«Infuse mathematical thinking in a lesson on Pythagoras Theorem»

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Mitchell Flowers
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«Infuse mathematical thinking in a lesson on Pythagoras Theorem» Background information: In year 2011, term 3, I was a relief teacher at NJC. The topic was on Pythagoras Theorem for year 2 students.

1 «Infuse mathematical thinking in a lesson on Pythagoras Theorem» Background information: In year 2011, term 3, I was a relief teacher at NJC. The topic was on Pythagoras Theorem for year 2 students. The original lesson material was similar to that commonly found in a textbook. After the introduction of the theorem, with geometrical representation, the lesson material was followed with numerous work examples and in-class exercises. As expected, the students had little difficulty in applying the theorem. In preparation, I had created this lesson to enrich the mathematical thinking and mathematical proof. Lesson Plan: (1) Compacting the original lesson material. Instead of working through the numerous examples, a teacher was to select only a couple of examples, with students active participation, to show the process of application of the theorem. Class practices were left as after class work. This had created 2 lessons worth of time for the enrichment activity. Pythagoras Theorem 2 2 a b c 2 (2) Enriching activity 1: Odd one out. The following is a list of the first few Pythagorean triples. Which one should not be included? ( 3, 4, 5 ), ( 5, 12, 13), ( 6, 8, 10), ( 7, 24, 25), ( 8, 15, 17), ( 9, 40, 41) Page 1 of 6 Learning & Teaching KLAng July 2012

2 Students were asked to find the odd one, and gave a reason to support their choice. Both classes of about 50 students in all, took a while to figure out the correct answer. The purpose of this activity is to prepare the ground for primitive Pythagorean triples. By the way the odd one is ( 6, 8, 10), which can be reduced to ( 3, 4, 5 ). A primitive Pythagorean triples has no common factor. The next activity was to list common properties of Pythagorean triples, base on these few examples. a. there is always one even, two odd numbers. b. the largest number is always odd. c. one of the 3 numbers is a prime. d. one of the 3 numbers is a multiple of 3. e. the even number is always a multiple of 4. f. one of the number is a multiple of 5. During this activity, many wild ideas might have been generated, but after the elimination process of checking against the others on the list, this was more or less what was left. Before the end of this lesson, teacher reminded the class to try to prove each of these. Making conjectures, doing guess and check, are key elements in a mathematician s work. Students should always be encouraged to make guesses, then, check or prove for correctness. (3) Enriching activity 2: Prove them all. a. there is always one even, two odd numbers. These are some parity properties of positive integers: (odd) + (odd) = (even) (1) (odd) + (even) = (odd) (2) (even) + (even) = (even) (3) The parity of a positive integer remains the same after squaring, i.e. squaring an odd integer results in another odd integer; squaring an even integer results in another even integer. The parity analysis of a b c will be similar to statement (1), (2) and (3). Page 2 of 6 Learning & Teaching KLAng July 2012

3 statement (3) is not possible. Primitive triples are co-prime, no common factor. If all the three numbers are even, then, there is at least a common factor 2, therefore the triples is not primitive, i.e. the triples can be simplified into a primitive one. So, a triples can only be (1) or (2). In other word, there will always be an even, two odd numbers. This is a direct proof by exhaustive listing. There was No indication that a student was capable of this proof. Teacher was to guide the class gradually to come to this conclusion. b. the largest number is always odd. There are only two possible arrangement of a triples: (odd) 2 + (odd) 2 = (even) (4) (odd) 2 + (even) 2 = (odd) (5) We need to show that statement (4) is impossible. Re-stating the statement to prove : the largest number cannot be even. Suppose the largest number is even, 2k. The two odds be 2m 1 and 2n 1. (2m 1) 2 + (2n 1) 2 = (2k) (4) 4m 2 4m n 2 4n + 1= 4k 2 4(m 2 m + n 2 n) + 2= 4k 2 The L.H.S. has a remainder 2 when divided by 4. The R.H.S. is divisible by 4. This is impossible. Contradiction! So the largest number is even, is impossible. The largest number cannot be even. That means the largest number is odd. This is an indirect proof by contradiction. For most of the students, this was the first time for them to witness this. Students found this hard to comprehend at first. c. one of the 3 numbers is a prime. By listing of a few more triples: ( 3, 4, 5 ) ( 5, 12, 13) ( 7, 24, 25) ( 8, 15, 17) ( 9, 40, 41) (11, 60, 61) (12, 35, 37) (13, 84, 85) (16, 63, 65) (16, 63, 65) has NO prime number. The conjecture was rejected. Page 3 of 6 Learning & Teaching KLAng July 2012

