Monday, January 18, 2016

[OlymLSec_20160118PPPC] A Square Proof by Contradiction


     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.  ©

     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

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