Sunday, April 19, 2015

Mathematics Education expert speaks

Article
Research tell us what to do with mathematics education crisis

Please visit my curation blog.

Tuesday, April 14, 2015

[AM_20150413RSD] Rationalising Denominators for #Surds

Question 

Introduction
     Expressions with surds in their denominators are cumbersome.  The good news is that we can make the denominators into rational numbers, which are nicer.  Rational numbers those that can be expressed as a ratio of integers i.e. they are (proper or improper) fractions or can be converted to fractions.  Whole numbers are also part of rational numbers because you can always put them upon a denominator of  1;  e.g. 2 = 2/1,  so  2  is a rational number.
     The standard trick for simplifying expressions with surds in their denominators is to rationalise the denominator by mutiplying the numerator and the denominator with its conjugate surd.  For example, the conjugate surd of   Ö5 + Ö2   is   Ö5 – Ö2.   Just change the  +  to  –  or the  –  to  +.  Let us see how the magic works.

Solution
Remarks
     Note that in the first step, I pulled out 2 as the common factor of the denominator, so that I get a simpler surd to work with.  Always try to work with simpler expressions.  This not only shortens your working, it reduces your chances of making a careless mistake.
     In mathematics, “rationalising” does not mean you give some reason or excuse for something that you know you have done wrong.  It means “make it into a rational number”.  Why does rationalising the denominator work?  This is because on the bottom (denominator) we have a difference-of-squares expression of the form
                                           (a + b)(ab)   which is equal to   a2b2.
Since squaring “gets rid” of square roots,  a2  and  b2  will give you rational numbers (whole numbers or fractions), you will end up with a nice number downstairs (on the denominator).  Pupils should make sure they have this technique in their repertoire of skills.

Suitable Levels
GCE ‘O’ Level Additional Mathematics
* revision for GCE ‘A’ Level H2 Mathematics
* revision for IB Mathematics HL / SL
* other syllabuses that involve surds
* precocious kids who always want to learn more

Monday, April 13, 2015

[OlymUSec_20150412BDP] Guessing Cheryl’s Birthday

Question

Introduction
     This “Primary 5 mathematics” (actually an upper secondary Olympiad) logic puzzle has gone viral.  It has been making its rounds in various forums in Singapore and overseas, stumping adults and children alike.  It is actually a parody of an old puzzle.  Can it even be solved?  It seems that there is no information given by each parties that we can exploit.  Actually there is!  In a subtle way ...

Solution
     In the beginning, everybody knows that Albert knows only the month and Bernard knows only the numerical day of the month.
     When Albert tells us “I don’t know when Cheryl’s birthday is, but I know that Bernard does not know too.” he is leaking out information (from his knowledge of the month) that the day of the month appears more than once and cannot be (June 18 or May 19).  Actually, the original phrasing is more like “If I don’t know when Cheryl’s birthday is, then Bernard does not know too.”.  The person who set this question merely changed the names of the people and the dates, without appreciating the subtle but crucial difference between a statement of fact and an implication (an “if ... then ... ” statement). 
     Ruling out June 18 and May 19, we also know that Albert knows that the birthday month is neither June nor May.  Otherwise, how would he have been so confident in saying that he knows Bernard would not know Cheryl’s exact birthday?  So we can eliminate those months.
     Bernard acknowledges the above state of affairs and the embedded hint.  With the choice narrowed down and with his knowledge of the numerical date, he now knows Cheryl’s birthday.  Since we know that Bernard knows Cheryl’s birthday, we know that it cannot be a numerical date that appears more than once (otherwise he would not have been able to know).  So we can cross out July 14 and August 14.

     Now Albert would telepathically thank Bernard for this helpful hint.  Because now he is able to deduce Cheryl’s birthday with his knowledge of the month.  That would mean that this cannot be a month with two candidate dates.  We blot out the August dates and see for ourselves the only remaining possibility.

Conclusion: Cheryl’s birthday is  July 16.

Remarks
     This puzzle was solved using the process of elimination and analysing our knowledge of what each party knows and can know.  Thus we successively narrow down the possibilities until the answer becomes obvious.  Here we learn that
     knowledge of other people’s knowledge can itself give us knowledge
This principle is actually employed in cryptology (the use of secret codes) which finds applications in fields like banking, the military (cf. interesting story of how the German Enigma code was broken in WorldWar II) and communications.  As an example, radio communication can tell the enemy of troop positions and warn of an impending attack, and that is why radio silence is imployed as a precaution.  Sensitive information in certain organisations is restricted on a “need to knowbasis.

