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IOSR Journal of (IOSR-JM) e-ISSN: 2278-5728, p-ISSN: 2319-765X. Volume 16, Issue 4 Ser. III (Jul.–Aug. 2020), PP 21-32 www.iosrjournals.org

A Reviewon the TheoryofContinued

M.Narayan Murty1 andPadala Navina2 1(Retired Reader in Physics, No.503, SR Elegance Apartment, Gazetted officers’ colony, Isukhathota, Visakhapatnam-530022, Andhra Pradesh, India.) 2(Mathematics Teacher, Sister Nivedita School, Ameerpet, Hyderabad – 500016, Telangana, India)

Abstract: In this article, the theory of continued fractions is presented. There aretwo types of continued , one is the finite and the other is the infinite continued fraction. A rational can be expressed as a finite continued fraction. The value of an infinite continued fraction is an . The ratio of two successive Fibonacci , which is a , can be written as a simple finite continued fraction. The can be expressed as an infinite continued fraction. The concept of golden ratio finds application in architecture. Using the convergents of finite continued fraction, the relation between Fibonacci numbers can be calculated and Linear Diophantine will be solved. Keywords:Continued fraction, Convergent, Fibonacci numbers, Golden ratio. ------Date of Submission: 17-07-2020 Date of Acceptance: 01-08-2020 ------

I. Introduction The credit for introducing continued fraction goes toFibonacci. The nickname of famous Italian mathematician Leonardo Pisano(1170-1250) is Fibonacci. In his book LiberAbaci, while dealing with the resolution of fractions 1 1 1 into unit fractions, Fibonacci introduced a kind of “continued function”. For example, he used the symbol 3 4 5 as an abbreviation for 1 1+5 1+ 1 1 1 4 = + + (1) 3 3 3.4 3.4.5 Continued fraction is two types, (a) Finite continued fraction and (b) Infinite continued fraction.A fraction of the form given below is known as finite continued fraction. 1 푎0 + 1 (2) 푎1+ 1 푎2+ 1 푎3+ ⋱ 1 1 푎 + 푛−1 푎푛 Where, the numbers 푎1, 푎2, … , 푎푛 are the partial denominators of the finite continued fraction and they all are real numbers. The number 푎0 may be zero or positive or negative. This fraction is denoted by the symbol 푎0; 푎1, 푎2, … . , 푎푛 and it is called simple if all of the 푎𝑖 are . The value of any finite simple continued fraction will always be a rational number. 1 If 푎 > 1 in the finite continued fraction (2), then 푎 = 푎 − 1 + 1 = 푎 − 1 + , where 푎 − 1 is a 푛 푛 푛 푛 1 푛 positive . Hence, every rational number has two representations 푎0; 푎1, 푎2, … . , 푎푛 and 푎0; 푎1, 푎2, … . , 푎푛−1, 1 as a simple finite continued fraction if 푎푛 > 1. If 푠 = 푎0; 푎1, 푎2, … , 푎푛 , where 푠 > 1, then 1 = 0; 푎 , 푎 , 푎 , … , 푎 (3) 푠 0 1 2 푛 Although due credit is given to Fibonacci for introducing continued fractions, most authorities agree that the theory of continued fractions began with Rafael Bombelli, the great algebraist of Italy. In his book L’ Opera (1572), Bombelliattempted to find the value of roots of integers by means of infinite continued fractions.He expressed 13 as an infinite continued fraction. 4 13 = 3 + 4 (4) 6+ 4 6+6+ ⋱ In general, an infinite continued fraction is written as 푏1 푎0 + 푏2 (5) 푎1+ 푏 푎 + 3 2 푎3+ ⋱

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Where 푎0, 푎1, 푎2, …. and 푏1, 푏2, 푏3, …. are real numbers. The expansion of an infinite continued fraction continues for ever. The infinite continued fraction in which there is 1 in all the numerators is called simple infinite continued fraction. Putting 푏1 = 푏2 = 푏3 = ⋯ = 1 in (5), the simple infinite continued fraction is written as 1 푎0 + 1 (6) 푎1+ 1 푎2+ 1 푎 + 3 푎4+ ⋱ The notation 푎0; 푎1, 푎2, … indicates a simple infinite continued fraction. The value of any infinite continued fraction is an irrational number.

II. Finite Continued Fractions Theorem2.1: Any rational number can be written as a simple finite continued fraction. Proof: Let 푎 푏, where 푏 > 0, is an arbitrary rational number. Now, let us write the following equations. 푎 = 푏푎0 + 푟1 0 < 푟1 < 푏 푏 = 푟1푎1 + 푟2 0 < 푟2 < 푟1 푟1 = 푟2푎2 + 푟3 0 < 푟3 < 푟2 (7) ...... 푟푛−2 = 푟푛−1푎푛−1 + 푟푛 0 < 푟푛 < 푟푛−1 푟푛−1 = 푟푛 푎푛 + 0 The above equations are rewritten as follows. 푎 푟 1 = 푎 + 1 = 푎 + 푏 0 푏 0 푏 푟1 푏 푟2 1 = 푎1 + = 푎1 + 푟1 푟1 푟1 푟2 푟1 푟3 1 = 푎2 + = 푎2 + 푟2 (8) 푟2 푟2 푟3 ...... 푟푛−1 = 푎푛 푟푛 Eliminating 푏 푟1 in the above first using the second equation, we get 푎 1 = 푎 + (9) 푏 0 1 푎1+푟1 푟2 Eliminating 푟1 푟2 in (9) using third equation of (8), we obtain 푎 1 = 푎 + 푏 0 1 푎1 + 푟2 푎2+ 푟3 Continuing in this way we get the following expression. 푎 1 = 푎 + 푏 0 1 푎1 + 1 푎2+ 1 푎3+ ⋱ 1 1 푎 + 푛−1 푎푛 Thus, the rational number 푎 푏is expressed as a simple finite continued fraction. Hence, the Theorem2.1 is 71 proved. As an example, let us apply this Theorem 2.1 to the rational number . 55 71 16 71 = 1 × 55 + 16 = 1 + 55 55 55 7 55 = 3 × 16 + 7 = 3 + 16 16 16 2 16 = 2 × 7 + 2 = 2 + 7 7 7 1 7 = 2 × 3 + 1 = 3 + 2 2 From the above equations, we obtain 71 1 1 1 = 1 + = 1 + = 1 + 55 55 7 1 3 + 3 + 2 16 16 2+ 7 71 1 ⇒ = 1 + 1 (10) 55 3+ 1 2+ 1 3+2 DOI: 10.9790/5728-1604032132 www.iosrjournals.org 22 | Page A Reviewon The Theory of Continued Fractions

