[MathEd] Critique: Problematising Mathematics Education

In Response to Article
The Politics of Math Education


My Comments

1. It is good to problematise mathematics education.  Certainly politics is involved in the choice of mathematics curricula.

2. However, is it good to argue and debate so much that nothing gets done?  Are you chasing down false dichotomies?  Are you assuming that you cannot have it all?

3. In Singapore, we do not argue so much and students go on to perform well in "mathematics".  Unfortunely, I feel, they get a very narrow and distorted view of what mathematics really is.

4. What is the answer the dilemma?  I think it's a question of identity.  I think students should make well-informed negotiated decision about the kinds of people they want to be and how a wholistic mathematics education serves to develop them not just in terms of skills and content, but also in terms of values, habits / dispositions, problem solving ability and critical thinking, ... etc.  I would be interested to learn of and even work with curriculum planners heading in this direction.

[NumTh Expository] Divisibility Tests for 7 and other digits

Introduction

     Let  n = ...ABCDEFG = ... + A´10⁶ + B´10⁵ + C´10⁴ + D´10³ + E´10² + F´10 + G. 

I invent the notation  ‘n ¸ d’  to mean  “n  is divisible by  d”  (as integers).  For a definition

n ¸ d   Û   d | n   Û   n = kd  for some integer  k

Many of you may have heard of the following divisibility tests:-

n ¸ 1   is always true   n ¸ 2   Û   the last digit  G ¸ 2   Û   the last digit  G Î {0, 2, 4, 6, 8}   n ¸ 3   Û   the sum of digits  ... A+B+C+D+E+F +G ¸ 3   n ¸ 4   Û   the last digits  FG ¸ 4   n ¸ 5   Û   the last digit  G ¸ 5   Û   the last digit  G Î {0, 5}   n ¸ 6   Û   n ¸ 2  and  n ¸ 3  (see above)   n ¸ 8   Û   the last digits  EFG ¸ 8   n ¸ 9   Û   the sum of digits  ... A+B+C+D+E+F+G ¸ 9 n ¸ 10   Û   the last digit  G = 0

What about divisibility by 7?

Modulo Arithmetic

     Divisibility tests are all based on modular arithmetic (a.k.a. clock arithmetic), as part of number theory.  This arithmetic was developed initally by a very clever mathematician

called Carl Friedrich Gauss.  The key idea is that numbers can be grouped into separate classes.  For example, when considering division by  3,  we can split the integers into three equivalence classes called residue classes

     [0] = { ..., -3, 0, 3, 6,   9, 12, ... }

     [1] = { ..., -2, 1, 4, 7, 10, 13, ... }

     [2] = { ..., -1, 2, 5, 8, 11, 14, ... }

The members of each class have the same remainder when divided by  3.  For example 

     10 ¸ 3 = 3r1,  7 ¸ 3 = 2r1,  4 ¸ 3 = 1r1  (all the remainders are equal to 1).

So  10, 7, 4, 1 ... etc are all members of the class  [1]  represented by  1.  Members of the same class are called congruent.  For example  10 º 4 (mod 3),  read as “10 is congruent to  4  modulo 3” and it means  10  and  4  have the same remainder when divided by  3.

a º b (mod m)   Û   (a – b) ¸ m   Û   a = b + km  for some integer  k

As you know,  “a ¸ m  gives a remainder of zero”  means  “a  is divisible by  m”.  Hence

a º 0 (mod m)   Û   a ¸ m

Congruency (for a given modulo m) is a type of equivalence relation.  We have these laws

Reflexivity:   For every integer a,  a º a Symmetry:    If  a º b,  then  b º a. Transitivity: If  a º b  and  b º c,  then  a º c.

These properties are similar to ‘=’.  We also have some very neat arithmetical laws.

