The associative property is one of those properties that does not get much talk about, because it is taken for granted, and it is used all the time, without knowing. The associative property has to do with what operands we process first when operating more than two operands, and how it does not matter what operands we operate first, in terms of the final result of the operation.

The associative property is a cornerstone point in Algebra, and it is the foundation of most of the operations we conduct daily, even without knowing. Doing Algebra without the associative property, although possible, it is rather hard. There are structures in Math in which associativity is not assumed to be true, but those are much more limited.

At the core of the associative property, we need to understand first the idea of operation. Without getting too technical, an operation "\(\circ\)" is simply a way of taking two elements \(a\) and \(b\) on a certain set \(E\), and do "something" with them to create another element \(c\) in the set \(E\).

So then, you take \(a\) and \(b\), you operate them and you get \(c\). Such action can be put mathematically as \(a \circ b = c\).

It is important to observe that you operated TWO elements, \(a\) and \(b\), to get \(c\). I make an emphasis again, you operate TWO elements, \(a\) and \(b\). So far, so good. So, question, what if you want to operate three elements. Well, you cannot, after all operations take TWO elements, so what would you do with the third one. Or can you?

Well, what if you operate two of them first, and then you operate the third one with the result of operating the first two elements? Yes, that can be done. So, say that you have three elements \(a\), \(b\) and \(c\) and you want to operate them. One way is to operate \(a\) and \(b\) first, and then operate the result of with \(c\). That would be \((a\circ b)\circ c\).

Notice the parenthesis there. It is there for a reason. By writing \((a\circ b)\circ c\) you are saying that you are operating \(a\) and \(b\) first, and THEN you operate \(c\). Fair enough. That looks like a satisfactory way of operating \(a\), \(b\) and \(c\). But it is that the only way? What if I operate \(b\) and \(c\) first, and THEN I operate \(a\) with the result of operatin \(b\) and \(c\). You would write that as \(a\circ (b\circ c)\).

Now the big question: is it the same if I operate those three elements in ways shown above. Do I get the same final result if I operate the first two and the result is operated with the third one, or if operate the first element with the results of operating the other two? Or simply, is \((a\circ b)\circ c\) the same as \(a\circ (b\circ c)\). Dear friends, the answer is depends on whether the operation is associative.

**Definition:** An operation \(\circ\) is associative if for any three elements \(a\), \(b\) and \(c\), we have that

Not all operations satisfy this associative property, majority do, but some do not. The most common operations, the ones that we know, do satisfy associativity, such as the sum or multiplication

### EXAMPLE 1

Check some numbers to convince yourself that associativity is met for the common sum "\(+\)".

### ANSWER:

For example, let us consider 3 numbers: \(8\), \(4\) and \(7\). Let us check whether or not associativity is met for this data. Notice that:

\[ \large (8 + 4) + 7 = 12 + 7 = 19 \]On the other hand, we have that

\[ \large 8 + (4 + 7) = 8 + 11 = 19 \]Hence, in this case \((8 + 4) + 7 = 8 + (4 + 7)\).

## The Associative Property Used to Define Operations with more than two Operands

So, not all operations are associative, but most of the ones we know are. When associativity is met, then we can define, without ambiguity the operation of more than two operands. To make it simpler, we simply write \(a \circ b \circ c\), without parenthesis because due to the associativity property, we know that it does not matter how we group the operands, we will get the same final result of the operation.

### EXAMPLE 2

Let us define the following operation:

\[ \large a\circ b = ab+a-b \]Is this operation associative?

### ANSWER:

Notice that

\[\left( a\circ b \right)\circ c=\left( ab+a-b \right)\circ c= \left( ab+a-b \right)c+ab+a+b-c\] \[= abc+ac-bc+ab+a+b-c\]On the other hand, we have that

\[a\circ \left( b\circ c \right) = a\circ \left( bc+b-c \right)=a\left( bc+b-c \right)+a+bc+b-c\] \[= abc - ac + bc + ab + a + b - c\]Hence, it is not always true that \(\left( a\circ b \right)\circ c = a\circ \left( b\circ c \right) \). Therefore, the operation "\(\circ\)" is not associative.

## More About Associativity

Associativity is one of those things you take for granted and basically you use it without knowing. For example, when you write \(1 + 2 + 3\), you are implicitly assuming that associativity is met, because otherwise you would need to specify if you mean \((1 + 2) + 3\) or you mean \(1 + (2 + 3)\). When there is associativity, the parenthesis don't matter because you get the same result, so you just write \(1 + 2 + 3\).

Please do not confuse associativity with commutativity. When we say associativity is met, then which pair you operate first does not matter. That is **not the same** as saying that the order of the operation does not matter, which is a different thing (and it is call the commutativity property).

### Why is the associative property important?

The associative property is very important because it allows flexibility to conduct operations of more than two operands, in a way that it does not matter which pair of operands is operated first, so parenthesis are not needed. For some operations associativity is not met, and that is fine, but lack of associativity makes everything more cumbersome.

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