
Why does this innovative method of subtraction from a third grader always work? - yesprabhu
https://math.stackexchange.com/questions/2667980/why-does-this-innovative-method-of-subtraction-from-a-third-grader-always-work
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tankenmate
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
> 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
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
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
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.

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srtjstjsj
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.

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jm__87
Interesting.. this is how I do subtraction in my head. Didn't realize it
wasn't the "standard" approach :).

~~~
Zecc
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".

