Just to reiterate the idea I was trying to bring across today in response to a couple of posts. I'm on the same page as you all on the constraints in the real classrooms. It's important to try and balance both teaching approaches in our classrooms. Let our students at least experience once the meaning making or sense making of certain topics. After that we should have all the creative ways to let our students learn the various procedure in solving Math problems and examination type of questions efficiently.
BTW, some of you are asking about how to access the resources in the tutorials. You can access them here, bear with me as it's still work in progress.
My reflections @ 16 Sep (Part 1 of lesson). Please give me your inputs so i can refine the explanations. Thank you very much.
ReplyDeleteLCM of 2 numbers is multiple of HCF and unique factors of 2 numbers.
this only works with 2 numbers: LCM * HCF = n * m
To begin, i think we need to explicitly say we are going to find common multiples. This is as opposed to finding common factors.
So students should now be reminded common factors are different from common multiples.
There may be confusion about the ladder method (when to stop on ladder) as it is used for both HCF and LCM.
We can try to use link between ladder method and factor tree to explain procedure of LCM.
Explanation 1 for LCM
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To make a product in factory, we need parts.
To make a product in maths, we need factors.
if we make a product out of 4 and 6, then we take parts from 4 and 6.
remind students, 4 is made up of product of 2 and 2
draw venn diagram for 4
remind students, 6 is made up of product of 2 and 3.
draw venn diagram for 6, overlapping at the common factor of 4 and 6.
now we can see that this combination of 4 and 6 is the smallest and also shares parts/factors from 4 and 6.
Activity with students
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Get them to fill up table with columns
a, b, HCF(a,b), LCM(a,b), a*b, HCF(a,b)*LCM(a,b)
ask them what they discovered.
consider two simple positive integers.
what is HCF(4,6) and LCM(4,6).
when we multiply HCF(4,6)*LCM(4,6),
what are we actually getting?
explanation 2
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the factors or parts of 4 and 6 are given by:
4 = 2 x 2 and 6 = 2 x 3
these parts of 4 and 6, we take to be building blocks for 4 and 6.
For the smallest product of 4 and 6, we compare the parts/factors of 4 and 6.
when we compare the factors on the trees,
we see factors pair (2) and (2) are same. so we only take one 2.
next what is left, factors (2) and (3). they are different => they are prime factors. so we take these parts (2) and (3).
LCM = 2 * 3 * 2
Explanation 3 for LCM
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i.e. we want a big cake, we take more eggs and more sugar, more flour.
when we want small cake, maybe 1 egg, a bit of sugar and flour
factors of 4 = 2 x 2
and 6 = 2 x 3.
show this in factor trees to illustrate the products made up of parts/factors.
Now, we first take common factor of 2 numbers.
After we take common factor, what else is left: 3 from 6 and 2 from 4.
To make sure the final product looks like both 4 and 6. we need unique parts of 4 and 6 also.
i.e. we are the children of our parents. we may have our father's nose and our mother's eyes.
so to make smallest product, we took the common part/factor (2),
now we take the remaining parts, (3) and (2).
Using the factor tree, we continue to cancel out the 3 as we put it on the ladder.
next, we cancel the other 2 on the tree for 4, as we put it on the ladder.
this will give us the smallest product combining the common part of 4 and 6 and the unique part of 4 and 6.
Using an example, 4 = 2 x 2; 6 = 3 x 2
We take common factor of 4 and 6, which is 2, and put it on outside of the ladder.
We are now left 3 and 2, factors of 4 and 6.
S0, the LCM is 2 * 3 * 2
Explanation 4
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4 * 6 = 24.
this is a multiple/product, but is it the smallest product we can make from parts/factors of 4 and 6?
4 * 6 = 2 * 2 * 3 * 2
the product we want must also be a product of 4, it must also be a product of 6.
so we take 2 (common factor), cancelling 2 on the factor trees of 4 and 6.
now left with prime factors (making sure the product can belong to the family of multiples of 4 and also 6),
we take the remaining factors 2 and 3 and cancel from the trees.