Suppose  a º b (mod m)  and  c º d (mod m).  Then      a + c º b + d  (mod m)      a – c º b – d  (mod m)          ac º bd      (mod m)

For a given number  n = ...ABCDEFG,  we can write  n = 1000x + y   where  x = ...ABCD,  y = EFG.  With the above laws, modular arithmetic helps simplify calculations and provide powerful insights.  For example, in modulo 8,

         10 º 2,  100 º 10² º 2² º 4,  1000 º 10 ´ 10²  º 2 ´ 4 º 0.  Hence

         n = 1000x + y  º  0×x + y  º  y.

This explains the divisibility test by  8:  To see if a number is divisible by  8, we just need to look at the last  3  digits  y = EFG.

Test for divisibility by 7

     Using modulo 7,  10 º 3,  100 º 3² º 2,  1000 º 3 ´ 2 º 6 º -1.

     n = 1000x + y º  -1×x + y º  y – x.  Thus

     n ¸ 7   Û   n º y – x º 0   Û   x º y   Û    x – y ¸ 7.  We have the rule

n ¸ 7   Û   ...ABCD º EFG (mod 7)

This means that a number is divisible  7  if (and only if) after breaking off the last three digits, both resulting pieces have the same remainder when divided by  7.

Example 1

       For 12 516:  12 º 5  and  516 º 5 (mod 7).  So  12 516 ¸ 7.

Example 2

       For 654 321:  654 º 3  and  321 º 6 (mod 7).  So  654 321 not ¸ 7.

Example 3

       For 1 494 787:  787 º 3 º 1494 (mod 7).  So  1 494 787 ¸ 7.

       Alternatively, 1 494 – 787 = 707 ¸ 7.  So  1 494 787 ¸ 7.

Works like magic, doesn’t it?

[OlympiadGRN20150406] Grid Ratio Ninja

Perhaps you are an expert at Fruit Ninja, slashing fruits mercilessly with precision using your much-feared finger.  Now can you do the mathematical equivalent below?

Question

Solution

     We would be getting nowhere fast if we took a random approach.  We are only allowed two slashes.  Let us look at the question for more clues.  We have a  5  by  6  grid, giving a total of  30  cells.  The total number of parts in the ratio is  1 + 2 + 3 + 4 = 10.  10  parts correspond to  30 cells,  so each part is  3  cells.  So the number of cells in each portion are  3,  6,  9  and  12.

     Note that if we pair up  3  with  12  and  6  with  9,  we have  15 cells each.  We can just slash the grid down the middle to achieve this.

     We now need ratios of  1 : 4  and  2 : 3  for the remaining parts.  We certainly can achieve this if we were allowed two more slashes and if we used horizontal lines, like this:

However, we only have one more slash.  Can we make this slash do the work of the two horizonal slashes?  Yes!  Replace the two horizontal slashes with a single slanting slash that passes through the mid-points of the horizontal slashes.  This gives four trapezia.  The reason why it works is because for each portion, an extra triangular area is added but it is offset by the removal of a triangle of the same area, like this:-

We can check that the ratios of the areas of the portions are as required.  Done!

Heuristics Used

H02. Use a diagram / model

H04. Look for pattern(s)

H05. Work backwards

H08. Make suppositions

H09. Restate the problem in another way

H11. Solve part of the problem

[OlympiadXPH20150405] Seats in a Hall as Pigeonholes?

Question

There are  25  rows of seats in a hall, each row having  30  seats.  If there are  680 people seated in the hall, at least how many rows have an equal number of people each?

Introduction

     This is a mathematics olympiad type of question for primary / elementary school, I believe.  Let us make sure we understand the question.  Each row of seats has a certain number of people, which I shall call the “headcount”.  We want all the headcounts to be as different as possible, and yet we do not want so many rows to have the same headcount.  The number of rows with the same headcount should be kept as low as possible.  How low is low?