4 To prove, will without fail, begins with finding a counter-example to disprove. As we may have failed to disprove it, then, we have more confidence that the statement to prove, or the conjecture is truth. Developing a habit to find counter-examples, has a very important role in one s thinking process. Possibly one of the great value in learning mathematics is in learning to find counter-examples. So, this conjecture is disproved by counter-example. d. one of the 3 numbers is a multiple of 3. A positive integer can be expressed as 3k, 3k 1 or 3k 2. The squared of these integers are 9k 2, 9k 2 6k + 1 or 9k 2 12k + 4 respectively. Sum of any two squared integers : 9n 2 + 9k 2 = 9(n 2 + k 2 ) (6) 9n 2 + 9k 2 6k + 1 = 3( 3(n 2 + k 2 ) 2k ) (7) 9n 2 + 9k 2 12k + 4 = 3( 3(n 2 + k 2 ) 4k +1 ) (8) 9n 2 6n k 2 6k + 1 = 3( 3(n 2 + k 2 ) 2n 2k ) (9) 9n 2 6n k 2 12k + 4 = 3(3(n 2 + k 2 ) 2n 4k +1 ) (10) 9n 2 12n k 2 12k + 4 = 3(3(n 2 + k 2 ) 4n 4k + 2 ) (11) A squared integer, when divided by 3, can only have zero or 1 as remainder. Equation (6) implies that all three integers are multiple of 3, it is not a primitive triple. This is not possible. Equation (9), (10) and (11) each has a remainder 2 when divided by 3, this is NOT possible as the remainder of a square number can only be 0 or 1. There cannot be such a triple. So, only equation (7) and (8) are possible to form a triple. In both cases, there is a number that is a multiple of 3. At this point, the lesson had come to an end. The last two conjectures were given, without discussion. As an introduction to mathematically proof as a class activity, students were clearly unfamiliar to the proving processes. But, the rigours in mathematics, is developed through the proving process. Instead of omitting this from the current mathematics education, more of such activities should have been introduced, for mathematics education to be meaningful. Page 4 of 6 Learning & Teaching KLAng July 2012

5 e. the even number is always a multiple of 4. A positive integer can be expressed as 2k or 2k 1. The squared of these integers are 4k 2, 4k 2 4k + 1 respectively. Sum of any two squared integers : 4n 2 + 4k 2 = 4(n 2 + k 2 ) (12) 4n 2 + 4k 2 4k + 1 = 4( n 2 + k 2 k ) (13) 4n 2 4n k 2 4k + 1 = 4( n 2 + k 2 n k ) (14) Equation (12) implies that all three integers are multiple of 2, it is not a primitive triple. This is not possible. A squared integer, when divided by 4, can only have zero or 1 as remainder. Equation (14) has a remainder 2 when divided by 4, this is NOT possible as the remainder of a square number can only be 0 or 1. There cannot be such a triple. Only (13) is possible to have triples. Let the even number be 2n, the two odd numbers are 2k 1 and 2m 1, with 2m 1 > 2k 1. (2n) 2 + (2k 1) 2 = (2m 1) 2, noted that the even number cannot be the largest. (2n) 2 = (2m 1) 2 (2k 1) 2 4n 2 = (2m 1 + 2k 1)(2m 1 2k + 1) 4n 2 = 4(m + k 1)(m k ) n 2 = (m + k 1)(m k ) (m + k 1) and (m k ) are of different parity, i.e. when (m + k 1) is odd, then (m k ) is even, vice versa. (m + k 1)(m k ) is even, therefore n 2 is even too. This means that n is even, n = 2p. As such the even number 2n = 2(2p) = 4p is a multiple of 4. For completeness, the first part of this proof has been included, as the original proof in a, using parity only. Page 5 of 6 Learning & Teaching KLAng July 2012

6 f. one of the number is a multiple of 5. In a similar way, a positive integer can be expressed as 5k or 5k ± 1 or 5k ± 2. The squared of these integers are 25k 2, 25k 2 ± 10k + 1, 25k 2 ± 20k + 4 respectively. Sum of any two squared integers : 25n k 2 = 25(n 2 + k 2 ) (15) 25n k 2 ± 10k + 1 = 5(5n 2 + 5k 2 ± 2k ) (16) 25n k 2 ± 20k + 4 = 5( 5n 2 + 5k 2 ± 4k ) (17) 25n 2 ± 10n k 2 ± 10k + 1 = 5( 5n 2 + 5k 2 ± 2n ± 2k ) (18) 25n 2 ± 10n k 2 ± 20k + 4 = 5( 5n 2 + 5k 2 ± 2n ± 4k + 1) (19) 25n 2 ± 20n k 2 ± 20k + 4 = 5( 5n 2 + 5k 2 ± 4n ± 4k + 1) (20) Equation (15) implies that all three integers are multiple of 5, it is not a primitive triple. This is not possible. A squared integer, when divided by 5, can only have zero or 1 or 4 as remainder. Only (16), (17) and (19) satisfy this requirement. In addition, (16) and (17) each has a number that is a multiple of 5, 5n. (19) implies the sum of two squared number is a multiple of 5. That implies that the largest number is a multiple of 5. Therefore, there will always be a number in a triple that is a multiple of 5. In conclusion: It wasn t easily for the class. However, I believe that through this activity, students had been challenged intellectually. It provided an opportunity to give students a better understanding of mathematics. This lesson had extended the properties of integers with parity, divisibility. The different methods of prove and disprove had been an eye opener for these students. This lesson is suitable for more mathematical-able sec 2 classes. By the way, abc is divisible by 60. Other proving methods are possible. Page 6 of 6 Learning & Teaching KLAng July 2012

PROBLEM SET 3: PROOF TECHNIQUES

PROBLEM SET 3: PROOF TECHNIQUES CS 198-087: INTRODUCTION TO MATHEMATICAL THINKING UC BERKELEY EECS FALL 2018 This homework is due on Monday, September 24th, at 6:30PM, on Gradescope. As usual, this homework