Are Subtraction and Division Commutative and Associative Operations?
Date: 05/15/2008 at 11:53:02 From: Uli Subject: subtraction and division: commutative and associative? Math sources (textbooks, teachers, even this website) always say subtraction and division are neither associative nor commutative. But defined properly, they plainly are both associative and commutative. So why does everyone say they're not? I'll use the commutativeness of subtraction as an example; the other situations are analogous. The expression "a - b" is defined as "a + (-b)" for any real number. see, e.g., Keedy/Bittinger/Beecher, "Algebra and Geometry," Sixth Ed'n, at p. 8, Theorem 2. That expression commutates beautifully: a + (-b) = (-b) + a 2 + (-3) = (-3) + 2 Thus, given the definition of subtraction, the statement that subtraction is not commutative is false. The statement appears to result solely from the obvious fact that the statements "a + (-b)" and "b + (-a)" are different. But isn't that just being willfully sloppy about the meaning of the "-" sign? In other words, even with a minimum of care about what the signs mean, the alleged non-associativeness and non-commutativeness of subtraction and division (with "a divided by b" defined as "a x 1/b", (see Keedy/Bittinger/Beecher at p. 9, Theorem 4) vanishes. So why does everyone insist that these operations are in fact non-commutative and non-associative? Incomprehensibly to me, this even includes the Keedy/Bittinger/Beecher textbook on p. 10, problems 75-78.
Date: 05/15/2008 at 15:42:30 From: Doctor Peterson Subject: Re: subtraction and division: commutative and associative? Hi, Uli. You wrote: >Math sources (textbooks, teachers, even this website) always say >subtraction and division are neither associative nor commutative. But >defined properly, they plainly are both associative and commutative. >So why does everyone say they're not? Because AS A BINARY OPERATION, they are not. In order to apply these properties to them, you have to rewrite them (at least in your head) as additions and multiplications. We say that an operation "#" is commutative if for any a and b, a # b = b # a If we take # as "-", then it is NOT true that a - b = b - a and that is all we are saying when we say that it is not commutative. It does not fit the definition of commutativity, applied directly. And this definition applies to the literal symbolism of an operation, not to some alternate way of thinking of it. >I'll use the commutativeness of subtraction as an example; the other >situations are analogous. The expression "a - b" is defined as "a + >(-b)" for any real number...that expression commutates beautifully: > >a + (-b) = (-b) + a >2 + (-3) = (-3) + 2 What you're doing here is commuting the ADDITION, not the subtraction. That's fine; this is what anyone who is good at math does automatically. But you have to call it what it is. Commuting the subtraction would mean commuting the minuend and subtrahend (a and b), not the two addends (a and -b). The fact that subtraction and division are not associative or commutative when treated (naively, one might say) as operations in themselves (simply taking the symbols as given) is the reason we teach that subtraction should be thought of as addition of the negative, and division as multiplication by the reciprocal. That is, we DEFINE subtraction as an addition so that we can use the nice properties of the latter operation, and not have to treat subtraction in practice as an operation in itself. So there is nothing inconsistent in a text saying both things. If you have any further questions, feel free to write back. - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/
Date: 05/20/2008 at 17:48:01 From: Uli Subject: Thank you (subtraction and division: commutative and associative?) Thank you!!! A little background clarifies everything. You wrote, "we DEFINE subtraction as an addition so that we can use the nice properties of the latter operation, and not have to treat subtraction in practice as an operation in itself." That's refreshingly honest and accurate, and extremely helpful as I help my daughter with her beginning algebra and in the process work through a number of questions I never got answered when I was in school. I really appreciate your taking the time to answer me so thoroughly. What a wonderful institution Dr. Math is!
Date: 05/20/2008 at 20:58:50 From: Doctor Peterson Subject: Re: Thank you (subtraction and division: commutative and associative?) Hi, Uli. Thanks! This is just what we're here for; often a perspective that is not commonly taught in schools can make a big difference, and the way we get questions outside of the classroom context makes it easier to see them in different ways. One thing I've learned from doing this over the years is that, as important as definitions are in math, they are important as a TOOL, which we design to meet a specific need, and consequently we can rework them as needed in order to "sharpen" them, rather than holding on to the first way they were defined. In this case, the idea of subtraction existed before the idea of negative numbers, but when it was rethought as the addition of the negative (opposite), it made a lot of work far easier than it had been previously. Please write any time you have questions like this! - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/
Search the Dr. Math Library:
Ask Dr. MathTM
© 1994-2015 The Math Forum