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 can 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 teacher. 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) since (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 http://mathforum.org/dr.math/ Date: 07/01/2002 at 01:46:06 From: Randy Nixon Subject: Thank you (Relationship between commutative and associative property?) Thanks for your answer! Yes it does make sense. :) |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2013 The Math Forum
http://mathforum.org/dr.math/