Suitable Levels
Upper Secondary Olympiad
* other syllabuses that involve knowledge or epistemology
* application of mathematical principles in real life
* for all people interested in logic puzzles

[OlymPri20150412LGC] Your #logic a House of Cards?

Question


Introduction
     This is an olympiad type of question that tests one’s use of logic.  Sadly, logic is seldom taught in schools, except maybe in courses like knowledge inquiry, theory of knowledge, or in selected topics like geometry.  Mathematics is actually very much connected with logic, but it seems to be regarded as difficult to teach.  Being able to use logic and think critically is very important in our lives in various aspects.
     Let us study how we can analyse conditional statements using logic.

Analysing Conditional Statement with Logic
     A conditional statement is a statement of the form “If  H  then  C.”  An example would be: “If the blue litmus paper turns red, then the liquid is an acid”.  The front clause H (the part about the litmus paper turning red) is called the hypothesis (or premise).  The latter clause  C  (the part about the acid) is called the conclusion.  Every conditional statement has three other related forms: the contrapositive, the converse, and the inverse.

     For example, let say there is a NC16-rated movie and you can watch it only if you are at least 16 years old.  So  H := “watch NC16”  and  C := “16 years old and above”.  The conditional statement means “If you want to watch the movie, then you must be 16 years old and above”.  The contrapositive is “If you below 16 years old, then you cannot watch the movie”.  The contrapositive always has exactly the same meaning as the original conditional statement.  If one is true, so is the other.  If one is false, so is the other.  The contrapositive is merely phrased in a different way.
     The converse says “If you are 16 and above, then you will watch the movie.”  The inverse would say “If you don’t want to watch the movie, then you are not 16 and above.”  Note that the inverse is the contrapositive of the converse.  The inverse always has exactly the same meaning as the converse.  If one is true, so is the other.  If one is false, so is the other.
     The converse (and the inverse) has a different meaning from the original statement.  If the original statment is true, the converse may or may not be true.  In our example, “If you want to watch the movie, then you must be 16 years old and above” is true, but the converse “If you are 16 years old and above, then you want to watch the movie.”  may not be true, as some people who are 16 years old and above may want to do something else.  Like reading a book.  Or playing golf.  So you see, there are really two camps: the original conditional statement and its contrapositive in one camp, and the converse and the inverse in the other.
     Let us analyse the given problem using logic.


Analysis
     The statement  ‘any card with a letter A on one side always has the number “1” on the other side’ can be rephrased as ‘If letter A appears, then the other side is “1”’.  Let us analyse each position in order.  For (a), since the letter A appears, we definitely need to know what is behind the card to test whether the claim is true.  For (b), note that the claim does not say ‘If the letter is not A, then the other side is not “1”’ (inverse)  nor  ‘If the number is “1”, then the other side must be an A.’ (converse).  It is possible that A is on the other side and we are fine.  But it is also possible to have a non-A with the number  “1”.  The given condition does not prohibit that.  So, if the other side is not an A, it is still OK.  So the other side can be anything, and it does not matter.  For (c) we definitely have to flip to check the other side in case the other side is an A, then the claim would be proven false.  For (d), since the card is not an A, it is irrelevant.  The claim is talking about  A.  It is not talking about B.  We can summarise the analysis in the diagram below:-

Ans:  We need to flip (a)  and  (c)

Suitable Levels
Primary School Olympiad

* syllabuses that involve logic and epistemology

[S2_AFMLCM_20150412] Conquering Algebraic Fractions

Question

Introduction
     This is an equation involving algebraic fractions, usually for secondary 2 (approximately grade 8) pupils in Singapore.  Many students (and teachers?) like to use the “cross-multiplying” method, as shown in Solution 1.  A usually more efficient method is to multiply every term by the Lowest Common Multiple (LCM) of all the denominators appearing in the equation, as shown in Solution 2.


Discussion
     Note that division by zero is not allowed.  Furthermore, in algebra, it is dangerous to cancel or divide by an unknown quantity, because there is a possibility that you are dividing by zero.  So any division or cancellation by an unknown quantity must be justified beforehand.  Mathematics is not a game of blind senseless manipulations.  If you look at the second solution, which is short and sweet (only 4 steps), multiplying through by the LCM of denominators not only avoids this awkwardness, but it clears all the fractions in one fell swoop.  The solution takes only  4  steps, and it is in fact the recommended method.  All students, whether “good” or “poor” in maths, should use the second method.  Teachers who refuse to use/teach the LCM method (out of habit, or because their own teachers taught them otherwise, or because this makes them or their pupils “uncomfortable”) are really doing the weaker students a huge disservice.  You are widening the achievement gap.  The better students are better, precisely because they use better methods.  The longer one’s working is, the higher the chances of making mistakes and the more time is wasted.  If the “weaker” pupils have to jump through lots of hoops to achieve a certain standard before they are allowed to learn this “advanced” method (actually it’s just the normal method), they will have to unlearn their old method and may get confused as they learn this method.  A triple whammy!  All learners need to practice anyway, so one might as well practice the correct thing right from the beginning and learn good habits (striving for efficient, effective, elegant solutions).  So please, please, please everyone: use the LCM method!



