3 ms·

Why does this innovative method of subtraction from a third grader always work?

14 points by yesprabhu 9y ago

tankenmate 9y ago
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.
jm__87 9y ago
Interesting.. this is how I do subtraction in my head. Didn't realize it wasn't the "standard" approach :).
- srtjstjsj 9y ago
  This is how it's taught at my local school, but it's a bit hidden, because the the 7 is "broken apart", not "subtracted": 61 - 17 = 60 + 1 - (10 + 1 + 6) = (60 - 10) + (-6) "Breaking apart" is equivalent to subtraction, but feels different when doing mental calculation.
- Zecc 9y ago
  As do I, sort of. To subtract mentally I adjust the numbers of both sides by the same amount so it rounds up nicely in either side. In this case I'd go: 61-17 =>(-1 on both sides)=> 60-16 =>(-10 on both sides)=> 50-6 =>("normal" borrow)=> 44 Had it been, say 148-34, I could go either: 148-34 =>(+2)=> 150-36 =>(-6 with borrow)=> 144-30 =>(-30) 114 or most likely 148-34 =>(+2)=> 150-36 =>(-30)=> 120-6 =>(-6) 114 though in this simple case I'd just 148-34 =>(set the 100 aside)=> 48-34 =>(split digits)=> 40-30 + 8-4 = 10+4 =>(join digits again)=> 14 =>(put the 100 back in)=> 114 This is in line with answers that were given to the Math Exchange question. Edit: @srtjstjsj put it nicely when they call it "breaking apart".
srtjstjsj 9y ago
It's interesting that the poster knows "subtrahend" and "minuend" and fully explains the algorithm, but can't solve the pre-algebra problem that shows that it is accurate.

14 points by yesprabhu 9y ago

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.

jm__87 9y ago

Interesting.. this is how I do subtraction in my head. Didn't realize it wasn't the "standard" approach :).

srtjstjsj 9y ago
This is how it's taught at my local school, but it's a bit hidden, because the the 7 is "broken apart", not "subtracted": 61 - 17 = 60 + 1 - (10 + 1 + 6) = (60 - 10) + (-6) "Breaking apart" is equivalent to subtraction, but feels different when doing mental calculation.
Zecc 9y ago
As do I, sort of. To subtract mentally I adjust the numbers of both sides by the same amount so it rounds up nicely in either side. In this case I'd go: 61-17 =>(-1 on both sides)=> 60-16 =>(-10 on both sides)=> 50-6 =>("normal" borrow)=> 44 Had it been, say 148-34, I could go either: 148-34 =>(+2)=> 150-36 =>(-6 with borrow)=> 144-30 =>(-30) 114 or most likely 148-34 =>(+2)=> 150-36 =>(-30)=> 120-6 =>(-6) 114 though in this simple case I'd just 148-34 =>(set the 100 aside)=> 48-34 =>(split digits)=> 40-30 + 8-4 = 10+4 =>(join digits again)=> 14 =>(put the 100 back in)=> 114 This is in line with answers that were given to the Math Exchange question. Edit: @srtjstjsj put it nicely when they call it "breaking apart".

srtjstjsj 9y ago

It's interesting that the poster knows "subtrahend" and "minuend" and fully explains the algorithm, but can't solve the pre-algebra problem that shows that it is accurate.