Teaching maths so it counts in real life

- Group B

As we have seen how we could prove that the sum of the interior angles of a regular polygon would sum up to (n-2) x 180 degrees based on the information that the sum of exterior angles always sums up to 360 degrees.

We thought it would be nice to share what an alternative proof (hopefully it holds for all cases). This alternative involves cutting any n-sided polygons into triangles with a vertex of the polygon arbitrarily chosen as a common vertex for all the triangles to be formed. Examples of 4-sided, 6-sided and 8-sided polygons are shown in Figure 1 below.

Figure 1

As we observe from these “cut” polygons, it is obvious that the resulting number of triangles formed is always 2 less than the number of vertices initially present, i.e., for a n-sided polygon, there will be (n-2) triangles being formed.

Tapping on the prior knowledge that the sum of the interior angles of a triangle is 180 degrees, the sum of interior angles of a n-sided polygon would thus be (n-2) x 180.

Group A

Pictures on Algetiles

Dear all,

From trying to brainstorm for a consistent approach using algetiles...

To our brilliant hypotheses!

Group D

Group A

Group C

Group E? F?

Group B?

Blurred shot...

Group A

And our ever photo-ready Liting. =)

It was fun! Looking forward to more! =)

God Bless!
Regards,
Daniel

Group Reflection on Expositional Teaching

It had been an interesting experience to be involved in such an activity, be it pretending to be a student or a teacher in the class.  Throughout this entire week, everyone of us had benefited a lot from each other in a way or another. This could be seen in two ways. We were able to know the common mistakes which we might made when teaching the class. For instance, the expression of –b(a+c). Many of us would have told the class that it is “minus b” instead of “negative b”. Without this expositional teaching, probably we would not have realised it.

Furthermore, we are able to learn the various approaches used to bring the different concepts of algebra across to students. Be it through the use of whiteboard or powerpoint or even the way that we try to engage the students. All these are good points that we are able to learn from one another. Don’t you agree?

Working as a group is different from working in a group in order to reflect the links within a common overall approach adopted by the whole team. For instance, some of us might want to call the distributive law as umbrella method while others might want to call it the rainbow method.  In order for the whole team to be able to show consistency in their teaching, the team members have to communicate well and come to a conclusion on the way which they would like to present to the class.  If we are just working in a group, we might only just come together and fix our “jigsaw” together to present instead of sharing personal views to result in the group to show consistency in the way to teach the class. As such, we guess this expositional teaching had introduced an element of collaborative learning. On a good note, all of the 6 groups had achieved that! Great job guys!

Just to summarise some points which we might want to take note for TP..

1.   Negative vs minus sign. Example: -2a(a+b) should be seen as negative 2a instead of minus 2a

2.   Always have to follow through with checking of solution. For instance, the part on solving simultaneous equation.

3.   Have to take note of the writing of  on whiteboard or even the use of powerpoint to prevent confusion with the multiplication x.

 And the list goes on... ....

- Group F

Formula for volume of a pyramid

It was really interesting to see how 3 pyramids could be put together to form a cube. We thought that the visual aid of seeing the three pyramids being combined to form the cube was a very good tool to help students understand why the formula for finding the volume of a pyramid is:
(1/3) x base x height of the pyramid.

There is another way to show the same result. This approach involves dividing a cube into 6 square pyramids. How we do this is to draw a line from every vertex to the vertex diagonally opposite. Connecting the four vertexes together, you will get 6 square pyramids as show in the picture below.

As you can see the vertex of the pyramids converge at the center of the cube. Thus we know that the height of the pyramid is half that of the height of the cube. On other words:
height of cube = 2 x height of square pyramid
base of the cube = base of the  square pyramid

Based on the information above we can arrive at this expression:

Volume of square pyramid 

= (length of cube x breadth of cube x height of cube) ÷ 6

= [length of cube x breadth of cube x (2 x height of square pyramid)] ÷ 6

= (1/3) x base of cube x height of pyramid

= (1/3) x base of square pyramid x height of square pyramid 

This can be easily seen in the animation below.

