Tuesday, March 31, 2015

[Pri20150330FBT] Open-ended Question on #Fractions

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


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?

Monday, March 30, 2015

[Pri20150303VSA] A way to Visualise Arithmetic Progressions

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

     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”.

Sunday, March 29, 2015

[AJC Promo 2012 Q5] Range of Composite Functions with Inverse

Diagram 1.  Whole Question

     This is a question taken from Anderson Junior College, one of Singapore’s above-average junior colleges (in terms of the calibre of student intake).  By now this college is reputed to set the most difficult examination questions in Singapore.  It seems that they are trying to give the top junior colleges a run for their money, so to speak.  The question is difficult because it really tests students’ understanding of the concepts.  If you do not understand what is happening, you would be totally lost – even your graphing calculator (GC), the student’s favorite psychological crutch, would not be of much help.
Diagram 2.  Part (i) of Question

     Part (i) of the question tests students’ understanding of 1-to-1 functions (a.k.a. one-one functions or injective functions).  A function is invertible if and only if the function is one-one.  [ For the current A level H2 syllabus, it is assumed that the codomain is always the same as the range, so there is no need to worry about survjectivity. ]  Using the GC to graph the function  f   and to obtain the local maximum point, one sees that the required domain is  -2 < x < 0.  
Ans:  k = -2.   [ I am using black for explanations and blue for written answers. ]
     The domain is highlighted in yellow in the diagram below.  If the yellow region were to extend to the left beyond this point, it would be possible for a horizontal line to cross two points in the yellow region.  That would make the function  f  not 1-to-1 and hence not invertible.
Diagram 3.  Graph of f

Diagram 4.  Part (ii) of Question

For part (ii), we recall that a composite function exists if and only if the range of the first function (read from right to left) is a subset of the domain of the second function.
We require
Range of g  Í  Domain of f-1
The domain of  f-1  (the second function) is actually the range of  f.  From the above diagram, the relevant part of  f(x)  goes from -¥  to  f(-2) = -2 + ln 4 = 2 ln 2 – 2 » -0.614.  The range of  g is everything from -1 downwards (see diagram below: imagine taking every possible point of  R, the domain of g on the x-axis and shooting them over to the y-axis).  We write
            Range of g     = (-¥, -1]
            Domain of f-1 = (-
¥, 2 ln 2 – 2]
Since Range of g Í Domain of f-1,  therefore  f-1g  exists.

Diagram 5.  Range of g

     Finding the range of composite functions is something that many students have difficulty with.  There are two methods: the direct method and the two-step method.  The direct method is usually difficult or infeasible.  In this case, finding range from the graph of  y = f-1g(x) is practically impossible, because there is no simple formula for  f-1.   
The two-step method:
Step 1.  Find the range of the first function.
Step 2.  Transfer this range to the x-axis of the graph of the second
             function and map every point therein over to the y-axis.

     Step 1 has been done already.  We have found that  Range of g = (-¥, -1] . 
Diagram 6.  The Two-step Method

     For step 2, although the formula for  f-1  is impossible to find (it’s a pretty nasty question, isn’t it?), we know that this graph is a reflection of the graph of  y = f(x)  (shown in blue on the diagram on the right) in the line  y = x, and we can sketch this (shown in red on the diagram on the right).  Now transfer the range of g from the y-axis of your first diagram over to the x-axis of this diagram on the right (shown in green).  Now imagine taking every possible point of this set and mapping it over to the y-axis of the second graph.  The problem is: how to find  f-1(-1)  when we don’t even know the formula for f-1?  (really evil problem, isn’t it?).  One way to deal with this is to make an educated guess for  f(what?) = -1. 
     Notice that f(-1) = -1.  Therefore  f-1(-1) = -1.
     What if your intuition really sucks and you cannot make a guess?  The GC can come to your rescue.  Set up the graph of  y = f(x)  with the restricted domain and then intersect that with the graph of  y = -1.  The intesection is at  x = -1,  which means f(-1) = -1,  or  f-1(-1) = -1,  as above.
Diagram 7.  Using Intersection on the GC

     Going back to diagram 6: From the range of g on the x-axis of the graph on the right, the points will land on every point from 0 down to f-1(-1) = -1, including -1 but excluding 0 (because x-axis is an asymptote for  y = f-1(x)).  We answer thus:-
     Range of  f-1g = [-1, 0)
Diagram 8.  Part (iii) of Question