This is the simple finite continued fraction of the rational number 71/55. Since, in general, the finite continued fraction (2) is denoted by the symbol 푎0; 푎1, 푎2, … . , 푎푛 , the above continued fraction is denoted by the symbol 1; 3,2,3,2 . As, 2 =1 + 1/1, this continued fraction can also be denoted by the symbol 1; 3,2,3,1,1 . That is, 71 = 1; 3,2,3,2 = 1; 3,2,3,1,1 55 Now, using (10), let us represent 55/71 as continued fraction. 55 1 1 = 71 = 1 (11) 71 1+ 55 1 3+ 1 2+ 1 3+2 Hence, the rational number 55/71 is represented as [0;1,3,2,3,2] or [0;1,3,2,3,1,1].The examples (10) & (11) 19 172 710 15 prove the statement (3).Similarly, the rational numbers , , − and − are denoted as given below. 51 51 457 23 19 = 0; 2,1,2,6 = 0; 2,1,2,5,1 51 172 = 3; 2,1,2,6 = 3; 2,1,2,5,1 51 710 − = −2; 2,4,6,8 = −2; 2,4,6,7,1 457 15 − = −1; 2,1,6,1 23 The sequence of numbers introduced by Italian mathematician Leonardo Pisano is known as Fibonacci sequence, given by 1,1,2,3,5,8,13,21,34,55,89,144,233, ...... (12) Each term in the sequence after the first two is the sum of the immediately preceding two terms. The 푛푡ℎ term, 푡ℎ denoted by 퐹푛 , is called 푛 . Note that 퐹3 = 퐹2 + 퐹1 = 1 + 1 = 2 and 퐹6 = 퐹5 + 퐹4 = 5 + 3 = 8 In general, we can write 퐹푛 = 퐹푛−1 + 퐹푛−2, (푛 ≥ 3)(13) Let us write the following equations for Fibonacci numbers using (13). 퐹푛+1 퐹푛−1 퐹푛+1 = 1 × 퐹푛 + 퐹푛−1 = 1 + 퐹푛 퐹푛 퐹푛 퐹푛−2 퐹푛 = 1 × 퐹푛−1 + 퐹푛−2 = 1 + 퐹푛−1 퐹푛−1 ...... (14) 퐹4 퐹2 퐹4 = 1 × 퐹3 + 퐹2 = 1 + 퐹3 퐹3 퐹3 퐹3 = 2 × 퐹2 + 0 = 2 퐹2 Using the above equations, we obtain 퐹 1 1 푛+1 = 1 + = 1 + 퐹푛 1 퐹푛 1 + 퐹푛−1 퐹푛−1 퐹푛−2 Continuing in this manner, one will get 퐹푛 +1 1 1 = 1 + 퐹푛 = 1 + 1 (15) 퐹푛 1+ 퐹푛−1 퐹푛−1 퐹푛−2 ⋱ 1 1+2 퐹 ⇒ 푛 +1 = 1; 1,1, … ,1,2 = 1; 1,1, … ,1,1,1 (16) 퐹푛 The above expression (16) shows that the ratio of two successive Fibonacci numbers 퐹푛+1 퐹푛 , which is a rational number, can be written as a simple finite continued fraction.

III. Convergentsof Finite Continued Fraction 푡ℎ Definition: The continued fraction made from 푎0; 푎1, 푎2, … . , 푎푛 by cutting off the expansion after the 푘 푡ℎ partial denominator 푎푘 is called the 푘 convergent of the given continued fraction and denoted by 퐶푘 . 퐶푘 = 푎0; 푎1, 푎2, … , 푎푘 (17) 퐶0 = 푎0(18) The convergent 퐶0 is called the zeroth convergent. Going back to the example (10), let us write the successive 71 convergents of = 1; 3,2,3,2 . 55 DOI: 10.9790/5728-1604032132 www.iosrjournals.org 23 | Page A Reviewon The Theory of Continued Fractions

