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She is applying the anti-commutative law properties of subtraction; a-b = - (b-a) Also note that you can do this with division. A lot of people are taught that

She is applying the anti-commutative law properties of subtraction; a-b = - (b-a) Also note that you can do this with division. A lot of people are taught that anti-commutativity is only applicable to subtraction but it also applies to any operator or function. a op b = Identity(op) op ( b op a ) Obviously that assume that a op b is defined and b op a is defined and that an identity exists under the operator op. Same goes for functions, f(a,b) = f(Identity(f()),f(b,a))

thaumasiotes 9y ago
> Also note that you can do this with division. A lot of people are taught that anti-commutativity is only applicable to subtraction Wouldn't the application of this to division be the identity a/b = 1 / (b/a) ? Because I'm pretty sure that's taught to everyone, and is in fact the only method taught for dividing rational numbers. > but it also applies to any operator or function Um... how's that? Can you apply it to the function f(x,y) = x^y? It looks like you're claiming that 2^3 is equal to 1^(3^2), but 8 is not actually equal to 1. (Then again, it isn't clear what you think Identity(f) would be here... there is no concept of "exponentiating 0 numbers" in the same way that the empty sum is 0 or the empty product is 1, because exponentiation is not associative and has no identity element.) Similarly, 0 is the additive identity, but it is not the subtractive identity because no subtractive identity exists.
- tankenmate 9y ago
  Not all operators or functions are anti-commutative; addition for example is commutative, not anti-commutative. Exponentiation is not commutative (2^3 != 3^2) and it is not anti-commutative because it has no identity (there are exponentiation equivalencies (sometimes labelled identities) but there is no symmetric identity like 0 for addition and subtraction or 1 for multiplication or division). An operator or function is called anti-commutative when it fits the necessary conditions, not the converse. You can't just label something as anti-commutative and then point out that is doesn't fit the conditions; that's putting the cart in front of the donkey.
  - taejo 9y ago
    You wrote: > it also applies to any operator or function. I understand what you mean (that anticommutativity is a property that any binary operator might have) but it did sound like you meant all operators are anticommutative.
    - thaumasiotes 9y ago
      He's wrong about anticommutativity, too, though. An operation ∘ is anticommutative if a∘b = -(b∘a). It happens that this is true of subtraction, but that has nothing to do with 0 being the "subtractive identity" or with the - in front of "-(b-a)" visually resembling the sign for subtraction. Vector cross product is anticommutative too, because a×b = -(b×a), but 0 is not an identity for vector cross product and negating a vector is not the same operation as taking the cross product of two vectors. The 0 and the negation come from the operation of adding vectors, not cross-multiplying them.
  - thaumasiotes 9y ago
    > there is no symmetric identity like 0 for addition and subtraction or 1 for multiplication or division 0 is an identity for addition, but not for subtraction. 1 is an identity for multiplication, but not for division. They are right identities for these operations, but 1 is also a right identity for exponentiation.

thaumasiotes 9y ago

> Also note that you can do this with division. A lot of people are taught that anti-commutativity is only applicable to subtraction Wouldn't the application of this to division be the identity a/b = 1 / (b/a) ? Because I'm pretty sure that's taught to everyone, and is in fact the only method taught for dividing rational numbers. > but it also applies to any operator or function Um... how's that? Can you apply it to the function f(x,y) = x^y? It looks like you're claiming that 2^3 is equal to 1^(3^2), but 8 is not actually equal to 1. (Then again, it isn't clear what you think Identity(f) would be here... there is no concept of "exponentiating 0 numbers" in the same way that the empty sum is 0 or the empty product is 1, because exponentiation is not associative and has no identity element.) Similarly, 0 is the additive identity, but it is not the subtractive identity because no subtractive identity exists.

tankenmate 9y ago
Not all operators or functions are anti-commutative; addition for example is commutative, not anti-commutative. Exponentiation is not commutative (2^3 != 3^2) and it is not anti-commutative because it has no identity (there are exponentiation equivalencies (sometimes labelled identities) but there is no symmetric identity like 0 for addition and subtraction or 1 for multiplication or division). An operator or function is called anti-commutative when it fits the necessary conditions, not the converse. You can't just label something as anti-commutative and then point out that is doesn't fit the conditions; that's putting the cart in front of the donkey.
- taejo 9y ago
  You wrote: > it also applies to any operator or function. I understand what you mean (that anticommutativity is a property that any binary operator might have) but it did sound like you meant all operators are anticommutative.
  - thaumasiotes 9y ago
    He's wrong about anticommutativity, too, though. An operation ∘ is anticommutative if a∘b = -(b∘a). It happens that this is true of subtraction, but that has nothing to do with 0 being the "subtractive identity" or with the - in front of "-(b-a)" visually resembling the sign for subtraction. Vector cross product is anticommutative too, because a×b = -(b×a), but 0 is not an identity for vector cross product and negating a vector is not the same operation as taking the cross product of two vectors. The 0 and the negation come from the operation of adding vectors, not cross-multiplying them.
- thaumasiotes 9y ago
  > there is no symmetric identity like 0 for addition and subtraction or 1 for multiplication or division 0 is an identity for addition, but not for subtraction. 1 is an identity for multiplication, but not for division. They are right identities for these operations, but 1 is also a right identity for exponentiation.