Sunday, November 6, 2016

[AM_20161105ITFF] False Friends in Integration (Calculus)

Question

Introduction
          False friends are words in two languages that look/sound alike, but differ significantly in meaning.  Do you know that there are also false friends in mathematics?  Can you distinguish and explain the difference between the two integrals?

Solution

          The integrand on the left has the variable  x  as the base and the constant  e  as the index.  So we integrate it using the Power Law.
          By contrast, for the integrand on the right, the base  e  is a constant whereas the index is the variable  x.  Integrating  e  to the power of  x  is the eeeeeeeeeeeeeeeasiest.  You just get back the same thing, plus the arbitrary constant of course.

Remark
          Many students make the mistake of trying to apply the Power Law for the exponential.  As a learner of mathematics, one needs to cultivate the habit of being observant and paying attention to detail.  This is part of developing one’s identity and character which is important in life.

Suitable Levels
GCE ‘O’ Level Additional Mathematics
GCE ‘A’ Levels (revision)
* revision for IB Mathematics HL & SL (revision)
* Advanced Placement (AP) Calculus AB & BC
* University / College Calculus
* other syllabuses that involve integral calculus

* whoever is interested

Thursday, November 3, 2016

[Enrich20161103NRP] The Napkin Ring Problem

Problem

     Two rings are made by drilling a cylindrical hole through a small sphere and a hole through the large sphere, such that the resulting rings have the same height (2h).
     Which ring has the larger volume of remaining material?

Solution
     The answer is: both rings have the same volume.  How can we know?
     There is a way to show this using integration.  But calculus is not necessary.

     Let  r  be the radius of any chosen sphere and let  a  be the radius of the cylindrical hole.  By Pythagoras’ Theorem,  h² = r² – a².  Consider a cross-section of the ring sliced a distance  x  from the centre of the sphere, perpendicular to the axis of the cylindrical hole.  The outer radius of this cross section is the square root of  r² – x².  Hence the area of the material in the cross-section is
               p [(r² – x²) – a²]  =  p (r² – a² – x²)  =  p (h² – x²)
Note that  r  does not appear in the formula.  That means the cross-section does not depend on  rA bigger (or smaller) sphere would have the same cross-sectional area for each distance  x  away from the centre.  By Cavalieri'sPrinciple, the other sphere will have the same volume!
     By the way what is this volume?  It is the same as that of a sphere without hole  i.e.  where  a = 0  and  r = h.  This works out to be  4/3 p h³,  where  h  is half the height of the ring.

H02. Use a diagram / model
H09. Restate the problem in another way
H11. Solve part of the problem
H13* Use Equation / write a Mathematical Sentence

Suitable Levels
GCE ‘A’ Levels H2 Mathematics (Number patterns, with algebra)
* revision for IB Mathematics HL & SL
* Advanced Placement (AP) Calculus AB & BC
* University / College Calculus
* other syllabuses that involve volumes and Pythagoras’ Theorem
* any learner who is interested






Saturday, October 22, 2016

[Pri1_20161021DIV] Labelling as a strategy for Division

Introduction
          The Singapore mathematics syllabuses are very well designed, especially the primary school syllabus.  Fundamental concepts and skills are introduced before going on to complex calculations and problem solving.  At primary 1, pupils learn the idea of multiplication and division of small numbers by grouping (or partitioning).  They are not made to recite the times tables meaninglessly.
          Division is easy if the number of things in each group is known.  You just keep on circling the known number of objects until everything is circled.  However, if the number of groups is required but the number of things in each group is not given, and if the objects are not arranged in a convenient way, the task can be a bit more challenging.  Remember: they have not memorised the multiplication tables yet.

Problem / Question


Solution (Suggested)
          One way to solve this problem is to label the fish 1, 2, 3, 1, 2, 3, ... in a cyclic fashion, assigning fish to each of the three friends one at a time, thereby ensuring that each person gets the same number of fish.  Start with “1” somewhere on the left, “3” on the right and “2” somewhere in the middle.  Assign the next “1” close to the previous “1”, the next “2” close to the previous “2” and the next “3” close to the previous “3”.  So all the 1s are close together, the 2s are close together and the 3s are close together.  After all the fish have been labelled, the partitioning (or grouping) becomes obvious.

H02. Use a diagram / model
H04. Look for pattern(s)
H09. Restate the problem in another way

Suitable Levels
Primary School / Elementary School Mathematics
* any precocious or independent learner who is interested




Wednesday, June 8, 2016

The passing of a Great Giant - Jerome Bruner

Jerome Bruner, a great psychologist, has died recently.  We thank him for his theories that guided the development of Singapore Mathematics education.