퐶0 = 1 1 4 퐶 = 1; 3 = 1 + = = 1.3333. .. 1 3 3 1 9 퐶2 = 1; 3,2 = 1 + 1 = = 1.2857 …(19) 3+ 7 2 1 31 퐶 = 1; 3,2,3 = 1 + = = 1.2916. .. 3 1 24 3 + 1 2+ 3 71 퐶 = 1; 3,2,3,2 = = 1.2909 … 4 55 The above values of convergents show that, except for the last convergent 퐶4, these are alternately less than or greater than 71/55, each convergent being closer in value to 71/55 than the previous one. The following are the three important properties of convergents of a continued fraction. (a) The convergents with even subscripts form an increasing sequence, that is, 퐶0 < 퐶2 < 퐶4 < ⋯ (b) The convergents with odd subscripts form a decreasing sequence, that is, 퐶1 > 퐶3 > 퐶5 > ⋯ (c) Every convergent with an odd subscript is greater than every convergent with an even subscript. That is, 퐶0 < 퐶2 < 퐶4 < ⋯ < 퐶2푛 < ⋯ < 퐶2푛+1 < ⋯ < 퐶5 < 퐶3 < 퐶1

푡ℎ Theorem3.1: The value of 푘 convergent 퐶푘 of the finite simple continued fraction 푎0; 푎1, 푎2, … . , 푎푛 is 푝푘 푞푘 . 푝푘 퐶푘 = (20) 푞푘 The numbers 푝푘 and 푞푘 (푘 = 0,1,2, … , 푛)are defined as 푝0 = 푎0 푞0 = 1(21) 푝1 = 푎1푎0 + 1 푞1 = 푎1(22) 푝푘 = 푎푘 푝푘−1 + 푝푘−2 푞푘 = 푎푘 푞푘−1 + 푞푘−2푓표푟푘 = 2,3, … , 푛(23) Proof: As per (17) and (18), the convergents of 푎0; 푎1, 푎2, … . , 푎푛 are given by 푎0 푝0 퐶0 = 푎0 = = [Using (21)] (24) 1 푞0 1 푎1푎0+1 푝1 퐶1 = 푎0; 푎1 = 푎0 + = = [Using (22)](25) 푎1 푎1 푞1 1 푎2 푎1푎0+1 +푎0 퐶2 = 푎0; 푎1, 푎2 = 푎0 + 1 = (26) 푎1+ 푎2푎1+1 푎2 For 푘 = 2, we have from(21),(22) & (23), 푝2 = 푎2푝1 + 푝0 = 푎2 푎1푎0 + 1 + 푎0(27) 푞2 = 푎2푞1 + 푞0 = 푎2푎1 + 1(28) Substituting (27) & (28) in (26), we obtain 푝2 퐶2 = (29) 푞2 Noting (24), (25) and (29), we can write in general that 푝푘 퐶푘 = 푞푘 Hence,the Theorem3.1 is proved. 71 Let us see how this theorem works in case of the example = 1; 3,2,3,2 . In this case 푎 = 1, 푎 = 3, 푎 = 55 0 1 2 2, 푎3 = 3 &푎4 = 2.Using (21), (22) and (23), we calculate 푝푘 and 푞푘 for 푘 = 0,1,2,3 &4. 푝0 = 푎0 = 1 푞0 = 1 푝1 = 푎1푎0 + 1 = 3 × 1 + 1 = 4 푞1 = 푎1 = 3 푝2 = 푎2푝1 + 푝0 = 2 × 4 + 1 = 9 푞2 = 푎2푞1 + 푞0 = 2 × 3 + 1 = 7 푝3 = 푎3푝2 + 푝1 = 3 × 9 + 4 = 31 푞3 = 푎3푞2 + 푞1 = 3 × 7 + 3 = 24 푝4 = 푎4푝3 + 푝2 = 2 × 31 + 9 = 71 푞4 = 푎4푞3 + 푞2 = 2 × 24 + 7 = 55 71 Using the above values, the convergents of = 1; 3,2,3,2 aregiven by 55 푝0 1 푝1 4 푝2 9 푝3 31 푝4 71 퐶0 = = = 1, 퐶1 = = , 퐶2 = = , 퐶3 = = and 퐶4 = = 푞0 1 푞1 3 푞2 7 푞3 24 푞4 55 The above values for convergents are same as those given in (19) for the continued fraction [1;3,2,3,2]. So, the Theorem3.1 is proved for the continued fraction [1;3,2,3,2]. 푝푘 푡ℎ Theorem3.2: If 퐶푘 = is the 푘 convergent of the finite simple continued fraction 푎0; 푎1. 푎2, … , 푎푛 , then 푞푘 푘−1 푝푘 푞푘−1 − 푞푘 푝푘−1 = −1 1 ≤ 푘 ≤ 푛(30)

DOI: 10.9790/5728-1604032132 www.iosrjournals.org 24 | Page A Reviewon The Theory of Continued Fractions

