not functions are binary and the most prominent elementary ternary relations are in fact binary functions.

"Elementary" shall mean "part of the signature of a first-order theory".

The most prominent ternary relation that comes to my mind is the betweenness relation in geometry.

I am looking for examples from all over mathematics of elementary ternary relations that are not binary functions (resp. the corresponding theories).

Examples of elementary quaternary relations are also welcome!

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Coprimeness, as in there is no $p$ dividing all three of $a$, $b$, and $c$. –  Dan Piponi Jan 31 '10 at 0:17

Geometry seems like a natural source, e.g., colinearity of three points. Or how about, for three non-colinear points, clockwise(p,q,r) if the path going through p, q, and then r runs clockwise on the circle whose circumference they lie on? It seems unintuitive to me to try to break up this relation into binary functions and relations.

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Majority/median functions such as $$f(x,y,z) = (x \vee y) \wedge (y \vee z) \wedge (z \vee x)$$ are very natural. (This also makes sense for more than three arguments, but it is degenerate in the case of two arguments.)

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Isn't this a 4-ary relation in 3-ary function's disguise? –  Hans Stricker Jan 30 '10 at 20:45
Every 3-ary function is a 4-ary relation. Is that a problem? –  François G. Dorais Jan 30 '10 at 21:00
Of course it's not a problem per se, but it's not what I am looking for: relations that are NOT functions. –  Hans Stricker Jan 30 '10 at 22:41
I'm sorry, I thought the word "binary" in your question was important. –  François G. Dorais Jan 30 '10 at 22:52

You might be willing to count this as a quaternary relation, from the Wikipedia article:

In other words, ideas like "x is closer to a than y is to b" make sense in uniform spaces.

http://en.wikipedia.org/wiki/Uniform_space

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This may be too specific for what you're going for, but this question immediately brought to mind to triality on $D_4$. Here's triality in a nutshell:

View $\mathbb{R}^8 = \mathbb{O}$ as the inner product space collection of all Cayley numbers (where $\{1,i,j,k,l,li,lj,lk\}$ forms an orthonormal basis.). The collection of all orientation preserving isometries of $\mathbb{O}$ is simply $SO(8)$ (whose Lie algebra is $D_4$).

For $A$, $B$, and $C$ in $SO(8)$ and $x$ and $y$ in $\mathbb{O}$, consider the equation $$A(x)B(y) = C(xy)$$

Triality is the following claim: Given $A$, there exists a $B$ and $C$ making this equation hold for all Cayley numbers. The choice of the pair (B,C) is ALMOST unique - the only ambiguity is in replacing $(B,C)$ with $(-B,-C)$. Likewise, given $B$ or $C$, the other two matrices exist and are unique up to simultaneously changing the sign of both.

This gives rise to a natural relation $R\subseteq SO(8)\times SO(8)\times SO(8)$ with $(A,B,C)\in R$ iff $A(x)B(y) = C(xy)$ for all Cayley numbers $x$ and $y$. The ambiguity in sign shows that this is not a 2-ary function.

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