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I understand that the Cartesian product operation is not associative if it is understood as a binary operation.

I.e. $(A \times B) \times C \ne A \times (B \times C)$.

However when mathematicians write, e.g. $A\times B\times C$ they actually mean a Cartesian product of arity 3 which is a different operation than the successive application of two binary Cartesian product operators. Since its a ternary or, in general, n-ary operator, associativity does not come into play. Using postfix notation would make that clear. However, since infix is traditionally used in Math, my questions are:

  1. What kind of notation can be used to signify that the "$\times$" in $A\times B\times C$ actually denotes a ternary operator and not a binary one? Essentially, I am looking for a notation to denote that we are using a single ternary operator "..$\times$..$\times$.." as opposed to two applications of the binary "..$\times$.." operator. I am sure connecting the two "$\times$"s with a curved underline would make that plain but I guess it's hard to do in typography.
  2. What are some other similar examples of n-ary operators that yield different results than the successive application of their binary counterparts?
Jukka Kohonen
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Marcus Junius Brutus
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    Although $(A\times B)\times C\neq A\times (B\times C)$, there is an incredibly obvious bijection between them that preserves every interesting property about them. As such we usually say $(A\times B)\times C\equiv A\times (B\times C)$ because for all intents and purposes they are the same to us. – JMoravitz Sep 27 '17 at 01:04
  • Anybody want to give a hint to the 'incredibly obvious bijection' between $(A\times B)\times C $ and $ A\times (B\times C)$? – john Mar 04 '20 at 15:39
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    For the second question, the logical NAND / NOR operators can be examples, if defined as NAND: "not (... and ... and ...)", NOR: "neither, ..., nor, ..., nor, ...". These are interesting because neither is associative, but (unlike the Cartesian product) both are commutative. Event the n-ary version does not depend on the order of the operands. Yet, the n-ary versions give different result than the successive application of the binary versions. – balage May 08 '20 at 15:30
  • @John: I would go for $f: ((a,b),c) \mapsto (a,(b,c))$. – Jukka Kohonen Sep 16 '21 at 00:20

2 Answers2

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For the first question, you could write $$ \times(A,B,C). $$ A more standard notation would be to consider the sets to be indexed by an ordered set, such as $A_1,A_2,A_3$ are indexed by $\{1,2,3\}$, and then to consider $$ \prod_{i=1}^3 A_i. $$

Jagriff
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For the second question, one can view a chain of $n-1$ inequalities as an $n$-ary operation from $\mathbb R^n$ to $\{0,1\}$, representing truth values as numbers. Then $x < y < z$ is quite different from the iterated binary operations $(x<y)<z$ and $x<(y<z)$.

This has practical relevance in computational contexts: see, for example, how $\cdot < \cdot < \cdot$ works in Julia:

julia> 0 < -5 < 5
false

julia> 0 < (-5 < 5)
true


julia> (0 < -5) < 5
true

Python and SageMath have a similar chained inequality.

Even simpler is the chained equality: $a=b=c$ does not mean $a=(b=c)$ or $(a=b)=c$. Many binary relations are chained in a similar way, for example the subset relation $S \subset T \subset U$, and could be seen as $n$-ary relations.

Jukka Kohonen
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