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Re: Mathematical understanding
Posted:
Nov 8, 2010 12:45 PM
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On Nov 8, 2010, at 10:50 AM, Phil Mahler wrote:
> On 11/8/10 10:22 AM, "Alain Schremmer" <schremmer.alain@gmail.com>
> wrote:
>
> Quite so! But I had assumed that Mahler had meant to write
>
>>> ... I still believe that practice is a good thing, and follows,
>>> not precedes, understanding.
>
> I'm dropping out for the rest of the day, I know people react to
> (what they perceive as) clutter.
> But I didn't mean what Alain says, though I appreciate his thoughts.
>
> When I teach why we add exponents to multiply (this is in the
> context of low level courses, whole number exponents), I think the
> students "understand" it when I draw the usual "picture", for
> example 3 x's followed by 4 x's is 7 x's for factors. Then they
> practice. But many will get it wrong on the next test. Where did
> the understanding go? It's the practice that helps them get it
> right, not the theory. Hopefully after a few weeks of practice, and
> my reiteration of the theory, it does click. Of course that still
> doesn't mean that next semester they won't get it wrong. Even
> knowledge of a reason requires refreshing.
[A] A variation of the above does work rather permanently:
(1) Get the students to compute 7?5?5 ? 4?5?5?5?5
Hopefully they will get 437500
Now tell them you are not really interested in the "answer" and ask
them something along the line of: "how many 5s there are in there and
what is the product of what's left "
Moaning and groaning, they will divide 437500 by 5, get 87500,
divide 87500 by 5, get 17500, etc and hopefully will end up with a
response along the line of "there are six 5s and a 28"
(2) Ask if there might not be a "better way". The probable response
will be a blank stare.
(3) Get the students to compute 4?5?5?5?5?5 ? 3?5?5?5?5?5?5?5?5
The students won't be happy so settle to asking them what the
product of what's left is going to be. After a few hesitations, they
will probably say 28. Now ask them to find how many 5s and
consternation will return. If not, raise the ante.
(4) The students are now ready for a "better way" and for the fact
that the hard part is the number of 5s. So, offer them the following
"code":
7x5^+3 is code for " 7 multiplied by 3 copies of 5"
that is 7?5?5?5
BUT the code for "multiplied" is the + before the 3 and NOT the x
between the first 5 which is merely a SEPARATOR. You should introduce
the metalanguage: 7 is the coefficient and +3 is the exponent.
Here a bit of practice in and out of the code is good, eventually up
to , say. 4x5^+13 ? 3x5^+17. The purpose is to familiarize the
students with the new language and for them really to accept that it
is a shorthand. So they must get impatient with the longhand.
(5) Now let the students rewrite the initial 7?5?5 ? 4?5?5?5?5 in
code. They will ask what to do about the middle ? ; tell them to do
what seems proper and they will eventually write 7x5^+2 ? 4x5^+4. Ask
them what the coefficient of the result is going to be and they will
say 28; ask them how many 5 there will be and, after some hesitation,
they will say 6.
[B] By far the better approach is, after addition of integers has
been done, to introduce the code
7x5^-3 is code for " 7 divided by 3 copies of 5"
that is 7/5?5?5 (written as a fraction)
AND the code for "divided" is the - before the 3.
Then, of course, the students eventually enjoy 7?5?5 / 4?5?5?5?5 etc.
[C] It works even better when it is not an isolated happening and, by
that time, using code to shorthand long stuff has become a routine
idea because then the students can concentrate on the case at hand.
Regards
--schremmer
P.S. to Marsh: I hope the above computations are correct as I did
them in my head and I always was pretty awful at mental arithmetic.
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