One of his greatest theories that is useful for children's learning is the Concrete-Pictorial-Abstract (Enactive-Iconic-Symbolic) approach.

It is impossible to use only words to explain simple mathematical concepts to a young child, since those concepts cannot be further explained using words.  You will run out of words to explain!  What definitely does not work is to start from use just words and they don't "get it", then scold them for being stupid.

They have to learn buy touching and playing with things, then from looking at pictures and then using abstract words or reasoning.

Saturday, February 27, 2016

[S1_20160227FZCK] Factorisation by Chunking

Problem / Question
 

Solution

Commentary
     Here I illustrate the usefulness of chunking to factorise (AmE: factor) an algebraic expression.  Observe that  3a – 2b  is a repeated part of the expression.  I call it a “chunk”.  To make it clear, I rewrite  (3a – 2b)²  as   (3a – 2b)(3a – 2b)  so that you can see it as two copies of the same chunk.  I highlight in yellow one copy of  (3a – 2b)  from each of 
(3a – 2b) (3a – 2b)   and  -3(3a – 2b).  The remaining stuff are highlighted in blue and green.  Take out the yellow chunk as common factor by writing it out on the left in the third line, shown in yellow.  You can pull out the common factor by writing it out to the right if you want, but here I chose to put it on the left.  The result would be equivalent anyway.  Once you have written out the common factor,  you write out the other stuff (shown highlighted in blue and green) into another other bracket.
     Once you understand how it works, you can actually do the second line mentally and write down the answer straightaway.  Chunking is a very useful technique in mathematics.  Here are some more examples of the technique of chunking: (1), (2), (3).

H04. Look for pattern(s)
H10. Simplify the problem
H11. Solve part of the problem

Suitable Levels
Lower Secondary Mathematics (Sec 1 ~ grade 7)
GCE ‘O’ Level “Elementary” Mathematics
* other syllabuses that involve algebra and factorisation (factoring)
* any learner who is interested





Tuesday, February 23, 2016

[Pri20160223FPDM] Pernicious Portion Problem? Shift Happens!

Problem / Question

Strategy
     This seems to be a confounding question on decimals.  What shall we do with the triangles?  Is there a short cut?

     Yes!  What you can do is to imagine putting the two triangles together to form a rectangle.  And then the solution becomes easy!  This is because the area is unchanged and hence the proportion of the shaded area is unchanged, is the same as before.  We can make use of fractions and convert it to a decimal.


Solution


H02. Use a diagram / model
H04. Look for pattern(s)
H09. Restate the problem in another way
H10. Simplify the problem

Suitable Levels
Primary School Mathematics
* other syllabuses that involve fractions and decimals
* any learner who is interested


Monday, February 22, 2016

[Maths Education] Mathematical Journalling

     Nowadays, I see some schools / textbooks asking students to search the internet and to write on a certain problem on their “mathematics journal”.  It's so very guided.  It's so artificial.  The questions should come from the learners themselves, out of their own curiosity.  The learn then seeks to answer their own questions.  The journal can serve as to document and summarise their process of learning.

     The best maths journals are self-initiated.  Great mathematician Karl Friedrich Gauss and renowned scientist Richard Feynman kept math journals on their own accord, not because some teacher told them to do it. 

     When I was a student, I borrowed books from the National Library on things out of the normal curriculum.  I kept notes of things I learned.  I also did my own investigations.  I accidently discovered quadratic equations when I was in Primary 4.  I read guidebooks, asked my friend's elder brothers and sisters, my Chinese teacher (!) and other people to find out more.  I did not like factorisation by trial-and-error.  Neither did I like completing the square nor using the quadratic formula.  So I did my own research to find a sure-fire way to factorise without trial-and-error.  I finally managed to find a way, but my method had an uncanny similarity to the quadratic formula.  It was a Pyrrhic victory, but it was fun!  I thoroughly enjoyed it.

     If students need to be told or goaded to write mathematics journals, then we as educators need to ask ourselves:  Why?  What is their conception of mathematics and education?  What experiences have they gone through that lead them to these beliefs?

     Some food for thought, eh?