Suitable Levels
* Secondary 2 Mathematics (“Algebraic Fractions”)
* GCE ‘O’ Level (“Elementary”) Mathematics Revision
* other syllabuses that involve algebraic Fractions
* precocious children who want to learn algebra


Sunday, April 12, 2015

[EM_20150412SCA] Is your Airport Design career “taking off”?

Question

An Airbus 380 has constant acceleration of  1 m/s2.  Its takeoff velocity is 280 km/h.  How long must the runway be at a minimum to allow the plane to take off?

Introduction
     A practical question for airport design, perhaps?  I present two solutions.  The first uses a graphical method (speed-time graph) which is in the (“Elementary”) Mathematics syllabus and the other uses formulas for motion under constant acceleration taught in Physics.  Whichever method is used, remember to convert from km/h to m/s.  The target velocity is  700/9 m/s,  and the time to achieve this is  700/9 s  starting from rest,  since the acceleration is  1 m/s2.

Solution 1 [“Elementary” Mathematics, speed-time graph]
     The speed-time graph is very useful because it is able show the acceleration (as the gradient or slope of a straight line) and at the same time the area under the graph gives the numerical value of distance travelled.  If the acceleration is constant, we usually we get a trapezium.  But since the aeroplane starts from rest, we get a triangle (see diagram below).  All we need to do is to calculate the area under the graph and get the answer.
     It is interesting to observe that if we used the average speed  700/18 m/s,  we would also get the answer because the area under the graph (yellowish green rectangle) is the same as the area of the triangle.  This trick works for constant acceleration, but it may not work in other situations.


Solution 2 [Physics, constant acceleration]
     We use the important formulas  v = u + at  and  s = ut + ½at2.  In our example,  u = 0  because the initial velocity is zero (the airplane starts from rest).  This makes our calculations very easy.  If we compare the two methods, you find that the calculations are very similar, and we get the same answer.  Remember that speed = |velocity|  the magnitude of velocity.  In this relatively easy problem, the velocity means the same as speed because we are going in a straight line and in one direction only.  In other situations, this may not be so.

Heuristics Used
H02. Use a diagram / model
H05. Work backwards
H09. Restate the problem in another way
H13* Use Equation / write a Mathematical Sentence

Suitable Levels
GCE ‘O’ Level “Elementary” Mathematics
GCE ‘O’ Level Physics
* other syllabuses that acceleration, speed and distance
* precocious kids who always want to learn more

[AM_20150412SQS] Perfect Squares Lurking Absurdly

Question


Introduction
     Surds are expressions involving roots like square roots.  They are usually irrational numbers.  If you try to put them into a ratio of integers, you are absurd!  Many students (and teachers?) are not sure of how to put square roots of numbers in a simple form.  The trick is to use square numbers or perfect squares.  These are squares of whole numbers.  For example:-
     12 = 1 ´ 1 = 1                      Ö1 = 1
     22 = 2 ´ 2 = 4                      Ö4 = 2
     32 = 3 ´ 3 = 9                      Ö9 = 3
     42 = 4 ´ 4 = 16                  Ö16 = 4
     52 = 5 ´ 5 = 25                  Ö25 = 5
A number like  4  can be represented by a real square whose sides have length  2  units.  Note also that if you take the square root of a perfect square, you always get a nice whole number.
     How do you deal with numbers that are not perfect squares?  You factor out as many perfect squares as possible.  This would eventually lead to surds with small numbers, which are more manageable.  Here are some examples:-
With this weapon in our hands, let us kick some butt.

Solution
Moral of the Story
     Using perfect squares reduces your square roots to surds involving square roots of prime numbers, which are easier to combine or cancel.  This gives a short and sweet solution.  In mathematics, always try to do things by the cleanest way (if you can).


H10. Simplify the problem
H11. Solve part of the problem

Suitable Levels
GCE ‘O’ Level Additional Mathematics
* revision for GCE ‘A’ Level H2 Mathematics
* revision for IB Mathematics HL / SL
* other syllabuses that involve surds
* precocious kids who always want to learn more