# How much commutativity in associativity

1. May 19, 2005

### Kocur

How much commutativity in associativity?

Please correct me if I am wrong.

By the definition, binary operation "+" on set S is associative if and only if, for all elements x, y, and z from S, the following holds:

x + (y + z) = (x + y) + z.

In other words, the order of operation is immaterial if the operation appears more than once in an expression.

Now, operation "+" may be either commutative or not. Let us consider the later case. If "+" is not commutative, we have the following:

x + (y + z) may be different from x + (z + y) and
x + (y + z) may be different from (y + z) + x and
x + (y + z) may be different from (z + y) + x
.

Thus, the result of x + (y + z), depending on the way the operation is performed may by different from the result of (x + y) + z.

The whole problem disappears whenever "+" is commutative.

So, can we really claim that "+" is associative without referring, maybe even not explicitly, to commutativity?

Kocur.

Last edited: May 19, 2005
2. May 19, 2005

### AKG

Huh? If your operation is associative, then no, the result will not be different. What does:

x + (y + z) may be different from x + (z + y) and
x + (y + z) may be different from (y + z) + x and
x + (y + z) may be different from (z + y) + x.

have to do with whether or not (x + y) + z = x + (y + z)?

3. May 19, 2005

### dextercioby

Let (G,+) be a group.

Proposition

(G,+) abelian--------->"+" associative.

In the following,the "x","y" & "z" are elements of the group.

x+y+(-(x+y))=0 (1) ("x+y" is an element of the group (closure) and it has an inverse wrt "+")

From (1) it follows that

-x+x+y+(-(x+y))=-x =>y+(-(x+y))=-x=>-y+y+(-(x+y))=-y+(-x)=>-(x+y)=-x+(-y) (2) (i used commutativity of addition & inverses)

Consider the expression

x+(y+z)+(-(x+y))+(-z) (3)

If i show that (3) =0,then,from the inverse property and commutativity (meaning i can add both to the left & to the right)

x+(y+z)=(x+y)+z ,which is what i should be proving.

x+(y+z)+(-(x+y))+(-z)=x+(y+z)+(-x)+(-y)+(-z)=x+(-x)+(y+z)+(-(y+z))=0

Q.e.d.

4. May 19, 2005

### AKG

dexter, the question was whether an operation can be associative without being commutative, not that being commutative implied being associative. Also, I don't think you mean to call G a group, since the group operation is associative by definition.

Also, here you have:

-x+x+y+(-(x+y)) = -x --> y+(-(x+y)) = -x

but this does not follow. I'm going to write xy instead of x + y, and I (for identity) instead of 0. You started with:

$$(xy)(xy)^{-1} = I$$

then you said:

$$x^{-1}[(xy)(xy)^{-1}] = x^{-1}(I) = x^{-1}$$

which is a correct inference. You then concluded that

$$x^{-1} = x^{-1}[(xy)(xy)^{-1}] = [x^{-1}x][(y)(xy)^{-1}] = I[(y)(xy)^{-1}] = [(y)(xy)^{-1}]$$

But the second equality presumes associativity, so the proof fails.

5. May 19, 2005

### arildno

Matrix multiplication is associative, but not in general commutative.

6. May 20, 2005

### Kocur

AKG wrote:

Huh? If your operation is associative, then no, the result will not be different. What does:

x + (y + z) may be different from x + (z + y) and
x + (y + z) may be different from (y + z) + x and
x + (y + z) may be different from (z + y) + x.

have to do with whether or not (x + y) + z = x + (y + z)?

Well, I just wanted to show, that if "+" is non-commutative, we can obtain different results for x + (y + z), depending on the order in which we perform operations.

The results for (x + y) + z may also be different, due to the same reason.

It follows that (x + y) + z might be different from x + (y + z).

Dextercioby:
I was thinking about something else. Namely, I was trying to show that the order in which non-commutative operations are performed is always important. Thus, I am not sure if we can talk about associativity in such a case .

AKG:
You got it right in your second post.

Arildno:
You are right.

7. May 20, 2005

### matt grime

Yes you *can* but you are not necessarily guaranteed to do so, and indeed if '+' is associative you are guaranteed *not* to. And? The commutativity or otherwise of * is neither here nor there.

Here is an operation that is commutative but not associative:

define * on {a,b,c} by a*b=b, a*a=c and c*b=a extending to commutativity by fiat and thence to any binary operation on {a,b,c}. Any choice of extension wil do

Then (a*a)*b=c*b=a, yet a*(a*b)=a*b=b

That can easily be adapted to give an operation that is neither associative or commutative by losing the "by fiat" bit and say decalring b*a=c

matrix multiplication is associative but not commutative

and addition on R is both associative and commutative, therefore all 4 possibilities can occur, and the properties are seen to be completely independent!

Note, Dexter, how come you 'prove' the operation in an abelian group is associative? That is part of the definition of group. Did you mean to say that in an abelian group some of the axioms are redundant?

8. May 20, 2005

### AKG

If "+" is non-commutative, then there's nothing to show. If "+" is non-commutative, then that means precisely that we can obtain different results depending on the order we perform operations. Well actually, it's not the order of operations (that's the idea behind associativity), it's the order that we put the elements that we're operating on. Associativity says that (x + y) + z = x + (y + z). The two plus signs in each expression of the equation are the operations, and associativity is saying that the order of doing the operations don't matter, you can go left to right or right to left. Commutativity says that x + y = y + x. That is, the order in which you put the elements you're operating on doesn't matter.
It absolutely does not if we're assuming "+" is associative, which is what your first post appears to do. It says that we cannot say that "+" is associative without referring to commutativity, that is, an operation cannot said to be just associative but non-commutative. However, that's wrong, and nothing you've done even suggests otherwise. You've basically demonstrated what non-commutativity means, that a sum written in different orders might be different. Where in the world you get from this that associativity might not hold, I have no idea.

As far as I can tell, you might be hoping to prove it based on something like this:

$$a \neq b \neq c$$

Therefore

$$a \neq c$$

Replace a and c with 5, and b with 3, and see if this is valid. If this is not what you're thinking, then I honestly have no clue why you believe that something in your posts shows that associativity requires commutativity.

9. May 21, 2005

### Kocur

I think that you are right AKG. Thank you for helping me out of confusion.

Kocur.

10. May 22, 2005

### matt grime

Did you not read my post where I gave expicit binary operations that show that commutativity and associativity are unrelated?