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?
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.