Part (iii) tests students understanding of increasing and decreasing functions.
f  is an increasing function means
     whenever   a > b,   f(a) > f(b)  
f  is an decreasing function means
     whenever   a > b,   f(a) < f(b)  
These are in fact the definitions of increasing and decreasing functions.  One can recognise an increasing function from its graph by the up slope (positive gradients) as you move from left to right.  For a decreasing function, the slope will be down as you move from left to right (negative gradients).  [An interesting note:  if  f  is a decreasing function,  then  f-1  will also be a decreasing function.]
   In our case,  the graph of  f  is down-sloping, so it is a decreasing function. 
          Since  f  is a decreasing function,  whenever   a > b,   f(a) < f(b).
Once again we seem to have the pernicious problem of not knowing formula for  f-1.  How to solve the inequality then?  Well, we can apply  f  to both sides of the inequality and the inequality reverses (because f is a decreasing function).  Note that  f  and  f-1  “cancel”  as functions  i.e.  ff-1(w) = w  for whatever the  w  is as long as it is well-defined.  Hence we proceed as follows
                                          f-1g(x) > -1
                                        ff-1g(x) < f(-1)
                                              g(x) < -1
                                          -1 - x2 < -1
                                               - x2 < 0
This latter inequality is the last trick on the question-setter sleeve desgined to unsettle the student.  How do you solve this inequality?  Do you need to equate or intersect with anything?  Anyway, what is the meaning of solve?  
To solve an inequality means to find all the possible values of  x  such that when you substitute each value into the inequality, the inequality becomes a true statement.
Note here that if we substituted  x = 0,  we would get  0 < 0, which is not true.  However, if we substituted any other real number,  x2  would always be a positive number, the LHS would always be a negative number, which is less than zero.  Conclusion:  x  can be any real number except 0.
Ans:  x Î R \ {0}

Note that the written solution (the parts typed in blue) is actually very short, although the explanation is rather long, because a lot of deep thinking is involved.

Let us think back on the lessons learnt while solving this particularly difficult problem.
*  the reason why this problem seems difficult is because it tests students’ understanding of
    concepts (which most are weak in).  From experience with many cohorts of students, the
    JC teachers know what concepts students are weak in and they like to set questions that
    exploit the chinks in students armours.
*  remember the horizontal line test and domain restriction to get a 1-to-1 function, so that the
    function is invertible.
*  remember the condition for the existence of composite functions
*  inverse functions swap the domain and range with the original functions
*  the two-step method is recommended for finding range of composite functions
*  increasing functions preserve inequalities, while decreasing functions reverse
    inequalities.  You can recognise a decreasing function from the downward
    slope of its graph as you go from left to right.
Although you do not have the formula for  f-1,
*  the value of  f-1(-1)  can be found by intersecting the graph of  y = f(x)  with the horizontal line
     y = -1.
*  f  and  f-1  “cancel” each other.
*  what is the meaning of “solve an inequality”?
*  How to solve inequalities like  -x2 < 0?  What about  -x2 > 0?   x2 > 0?

Tuesday, March 17, 2015

[Maths History] #Newton's #proof that #pi is #transcendental ?

Article (from Quora, re: Alejandro Jenkins)
How do you prove that a number is a transcendental number?

Although the concept of "transcendental number" was not defined during Newton's time, Newton's Principia contained the germ of idea that seems to lead to a relatively easy proof of the idea that p  is transcendental.  Arnol'd described how Newton showed that the trajectory of planets cannot be expressed in terms of any polynomial function of time.

[#Critique] Do pediatricians suck at #maths (#math) ?

It looks like some pediatrician is complaining that maths (that's the way we spell it in Singapore) is too hard.

Paul has two-thirds as many postcards as Sean has and the number of postcards Sean has is three-fifths the number of postcards Tim has. If the boys have a total of 280 postcards, then how many more postcards does Tim have than Paul?

He says “Obviously it’s complex, but not impossible, ...  I would use a complex algebra equation to solve it and I’d guess it’s more appropriate for an eighth and ninth grade student than one still in the fifth grade.”

Oh really?  Here's an easy solution.  No complicated ("complex" can mean a few other things) algebra, nor any algebra for that matter is needed.

To answer the question "Do pediatricians suck at maths?"  My answer is: I don’t know.

But remember: Don’t use a cannon to shoot a fly, or as the Chinese say: Why kill chicken with an ox cleaver?.  Thanks for reading this blog post.  J

Monday, March 16, 2015

[#Critique] Did #Seth #Godin know what #transcendental means in #maths?