Cheers,

Group F

Approximation: An important concept

I am sure many of you are pretty amazed by the proof that we saw on the formula for surface area of sphere. This approximation approach proposed by the famous Mathematician Archimedes is simply ingenious. This concept of approximation or estimation has far-reaching application besides the surface area of a sphere. Approximation is used in  math, science, economics, finance and innumerable subject areas to solve problems. Consequently, it's a concept that can be applied in scientific research and problem-based learning to make the study of a complex problem simpler using simple process or model. Many of our real-life problems are solved through simplified models derived from approximation. One such famous model is the supply and demand model in economics, this model is so valuable in the study of both micro and macro economics! 







This is just one example that illustrates the usefulness of solving a problem using models. Essentially, solving problems through models is an important skill that students need to acquire in school to enable to solve problems efficiently. Also, It will be beneficial for students to see the applications of math. It is crucial that we do not reduced math into something that is impractical to the real world that we live in. To provide extension in learning is one way to stir the interest and motivation of students in the subject.  This opinion is an extension of what I have learn through a series of reflection. Enjoy and feel free to post your sentiments and views with regards to this post :) 


Cheers,


Wei Chuan (Matthias)

The Cavalieri's Principle

Although one could easily understand the logic as expressed by the wikipedia’s coin analogy of the principle, our group actually faced difficulties in trying to understand the principle analytically from the theory statements.

Initially, we had the interpretation that as long as we have 2 regions bounded between two parallel lines, any other line parallel to these lines would cut the regions in equal lengths and the area would subsequently be the same, as show in Figure 1 below. (i.e. EF = GH, and the purple and orange areas would be the same).

Figure 1

Subsequently, we thought of several counter examples (one of which shown in Figure 2) that this would not be the case. By having an oval stretched horizontally, we see that the shape is still bounded by the two parallel lines, but the observation of AB = CD no longer holds, and it is obvious that the areas of these two figures are definitely not the same.

Figure 2

Next, we also observed that the areas of two regions bounded by the two parallel lines could be identical without having the two lengths being the same – as shown in Figure 3.

Figure 3

Conclusively, based on these observations, we decided that the restrictions for the Cavalieri’s principles are:

• The shape need not be the same 
• Individual Lengths need not be the same
• But the sum of the lengths need to be the same = > area will be the same

In general, it would be easier to visualise the principle as a form of rearrangement of the basis set and,

1) If the sum of lengths are the same => the area will be the same. (2D case)

2) If the sum of areas are the same => the volume will be the same. (3D case)

As such, after much struggle in understanding the point of the principle, we decided that the visualising it with a rope works best for our group. A long rope can be cut into several pieces of shorter ropes and no matter how it is rearranged, the sum of lengths would be the same => resulting in the sum of areas being the same (2D) => the volume being the same. (3D)

Figure 4

Through this post, we hope that it will reaffirm the understanding of the principle.

Group A

Volume of Sphere by Cavalieri's Principle

Hello everyone,

Here is the proof that I shared in class just now. In a much clearer sense. Hope it helps =)
I recommend saving the pictures below and reading in actual size.

Link on video to find surface area of sphere

Hi everyone

Attached is the link to the video shown in class today,
http://www.rkm.com.au/animations/animation-Sphere-Surface-Area-Derivation.html

Huanlong

Volume of Hemisphere equals volume of doughnut Qn.

Prem -
Hee folks - i came across this to see if it helps.


This proof was known to ancient greeks and does not involve calculus or integration.

Take a hemisphere.

Surround it by a cylinder of the same radius as the hemisphere, and the same height as the height of the hemisphere. We assume you know the volume of this cylinder: volume is area of the base multiplied by height. Note that the height is the same as the radius of the base:


Take an upside down right circular cone in the cylinder. The 'base' of the cone will be at the top of the cylinder, and the point at the bottom will be at the center of the hemisphere. The volume of a cone is .

Proposition

On any horizontal slice of this configuration, the area of the cross section of the hemisphere
is the DIFFERENCE between the area of the cross section of the cylinder MINUS the area of the cross section of the inverted cone.

