Why does this innovative method of subtraction from a third grader always work?
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Why does this innovative method of subtraction from a third grader always work?
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Re: Why does this innovative method of subtraction from a third grader always work?
#2She 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))
Re: Why does this innovative method of subtraction from a third grader always work?
#3She 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…
> 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.
Re: Why does this innovative method of subtraction from a third grader always work?
#4She 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…
> 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…
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.
Re: Why does this innovative method of subtraction from a third grader always work?
#5Interesting.. this is how I do subtraction in my head. Didn't realize it wasn't the "standard" approach :).
Re: Why does this innovative method of subtraction from a third grader always work?
#6Earlier quoted context omitted.
> 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…
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 functio…
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.
Re: Why does this innovative method of subtraction from a third grader always work?
#7Earlier quoted context omitted.
> 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…
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 functio…
> 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.
Re: Why does this innovative method of subtraction from a third grader always work?
#8It'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.
Re: Why does this innovative method of subtraction from a third grader always work?
#9Interesting.. this is how I do subtraction in my head. Didn't realize it wasn't the "standard" approach :).
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.
Re: Why does this innovative method of subtraction from a third grader always work?
#10Earlier quoted context omitted.
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 functio…
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.
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.