Magic and irrational (by Seth Godin)

My Comments

This is an interesting piece of inspirational waffle about p to celebrate "p day".  Hopefully this special day helps to keep this important piece of knowledge that has a long cultural history fresh in our collective human consciousness, and perhaps stem the decline in mathematical literacy of some of us.

However, his article made me wonder whether he knew what transcendental means.  It is a pretty technical concept, and even if one knew what it means, it's a challenge to put it in layman's terms and say something funny with it.  If he knew the meaning, he certainly did not show it.

But I do not blame him.  Seth Godin is known for his marketing, not for his maths.  So this is not a personal attack.  I'm just following his advice, trying to be a purple cow.  Perhaps after some "moo moo" here and "moo moo" there, and transcendental meditation, this zero of a cow could break off from its roots of limited degree, eat the pi in the sky and jump over the Mooooooooon to become a hero!


[#Critique] Is #pi #infinite?

Article / Resource
The infinite life of pi - Reynaldo Lopes

My Comments
This is a well-produced video in which the team took great care in producing an interesting narrative.  The circles were drawn imperfect and shaky, so as to catch the attention of viewers and to highlight them.  A couple of points they missed out:-

1) p being irrational, does not only have an infinite number of digits in its decimal expansion, the digits are also non-repeating.  A rational number (fraction) like 1/7 also has an infinite number of digits, but they repeat: 1/7 = 0.142857 142857 142857 ...

2) near the end of the video, the animators depicted universe < p.  I find the artistic licence disturbing.  I am sure there is some way to artistically depict the known universe having less number of atoms than number of digits of p.

To answer the question in the title of this article "Is p infinite?": No, but her number of decimal digits is.

[Technology] #Calculator that makes you #Think

Article / Resource
QAMA Calculator Examples

My Comments
This calculator aims to enforce human estimation, and cultivate number-sense.  More designers / manufacturers / inventors should do likewise by at least trying to make students think.

Monday, March 9, 2015

[IB-HL H&H_Rev6C Q11] Factors of a Complex Polynomial


     This question is taken from the Haese & Harris textbook for IB HL Mathematics and is rather challenging.  Here I present two solutions.  In the first solution, I use substitution to make a variable “disappear”.  [ heuristics: H10. Simplify the problem, H11. Solve part of the problem]  And that allowed me to crack the rest of them problem.  In the second solution, I rephrased the problem in terms of tangents to curves. [ heuristic H09. Restate the problem in another way ] This gives another angle from which to tackle the problem.

Solution 1

Solution 2

Thinking Back
     Both solutions are related in the sense that they hinge on some form of the relation maked as [*].  That led to an equation in  a.  Once a  is found,  k  can be found, and the solutions proceed similarly.  The equation [*] is a manifestation of the fact that for a repeated root, both P(x) and P’(x) share a common factor.

[Pri20150308CSR] Charitable Savings Ratios?


     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.

     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.

Before    $240   $192      5 : 4
After      $160   $112     10 : 7

Solution makes sense.

H02. Use a diagram / model
H04. Look for pattern(s)
H05. Work backwards
H10. Simplify the problem

Sunday, March 8, 2015

[Pri20150306APC] Apples-with-Pears Comparison


     This question involves money and looks rather challenging, because there are a few unknown quantities.  To solve it, we use the concept of unit costs, so that we can make an “apples-to-apples” ... er ... I mean “pears-to-apples” J comparison of the prices.

     Since absolute dollar amounts are given, we can quickly solve [H11. Solve part of the problem] for the total costs of pears and apples as follows [H02. Use a diagram / model]:-

     Now we know that the total cost (in dollars) of pears is 45 and for the apples it is 50 (5 more than 45).  Although we do not know the absolute numbers of pears and apples, we know their ratio.  So let us write these down as, say, ‘square’ units. 

     Dividing the total costs by the numbers gives the unit costs, which we know only in ratio terms.  So let us use, say, ‘circle’ units to denote these.  But we know that the unit cost (in $) of an apple is 0.50 less than that of a pear.  And that is equivalent to 5 ‘circle’ units.  From here [H05. Work backwards], we quickly work out the cost of a pear (15 ‘circle’ units) as $1.50.

     Ta da!

H02. Use a diagram / model
H05. Work backwards
H11. Solve part of the problem

[Pri20150306COH] The Cards of Hearts?



