Right way to show associativity.

Question

Authors usually write that $*$ is associative on a set $S$ if,

$(a*b)*c=a*(b*c)$ $\forall a,b,c \in S$

I think it should have been,

$(a*b)*c=a*(b*c)=(a*c)*b$ $\forall a,b,c \in S$

I made all possible groups i.e. $(a,b)$, $(b,c)$ and $(a,c)$ so that for example if set were $\mathbb R$ and operator were "$.$" then I would be able to re write this expression in 3 ways,

$$1.(2.3)=(1.2).3=(1.3).2$$

, which is actually true.

Answer 1

The problem with binary operations is that they are defined on just two elements, i.e. $a*b$. Therefore, when it comes to calculating $a∗b∗c$, we don't know how to apply the operation because there are now three elements. This is why we use the brackets, and say "we do $a*b$ first, then we'll apply $c$ to the result", i.e. $(a*b)*c$. Then, in a set where associativity of the binary operation holds, we know it's okay to do $a*(b*c)$ instead of $(a*b)*c$.

The reason why $(a*c)*b$ isn't the same, is because that comes down to helping us answer the question of "what is $a*c*b$?", and this is only ever the same question as "what is $a*b*c$?" when we also have commutativity of the elements $b$ and $c$.

Hope this helps.

Answer 2

Your proposed reformulation assumes commutativity of $*$, which isn't and shouldn't be required for associativity.

Answer 3

"Associativity" just means "it doesn't matter how we bracket terms and the order in which we simplify, we'll still get the same answer". It says nothing about the order in which the terms appear. Notice the difference between "the order in which we evaluate the expression" and "the order of the terms in the expression".

Associativity says $(x * y) * (b * c)$ can unambiguously be written as $x*y*b*c$, but it doesn't have anything to do with commutativity. For example, if $*$ is the "concatenation" operation on strings, $x*a*b = xab$ however we perform the concatenation, but it certainly is not equal to $xba$.

Answer 4

A binary operation ☆ is associative on a set $\mathbb S$ if,

$$(a☆b)☆c=a☆(b☆c) \forall a,b,c \in \mathbb S$$

Suppose you are verifying that whether $\mathbb S$ is associative w.r.t ☆ or not.

Let $x,y,z$ be some definite elements of $\mathbb S$.

Assume you find out that,

$$(x☆y)☆z=x☆(y☆z)$$

Since you have to check associativity according to usual definition and for all $a,b,c \in \mathbb S$, you find out associativity holds for another triplet $(x,z,y)$ i.e.

$$(x☆z)☆y=x☆(z☆y)$$

If ☆ is commutative, you can unambigously write;

$$(x☆y)☆z=x☆(y☆z)=(x☆z)☆y$$

, otherwise not.

With your definition, we would have to verify $$(x☆y)☆z=x☆(z☆y)$$

and suppose we had verified

$$(x☆z)☆y=x☆(z☆y)$$

then the following holds if and only if $y☆z=z☆y$

$$(x☆y)☆z=(x☆z)☆y$$

and that would be equivalent to say that if $☆$ is not commutative, then it's not associative.