Proof of the proposition

Suppose that we made out "slice" at certain height h.
Radius of the slice of hemisphere would be . The area will be, therefore, .
Area of the slice of the cylinder would be , since the cylinder has radius r regardless of where we make the slice.
Radius of the cone would be h, so the area of the slice would be .
Subtracting the area of the cone slice, from area of the cylinder slice, we get

Since is the same as the formula for the area of the hemisphere slice, the proposition is proven.

Principle of volume

If two solids have cross sections of equal area for all horizontal slices, then the have the same volume.

Final result

Therefore,
The volume of the sphere is twice that, .

Some ICT Tools

Hi there everyone, 

Here are some ICT tools you all might wanna explore for lesson, teaching & also learning. 

http://www.geogebra.org/cms/

http://vyew.com/s/

http://sketchup.google.com/

http://maps.google.com.sg/

http://www.mindmeister.com/

http://www.mindomo.com/

http://info.mindjet.com/MindManagerB.html?cmpg=APAC_-_Google_AS_Mindjet&gclid=CJGx6dz9saoCFcwa6wod1R5E7g

http://www.wallwisher.com/
http://en.linoit.com/
http://prezi.com/

They are all FREE Sites!!! 

I personally had used google sketchup and geogebra for math teaching during my ESE. It is really fun and there are many ways we can use it. 

So yup, do give them a try if you have not tried them before. =) 


Cheers 
Suvie =) 

YES! We are INTERESTED(:

Out of curiosity, my group tried the LCM question which was given by Shi Wei. We tried both methods and we had come to a similar conclusion as what she gets! 

Hence, we conclude that we should use prime factors to do the ladder method. This will be a good example for teachers to show that one should not use composite factors while doing the ladder method.

(Group F)

Hi everyone!

Looking at the prime factorisation method for solving HCF and LCM, we thought of a question that would require us to work backwards to find a number N of a pair, given the HCF and LCM. Here’s an example and solution that we would like to share.

Example: Given that 252 and a number N have a LCM of 1260 and HCF of 36, find the number N.

Method:

Express 252, 1260 and 36 as a product of prime factors

We can also derive the answer using our famous Dina's Venn Diagram

We feel that this example will be helpful for students to have an overall understanding of HCF and LCM.

Hopefully, it is a good take away from us.

Cheers,

Subashini (Group F)

Caution! Beware of how you use Ladder method when finding LCM

Hi everyone!

My group basically had two questions in mind:

(1) Can we divide by composite numbers when using the Ladder method?

(2) Will the order of division affect the final answer for LCM when using the Ladder method.

For efficiency sake, some of us will do a ‘short cut’ when doing the Ladder method by dividing composite number that is common to at least 2 of the numbers instead of dividing by prime numbers. This is a mentality that is common to students as well.

However, my group vaguely remembered that doing the sums this way will result in an answer bigger than the LCM possible. Thus this led us wonder if we can divide using composite numbers in Ladder method and that perhaps the order of division is important when using the Ladder method. To answer our hypothesis, we begin to dig through the different sums and here is an example we found!

Find the LCM of 60, 75 and 100.

We learnt that when finding LCM using the ladder method, we have to divide it by a factor that is common to at least 2 of the numbers.

In this question, 5 is a factor common to all the three numbers. 10 is also a factor that is common to 2 of the numbers. Both division by 5 or 10 satisfy the procedure’s requirement as stated above. Hence, we should technically get the same answer if we begin by division of 5 or 10.

Method 1: Begin by dividing by a factor that is common to only 2 numbers

10	60, 75, 100
5	6  , 75, 10
3	6  , 15, 2
2	2  , 5  , 2
	1  , 5  , 1

Therefore, LCM = 10 x 5 x 3 x 2 x 5 = 1,500

Method 2: Begin by dividing by a factor that is common to all the factors

5	60, 75, 100
4	12, 15, 20
5	3  , 15, 5
3	3  , 3  , 1
	1  , 1  , 1

Therefore, LCM = 5 x 4 x 5 x 3 = 300

Comparing the answer of method 1 and 2, we see that we have multiplied an additional factor of 5 in method 1, resulting in a bigger answer.

