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From: Pat Ballew <firstname.lastname@example.org> To: Teacher2Teacher Public Discussion Date: 2000060308:56:51 Subject: Dividing Fractions Disclaimer: I am not an elementary teacher, and my views of teaching elementary math are formed from working with some very good elementary teachers at times, both in the US and Japan, and the responses of students in a few classrooms where I have taught demonstration lessons. All in all a very limited experience. That being clear, I would still like to offer a comment here. It seems that dividing a fraction by a fraction is made more difficult by american teachers and students because they almost always perceive division as a partitioning process (how many of those in this) rather than a missing product model (what must that be multiplied by to get this). It seems when I talk to teachers in the Orient they are much more likely to look at both models for examples. When trying to explain the "flip and multiply" algorithm, it seems that the missing product model is much more direct. If that is not clear, an example will perhaps help. The division of 3/4 by 1/2 can be written as a missing product equation 1/2 * ? = 3/4 . If students are good at multiplying fractions, they will, I would think, be much less likely to get a really bad answer for this (I think 3/8 is the most common wrong answer by students struggling to understand division by fractions, when students freqently multiply by 1/2 instead of dividing). It may be that no algorithm other than guess and test is really necessary for this... Imagine the most difficult case, all prime values for numerators and denominators of both cases... (2/3) / (5/7) First it is rewritten as a missing product 5 ? 2 --- * ----- = ----- 7 ? 3 I would think any student who has learned to pre-cancel (I'm not sure what word is in vogue for this process in elementary schools today, so forgive me if the language is, at times, arcane) then she should see that We need a 7 in the numerator to get rid of the other 7, and a 2 to create a 2 in the product, so the quotient must have 14 in the numerator... and... we need a 5 in the denominator to get rid of the other five and a 3 to make a three in the product, so we need 15 in the denominator, giving us a quotient of 14/15.
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