Mathematical ideas which get developed..

[apple2357]

Here is something i have been thinking about..

Take exponentiation - it is introduced as repeated multiplication: So 5^3 is another way of writing 5x5x5 and that makes sense.
However, when we introduce negative powers or rational powers this way of looking at it doesn't work. So we define ( do we define?) negative powers to mean one thing, rational powers another and power of zero too. These are entirely consistent with the rules which come out of 'repeated multiplication' idea. But it can be a stumbling block for young children to learn these ideas as they are essentially two different ones.

Can anyone offer me any other examples in mathematics where we start by thinking in one way but then develop it to work for other contexts? The only other one i can think of is factorial and how we define 0!, and that doesn't feel consistent with the definition of n!.

 

[Harry_the_cat]

Trigonometry. We often introduce trigonometry by referring to right angle triangle, ie sin A = opp/hyp. This works for acute angles only. Then we extend into angles of any size, but these definitions no longer hold.

 

[apple2357]

Ah yes. Good one

 

[topsquark]

One of my favorites is "clock math." We have developed basic algebraic techniques to work on the real number system. These only need to be rewritten slightly in order to be used to do problems on a 12 (or 24) hour clock. So we have things like [math]9 + 6 = 15[/math] for the usual algebra but we have [math]9 \oplus 6 = 3[/math] in clock math. The structures are easy to extend as an example and this leads to more complicated algebraic systems.

-Dan

 

[apple2357]

One of my favorites is "clock math." We have developed basic algebraic techniques to work on the real number system. These only need to be rewritten slightly in order to be used to do problems on a 12 (or 24) hour clock. So we have things like [math]9 + 6 = 15[/math] for the usual algebra but we have [math]9 \oplus 6 = 3[/math] in clock math. The structures are easy to extend as an example and this leads to more complicated algebraic systems.

-Dan

Interesting. Can't say i have ever heard of clock math as such? I recognise what you are doing in terms of modular arithmetic but that might be something different?

 

[JeffM]

Actually I prefer to define exponentiation recursively for a > 0.

[MATH]a^0 = 1; \text { integer } n > 0 \implies a^n = a * a^{(n-1)}.[/MATH]
Thus, repeated multiplication is a consequence of the definition for n > 1. Obviously multiplication is a binary operation and a¹ makes no sense in terms of repeated multiplication.

We then can prove that, for non-negative integers b and c with a > 0

[MATH]a^b * a^c = a^{b+c} = a^{c+b} \text { and } (a^b)^c = a ^{bc} = (a^c)^b.[/MATH]
It then becomes simple to show that the only way to maintain the first result
while extending exponents to all integers is

[MATH]a > 0, \ b \text { and } c \text { any integers, and } a^b * a^c = a^{b + c} \implies[/MATH]
[MATH]1 = a^0 = a^{\{c + (-c)\}} = a^c * a^{-c} \implies a^{-c} = \dfrac{1}{a^b} = \left ( \dfrac{1}{a} \right )^b.[/MATH]
That gets us all integer exponents without any nonsense about repeated multiplication with negative numbers.

From here, we can consistently extend to reciprocals of integers only by this

[MATH]a > 0, \ b \text { and } c \text { integers with } c \ne 0, \text { and } a^{bc} = (a^c)^b \implies[/MATH]
[MATH]a = a^1 = a^{\{c * (1/c)\}} = (a^{(1/c)})^c \implies a^{1/c} = \sqrt[c]{a}[/MATH]
cth root not rendering correctly ................Now it is! - just the touch of right finger.....

It is not that we have to keep changing the initial definition. Instead, we are expanding a basic definition consistently with our initial results.

We can do the same kind of recursive definition with multiplication and avoid nonsense about repeated addition or with the factorial function. Start with a good definition, and everything becomes more logical.

 

[Dr.Peterson]

Of course, the very idea of number starts out as just 1, 2, 3, and gets extended to zero, negative, rational, real, complex; and at each step all the operations have to be extended, so that multiplication which started as repeated addition, turns into scaling, and eventually involves rotation.

 

[apple2357]

Actually I prefer to define exponentiation recursively for a > 0.

[MATH]a^0 = 1; \text { integer } n > 0 \implies a^n = a * a^{(n-1)}.[/MATH]
Thus, repeated multiplication is a consequence of the definition for n > 1. Obviously multiplication is a binary operation and a¹ makes no sense in terms of repeated multiplication.

We then can prove that, for non-negative integers b and c with a > 0

[MATH]a^b * a^c = a^{b+c} = a^{c+b} \text { and } (a^b)^c = a ^{bc} = (a^c)^b.[/MATH]
It then becomes simple to show that the only way to maintain the first result
while extending exponents to all integers is

[MATH]a > 0, \ b \text { and } c \text { any integers, and } a^b * a^c = a^{b + c} \implies[/MATH]
[MATH]1 = a^0 = a^{\{c + (-c)\}} = a^c * a^{-c} \implies a^{-c} = \dfrac{1}{a^b} = \left ( \dfrac{1}{a} \right )^b.[/MATH]
That gets us all integer exponents without any nonsense about repeated multiplication with negative numbers.

From here, we can consistently extend to reciprocals of integers only by this

[MATH]a > 0, \ b \text { and } c \text { integers with } c \ne 0, \text { and } a^{bc} = (a^c)^b \implies[/MATH]
[MATH]a = a^1 = a^{\{c * (1/c)\}} = (a^{(1/c)})^c \implies a^{1/c} =[/MATH] \sqrt[c]{a}.[/MATH]

cth root not rendering correctly

It is not that we have to keep changing the initial definition. Instead, we are expanding a basic definition consistently with our initial results.

We can do the same kind of recursive definition with multiplication and avoid nonsense about repeated addition or with the factorial function. Start with a good definition, and everything becomes more logical.

Thats really interesting to read. Thanks

 

[Cubist]

An interesting question, and very interesting responses!

I guess that unless an advancement completely breaks the previous knowledge, then it makes a lot of sense to leave the previous knowledge in place as a "stepping stone" for the next student's learning.

 

[Dr.Peterson]

But it can be a stumbling block for young children to learn these ideas as they are essentially two different ones.

Then there are the classic "Well, no, actually we can ..." statements:

"You can't subtract a larger number from a smaller one."​

"You can't take the square root of a negative number."​

I try to reword those so they can be extended later.

 

[topsquark]

Interesting. Can't say i have ever heard of clock math as such? I recognise what you are doing in terms of modular arithmetic but that might be something different?

Yes, clock math is Algebra a modulo 12. I didn't know that until my 3rd year in college.

-Dan