Proof: (a) For 푘 = 1, the LHS of (30) = 푝1푞0 − 푞1푝0 = 푎1푎0 + 1 × 1 − 푎1 × 푎0 [Using (21) & (22)] = 1 = −1 1−1 = RHS of (30) for 푘 = 1 So, the given theorem is proved for 푘 = 1. (b) Let us assume that the formula (30) is true for 푘 = 푚, where 1 < 푚 < 푛. For 푘 = 푚 + 1, LHS of (30) = 푝푚 +1푞푚 − 푞푚+1푝푚 = 푎푚+1푝푚 + 푝푚−1 푞푚 − 푎푚+1푞푚 + 푞푚−1 푝푚 [Using (23) for 푘 = 푚 + 1] = − 푝푚 푞푚−1 − 푞푚 푝푚−1 = − −1 푚−1 [Using (30) for 푘 = 푚] = −1 푚 = RHS of (30) for 푘 = 푚 + 1 Hence, the theorem is true for 푘 = 푚 + 1, whenever it holds good for 푘 = 푚.It follows by induction that the theorem is valid for all 푘 with 1 ≤ 푘 ≤ 푛. 푡ℎ Corollary 3.1: If 퐶푘 = 푝푘 푞푘 is the 푘 convergent of the simple finite continued fraction 푎0; 푎1, 푎2, … . , 푎푛 and 푎0 > 0, then 푝푘 = 푎푘 ; 푎푘−1, … … , 푎1, 푎0 (31) 푝푘−1 and 푞푘 = 푎푘 ; 푎푘−1, … … , 푎2, 푎1 (32) 푞푘−1 Proof: Using (21),(22) and (23), we obtain 푝푘 푝푘−2 푝푘 = 푎푘 푝푘−1 + 푝푘−2 = 푎푘 + 푝푘−1 푝푘−1 푝푘−1 푝푘−3 푝푘−1 = 푎푘−1푝푘−2 + 푝푘−3 = 푎푘−1 + 푝푘−2 푝푘−2 푝푘−2 푝푘−4 푝푘−2 = 푎푘−2푝푘−3 + 푝푘−4 = 푎푘−2 + 푝푘−3 푝푘−3 ...... 푝1 1 1 푝1 = 푎1푎0 + 1 = 푎1푝0 + 1 = 푎1 + = 푎1 + 푝0 푝0 푎0 푝 1 1 1 푘 = 푎푘 + 푝푘−1 = 푎푘 + 푝푘−3 = 푎푘 + 1 푝푘−1 푎 + 푝 푘−1 푝 푎푘−1 + 푝푘−4 푘−2 푘−2 푎푘−2+ 푝푘−3 ⋱ 1 1 푎1 + 푎0 푝푘 ⇒ = 푎푘 ; 푎푘−1, … . , 푎1, 푎0 푝푘−1 Hence, the relation (31) is proved. Similarly, using (21),(22) & (23), we can write 푞푘 푞푘−2 푞푘 = 푎푘 푞푘−1 + 푞푘−2 = 푎푘 + 푞푘−1 푞푘−1 푞푘−1 푞푘−3 푞푘−1 = 푎푘−1푞푘−2 + 푞푘−3 = 푎푘−1 + 푞푘−2 푞푘−2 푞푘−2 푞푘−4 푞푘−2 = 푎푘−2푞푘−3 + 푞푘−4 = 푎푘−2 + 푞푘−3 푞푘−3 ...... 푞2 푞0 1 푞2 = 푎2푞1 + 푞0 = 푎2 + = 푎2 + 푞1 푞1 푎1

푞 1 1 1 푘 = 푎푘 + 푞푘−1 = 푎푘 + 푞푘−3 = 푎푘 + 1 = 푎푘 ; 푎푘−1, … . , 푎2, 푎1 푞푘−1 푎 + 푞 푘−1 푞 푎푘−1 + 푞푘−4 푘−2 푘−2 푎푘−2+ 푞푘−3 ⋱ 1 1 푎2 + 푎1

DOI: 10.9790/5728-1604032132 www.iosrjournals.org 25 | Page A Reviewon The Theory of Continued Fractions

Thus, the expression (32) is proved.

푡ℎ Corollary 3.2:If 퐶푘 = 푝푘 푞푘 is the 푘 convergent of the simple finite continued fraction 푎0; 푎1, 푎2, … . , 푎푛 , then (−1)푘−1 퐶푘 − 퐶푘−1 = (33) 푞푘푞푘−1

푝푘 푝푘−1 푝푘푞푘−1−푞푘푝푘−1 Proof: 퐶푘 − 퐶푘−1 = − = 푞푘 푞푘−1 푞푘푞푘−1 Using (30) in the numerator of RHS of the above expression, we get (−1)푘−1 퐶푘 − 퐶푘−1 = 푞푘 푞푘−1 Hence, the relation (33) is proved.

IV. Relation Between Fibonacci Numbers UsingConvergents

In modern usage, the Fibonacci sequence begins with 0. The Fibonacci sequence is given by 0,1,1,2,3,5,8,13,21,34,55,89,.....(34) 푡ℎ If 퐹푘 with 푘 ≥ 0 denotes 푘 Fibonacci number, then 퐹0 = 0, 퐹1 = 1, 퐹2 = 1, 퐹3 = 2, 퐹4 = 3, 퐹5 = 5, 퐹6 = 8, … ..(35) The Fibonacci sequence (34) satisfies the relation 퐹푘 = 퐹푘−1 + 퐹푘−2for 푘 ≥ 2(36) Consider the continued fraction 0; 1,1,1, … ,1 in which all the patial denominators are equal to 1. The first few convergents of this continued fraction are written in terms of Fibonacci numbers as given below. 0 퐹0 1 1 퐹1 1 1 퐹2 퐶0 = 0 = = , 퐶1 = 0; 1 = 0 + = = , 퐶2 = 0; 1,1 = 0 + 1 = = , 1 퐹 1 1 퐹 1+ 2 퐹 1 2 1 3 2 퐹3 3 퐹4 퐶3 = 0; 1,1,1 = = , 퐶4 = 0; 1,1,1,1 = = 3 퐹4 5 퐹5 Looking at the above relations, we can write in general 퐹푘 퐶푘 = for 푘 ≥ 2(37) 퐹푘+1 Using the above expression in (20), we obtain 푝푘 퐹푘 퐶푘 = = (38) 푞푘 퐹푘+1 Hence, we can take 푝푘 = 퐹푘 and 푞푘 = 퐹푘+1(39) Taking the relation (30), we have 푘−1 푝푘 푞푘−1 − 푞푘 푝푘−1 = (−1) Using (39) in the above expression,we obtain 푘−1 퐹푘 퐹푘 − 퐹푘+1퐹푘−1 = (−1) 2 푘−1 ⇒ 퐹푘 − 퐹푘+1퐹푘−1 = (−1) (40) Thus, the relation between Fibonacci numbersis derived from convergents of continued fraction. Forexample, if 2 2 4−1 푘 = 4, LHS of (40) is퐹4 − 퐹5퐹3 = 3 − 5 × 2 = −1 = (−1) , which is the RHS of (40).

