d 2 b 2 213 Let y b d Dividing 212 and 213 by d 2 we obtain p 5 y y 1 y 1 2 y 2

D 2 b 2 213 let y b d dividing 212 and 213 by d 2 we

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d ) 2 + b 2 . (2.1.3) Let y = b d . Dividing (2.1.2) and (2.1.3) by d 2 , we obtain p 5 y ( y + 1 ) ( y + 1 ) 2 + y 2 and p 5 y y 2 + 1. (2.1.4) The first of these inequalities simplifies to p 5 ( y 2 + y ) 2 y 2 + 2 y + 1, which yields ( 2 - p 5 ) y 2 + ( 2 - p 5 ) y + 1 0. Dividing by 2 - p 5, which is negative, we obtain y 2 + y - ( 2 + p 5 ) 0. Solving for the roots, we obtain y - 3 - p 5 2 or y 1 + p 5 2 . On the other hand, the second inequality in (2.1.4) can be written y 2 - p 5 y + 1 0, which yields p 5 - 1 2 y p 5 + 1 2 . Therefore the only possibility is y = 1 + p 5 2 , which is a contradiction since y is rational. Case 2: Similar to Case 1 and left to the reader. Theorem 2.1.9. p 5 is the best (i.e. largest) constant possible in Hurwitz Theorem. In other words, for every c > p 5 there exists an irrational number x such that there exist only finitely many rational numbers h k such that x - h k < 1 ck 2 . In fact x = 1 + p 5 2 satisfies this property for all such c. Proof. Suppose there exists infinitely many rational h k with x = 1 + p 5 2 such that h k - x < 1 ck 2 . (2.1.5) 19
We will show that c p 5. By the triangle inequality, h k - 1 - p 5 2 = h k - 1 + p 5 2 + p 5 h k - 1 + p 5 2 + p 5 < 1 ck 2 + p 5. (2.1.6) If we multiply (2.1.5) and (2.1.6), then the left hand side is: h k - 1 + p 5 2 · h k - 1 - p 5 2 = h 2 k 2 - h k - 1 = 1 k 2 h 2 - hk - k 2 1 k 2 . Therefore we obtain 1 k 2 < 1 ck 2 1 ck 2 + p 5 . Multiplying by ck 2 we obtain c < 1 ck 2 + p 5. As k goes to infinity, the right side approaches p 5. Therefore c p 5 as desired. 2.2 Continued Fractions Let a 0 , a 1 ,... be real numbers. Assume that a i > 0 for i > 0. Define a 0 , a 1 ,..., a n = a 0 + 1 a 1 + 1 a 2 + 1 . . . + a n - 1 + 1 a n . Example 3 . 2,3,4 = 2 + 1 3 + 1 4 = 2 + 4 13 = 30 13 . Definition 2.2.1. A simple continued fraction is a continued fraction a 0 , a 1 ,..., a n where each a i is an integer. Clearly, a simple continued fraction is a rational number. The converse is also true: Proposition 2.2.2. Any rational number can be written as a simple continued fraction. Proof. We are given a rational number p q in reduced form. If q = ± 1, then p / q = ± p = ⟨± p , and we are done. Therefore suppose that | q | 6 = 1. We assume without loss of generality that 20
p > 0. We let u 0 = p and u 1 = q . By the Euclidean Algorithm u 0 = u 1 a 0 + u 2 with 0 u 2 < | u 1 | u 1 = u 2 a 1 + u 3 with 0 u 3 < | u 2 | . . . u j = u j + 1 a j . Claim: a 0 , a 1 ,..., a j = p q . Define r i = u i u i + 1 . Dividing the equation u 1 = u i + 1 a i + u i + 2 by u i + 1 , we see that r i = a i + 1 r i + 1 . Starting from r 0 = a 0 + 1 / r 1 , and applying this equation recursively for r 1 , r 2 ,..., until r j = a j , we see that So r 0 = a 0 , a 1 , a 2 ,..., a j . Since r 0 = p / q , this proves the desired result. Note that there is a simple way to alter a given simple continued fraction representation of a given rational number. Namely, suppose we have x = a 0 , a 1 ,..., a j with a j > 1. Then we also have x = a 0 , a 1 ,..., a j - 1,1 , since ( a j - 1 ) + 1 / 1 = a j . To eliminate this ambiguity, we impose the condition on our simple continued fractions that they do not end with 1 except for the case 1 = 1 . We formalize this with the following definition: Definition 2.2.3. We say that a simple continued fraction x = a 0 , a 1 ,..., a j is in simplest form if j = 0 or if j > 0 and a

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• Spring '14
• Math, Number Theory, Prime number, Rational number, kN, Continued fraction

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