Thursday, February 18, 2016

[P6_20160217RTTU] Books on Bookshelves

Problem


Introduction
     Here we have a numerically challenging problem that involves ratios, and it ultimately reduces to an algebraic problem with two unknowns.  Nevertheless, we are spoilt for choice as regards to methods of solution:-
     (1)   Bar Diagram Modelling
     (2)   explicit letter-symbolic Algebra
     (3)   “p” and “u”  (parts and units)
     (4)   Distinguished Ratio Units
     Despite the fact that Bar Diagram Modelling made “Singapore mathematics” famous, let us remember that it is only one of the ways of solving problem by diagramming, which is just one of the eleven Primary School heuristics recommended by the Singapore Ministry of Education.
     The methods have a lot in common, and they differ mainly in the form of presentation.  However, standard Bar modelling is impractical under high-stakes high-stress examination conditions for this problem, not least because one would have to cut the bars into many pieces.  One should not cut off one’s feet just so as to fit the shoes (削足适履), as one Chinese saying goes.  We need to be flexible and open-minded.  I present a solution using my own Distinguished Ratio Units.

Solution
Ans:  735 books

Commentary
     First off, we need to equalise the numerators of  2/5  and  11/4 = 5/4  and put them ratio form.   This is because the  “2”  in the  2/5  represents the same quantity as the  “5”  in  5/4.
We do this adjustment by multiplying the former through by  5  and the latter through by  2.  Thus we deduce that the original number of books in A and in B are  25  and  8  “heart” units respectively. 
     Next, we add on the  2  and  3  “triangle” units.  By doing a comparison, we can figure out that  1  “triangle” unit must be  45  more than  17  “heart” units.  So  2  “triangle” units must be equal to  34  “heart” units plus  90.  Replacing the  2  “triangle” units (shown in yellow) with their equivalent, we now know that  59  “heart” units plus 90 gives  444.  This allows us to figure out that  1  “heart” is actually  6.  Thus, we can work out what  1  “triangle” unit, and then what  5 “triangle” units are worth.

Final Remarks
     Due to the difficulty of the numbers, the solution presented above is about as streamlined as I can make it to be.  
     There is another variation that can be used – equalising the “triangle” units (akin to the technique of elimination in standard algebra).  What we do is we multiply the group with total  444  by  3  and to multiply the group with total  489  by  2.  This would give  6  triangle units on each side.  Then we can compare the “heart” units and continue from there.  This way of proceeding is not for those who fear 4-digit numbers.
     If there are nicer or more elegant ways to tackle this question, I would definitely love to hear from you.

H01. Act it out
H04. Look for pattern(s)
H05. Work backwards
H06. Use before-after concept
H09. Restate the problem in another way
H11. Solve part of the problem

Suitable Levels
Primary School Mathematics
* other syllabuses that involve whole numbers and ratios
* any problem solver who loves a challenge






Wednesday, February 17, 2016

[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.

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





Monday, February 8, 2016

Happy Chinese New Year 2016 (#LunarNewYear)


2016
= 12 × 168


     Wishing all my readers a Happy Chinese New Year ... or more accurately a Lunar New Year – many other Asians (e.g. the Japanese and the Koreans) also celebrate this festival.  This year  2016  is mathematically special, because it is the product of  12  and  168.  12  is a lucky number for Western people (e.g. 12 signs of the zodiac and 12 days of Christmas)  and “8” is especially auspicious in Chinese because it sounds like /  [fa in Mandarin,] which means to prosper or to grow.  168 sounds like 一路发 [yaad lou faat in Cantonese] which is to prosper all the way through life.

     I think this year will be challenging, because the Fire Monkey could be monkeying around even more with the economy and world peace.  Nevertheless if all human beings can unite together and learn to think critically, creatively and logically (and mathematics is about all these), the planet Earth can be a better place.  So best wishes to one and all!



Thursday, February 4, 2016

[EM_20160204PBIE] “Hillarious” Mathematics of Politics?

Problem
Six coins are tossed to decide a result for either “C” or “S”.  Assuming that the coins are fair, and that the results of the tosses are independent, calculate the probability that all the tosses are in favour of “C”.

Introduction
     Hot in recent news is the story of the purported six coin tosses that were needed to determine certain county delegates in the race between Hillary Clinton and Bernie Sandersin the state of Iowa.  All six coin tosses were in favour of Hillary Clinton, and the result is so improbable that some people said it was “hillarious”.
     How is the probability calculated?  This is an example of mathematics in real life events that has the potential to affect the United States of America, and the whole world (including Singapore).

