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

Why does this innovative method of subtraction from a third grader always work?(math.stackexchange.com)

14 points by yesprabhu 8 years ago | 10 comments

tankenmate 8 years 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 8 years 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 8 years 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.

srtjstjsj 8 years 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.

jm__87 8 years ago |

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

Zecc 8 years 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 8 years 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.