     For this question, I shall illustrate my technique of Distinguished Ratio Units to model the situation.  Read the question carefully and translate the information into a diagram [ using heuristic H02 ] like below:-

     I used three different shapes (circle, triangle and square) to envelop the numerical counts of the different kinds ratio units. Although we do not know how many circle units’ worth of cards Kelly had at first, we quickly notice that  9 circle units are equivalent to 3 triangle units, so that one triangle unit is the same as 3 circle units.  So Kelly had 2 circle units’ worth of cards [ heuristic H05 ], as depicted below:-

This allows us to answer part (a) of the question already, namely that the required ratio is 10 : 2 i.e. 5 : 1.  With different types of units, it is difficult to compare things.  However, note that in the exchange of cards, the total number of cards remains constant.  Taking the LCM of 12, 4 and 11 which is 132, we can change all the ratio units to a common type of unit [ H09], say ‘heart’ unit, based on the total being 132 ‘heart’ units.  To do that, we can multiply the numerical counts in columns 1 & 2 by 11, multiply column 3 by 33 and multiply column 4 by 12.  This is what we would get:-

     To answer part (b) of the question, we actually do not need to bother about columns 2 and 3.  Just focus on columns 1 and 4.  From column 4 we observe that 84 minus 48 which is 36 ‘heart’ units gives 72, so one ‘heart’ unit corresponds to 2.  The number of cards won by Kelly can be found by comparing the 84 ‘heart’ units and 22 ‘heart’ units highlighted in yellow.

  That means 62 ‘heart’ units and that corresponds to 134.  And we are done!

Answer (a)   5 : 1
              (b)   134

H02. Use a diagram / model
H05. Work backwards
H09. Restate the problem in another way

Thinking Back

     In this question, we have used Distinguished Ratio Units to model the given situation.  We worked backwards to find that Kelly’s initial holdings were worth 2 circle units.  Then we converted everything to a common unit (‘heart’ unit) based on the constant total of 132 heart units.  Once again, I © hearts!

Saturday, March 7, 2015

[Pri20150306PPP] A Perspective for Perplexing Parallelograms


     Let us first be clear about what is given in the problem.  I have indicated the numerical values of the areas in the diagram below.

     First we construct  HK  parallel to  AB  passing through  E[ Heuristics H02 & H09 ]  This makes the problem easier to solve because there are congruent triangles and there are connecting ratios along the sides of the parallelogram as well as along the diagonal.  Namely  BF : FA = BE : ED = BK : KC.  We denote the  Area of DBKE by  x, and observe that it is the same as  Area of DBFE  by congruency (you can shift and rotate  DBKE  to get  DBFE).   Area of DAHE = Area of DAEF = 1 cm2  because of congruent triangles.  [ H04 ]

     Let us try to figure out  x [ H11 ].  Making further observations [ H04 ], note that  Area of DADB = Area of DCDB (big congruent triangles) and Area of DHDE = Area of DGDE  (congruent triangles).  See diagram below.

     Because of all these congruent triangles, the parallelogram AHEF and EGCK are forced to have the same areas (area measures), even though they may not be congruent:-
      Area EGCK = Area DCDB – Area DGDE – Area DBKE
                           = Area DADB – Area DHDE – Area DBFE 
                           = Area AHEF = 2 cm2.
We conclude that  Area DBKE = Area DBFE = 13/5 – 2 = 3/5  [cm2].  [ H05 ]

Here is the important thing:-

If triangles/parallelograms have the same height, their area ratio equals their base ratio.
     [ This is different for similar triangles/parallelogram. Do not confuse these two situations. ]

     Area DBFE : Area DAEF = BF : FA = BE : ED = BK : KC  = 3 : 5.
Now we can quickly fill in the areas of the remaining pieces as follows:-

      Area ABCD = 2 ´ Area DABD = 2 ´ (3/5 + 2 + 5/3) = 8 8/15  [cm2]   Done!

H02. Use a diagram / model
H04. Look for pattern(s)
H05. Work backwards
H09. Restate the problem in another way
H11. Solve part of the problem

Thinking Back

     After solving a problem (especially a difficult one), it is always good to think back and recollect what we have learnt by solving the problem.  We have used the following important geometrical facts

#1   If two shapes are congruent (that means you can shift, rotate and/or reflect so
       that they coincide), then their areas are equal.
#2   If triangles/parallelograms have the same height, their areas’ ratio equals their
       bases’ ratio.
#3   Sometimes, two shapes can have the same area even if they are not congruent.

      Drawing a construction line splits the diagram into various parallelograms and triangles with the same height, for which we can compare the ratios.  Using congruent triangles, we realise that two of the smaller parallelograms have the same area even though they are not congruent.  With the right perspective, we can deal with these perplexing parallelograms.  Using comparison, we found the area of the small triangular piece  x, and from there we use fact #2 above to work out the rest.