Solution 1

     One strategy to solve this is to start filling up the empty seats with as many as possible.  Indeed this is the approach taken by the official “model answer”, which for copyright reasons I cannot show.  However I am going to do something even better: to explain the solution visually.  Recall that the sum of an arithmetic progression is 

                   ½ ´ number of terms ´ (first term + last term)

     After filling up the empty seats with different numbers of people, we would have filled up  30 + 29 + ... + 7 + 6 = ½ ´ 25 ´ (30 + 6) = 450  seats.  This is stage 1, shown in orangey-yellow in the diagram below.  We have now  230  people remaining to be seated. 

     For stage 2, we try to fill in the remaining seats with as many people as possible but keeping the headcounts all different.  So we ramp up  6 to 30,  7 to 29,  8 to 28, ... etc.  The additional number filled up is  24 + 22 + ... + 2 = ½ ´ 12 ´ (24 + 2) = 156.  This is stage 2, shown in green.  We have  74  seats remaining.

     For stage 3, we ramp up  18 to 30,  19 to 29,  20 to 28, ... etc.  The additional number filled up is  12 + 10 + ... + 2 = ½ ´ 6 ´ (12 + 2) = 42.  This is stage 3, shown in blue.  We have  32  seats remaining.

     For stage 4, we ramp up  19 to 30,  20 to 29,  21 to 28, ... etc.  The additional number filled up is  11 + 9 + 7 + 5 = 32.  We are done.  This is the final stage, shown in pink. 

Discussion

     It turns out that there is another way to look at the problem.  The above solution can be depicted in a “ball and bin diagram” as shown below.

There are  25  balls in the diagram, each representing a row’s headcount.  There are  4  rows with 30 people,  4 rows with 29 people, ...,  3 rows with 26 people,  3 rows with 25 people,  2 rows with 24 people, and  1 rows with 26 people.  We can actually calculate the totals for rows with repeated headcounts.  For example in the diagram, there are  4  layers of balls with each layer representing 30 + 29 + 28 + 27 (shown in dark turquoise) and the sum is = 4 ´ [½ ´ 4 ´ (30 + 27) ] = 456.  Those with rows with three of the same headcounts total up to  153  (shown in pink).  The rows with two of the same headcounts total up to  48  (shown in yellowish green).  There is one and only one row with  23  (shown in dirty green).  All this give a grand total of  680.  We can calculate the grand total  for every ball and bin diagram in this fashion.  We want a grand total of  680 and there must be 25 balls.  Using this diagram and a generalised version of the Pigeonhole Principle, we can have a very short and sweet solution.  Before that, let me briefly explain the Pigeonhole Principle.

Pigeonhole Principle

     Let us say there are 30 pigeonholes and 31 envelopes.  We can represent this with a ball and bin diagram using 30 columns (for pigeonholes) and 31 balls (representing the envelopes).  If you try to put one envelope into each pigeonhole it is impossible.  One of the pigeonholes will have two envelopes.  In the ball and bin diagram, there will be at least column with two balls.  If you try to put everything flat to one layer, that is 30 balls and you still have one ball left.  You will have to put this remaining ball somewhere on the second layer.  This illustrates the Pigeonhole Principle.

     Likewise if you have  91  balls and  30 columns, there must be one column with  4  balls.  If it is all  3  balls, that fills with only 90 balls.  Your remaining ball has to go somewhere on the fourth layer.  This illustrates the Extended (or Generalised) Pigeonhole Principle.

Solution 2

     I am going to further extend the Pigeonhole Principle to Pigeonholes with Valuations (with some number attached to each configuration e.g. grand total).  We have 25 balls and 30 bins (columns) and we want to fill up the grand total number as quickly as possible achieving a grand total of  680, using as few layers as possible.

     Starting from  30  downwards and, using  3  layers,  we have  8  balls for each layer, filling all the colums for  30,  29,  ... ,  23.  We now have used up  3 ´ 8 = 24 balls that represents a grand total of  3 ´ [½ ´ 8 ´ (30 + 23) ] = 636.  There are 44 seats left over, but only one remaining ball.  The best we can do is to put the remaining ball at  22  for one layer.  That leaves  22 unallocated seats.  So  3  layers are not enough.  We need  4  layers.