Solution

Discussion
     In order to make the calculation, we make two assumptions: 
          (1) that the coins were fair, and
          (2) that the coin toss results are independent.
     So, what does it mean that the coins are “fair”?  It means that the probability of getting a “heads” is the same as the probability of getting a “tails”, which means ½ for each.
     Coin toss results can be “heads” or “tails”.  These are examples of events.  An event is something that can happen or not happen, and we associate a probability with it.  The probability is a number that indicates how likely the event happens.  It is between  0  and  1  inclusive.  Zero probability means a practically impossible event.  A probability of  1  means a practically certain event.  [The reason for me using the word “practically” is technical, which I shall not discuss.]  If the events do not affect one another (i.e. in our case, the coin tosses are not affected by the other coin tosses) then the events are said to be independent.  If the events are independent, then we can simply multiply the individual probabilites together, as above.  If the events are dependent, the calculation would be more complicated.

     So are the coin toss results valid?  I do not know.  All I can say is: improbable does not mean impossible.  Mathematics cannot tell whether the above assumptions (1) and (2) are correct.  But at least I “lay all the cards on the table”, so that astute students of probability know the basis of these calculations.  It is up to you to decide, but at least you would have made a mathematically-informed decision.  There could be other twists to the story, which is beyond the scope of this article.  This is one of the reasons why you need to learn mathematics carefully and think critically, whether or not you would become a mathematician,engineer, teacher or have a mathematics-intensive career.

H08. Make suppositions
H10. Simplify the problem
H11. Solve part of the problem

Suitable Levels
GCE ‘O’ Level “Elementary” Mathematics (Number patterns, with algebra)
* other syllabuses that involve probability
* anybody in the whole wide world!









Monday, February 1, 2016

[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.

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




Tuesday, January 26, 2016

[S1_20160126FMNT] Largest Common Remainder for Three Divisors

Problem / Question

Find the greatest 4 digit number that will divide 63, 45 and 69 so as to leave the same remainder.

Solution
     LCM + (smallest number – 1) = 7245 + 44 = 7289   J

Remarks
     This question is from National University of Singapore High School, which caters to students who are very interested in science and mathematics, and possibly an academic career.  Do not be fooled by my short and sweet solution.  The question is actually quite challenging, and I went by a long way before coming up with this elegant solution.  This reminds me of  Human Resource managers who think that more lines of code written by programmers means more work is done.  Actually, a lot of hard thinking could be involved in writing a one-line code that does the same job.
     Research shows that when expert problem solvers realise that they are stuck, they change tactics.  They go back to the drawing board.  “Insanity is doing the same thing over and over again and expecting the different results.” said Albert Einstein.  The five stages of mathematical problem solving are
     1. Understanding the problem
     2. Planning a strategy
     3. Executing the plan
     4. Evaluation
     5. Reflection (which can even include blogging about it!)
At the evaluation stage, if one finds that one is not getting the results, or if the approach is not elegant, one goes back to stage 2 to devise a new strategy.  I had tried using a complicated Chinese Remainder Theorem approach, got the solution after one page of work, and realised that the problem can be solved very simply.

How does the solution work
     The set of remainders dividing by  63,  45  and  69  repeat themselves in a cycle and the length of the cycle happens to be the Lowest Common Multiple (LCM).  We can see this quite easily: suppose  x  and  y  both give a remainder  R  dividing by  63,  45  and  69,  then  xy  gives a remainder of  0  when divided by  63,  45  and  69,  which means  xy  is a common multiple of  63,  45  and  69, of which the lowest is the LCM.
     It is a routine matter to get the LCM via prime factorisation as follows:   63 = 32 × 7,   45 = 32 × 5,  and   69 = 3 × 23.  We pick the highest power for each occurring prime and we obtain  LCM = 32 × 5 × 7 × 23 = 7245.
     The next thing to note is that remainders must be less than the divisors.  Hence the largest common remainder must be  44,  one less than the smallest divisor.  You cannot have a remainder of  45  when divided by  45,  because you could simply have bumped up the quotient (the result of division) by 1 and get zero remainder.  So from 7245, 7246, 7247, ... to 7289,  you get common remainders of  0, 1, 2, ..., 44.  Once you hit  7290,  the remainder for division by  45  will hit  0  while the remainders for  63  and  69  will be  45  but this will not be a common remainder.  The next time we get a common remainder will be  2×LCM = 2×7245 = 14490  which is  5 digits long.  So within  4  digits, the highest number with a common remainder is  7289,  which gives a common remainder of  44.

