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Re: 1 3/4 : 1/2 [WAS Re: Data from Ma's study]
Posted:
May 14, 2000 6:43 AM
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At 09:36 AM 5/12/00 -0400, Paul S. Ache III wrote:
>There appears to be two different situations under which division of
>fractions might occur in the classroom. The first one is what, I think,
>Darrel is referring to; a situation in which an answer to the problem 1
>1/2 divided by 1/2. In my opinion, which means little to most, the
>processes described by Darrel are appropriate.
I was addressing only the problem at hand. The fact that people could not
solve the problem may imply that not only do they not know fraction
division and have an incomplete understanding of fractions, they probably
also have an incomplete understanding of division.
>The second situation is one, I think, Wayne is referring to; getting
>students ready for algebra. In this case, I don't see a connection
>between Darrel's processes and the one used in algebra, but the process of
>invert and multiply will easily lead to an answer.
>One of the major objectives of elementary school mathematics education
>should be to move students from the concrete to the abstract, from what is
>familiar to that which is not. I have asked my students more than once to
>provide a concrete model for the division of fractions and after having
>spent time on operations on whole numbers, they have little difficulty
>providing a mathematically correct response. The difficulty arises when I
>ask them to make a connection to invert and multiply. (I have yet to see
>a connection.) So, at this point, the use of concrete manipulatives seems
>to break down and we must apply the logic of mathematics. Without proper
>understanding and proper training, I think most teachers revert to a "just
>do it" approach. Comments?
Paul is correct; if we want to have students have a complete understanding
of why invert & multiply (I&M) works, they have to understand the logic of
mathematics and the relationship between multiplication and division.
However, we often tend to focus on I&M as the only way to solve fraction
division. I&M does not solve the problem that you are trying to solve--it
solves a slightly different problem.
For example, if I have five mini-pizzas for a birthday party and I want to
give 1/2 pizza to each person, the maximum number of people I can have at
the party is determined by the division problem of 5 pizzas divided by 1
pizza per two people, or 5 divided by 1/2. However, if I have 5 mini-pizzas
and I know that two people will share each pizza, the maximum number of
people I can have at the party is 5 pizzas multiplied by two people per
pizza or 5 x 2. In both cases the answer is 10 people. The difference
between the problems is subtle but it is there.
An interesting observation is that if you ask most people a question that
will require them to divide by a fraction they are familiar with, they will
tend to solve it as a multiplication problem and don't see it as a division
problem. Yet if you ask a similar question using whole numbers, the same
people will solve it as a division problem. ("I have 20 cupcakes for a
party and I want to give two to each person. How many people can I invite?"
has an answer of 10, yet nobody has told me they multiplied 20 by 1/2.)
This is what led to my question about why we perceive fraction division as
different from "standard" (an admittedly sloppy term I intended to mean
"whole number") division.
So, rather than I&M, I prefer to see the terms divided straight across
because it is similar to multiplying straight across. In this case, 1 3/4
(aka 7/4) divided by 1/2 gives [7/1] / [4/2] = 7/2, or 3 1/2. Obviously, if
you want a simple fraction, then it's best to use common denominators for
the general case so that the denominator of the resultant answer is "1".
Students will see, as they try drawing, or using manipulatives, to solve
various division problems, that it is much easier if the pieces are the
same size.
<snip>
>On Thu, 11 May 2000, Wayne Bishop wrote:
>
> > Looking ahead to algebra competence, as I hope we're all doing, would you
> > please explain this idea so that it would make sense to someone who is
> > learning
> > to divide x/(x - 1) by x/(x^2 - 1)? Thanks in advance,
> >
> > Wayne.
> > ---------------------------------------------------
> >
Lou answered this quite well and, as I said, I was only addressing the
fraction problem and hand. However, here's another way: To figure this out
it really helps if we deal with the same size pieces, aka common
denominators. We see that (x^2 - 1) is (x - 1)(x + 1). Since it looks like
this is the LCM of the two denominators, we will go ahead and find the
equivalent fraction for the first term with the LCM as the denominator,
then divide straight across. We end up with
[x(x + 1)/(x^2 - 1)] / [x/(x^2 - 1)] = [x(x + 1)/x] / [(x^2 - 1)/(x^2 - 1)]
= x + 1
It's a little longer than I&M, but it does solve the problem we asked and
not a slightly different one.
Derrel
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