Suppose we have the set A{a,b}.

Can this set have the associative property?

All the definitions talk about 3 element sets, and not about a two element set.

A={a,b}. Suppose that *is a binary operation, can a set have that property with only two elements belong to it? Thank you.

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    $\begingroup$ What do you mean by associative property? The way I know it, associativity is not a property of a set, but of a binary operation. $\endgroup$ – tomasz Jul 28 '13 at 14:39
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    $\begingroup$ @Alan You need more than sets to talk about associativity. You need a binary operation and you mention none. $\endgroup$ – Git Gud Jul 28 '13 at 14:39
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    $\begingroup$ @Alan Plus I don't understand what you mean with 'the set $A\{a,b\}$'. Do you mean $A=\{a,b\}$? $\endgroup$ – Git Gud Jul 28 '13 at 14:46
  • $\begingroup$ @Git Gud edited my question. $\endgroup$ – Alan Jul 28 '13 at 14:58

You may still need properties like $$(a*b)*a=a*(b*a).$$

Even in a one-element set you may need $$(a*a)*a=a*(a*a).$$

  • $\begingroup$ On a one element set there is only one binary operation and it is associative... $\endgroup$ – MichalisN Jul 28 '13 at 15:01
  • $\begingroup$ I think his problem is not with whether the property holds but how does the property looks like. $\endgroup$ – OR. Jul 28 '13 at 15:04
  • $\begingroup$ Then I missinterpreted your words 'you may need' :) $\endgroup$ – MichalisN Jul 28 '13 at 15:13

The definition of associativity, being a sentence in first order logic, has three variables, but has no requirements on the size of the set it operates on. In fact, the empty function $\emptyset\to\emptyset$ is trivially associative (and commutative and many other things). A binary operation on a singleton set will be, of necessity, associative, commutative, unital, etc. These cases are trivial, and we would hardly need to even conceptualize these properties if they were all we deal with. But since we do deal with bigger domains, we do need three variables to express the full associative property over larger sets.


As per wikipedia (read closely):

"In mathematics, the associative property is a property of some binary OPERATIONS (multiple)..."

"Within an expression containing two or more occurrences in a row of the same associative operator, the order in which the operations are performed does not matter as long as the sequence of the operands is not changed. That is, rearranging the parentheses in such an expression will not change its value. Consider the following equations..."

"Associativity is not the same as commutativity, which addresses whether or not the order of two operands changes the result. For example, the order doesn't matter in the multiplication of real numbers, that is, a × b = b × a, so we say that the multiplication of real numbers is a commutative operation."

So the answer is no, that would be commutativity for two operators.

The very reason for the property is that you are testing if those three (or more) elements would have two (or more) binary operators where you could alter which operation comes first and still get the same result.

This also extends that you can not perform this test for property on a one element set because there is no such thing as a binary operator for a single element. A unary operator operates on a one element set. This is not a property of the operator itself but on the set because it tests against more than one operator (remember it takes two operators to make this test for this property.)

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    $\begingroup$ I appreciate your taking time to respond to this four year old Question. However you seem to be making a major mistake in reasoning. There is no necessary relationship between the number of operators, the number arguments of an operator, and the number of elements in the domain of an operator. In particular you can have a binary operator "on a one element set". It doesn't take "two operators" to define associativity (or commutativity). These properties refer to a single binary operation. $\endgroup$ – hardmath Sep 4 '17 at 4:55

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