     To show that  4  layers are enough, we can imagine filling up  4  layers of  30 + 29 + ... + 25.  That gives a grand total of  4 ´ [½ ´ 6 ´ (30 + 25) ] = 660,  with  20 left over.  This can be handled by putting the remaining ball on  20  for one layer.  And we are done.  The latter configuration is an alternative configuration to the one given in the model answer.  But most importantly, it works.  We have shown that at there must be least  4   rows with the same number of people each.  

Note

1.  There can be more than one set of  4  rows with the same number of people each, but we just need to show that there exists (one set or more of)  4   rows with the same headcounts.

2.  One can imagine configurations that have  5  rows with the same number each, but it would not be “least”.

3.  To answer the question in the title of this article, the seats are not the pigeonholes.  Rather, the pigeonholes (or bins, or columns) represent possible numbers of people in a row.  Each ball in bin  j  represents a row that has  j  people.

Heuristics Used

H02. Use a diagram / model

H03. Make a systematic list

H04. Look for pattern(s)

H05. Work backwards

H06. Use before-after concept

H07. Use guess and check

H08. Make suppositions

H09. Restate the problem in another way

H11. Solve part of the problem

[Pri20150330FBT] Open-ended Question on #Fractions

Question

Give a fraction with its value in between 2/5 and 1/2.  (Note: the denominator cannot be greater than 20)

Discussion

This is an open-ended question for primary (elementary) school, which ought to be easy because there are many possible answers and you just need to supply one that meets the requirements.  However, most people are used to closed-ended questions, which have only one correct answer.  This is what prompted a parent to pose this question on a parent-support forum on Facebook. 

Many people, including parents and even some private tutors, began to supply their answers to this question.  Much of the discussion clustered around the idea of the mid-point or the average of the two given numbers, namely ¹/₂ (²/₅ + ¹/₂) or  ⁹/₂₀, and why that gives an answer.

Then Dr Kho Tek Hong (retired curriculum specialist, the “father” of Singapore mathematics) chipped in.  He said that there are many possible answers e.g. ³/₇, ⁴/₉, ⁵/₁₁, ⁵/₁₂, ⁶/₁₃, etc.  He suggests that students be asked to justify their answers.  [ By the way, here is an interesting method to compare fractions. ]

And I thought that was a stroke of educational brilliance from The Guru!  It shows the spirit of the Singapore mathematics curriculum – to get students to think, to be open-minded, to use logic and to be able to communicate mathematically.  The Singapore curriculum is not meant to torture students, nor to get them to toil around a tortuous path around a high mountain seeking the elusive holy grail – although it often seems that way.  Teachers, tutors and parents would do well to help the learners it in the right spirit.  Mathematics is not always about calculations (although you need to do some calculation).  We also need to open our minds to multiple answers as well as to embrace various methods and concepts.

[Pri20150330CPF] How to #Compare #Fractions

Comparing Positive Fractions

     How do you compare positive fractions, which are taught in primary (elementary) school?  For example, which is bigger:  ⁵/₆  or  ³/₄ ?

The “orthodox” method is to put them both to a common denominator.  The Lowest Common Multiple (LCM) of  6  and  4  is  12.  Multiplying the left fraction by  ²/₂  and the right fraction with  ³/₃  gives, respectively,  ¹⁰/₁₂  and  ⁹/₁₂.

Since  ¹⁰/₁₂  >  ⁹/₁₂,  we conclude that  ⁵/₆  >  ³/₄.

Another Method

     Here is a “short-cut” that I learned from a schoolmate in primary school.  Basically you “cross-multiply”: multiply the left numerator with the right denominator, and multiply the right numerator with the left denominator, and then compare the products so formed.  That will give you the correct inequality or equality sign (viz. ‘<’, ‘=’ or ‘>’).

