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## HISTORY OF MATHEMATICS

by: Kennith Herman

15

0

2

# HISTORY OF MATHEMATICS MA 330

Marketplace > University of Kentucky > Mathematics (M) > MA 330 > HISTORY OF MATHEMATICS
Kennith Herman
UK
GPA 3.54

Staff

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COURSE
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KARMA
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## Popular in Mathematics (M)

This 2 page Class Notes was uploaded by Kennith Herman on Friday October 23, 2015. The Class Notes belongs to MA 330 at University of Kentucky taught by Staff in Fall. Since its upload, it has received 15 views. For similar materials see /class/228149/ma-330-university-of-kentucky in Mathematics (M) at University of Kentucky.

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Date Created: 10/23/15
MA330 Satrage Note on Chakrav la Method of Bh skara We describe below a new technique inspired by the historical works of BhaskaraBrahmagupta for the solution in positive integers of the equation 2 7 Dy2 i1 where D is a given positive integer which is not a complete square Notation We shall work in the eld F Q where Q is the usual eld of rational numbers For convenience7 we shall write a typical element a bD E F where ab E Q as a pair 171 and by its norm we shall mean a2 7 Dbz and we shall write it as HabH Thus7 our problem is to nd all elements of norm i1 and this set is known to form the group of units in the ring of elements 17 b with 17 b E Z the ring of integers A reader unfamiliar with this terminology can still follow the rest of the work Method We now describe the steps Choose start Choose second Choose a conveninent pair 11 such that the norm 12 7 D is as small as possible in the absolute value You simply need to compare the oor and the ceiling of Thus for D 41 we try a 67 The resulting norms are 62 7 43 77 and 72 7 43 6 Since 6 lt l 7 7 we start with 71 As another example7 for D 53 we try 78 and pick 7 with norm 74 rather than 8 with norm 11 We record the rst pair and its norm For example for D 43 we write 71 6 Let h be the absolute value of the norm of the rst pair 11 Choose a b such that a b E 0 mod h and b1 has minimal norm Thus7 for our chosen D 43 we get that b E 5 mod 6 The value of b is chosen from the sequence 51117 and clearly 5 is a winner The norm of 51 is 25 7 43 718 It is no accident that it is divisible by 6 This can be provedlC39hallenge How We shall record the leftover part of its norm7 namely 7 as the next bottom number We x the sign such that the norm of the pair on top is the product of the number below and the number to its left So our record is now 77 1 57 1 6 73 Continue Stop Conclusion Modi cation Now we repeat the above step by setting a 5 and h 3 The choices for b are 14 7 10 and the winner again will be the one closest to the D Now b 7 is the winner with the norm of 71 being 72 7 43 6 Again it is divisible by 3 and the division is 2 So our record is now 771 571 771 6 i3 72 To repeat the step we take a 7h 72 and deduce b 13 5 7 9 with the 777 clear winner 7 So the new display is 7 1 571 771 7 771 6 73 i3 72 Verify that by continuing this way we get 771 571 6 73 771 i2 771 73 571 6 wu 1 When h becomes i1 we stop We have the necessary information to nd our Ly We rst multiply the various numbers in our rst row in the eld F and notice that at each stage certain integers can be factored We throw away these common factors Here are the explicit steps number product Factor out Explanation 7 D 7 D 5 D 78 12D 13 2D Factored the bottom number 6 7 D 177 27D 759 9D Factored the bottom 3 7 D 7800 122D 400 61D Factored out 2 5 D 4623 705D 71541 235D Factored out 3 7 D 720892 3186D 73482 531D Factored out 6 Our answer is z 73482y 7531 and hence also z 3482y 531 How do we know that we are done We multiplied the pairs in the top row whose norms were at the bottom in two parts Thus the norm of the whole product is 663346233661 6 3 2 393 62 But we clearly divided by the same amount due to factoring so the resulting norm is 1 Clearly it is wasteful to write the pairs on top since the second number is always 1 So we shall shorten our display to 757757 673727361

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