V. Solutionof Linear Diophantine Equation UsingConvergents The linear Diophantine equation is 푎푥 + 푏푦 = 푐(41) Where 푎, 푏 &푐 are given integers. A solution of the above equation is obtained by solving the Diophantine equation 푎푥 + 푏푦 = 1 (42) Putting 푥 = 푥0 and 푦 = 푦0 in (42), we get 푎푥0 + 푏푦0 = 1 (43) Multiplication of both sides of above equation with 푐 gives 푎 푐푥0 + 푏 푐푦0 = 푐(44) The solution of Diophantine equation (41) is given by comparing it with (44). So, the desired solution is 푥 = 푐푥0 and 푦 = 푐푦0(45) 푎 Let the rational number is expanded into simple finite continued fraction as given below 푏 푎 = 푎 ; 푎 , 푎 , … , 푎 푏 0 1 2 푛 DOI: 10.9790/5728-1604032132 www.iosrjournals.org 26 | Page A Reviewon The Theory of Continued Fractions

According to (17) and (20), the last two convergents of the above continued fraction are 푝푛−1 푝푛 푎 퐶푛−1 = and 퐶푛 = = (46) 푞푛 −1 푞푛 푏 From the above equation, we can write 푝푛 = 푎 and 푞푛 = 푏(47) By changing the index푘 to 푛, the Eqn.(30) can be written a 푛−1 푝푛 푞푛−1 − 푞푛 푝푛−1 = −1 Using (47) in the above expression, we have 푛−1 푎푞푛−1 − 푏푝푛−1 = (−1) (48) Let us now consider two cases for finding the solution of linear Diophantine equation. (a) If 푛 is odd, the Eqn. (48) is푎푞푛−1 − 푏푝푛−1 = 1.Then,the Eqn.푎푥 + 푏푦 = 1 has a particular solution 푥0 = 푞푛−1and 푦0 = −푝푛−1. Hence, according to (45), the solution of linear Diophantine equation is 푥 = 푐푞푛−1 and 푦 = −푐푝푛−1(49) (b) If 푛is even, the Eqn. (48) is (−푎푞푛−1 + 푏푝푛−1) = 1. So, the Eqn. 푎푥 + 푏푦 = 1 has a particular solution 푥0 = −푞푛−1 and 푦0 = 푝푛−1. Then, the solution of linear Diophantine equation is 푥 = −푐푞푛−1 and 푦 = 푐푝푛−1(50) The general solution of Linear Diophantine equation (41) is given by the equations 푥 = 푐푥0 + 푏푡 , 푦 = 푐푦0 − 푎푡, for 푡 = 0, ±1, ±2, … ..(51) Example: Let us solve the linear Diophantine equation 18푥 + 5푦 = 24. In this example 푎 = 18, 푏 = 5 and 푐 = 24. 18 3 18 = 3 × 5 + 3 = 3 + 5 5 5 2 5 = 1 × 3 + 2 = 1 + 3 3 3 1 3 = 1 × 2 + 1 = 1 + 2 2 18 1 = 3 + = 3; 1,1,2 5 1 1 + 1 1+ 2

In this continued fraction 푛 = 3 and the convergent 퐶2 is given by 1 7 푝2 퐶2 = 3; 1,1 = 3 + 1 = = 1 + 2 푞2 1 Therefore, 푝2 = 7 and 푞2 = 2. For 푛 = 3, the Eqn. (48) is 푎푞2 − 푏푝2 = 1 ⇒ 18 × 2 − 5 × 7 = 1 So, as per the above expression, the particular solution of 18푥 + 5푦 = 1 is given by 푥0 = 2 and 푦0 = −7 According to (45), the particular solution of the Diophantine equation 18푥 + 5푦 = 24 is 푥 = 푐푥0 = 24 × 2 = 48 푦 = 푐푦0 = 24 × −7 = −168 Using (51), the general solution of the givenDiophantine equation is 푥 = 48 + 5푡 , 푦 = −168 − 18푡, 푡 = 0, ±1, ±2, … …

VI. Infinite Continued Fraction

An irrational number can be expressed as an infinite continued fraction.The infinite continued fraction in which there is 1 in all the numerators is known as simple infinite continued fraction. Two distinct infinite continued fractions represent two distinct irrational numbers. An early example of an infinite continued fraction is found in the work of William Brouncker. In 1655, hehad converted Wallis’s famous infinite product 4 3 × 3 × 5 × 5 × 7 × 7 × … … = 휋 2 × 4 × 4 × 6 × 6 × 8 × … … into an infinite continued fraction 4 12 = 1 + 휋 32 2 + 52 2+ 72 2+2+⋱