As we can see, since  20 > 18,  we conclude that  ⁵/₆  >  ³/₄. 

     Does this method work?  Yes, definitely.  You can try it out with a few pairs of fractions and you can see for yourself that it is so.  Is this method legit?  Why does it work?    I give a formal proof of the method below.

Further Discussion

     Use the above method with care.  Some school teachers may not accept the method not because it is not correct, but it sounds “dubious” to them because they have not heard of it or they are not able to prove it for themselves.  Pupils can use this short-cut to give them a quick look-ahead to certain questions, and as a back-up to check their answer after using the Lowest Common Denominator method.  In questions that ask pupils to arrange a few fractions in ascending and descending order, this “crossing method” may give some speed advantage if done carefully.

     In primary school, pupils focus on positive fractions.  In secondary school, negative numbers and fractions are introduced.  Does the above trick work for negative fractions?  Was my theorem and proof above carefully phrased enough to cover the negative fractions? 

     Note that this method works for comparing two fractions at a time only.  Sometimes this cross-multiplying gives rather big numbers.  In that case, it is better to multiply each fractions by the LCM of their denominators.  Essentially this is the same as the orthodox method, except that we do not write the denominators.  Can the above proof be extended to cover this new short-cut?  What do you think?

[Pri20150303VSA] A way to Visualise Arithmetic Progressions

Question 

What is the sum of 25 + 26 + 27 + ......... + 189 ?

Introduction

     This is a problem meant to stretch the minds of Singapore primary (elementary) school pupils.  Adults (e.g. parents, teachers and tutors) trying to help out usually recommend doing this using the “rainbow” method (where a bunch of arcs are drawn joining 25 and 189, 26 and 188, 27 and 187 ... and so on, forming something that looks like rainbow), or the sum of arithmetic progression formula

          ½ ´ number of terms ´ (first term + last term)

which is usually taught at the junior college (pre-university) level.  Although correct, do the learners understand the logic behind them?

A Visual Method

     In my visual representation below, the answer pops out almost immediately and the reasoning is made apparent to the student.

     Imagine a series of vertical bars representing 25, 26, 27 ... up to 189 joined together forming a staircase (shown in orange).  Make a copy of this (shown in blue) and flip it around and join the two shapes together to form a rectangle.  Note that the width of this rectangle represents the number of terms 189 – 25 + 1 = 165, while the uniform height is in fact first term + last term = 189 + 25 = 214.  Taking the “area” of the rectangle and dividing by two, we get  17 655, an answer that would agree with those found using the previously mentioned methods.  The diagram above is a kind of “proof without words”.

[Pri20150308CSR] Charitable Savings Ratios?

Question

Introduction

     Here is another one of those Singapore Mathematics problems that are two-variable simultaneous equations in disguise.  The key to solving this question quickly is to exploit the fact that the amount of donations are the same in this case.

Solution

     First, read the question and translate the information into a diagram [H02. Use a diagram / model].  I use different shapes (e.g. circle and square) to envelop the different types of units.

     Since we have ‘-80’  for both Sharon and Ryan [H04. Look for pattern(s)], we may deduce that 1 ‘circle’ unit  (5 minus 4 ‘circle’ units) is equal to 3 ‘square’ units (10 minus 7 ‘square’ units).  That means 5 ‘circle’ units is 15 ‘square units’.  [ H10. Simplify the problem]

     By comparison again, we realise that 5 square units is 80 [H05. Work backwards] and hence 15 ‘square’ units is 240.

Ans: Sharon’s savings was $240 at first.

Check

Before    $240   $192      5 : 4

After      $160   $112     10 : 7

Solution makes sense.

Heuristics Used

H02. Use a diagram / model

H04. Look for pattern(s)

H05. Work backwards

H10. Simplify the problem