I'm studying the Cartesian product, which is bound to the idea of a binary relation. Even with Cartesian products of several sets, n-ary Cartesian products, we have to think combinatorically as two sets at a time, recursively. Is there any sort of operation where there is, say, a triary operation happening. Obviously, when we have 3 − 2 − 1 we have to think of associative rules, forcing subtraction to be a binary operator "step-through" affair. There is no 3 ⊕ 2 ⊕ 1 operator that does something in one fell swoop to all three numbers is there? Lisp has (+ 1 2 3) and grade-school math has vertical addition with the plus-sign and a line under the lowest number, but these are not "internally" non-binary. The only thing that takes "three at once" is, yes, a product-based operator, again, an n-ary Cartesian product. Correct? Another example would be playing poker and being dealt five cards. The cards were shuffled, which is a combinatoric permutation of the stack of cards, which are then dealt into hands. Is there anything from probability (or anywhere) that doesn't start with a shuffle permutation therefore binary, rather, just looks at the five cards coming together non-permutation-, non-binary-wise?

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I&#039;m studying the Cartesian product, which is bound to the idea of a binary relation. Even with Cartesian products of several sets, n-ary Cartesian products, we have to think combinatorically as two sets at a time, recursively. Is there any sort of operation where there is, say, a triary operation happening. Obviously, when we have   3  −  2  −  1 we have to think of associative rules, forcing subtraction to be a binary operator &quot;step-through&quot; affair. There is no   3  ⊕  2  ⊕  1 operator that does something in one fell swoop to all three numbers is there? Lisp has (+ 1 2 3) and grade-school math has vertical addition with the plus-sign and a line under the lowest number, but these are not &quot;internally&quot; non-binary. The only thing that takes &quot;three at once&quot; is, yes, a product-based operator, again, an n-ary Cartesian product. Correct? Another example would be playing poker and being dealt five cards. The cards were shuffled, which is a combinatoric permutation of the stack of cards, which are then dealt into hands. Is there anything from probability (or anywhere) that doesn&#039;t start with a shuffle permutation therefore binary, rather, just looks at the five cards coming together non-permutation-, non-binary-wise?

Giancarlo Brooks · Accepted Answer

How about a 3-number average? (Or more numbers if you want an operator that takes more than 3 arguments.) You don&#039;t get the correct value by averaging the first two numbers and then averaging that result with the third.

I'm studying the Cartesian product, which is bound to the idea of a binary relation. Even with Carte

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