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Srinivasa Ramanujan (22 December 1887 – 26 April 1920) had contributed many problems on continued fractions to the Journal of the Indian Mathematical Society and his note books contain about 200 results on such fractions. The following are twoinfinite continued fractions given by Ramanujan. 2휋 5 5+ 5 1+ 5 1  푒 − = 푒−2휋 2 2 1+ 푒−4휋 1+ 푒−6휋 1+ 1+⋱ ⋱ 4  휋 = 12 1+ 32 2+ 52 2+ ⋱ ⋱

(1) for expanding an irrational number into a simple infinite continued fraction: Let 푥0 is an arbitrary irrational number and the sequence of integers 푎0, 푎1. 푎2, …. is defined as given below. 1 1 1 푥1 = , 푥2 = , 푥3 = , … …. (52) 푥0−푎0 푥1−푎1 푥2−푎2 In general, the above sequence can be written as 1 1 푥푘+1 = ⇒ 푥푘 = 푎푘 + (푘 ≥ 0) (53) 푥푘 −푎푘 푥푘+1 Let us now expand 푥0 into an infinite continued fraction using the above expression. 1 푥0 = 푎0 + 푥1 Then, by successive substitutionsof (53), we obtain 1 1 푥0 = 푎0 + 1 = 푎0 + 1 푎1 + 푎1 + 1 푥2 푎2+ 푥3 ⋮ 1 ⇒ 푥0 = 푎0 + 1 = 푎0; 푎1, 푎2, … , 푎푛 , 푥푛+1 (54) 푎1+ 1 푎 + 2 푎3+⋱ 1 1 푎푛 + 푥푛+1 Applying the definition (17) and the theorem (20) for the convergent to the above expression, we can write 푥0 = 푎0; 푎1, 푎2, … , 푎푛 , 푥푛+1 = 퐶푛+1 푝푛+1 ⇒ 푥0 = (55) 푞푛+1 Using (23) in the above expression, we have 푥푛+1푝푛 + 푝푛−1 푥0 = [∵ 푎푛+1 = 푥푛+1𝑖푛 (54)] 푥푛+1푞푛 + 푞푛−1 Subtracting the convergent 퐶푛 from 푥0, we get 푥푛+1푝푛 + 푝푛−1 푝푛 푥0 − 퐶푛 = − 푥푛+1푞푛 + 푞푛−1 푞푛 푞 푝 − 푝 푞 = 푛 푛−1 푛 푛−1 푥푛+1푞푛 + 푞푛−1 푞푛 −(−1)푛 −1 = [ Using (23) & (30)] 푞푛+1푞푛 (−1)푛 ⇒ 푥0 − 퐶푛 = 푞푛+1푞푛 71 Since, 퐶 = 푝 푞 ,the values (19) for convergents of = 1; 3,2,3,2 show that, as 푛 increases, the integers 푛 푛 푛 55 푞푛 are increasing. Hence, the above expression gives that lim 푥0 − 퐶푛 ≈ 0 푛→∞ ⇒ 푥0 = lim푛→∞ 퐶푛 = 푎0; 푎1, 푎2, … … . (56) Thus, the irrational number푥0 is expanded into an infinite continued fraction 푎0; 푎1, 푎2, 푎3 … . .The following are four examples for expressing a given irrational number as an infinite continued fraction using this algorithm.

Example 1: Let 푥0 = 5 ≈ 2.236. The calculations for finding the simple infinite continued fraction expansion of 5 using (53) are given below. 1 푥0 = 5 = 2 + 5 − 2 = 푎0 + . i.e., 푎0 = 2 푥1 1 5+2 1 푥1 = = = 5 + 2 = 4 + 5 − 2 = 푎1 + . i.e., 푎1 = 4 5−2 5−2 5+2 푥2

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1 So, 푥 = . According to the above expression, 푎 = 4.similarly, 푎 = 푎 = ⋯ = 4. Hence, the simple 2 5−2 2 3 5 infinite continued fraction representation of 5 is 5 = 2; 4,4,4,4, … … . (57)