H04. Look for pattern(s)
H05. Work backwards
H09. Restate the problem in another way
H10. Simplify the problem
H13* Use Equation / write a Mathematical Sentence

Suitable Levels
Lower Secondary Mathematics (Sec 1 ~ grade 7) challenge
* other syllabuses that involve factors and multiples or number theory
* any precocious or independent learner who loves a challenge






Monday, January 25, 2016

[AM_20160125DAHT] Horizontal Tangents via Quadratic Discriminants

Problem

Introduction
     This is a Additional Mathematics textbook problem.  This question is of an intermediate level of difficulty.  The general method is by differentiation.  The equation of the curve happens to be capable of being put into a quadratic equation in  x.  Hence we can also use the theory of quadratic discriminants.  I present both methods of solution.

Method 1 (Using differential calculus)


Method 2 (Using quadratic discriminants)

Heuristics Used
H04. Look for pattern(s)
H05. Work backwards
H09. Restate the problem in another way
H11. Solve part of the problem
H12* Think of a related problem
H13* Use Equation / write a Mathematical Sentence

Suitable Levels
GCE ‘O’ Level Additional Mathematics
* other syllabuses that involve differentiation or quadratic discriminants
* any independent learner who is interested






Friday, January 22, 2016

[Maths_Education] The Need to Harness and Transcend Technology

Problem / Question

Help me please!  My teacher needs it tomorrow!
How many millilitres are there in 3.4 litres?

Answer
     LOL!  Well, if she really needs it, ask her to type the question into Google!

Remarks
     Actually, this is not a joke.  Many easy questions in school mathematics are now answerable by Google.  In fact, Wolfram Alpha is able to answer mathematically more difficult but routine questions.  Wikipedia and YouTube are also useful for learning mathematics.  There are many other good resources available on the Internet and public libraries.  The sad thing is,
1)
It seems that our current generation of kids views homework as a chore to be done for the teacher, not as an experience to be used for their own learning.  OK, maybe the homework task should have been designed better, to ask non-Googleable questions, but educators need to be aware of that is happening to our kids.
2)
Our kids do not know how to choose and use the abundantly available technology and resources
3)
Neither are they taught how to do this in school (do the teachers know it themselves?)
4)
This type of question merely targets the lower levels of Blooms Taxonomy
5)
Even for these easy questions, children are unable or unwilling to make the effort to find their own answers, and how are they to engage in Higher Order Thinking, creative thinking, reasoning etc.?  
     As technology evolves and improves and replaces many jobs, and as our children de-evolve and slacken, when this generation grows up they will not only face an even more challenging environment for their careers than today, they may not have the right values, attitudes and dispositions for living life.
     The purpose of learning mathematics in school should be to learn how to think and to serve others.  Human students must be educated to use technology and go beyond technology, to seek answers and to help other people instead of merely relying from other people for “help”.  This is one of the reasons why an identity (“learning to be a type of person”) approach to learning mathematics is important.

Tuesday, January 19, 2016

[S1_Expository] Exploring the HCF and LCM with a calculator


How does it work?
     When we reduce a fraction to its lowest terms, we actually cancel out as many common factors as possible.  So eventually the numerator and the denominator of original fraction get cancelled by their Highest Common Factor (HCF), a.k.a. “Greatest Common Divisor (GCD) ” in America.  Just as England and the United States are divided by a common language(*), we can divide  756  by  6  to get the HCF.  Why?  That is because  756  was divided by the HCF to get  6.  You may try a similar trick with the denominators.  You get the same conclusion.
     It is useful to know that the product of two numbers is equal to the product of their  HCF  and  LCM.  So   756 × 1386 = HCF × LCM.  Hence we have
Observe that the bracketed number  (756/126)  is equal to  6  which is the numerator of the reduced fraction.  So we do not even need to make that calculation.  Just take the reduced numerator  6  and multiply that with the original denominator  1386  to get the LCM.  You may try a similar trick with  1386  and  126.  You end up multiplying  756  by  11,  which gives the same answer.

Further Exploration
     Try the above with different pairs of numbers.  How can you extend this to find the HCF and LCM of three or more numbers?

(*) OK, just kidding.  In Singapore, we sort of follow British English, but we are flexible.

Suitable Levels
Lower Secondary Mathematics (Sec 1 ~ grade 7)
* other syllabuses that involve factors, HCF (GCD) and LCM
* any interested learner


Monday, January 18, 2016

[OlymLSec_20160118PPPC] A Square Proof by Contradiction

Question

Explanation
     If  a + b = 11,  then  2ab = (a + b (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= 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.

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