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  • Level: GCSE
  • Subject: Maths
  • Word count: 4016

Beyond Pythagoras

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Introduction

Beyond Pythagoras Introduction Believed to have been born in 582BC and died in 500BC approximately, Pythagoras was a Greek philosopher and mathematician. He discovered some of the most influential theories of number, geometry and proportion which are still frequently used in modern mathematics. Pythagorean triples, on which this investigation is based, are sets of three numbers as seen in the above diagram. Here it is the simplest triple; 3,4,5. They will be written in the following form throughout this paper; a,b,c. There are other Pythagorean triples such as 5,12,13; which is the 2nd odd triple and 361,65160,65161; which is the 180th odd triple. His theorem is this: the square of the hypotenuse (longest side (c (5 (25)))) of a right-angled triangle is equal to the sum of the squares of the other two sides (a (3 (9)) and b (4 (16))). This can be expressed using the following equation: a2+b2=c2 Aim The aim of the investigation is to explore and mathematically express the relationships between the number of the triple (n), the length of each side (a; the shortest), (b; the intermediate), (c; the longest), the perimeter (P) and area (A) of a right-angled triangle. Please note: Throughout this investigation, the sum on the bottom line is the final answer to each set of sums. For example: y=2+2 y=4 In this example, the bottom line (y=4) should be interpreted as the answer. The main rule of the investigation is that the values of the above (n, a, b, c, P and A) must all be positive integers (i.e. whole numbers of greater value than zero). If any of the numbers concerned do not match these criteria (i.e. they are decimals), then they and their respective triple will be excluded from the sequence of triples and will not be considered in this investigation, as this would cause the investigation to be infinitely complex. For example, if in the first term of the sequence, a equalled 1, then b and c would not be integers (1�3 recurring and 1.6 recurring respectively), as an integer has to be a whole number by definition. ...read more.

Middle

was worked out in the following way: The terms of b that had been previously collected were written down and the relationships between them were present in their most basic form at L2 as +4. b L1 L2 4 >8 12 >4 >12 24 >4 >16 40 >4 >20 60 >4 >24 84 >4 >28 112 >4 >32 144 From previous work, it is known that if the lowest level is L2, you halve the number on this level (4/2=2) and this forms the first part of the formula. Because the extracted value (4) is from L2, it shows that the relationship between the terms of b is exponential. In maths, exponents are displayed as indices (n2). It can be noted that the L1 (level 1) relationship between the terms was steadily increasing in value (by 4 each time), and L2 said this. This suggests that the first component of the formula will be quadratic, because the L1 relationship is exponential. tn=2n2 However, this does not yet work and requires further refinement. For example, the first term is not: t1=2n2 t1=2(1x1) t1=2 it is t1=4 The difference between 2 and 4 is 2, so the formula may be: t1=2n2+2 t1=2x12+2 t1=4 THIS IS CORRECT FOR THE FIRST TERM This can be checked by implementing it in the next terms of the sequence: t2=2n2+2 t3=2n2+2 t4=2n2+2 t2=2x22+2 t3=2x32+2 t4=2x42+2 t2=10 t3=20 t4=34 Note that in the above formula, the number which replaces n has been written in bold. The formula does not work, as the terms except for the first are not true using this formula. This problem can be solved by comparing the above values for t2, t3 and t4 with the true values. Because this formula does not work, the +2 at the end of tn=2n2+2 can be disregarded. b difference current formula t1=4 0 t1=4 t2=12 2 t2=10 t3=24 4 t3=20 t4=40 6 t4=34 It can be seen that as each term passes, the difference increases by two. ...read more.

Conclusion

For example, Pythagoras' theorem states that a2+b2=c2, and so by substituting a, b and c for their respective formulae the same should be true; that by adding the square of the formula of the relationship between n and a to the square of the formula of the relationship between n and b, this will equate the square of the formula of the relationship between n and c. a2+b2=c2=(2n+1)2+(2n(n+1))2=(2n(n+1)+1)2 a2 was calculated: a2=(2n+1)2 a2=(2n+1)(2n+1) a2=4n2+2n+2n+1 a2=4n2+4n+1 b2 was calculated: b2=(2n(n+1))2 b2=(2n2+2n)2 b2=(2n2+2n)(2n2+2n) b2=4n4+4n3+4n3+4n2 b2=4n4+8n3+4n2 c2 was calculated: c2=(2n(n+1)+1)2 c2=(2n2+2n+1)2 c2=(2n2+2n+1)(2n2+2n+1) c2=4n4+4n3+2n2+4n3+4n2+2n+2n2+2n+1 c2=4n4+8n3+8n2+4n+1 Using the above, it is now possible to see if the formulae are true: a2+b2=c2 (4n2+4n+1)+(4n4+8n3+4n2)=4n4+8n3+8n2+4n+1 4n2+4n+1+4n4+8n3+4n2=4n4+8n3+8n2+4n+1 4n4+8n3+8n2+4n+1=4n4+8n3+8n2+4n+1 The above calculation proves that the formulae are true because they obey Pythagoras' theorem. The whole process was repeated for even numbers. When all of the formulas had been calculated, a spreadsheet was built to work out the terms of the sequence after 8 which utilised the discovered formulae. The values of b and c in the even triples can be discussed. The "true" triples are the ones that satisfy the statement n-1<b<n+1: the value of b is greater than the previous term and less than the next term and the same is true for c. Also to be a "true" triple, the difference between b and c must be 2 in any one term. These two criteria directly affect each other. PROVE THE FORMULAE. EVEN The first aim of the investigation is to collect data so that it can be analysed. I wanted to calculate the first 8 Pythagorean triples using trial and error at first, but realised that using this time to devise a method to find these numbers would be more effective. After much trial and error, I came to that if you square a and divide it by 2, you end up with the base number for the next two sides, b and c. By subtracting and adding 0�5 to the answer, you can find b and c respectively. For example, in the first Pythagorean triple: ?? ?? ?? ?? 1 ...read more.

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