Example 2: Let us now expand 7 ≈ 2.6457 as a simple infinite continued fraction. Taking 푥0 = 7, let us make the following calculations using (53). 1 푥0 = 7 = 2 + 7 − 2 = 푎0 + . i.e., 푎0 = 2 푥1 1 7+2 7+2 3+ 7−1 7−1 1 푥1 = = = = = 1 + = 푎1 + . i.e., 푎1 = 1 7−2 7−2 7+2 3 3 3 푥2 3 3( 7+1) 3 7+3 6+ 3 7−3 7−1 1 푥2 = = = = = 1 + = 푎2 + . i.e., 푎2 = 1 7−1 7−1 7+1 6 6 2 푥3 2 2( 7+1) 2 7+2 6+ 2 7−4 7−2 1 푥3 = = = = = 1 + = 푎3 + . i.e., 푎3 = 1 7−1 7−1 7+1 6 6 3 푥4 3 3( 7+2) 12+ 3 7−6 1 푥4 = = = = 4 + ( 7 − 2) = 푎4 + . i.e., 푎4 = 4 7−2 3 3 푥5 1 푥 = = 푥 . Then, we get 푥 = 푥 , 푥 = 푥 &푥 = 푥 . So, 푎 = 푎 = 1, 푎 = 푎 = 1, 푎 = 푎 = 1and 5 7−2 1 6 2 7 3 8 4 5 1 6 2 7 3 푎8 = 푎4 = 4. Then, we obtain 푥9 = 푥5 ,푥10 = 푥6, 푥11 = 푥7&푥12 = 푥8. So, 푎9 = 1, 푎10 = 1, 푎11 = 1and 푎12 = 4. This shows that the block of integers 1,1,1,4 repeat indefinitely. Thus, the simple infinite continued fraction expansion of 7 is given by 7 = 2; 1,1,1,4,1,1,1,4,1,1,1,4, … … (58) Example 3: Let us find out the infinite continued fraction expansion of 휋 = 3.14159265 …. 푥0 = 휋 = 3 + (휋 − 3) i.e., 푎0 = 3 1 1 푥 = = = 7.06251330 … = 7 + 0.06251330 … i.e., 푎 = 7 1 휋−3 0.14159265 … 1 1 푥 = = 15.99659440 … = 15 + 0.99659440 … i.e., 푎 = 15 2 0.06251330 … 2 1 푥 = = 1.00341723 … = 1 + 0.00341723 … i.e., 푎 = 1 3 0.99659440 … 3 1 푥 = = 292.63467 … = 292 + 0.63467 … i.e., 푎 = 292 4 0.00341723 … 4 ...... Thus, the simple infinite continued fraction for 휋 is written as 휋 = 3; 7,15,1,292,1,1,1,2,1,3,1,14,2,1,1,2,2,2,2 … . . (59) Example 4: Let us express 푒, the base of natural , as an infinite continued fraction. 푒 = 2.718281828 …. 푥0 = 푒 = 2 + 푒 − 2 푎0 = 2 1 1 푥 = = = 1.39221119 푎 = 1 1 푒 − 2 0.718281828 1 1 푥 = = 2.549646785 푎 = 2 2 0.39221119 2 1 푥 = = 1.819350221 푎 = 1 3 0.549646785 3 1 푥 = = 1.220479319 푎 = 1 4 0.819350221 4 1 푥 = = 4.535572789 푎 = 4 5 0.220479319 5 ...... Thus, the pattern of infinite continued fraction expansion of 푒 is given by 푒 = 2; 1,2,1,1,4,1,1,6,1,1,8,1,1, … … . (60) The continued fraction representation of 푒 was found by Euler. In 1737, Euler showed that 푒−1 = 0; 2,6,10,14,18, … . (61) 푒+1 푒2−1 and = 0; 1,3,5,7,9, … … (62) 푒2+1 In the above two infinitecontinued fractions, the partial denominators form an progression.Following the given procedure, the value of 푥 in equation 3 2 = 2푥 can be represented by an infinite continued fraction as shown below. log 3 2 푥 = = 0.584962500721 … log 2 ⇒ 푥 = 0; 1,1,2,2,3,1,5,2, , … . . (63)

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Corollary 6.1: We have shown that an irrational number is represented by an infinite continued fraction. The converse is that an infinite continued fraction represents an irrational number. This statement is proved by taking the infinite continued fraction 푥0 = 1; 2,3,1,3,1,3,1, … . 1 ⇒ 푥0 = 1 + 1(64) 2+ 푦 Where, 푦 = 3; 1,3,1,3,1,3,1, … . 1 ⇒ 푦 = 3 + 1 1 + 푦 Solving the above equation, we obtain 푦2 − 3푦 − 3 = 0 3± 21 The solutions of this are 푦 = . Since 푦 > 3, we have to take 푦 = (3 + 21) 2 . 2 Substituting this value of 푦 in (64), we get 1 19− 21 푥0 = 1 + 2 = , which is an irrational number. 2+ 10 3+ 21 Thus, the value of an infinite continued fraction 푥0 is an irrational number. (2) simple periodic infinite continued fraction: If a simple infinite continued fraction contains a block of partial denominators 푏1, 푏2, … , 푏3that repeats indefinitely, then the continued fraction is known as periodic. A simple periodic infinite continued fraction is denoted as 푎0; 푎1, 푎2, … , 푎푚 , 푏1, 푏2, … , 푏푛 , 푏1, . . , 푏푛 , … . .

In short, it is represented as 푎0; 푎1, 푎2, … , 푎푚 , 푏1, 푏2, … . , 푏푛 , where the over bar indicates that this block repeats over and over. The block 푏1, 푏2, … , 푏3 is known as the period of the infinite continued fraction expansion and 푛 gives the length of the period. The following are few examples of simple periodic infinite continued fraction.

(a) 23 = 4; 1,3,1,8,1,3,1,8, … … = 4; 1 , 3 , 1 , 8 In this example, the period is 1,3,1,8 and the length of the period is 4. (b) 2 = 1; 1,2,2,2,2, … … = 1; 1, 2 (c) 3 = 1; 1,2,1,2,1,2, … … = 1; 1 , 2 (d) 5 = 2; 4,4,4,4, … … . = 2; 4 (e) 6 = 2; 2,4,2,4,2,4, … … = 2; 2 , 4 (f) 7 = 2; 1,1,1,4,1,1,1,4, … … = 2; 1 , 1 , 1 , 4 (g) 8 = 2; 1 , 4 (h) 10 = 3; 6 The following are the three general expressions for finding simple periodic infinite continued fractions of irrational numbers. For any positive integer 푛, (i) 푛2 + 1 = 푛; 2 푛 (65) (ii) 푛2 + 2 = 푛; 푛 , 2 푛 (66) (ii) 푛2 + 2푛 = 푛; 1 , 2 푛 (67) Proof of (65): 푛 + 푛2 + 1 = 2푛 + 푛2 + 1 − 푛 1 ⇒ 푛 + 푛2 + 1 = 2푛 + 푛 + 푛2 + 1 Substituting the above expression successively in RHS, we obtain 1 1 2 푛 + 푛 + 1 = 2푛 + 1 = ⋯ = 2푛 + 1 2푛 + 2푛 + 2 1 푛+ 푛 +1 2푛+2푛+⋱ ⋱ ⇒ 푛2 + 1 = 푛; 2푛, 2푛, … = 푛; 2 푛 Thus, the expression (65) is proved. For 푛 = 1,2,3, … 푒푡푐.,this expression gives the infinite continued fractions of rational numbers 2, 5, 10, 17, … , 푒푡푐. Proof of (66): 푛 + 푛2 + 2 = 2푛 + 푛2 + 2 − 푛 (68) 2 ⇒ 푛 + 푛2 + 2 = 2푛 + 푛 + 푛2 + 2 1 = 2푛 + 푛 + 푛2 + 2 2

