[S1_20160117AFNS] Much Ado About Nothing?

Problem

 

Introduction

     Assuming no typing errors, this is a tricky Secondary 2 question involving an equation with algebraic fractions.  How to solve it?  How to present the solution?

Solution

Remarks

[1]   Once the LHS expression has no meaning, it would not even make sense to continue.

[2]   This is a proof by contradiction type of argument.

It turns out that not all algebraic equations are soluble (or solvable).  This problem is one case in point.  The “unknown”  m  cannot be ⁵/₂ because that would make the expression undefined.  But if you substitute any other value, you always end up with nonsense like “15 = 0”.  So no matter what, there is no solution.  In other words, there is no value of  m  that you can substitute into the equation that makes it a true statement.

Heuristics Used

H05. Work backwards

H08. Make suppositions

Suitable Levels

* Lower Secondary Mathematics (Sec 2 ~ grade 8)

* GCE ‘O’ Level “Elementary” Mathematics

* other syllabuses that involve algebra

* any learner who is interested in algebra

[OlymLSec20160201PHHE] Pigeonhole Principle and Harry’s emails

Problem / Question

Handsome Harry has a secret email account that only four friends know.  Today he received 8 emails in that account. Which of the following is certainly true? (A)  Harry received two emails from each friend. (B)  Harry cannot have received eight emails from one of his friends. (C)  Harry received at least one email from each friend. (D)  Harry received at least two emails from one of his friends (E)  Harry received at least two emails from 2 different friends.

Introduction

      This question is from some Kangaroo Mathematics Competition, which tests students on logic and not necessarily things from Singapore Mathematics syllabus. 

Solution

      (D)  Harry received at least two emails from one of his friends

Explanation

      This is an example of the Pigeonhole Principle.  Perhaps the easiest way to understand this is to imagine an array of pigeonholes with four columns (one for each of Harry’s friends) and pigeons (representing individual emails sent from the friends).  In the diagram below, I draw dots instead of pigeons.

As you can see, no matter how the eight dots / pigeons are placed, at least one of the friends will have at least two dots.  It is not possible for all the friends to have less than two emails.

Formal Proof

     We can use a proof by contradiction argument.  Suppose it were not true that Harry received at least two emails from one of his friends.  That would mean each of his  4  friends sent at most one email.  But then the total number of emails would be  4  or less.  This contradicts the given fact that Harry received  8  emails.  So this state of affairs is not possible.  Therefore, the opposite is true.  We conclude that Harry received at least two emails from one of his friends.

Final Remarks

      The Pigeonhole Principle is very useful in many situations, including computer science.  In general, if you have more objects (“pigeons”) than there are containers or slots (“pigeonholes”), one of the containers must have at least two of those objects.

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

Suitable Levels

* Lower Secondary Mathematics Competition / Olympiad

* other syllabuses that involve logic, combinatorics or Pigeonhole Principle

* any precocious or independent mathematics problem solver who is interested

[OlymLSec_20160118PPPC] A Square Proof by Contradiction

Question

Explanation

     If  a + b = 11,  then  2ab = (a + b)² – (a² + b²) = 121 – 100 = 21.  But  2ab  is an even number, whereas  21  is odd.  This is a contradiction.  So (B) is impossible.  ©

Remarks

     Short and sweet isn’t it?  This uses the square-of-sum identity   (a + b)² = a² + 2ab + b².  I used the tactic of assuming the answer is correct  [H08]  and showing that this leads to something nonsensical [H05].  So the original assumption must be wrong.  This is called “proof by contradiction” or reductio ad absurdum (in Latin).

     By the way, the correct answer option is (E) from the Pythagorean Triplet   8² + 6² = 10²  with  {a, b} = {8, 6}.  The question seems to be taken from some Kangaroo mathematics competition.

Heuristics Used

H05. Work backwards

H08. Make suppositions

H09. Restate the problem in another way

H13* Use Equation / write a Mathematical Sentence

Suitable Levels

* Lower Secondary Mathematics competition

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

* other syllabuses that involve whole numbers and Pythagorean triplets

* any precocious or independent learner who loves a challenge