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Relationship Between Commutative and Associative Property?

Date: 06/30/2002 at 22:07:13
From: Randy Nixon
Subject: Relationship between commutative and associative property?

I'm attaching a transcript of a discussion on math I'm having with 
these 2 people.  I'd like your thoughts on the discussion if you 
have the time.

From S:
PEMDAS - Please Excuse My Dear Aunt Sally (Parenthesis, Exponents, 
Multiplication, Division, Addition, Subtraction)

Simply put, this is the order in which you do algebra, working from 
inside out, solve all the problems in the parentheses, removing each 
one. Then do all the exponents. Then from left to right do the M's 
and D's, and the A's and S's.

Re: from J:
Or from right to left, or from the middle outwards, or from the edges 
inwards. You'll get the exact same result. (This is known as the 
Asociative Property.)

Re: from S:
It's actually the commutative property. But you're right with
addition. With the S's, left to right is the the only way, or you'll
get drastically different answers.

Re: from D:
Actually, in an addition or multiplication unbounded as to the number 
of factors or addends, the ability to parse it in either direction 
depends on both the associative and commutative properties.

Re: From S:
Elaborate with an example?

Re: From D:
Take x + y + z.
We want to establish that a left-first parse equals a right-first 
parse, i.e. (x + y) + z = (z + y) + x.
The associative property tells us that (x + y) + z = x + (y + z)
Then the commutative property tells us that x + (y + z) = (z + y) + x
So both properties are needed to prove the equivalence.

Re: from S:
Ahhhh, I kind of see what you're saying...
I'm still unclear as to what you mean by 'needed', though....

The commutative property:

a + b + c = c + b + a

The associative property does have a factor in the order when you 
choose it to, i.e.:

a + (b + c)

You would do what is in parenthesis first, and then the rest. However
these are 2 different tools to get to the same solution. Yes, applying
both properties to the equation can prove the equation true, but I
don't see how one is required for the other. 

From what J said, I thought it fit the definition of the commutative 
property to a tee. Grouping terms and doing them out of order are 
examples two different methods. However in a broad sense the 
statement could also apply to the associative property, when you 
group them the right way.

Re from D:
I was defining the commutative property as
a + b = b + a (commutative property)
So, the associative property is required to extend that to
a + b + c = c + b + a (theorem)
If you instead define the commutative property as
a + b + c = c + b + a (commutative property)
then all I have to do is extend the test case by one variable 
(remember, I stated that the number of addends was unbounded):
a + b + c + d = d + c + b + a (theorem)
Once again, you need the associative property.

Re from S:
Again, I don't see it. This is what I was taught commutative property 
to be; That as long as you add, it doesn't matter what order you do 
them, that as long as you multipy it doen't matter what order they 
are in. Your statement regarding my adding a new variable to the 
equation is a bit unclear.

a + b = b + a is an example of the commutative property
so is
a + b + c + d + e + f = f + e + d + c + b + a

All the commutative property says that you can add up these terms out 
of order. 

The associative property states that you can group them any way you 

a + b + c + d + e + f= a + (b + c) + d + (e + f)

Perhaps you're stating a relationship between the associative and 
commutative properties that I am unfamiliar with. At any rate 
despite my enthusiasm for it, I don't pretend to know everything 
about math. 

With your permission I'd like to send this conversation to a math 

Date: 06/30/2002 at 23:22:22
From: Doctor Peterson
Subject: Re: Relationship between commutative and associative property?

Hi, Randy.

I assume you recognize that left-to-right and right-to-left are 
equivalent ONLY for an expression involving only additions or only 
multiplications, and that the order-of-operation rules are needed 
whenever operations are mixed together.

The problem is that without the associative property, you can't even 
WRITE something like

    a+b+c = c+b+a

and be sure of its meaning! Each side would have two different 
interpretations. So in order to talk about the commutative property 
applied to more than two operands, you have to assume the associative 
property. Or rather, I should say, you would have to depend strongly 
on the left-to-right order of operations, so that

    a+b+c means (a+b)+c and NOT a+(b+c)

In the absence of associativity, you could say that

    (a+b)+c = c+(b+a)


    (a+b)+c = c+(a+b) = c+(b+a)

by two applications of the commutative property.

But this is not the commutativity you were looking for; the right 
side is not adding right-to-left in the original expression, but 
first adding b+a and then adding that to c.

When one of you wrote

    a+b+c = c+b+a

you presumably meant

    (a+b)+c = (c+b)+a

and this requires associativity for its proof, just as D said:

    (a+b)+c = c+(a+b) = c+(b+a) = (c+b)+a

Does that make sense?

- Doctor Peterson, The Math Forum 

Date: 07/01/2002 at 01:46:06
From: Randy Nixon
Subject: Thank you (Relationship between commutative and associative

Thanks for your answer! Yes it does make sense. :)
Associated Topics:
High School Basic Algebra
Middle School Algebra

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