[AM_20151226EIQR] Looking for a Pea among Quadratic Roots?

Question

Introduction

     This question is about finding the parameter  p, and not about solving for the “unknown”  x.  It is heavy on algebra, one has to be patient, careful and meticulous.  Please refer to this article for a recapitulation of (Vieta’s) theory of Quadratic Roots.

Solution

Heuristics Used

H04. Look for pattern(s)

H05. Work backwards

H13* Use Equation / write a Mathematical Sentence

Suitable Levels

* GCE ‘O’ Level Additional Mathematics

* other syllabuses that involve quadratic roots
* any learner who is interested

[Pri20150523WNCA] Trees arranged in a Hexagon Outline

Question

Introduction

     This is real eeeasy peasy lemon squeezy, isn’t it?  54 ¸ 6 = 9  Ta da!  The answer, right? Wrong!  You got tricked!  Ha!  Ha!

     Always tryto understand the question and do the planning first.  Never be in a hurry and jump to thecalculation stage.  So what went wrong?  Well, the tree at each vertex is counted twice.
     Huh?
     Sometimes to understand the situation, it may be easier to consider a simpler problem.  Let us say there are four trees per side.  This is how it looks like from above.

     You can see that the corner trees (coloured in orange instead of brown) are counted twice, because they each serve as an extreme marker of two of the sides of the hexagon.  There are 18 trees and if you divide by  6,  you get  3  and not  4.  One way to count properly is to start from one corner tree and count groups of three trees, either in a clockwise or anti-clockwise (American: counter-clockwise) direction.

     Notice that the number of trees on one edge of the hexagon is equal to the number of trees in one group plus one (the corner tree for the next group).  So for  18  trees, the correct calculation is  18 ¸ 6 + 1 = 3 + 1 = 4  for the number of trees along one edge.  We use the same procedure for  54  trees.

Solution

     number of trees on each side = 54 ¸ 6 + 1 = 9 + 1 = 10

Final Remarks

     You may want to generalise it into a formula

                    # trees on each side = total # trees ¸ #sides + 1

However, I do not recommend that you purposely memorise this formula.  Mathematics is not about memorisation.  It is about understanding.  Once you understand it, the formula comes out automatically.  You may test yourself or get a friend to test your understanding by setting a similar question but changing the number of trees and number of sides.

Heuristics Used

H02. Use a diagram / model

H04. Look for pattern(s)

H10. Simplify the problem

Suitable Levels

* Primary School Mathematics

* other syllabuses that involve whole numbers

[S1_AFMLCM_20150412] Conquering Algebraic Fractions

Question

Introduction

     This is an equation involving algebraic fractions, usually for secondary 1 (approximately grade 7) 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 1 Mathematics

* GCE ‘O’ Level (“Elementary”) Mathematics Revision

* other syllabuses that involve algebraic Fractions

* precocious children who want to learn algebra

[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?