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Applying (68)successively in RHS ofabove expression, we get 1 1 1 2 푛 + 푛 + 2 = 2푛 + = 2푛 + = 2푛 + 1 2푛 + 푛2 + 2 − 푛 2 푛2+2−푛 푛 + 푛 + 푛+ 푛2+2 2 1 1 1 = 2푛 + 1 = 2푛 + 1 = 2푛 + 1 =..... 푛+ 푛+ 1 푛+ 1 2푛+ 푛2+2−푛 2푛+ 2푛+ 푛+ 푛2+2 2 푛2+2−푛 푛+ 2 1 = 2푛 + 1 푛 + 1 2푛+ 1 푛+ 1 2푛+ 1 푛+⋱ ⋱ ⇒ 푛2 + 2 = 푛; 푛, 2푛, 푛, 2푛, … = 푛; 푛 , 2 푛

Hence, the expression (66) is proved. Using this expression, we can write the infinite continued fractions of rational numbers 3, 6, 11, 18, … , 푒푡푐., taking 푛 = 1,2,3, … 푒푡푐.

Proof of (67): 푛 + 푛2 + 2푛 = 2푛 + 푛2 + 2푛 − 푛 (69) 2푛 ⇒ 푛 + 푛2 + 2푛 = 2푛 + 푛 + 푛2 + 2푛 1 = 2푛 + 2푛 + 푛2 + 2푛 − 푛 2푛 1 = 2푛 + 1 1 + 푛+ 푛2+2푛

By successive substitutions of (69) in RHS of above expression, we obtain

1 1 2 푛 + 푛 + 2푛 = 2푛 + 1 = 2푛 + 1 1 + 1 + 2푛 2푛+ 푛2+2푛−푛 2푛+ 푛 + 푛2+2푛 1 1 = 2푛 + 1 = ⋯ = 2푛 + 1 1 + 1 1 + 1 2푛+ 1 2푛+ 1 1+ 1+ 2 1 푛+ 푛 +2푛 2푛+ 1 1+⋱ ⋱ ⇒ 푛2 + 2푛 = 푛; 1,2푛, 1,2푛, … = 푛; 1 , 2 푛 Thus, the expression (67) is proved. This expression gives the infinite continued fractions of rational numbers 3, 8, 15, 24, … , 푒푡푐., for 푛 = 1,2,3, … 푒푡푐.

(3) Golden ratio as infinite continued fraction:

Consider the infinite continued fraction 1 푥 = 1 + 1 = 1; 1,1,1, … . . (70) 1+ 1 1+ 1 1+ 1 1+⋱ The above fraction shows that 1 푥 = 1 + or 푥2 − 푥 − 1 = 0 푥 1± 5 The solutions of this quadratic equation are 푥 = . Since, 푥 > 1, we must take the solution having positive 2 1+ 5 sign in the numerator of its RHS. Therefore,푥 = = 1.61803 …, which is the golden ratio ø. Hence, the 2 infinite continued fraction (70) represents the golden ratio. That is, 1+ 5 ø = = 1; 1,1,1, … = 1; 1 (71) 2

Importance of golden ratio: A rectangle, whose length and breadth are in the ratio,∅: 1, is Known as the golden rectangle. The ratio of the diagonal of a regular pentagon to its side is equal to the golden ratio ∅.Golden ratio

DOI: 10.9790/5728-1604032132 www.iosrjournals.org 31 | Page A Reviewon The Theory of Continued Fractions finds application in architectural designs.It has been observed that the Greeks have used the concept of golden ratio in the construction of temples. The Parthenon in Athens is the classic example of the golden ratio being used in architecture. It was constructed between 448-432 BC as a temple for the Goddess Athena. The great pyramid in Giza, Egypt, is another example of an ancient structure where golden ratio is used in its design. The ratio of the height of its triangular face to half of the side of its square base approximates to golden ratio. Some ratios in the human body, like the ratio of the height of a person to the distance between the naval point and the foot, are very close to golden ratio. If the ratios between two different parts of human body are close to golden ratio, the body appears beautiful. VII .Conclusion

The theory of finite and infinite continued fractions was discussed in this article. A rational number can be expressed as a finite continued traction and an irrational number can be expressed as an infinite continued fraction. Linear Diophantine equations were solved using convergents of finite continued fraction. The golden ratio was represented as an infinite continued fraction.

References [1]. Burton, David M., Elementary , Tata Me-Graw Hill Publishing Company Ltd. (2007) [2]. Narayan Murty M., Some facts about Fibonacci numbers, Lucas numbers and goldenratio, TheMathematics Student, Indian Mathematical Society, 87 No.1-2, (2018), 133 -144. [3]. Ramaswamy A.M.S., Contributions of to number theory, Research Gate, (2016), 1-6. [4]. SuryB., On the front cover, Resonance, Indian Academy of Sciences, 23 No.7,(2018),729.

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