- Level: International Baccalaureate
- Subject: Maths
- Word count: 3562
Investigating Divisibility
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Introduction
1. Factorise the expression P(n) = nx - n for x ? {2, 3, 4, 5}. Determine if the expression is always divisible by the corresponding x. If divisible use mathematical induction to prove your results by showing whether P(k+1) - P(k) is always divisible by x. Using appropriate technology, explore more cases, summarize your results and make a conjecture for when nx - n is divisible by x. 1.1 Factorise the expression P(n) = nx - n for x ? {2, 3, 4, 5}. When n is real numbers; x = 2 P(n) = n2 - n = n (n - 1) x = 3 P(n) = n3 - n = n (n2 - 1) = n (n + 1) (n - 1) x = 4 P(n) = n4 - n = n (n3 - 1) = n (n - 1) (n2 + n + 1) x = 5 P(n) = n5 - n = n (n4 - 1) = n (n2 + 1) (n2 - 1) = n (n2 + 1) (n - 1) (n + 1) 1.2 Determine if the expression is always divisible by the corresponding x. 1.2.1 x = 2 P(n) = n2 - n = n (n - 1) divisible by 2? When n = 1 n (n - 1) = 1 (1 - 1) = 1 (0) As 0�2 = 0, P(n) is divisible by 2 when x = 2 and n = 1 Assume n = k is correct k (k - 1) = k2 - k = 2M (where M is any natural number) Then considering n = k + 1 (k + 1) (k) = k2 + k = (k2 - k) +2k = 2M + 2k = 2(M + k) Therefore, P(n) = n2 - n is divisible by 2 when x = 2 and n = any natural numbers. 1.2.2 x=3 P(n) = n (n + 1) (n - 1) divisible by 3? When n = 1 n (n + 1) ...read more.
Middle
495k4 + 220k3 + 66k2 + 12k + 1 - k - 1 - k12 + k = 12k11 + 66k10 + 220k9 + 495k8 + 792k7 + 924k6 + 792k5 + 495k4 + 220k3 + 66k2 + 12k Therefore, as P(n) = nx - n and x = 12, P(k+1) - P(k) is not divisible by x. x = 13; P(k+1) - P(k) = (k + 1)x - (k + 1) - kx + k = (k + 1)13 - (k + 1) - k13 + k = k13 + 13k12 + 78k11 + 286k10 + 715k9 + 1287k8 + 1716k7 + 1716k6 + 1287k5 + 715k4 + 286k3 + 78k2 + 13k + 1 - k - 1 - k13 + k = 13k12 + 78k11 + 286k10 + 715k9 + 1287k8 + 1716k7 + 1716k6 + 1287k5 + 715k4 + 286k3 + 78k2 + 13k = 13(k12 + 6k11 + 22k10 + 55k9 + 99k8 + 132k7 + 132k6 + 99k5 + 55k4 + 22k3 + 6k2 + k) Therefore, as P(n) = nx - n and x = 13, P(k+1) - P(k) is divisible by x. x = 14; P(k+1) - P(k) = (k + 1)x - (k + 1) - kx + k = (k + 1)14 - (k + 1) - k14 + k = k14 + 14k13 + 91k12 + 364k11 + 1001k10 + 2002k9 + 3003k8 + 3432k7 + 3003k6 + 2002k5 + 10014 + 364k3 + 91k2 + 14k + 1 - k - 1 - k14 + k = 14k13 + 91k12 + 364k11 + 1001k10 + 2002k9 + 3003k8 + 3432k7 + 3003k6 + 2002k5 + 10014 + 364k3 + 91k2 + 14k Therefore, as P(n) = nx - n and x = 14, P(k+1) - P(k) is not divisible by x. 1.5 Summarize your results and make a conjecture for when nx - n is divisible by x. ...read more.
Conclusion
Hence, it forms = c x n. Therefore, n divides when n is prime. 4. State the converse of your conjecture. Describe how you would prove whether or not the converse holds. Converse means words so related that one reverses the relation denoted by the other. Hence, stating the converse of my conjecture simply means to rewrite my conjecture backwards. My original conjecture was in Pascals' triangle, when x is a prime number in and r is any number between 1 and x - 1, any term in that row is a multiple of designated prime number x. Converse of this would be, if every term in a row in Pascal's triangle, excluding first term and last term, are able to be divisible by a certain number, then that certain number is 1 or a prime number. To investigate whether this converse of my conjecture holds, firstly I would examine each term of each row whether it has a common factors. Then, by listing out all the common factors, it then could be determined whether the common factors are 1 or a prime number or not. If all the common factors are 1 or prime number, then the converse holds. However, if the common factors are not 1 or prime number, the converse of my original conjecture is false. For example, row 6 contains; 6, 15, 20, 15, 6 (excluding 1st and last term). The common factors of these terms would be 1. Hence, the converse holds. Examine row 4 which contains; 4, 6, 4 (excluding 1st and last term). The common factors of these terms would be 1 and 2. Hence, as 2 is a prime number, the converse holds. Another example would be row 9. Row 9 contains; 9, 36, 84, 126, 126, 84, 36, 9 (excluding 1st and last term). The common factors of these terms would be 1, 2 and 3. All of these three common factors are prime number, hence the conjecture holds. By considering these examples, it could be stated that the converse of my original conjecture is proved that it holds. ?? ?? ?? ?? 1 ...read more.
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