Explaining the difference

Prime factorization method:

60   = 2 x 2 x 3 x 5

75   =             3 x 5 x 5

100 = 2 x 2 x       5 x 5

Therefore, LCM = 2 x 2 x 3 x 5 x 5= 300

We can view this prime factorization method as “matching” the factors to find all the different factors that makes up the 3 numbers (as demonstrated by the different colour coded numbers)

Let’s try to understand the reason for the difference in method 1 by examining the effects of our steps used on its prime factors.

 If we are using method 1, we begin by division of 10.

60   = 2 x 2 x 3 x 5

75   =             3 x 5    x 5

100 = 2 x 2 x       5   x 5

When we begin by division of 10, we are matching off factors 2 and 5 of 60 and 100 respectively. This ignores the factor of 5 that also exists in 75. When we do that, we have only matched off part of the prime factor 5 that is common to all the three numbers. Thus, this resulted in an additional factor of 5 being multiplied for the final answer.

Significance of finding

2 significances are surfaced from our above observations.

1. Order of division is important when applying ‘short cut’ of division by composite number

Suppose we are asked to find the LCM of P₁, P₂ P₃ …P_n, division by composite number will only work if the user starts by dividing by a factor that is common to all the n numbers. Only when there are no more factors common to all the n numbers should we divide by a factor that is common to (n-1) number and so forth.

Coming back to the example of LCM of 60, 75 and 100, this would mean that we should first divide by 5 instead of 10. Cause 5 is a factor that is common to all the numbers. The numbers will then be reduced to 12, 15 and 20. We know that there are no more factors that are common to all the numbers in this set, so we should divide by the next factor that is common to 2 of the numbers. Factors common to 2 of the numbers at this stage are 2, 4 and 5. Hence, we can divide using composite number 4 because all the other factors are also common to maximum of only 2 numbers.

5	60, 75, 100
4	12, 15, 20
5	3  , 15, 5
3	3  , 3  , 1
	1  , 1  , 1

In the end, this “short cut” requires more strategizing and may not be such an efficient method after all the planning required. This should only be recommended to students who are advance.

2. Order of division is not important when we divide using only prime numbers

Dividing by prime numbers is a sure way of getting to the right answer though sometimes it is a tedious process. Since all numbers greater than 1 are actually product of prime factors, division by prime numbers ensures that we don’t accidentally match of part of the prime factors that is also common to others (like the case that happens when we divide by composite numbers). This is the beauty of the Fundamental theorem of arithmetic.

This is a straight-forward fool proof method highly recommended to all!

After going one big round, now I finally understand why my teachers always emphasised on dividing with only prime numbers when using the Ladder method. -___-''

That is all from my group now!

Group A

P.S.: If you are interested, you can also try finding the LCM of 120, 180 and 24. If not done correctly, it will also result in an answer bigger. =)

Some thoughts...

Being educated in an era when teachers taught using sedentary methods and classroom teaching was very much teacher-centred. We relentlessly practise in hope to do well in the national exam. This was the past. In the PGDE tutorial, I feel that I am constantly challenged to think of the ‘whys’ of math instead of the ‘hows’ of math. Currently, we are exposed to fundamental questions which challenge our understanding of various topics. At times, I admit that I have difficulty in comprehending some of these questions but the group discussion has definitely helped me to overcome some of these problems. As I attend the tutorial, I begin to realise that the teaching of math require teachers to be able to appreciate math rather than view math as a standard set of procedures required to derive standard solutions. It is beneficial for trainee teachers because a strong understanding in the fundamentals is crucial for us to articulate concepts clearly to student. Also, the brainstorming of approaches for teaching various topics is particularly useful as we are able to see the pros and the cons of different approaches. In addition, we were able to identify how the different approaches appeal to different groups of students. I am looking forward to more of such stimulating exercises!

- Matthias Low Wei Chuan