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Erick Kory Lawson
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1 IAP 07 Daniel W Stroock Traditional method for computing square roots: The traditional method for computing square roots is based on the identity (a + b) a + (a + b)b For example to compute one begins by finding the largest integer 0 a 0 < 0 for which a 0 Obviously a 0 Next one wants to find the largest integer 0 a < 0 such that (a 0 +a 0 ) or equivalently (a 0 +a 0 )a 0 a 0 and one finds that a One then looks for the largest integer a such that ( (a 0 + a 0 ) + a 0 ) a 0 (a 0 +a 0 ) and finds that a One continues in this way until one runs out of energy In decimal notation this procedure takes the following form: Thus equals apart from an error of at most 0 6 More generally if 0 l N < 0 (l+) then after n steps this procedure will produce the decimal representation of the number M such that 0 N M < 0 l n An alternative method: The traditional method is fine but somewhat tedious In addition from a mathematical standpoint it is not natural To understand the second of these suppose that N is an integer which is not the square of an integer Then N is an irrational number To check this suppose it were rational One could then find integers a and b such that N a b and one could assume that a and b are relatively prime (ie share no common factors greater than ) But then a and b would be relatively prime and N a b which since N and are relatively prime would mean that N a and b Irrational numbers are hard to understand and so when dealing with them one tries to find good rational approximations That is what decimal expansions do but they restrict themselves to rational numbers whose denominators are powers of 0 Other than the fact that most of us have ten fingers there is nothing sacrosanct about the number 0 or its powers and from a mathematical perspective this restriction is arbitrary and foolish To see how removal of the restriction to powers of 0 can improve the computation of square roots again consider interested in is u It is easy to verify that u u+ Obviously the integer part of is and so what we are and because 0 y + x + (y x) (x + )(y + )

2 Now define r 0 0 and r n r n + for n Then r and u r u+ u (u+) u r and u r n n (r n +)(u+) Hence u r n u n Writing r n as a n bn one has that a 0 b 0 and a n b n and b n a n + b n for n In particular r whose decimal expansion begins which is the first five places in the expression for as a decimal For a second example set u and observe that u u+6 u+ In addition note that because 6 y + 6 y + x + 6 x + y x (y + )(x + ) Hence if r 0 0 and r n r n +6 r n + for n then u r n u() n In particular r 0 7 whose decimal expansion begins 66 and coincides with the four places in the decimal expansion of The preceding examples suggest that given any N that is not a square one should set u N N and look for non-negative integers a b c d for which u au+b cu+d Indeed given such coefficients observe that ay + b cy + d ax + b cx + d y x (ac bc) (cx + d)(cy + d) and therefore if r 0 0 and r n ar n +b cr n +d then u r n u ad bc n d Clearly one wants ad bc n to be as small as possible but one can t take it to be 0 since ad bc ax + c bx + d b d for all x Hence the best that one do is have ad bc ± and so we want a b c d for which () u au + b cu + d and ad bc ± It turns out that finding the coefficients a b c d entails a good deal of interesting mathematics In fact it is not at all obvious that they will always exist In the first place c 0 since if it were 0 u would be rational Secondly consider the quadratic f(x) cx + (d a)x b Since f(u) 0 we know that f(x) c(x u)(x r) for some real number r Thus d a c(u + r) and b cur The first of these implies that u + r is a rational number and therefore that r N + s where s is a rational number Combined with the second of these we see that b c N + ( N + s) N + s N

3 which is possible only if s N Thus b c ( N N ) and d a + c N After multiplying the second of these by a and combining it with ad bc ± we obtain a + ac N c ( N N ) ± which means that ( a + c N ) c N ± Hence c N ± is the square of an integer β and () a c N + β b c ( N N ) d c N + β We now know that in order for () to hold β c N ± must be an integer in which case a b and d are given by () In the number theory literature an equation of the form c N ± Z + is called a Pell equation When N one choice is c since and β When N one can take c in which case + 0 and β 0 However finding a c by hand can take a lot of time For example when N the smallest c is and when N 67 it is 567 Thus it is important to develop a systematic procedure for finding c Continued fractions: Given a set {a m : m } Z + define the continued fraction Equivalently () [a n ] a n and [a a n ] a + a + + a n + a n [a m a n ] a m + [a m+ a n ] for m < n Determine p n and q n to be the relatively prime integers such that p n qn [a a n ] The rational number p n qn is called the nth convergent for {a m : m } The crucial equation that allows us to deal with the convergents is ( ) ( ) ( ) pn p () n 0 0 a a n q n q n To prove () one proceeds by induction on n When n there is nothing to do since p 0 0 p q 0 and q a Now assume } that n and that () holds for all choices of sequences {a m : m } and let : k be the convergents for {a +m : m } Then { p k q k ( ) ( p n p n 0 a q n q n ) ( ) 0 a n+ Z denotes the set of all integers N is the subset of non-negative integers and Z + is the subset of strictly positive integers

4 At the same time by () for all m p m q m a + p m q m q m a q m + p m and so since q m and a q m +p m are relatively prime p m q m and q m a q m +p m Hence ( ) ( ) ( ) ( ) ( ) pn p n+ 0 p n p n 0 0 q n q n+ a a a n+ and so From () we have ( pn q n ) ( 0 a q n ) ( 0 a n q n ) ( ) ( ) ( ) 0 pn p n q n q n a n (5) p n a n p n + p n and q n a n q n + q n for n In addition by taking the determinant of both sides of () one finds that (6) p n q n p n q n ( ) n From (6) and (5) one has (7) p n q n p n q n ( )n q n q n and p n q n p n+ q n+ ( )n a n+ q n q n+ From these it follows that there exists a u (0 ) such that (8) p n q n u and p n+ q n+ u We will use [a a n ] to denote the number u in (8) In order to connect these considerations with those in the preceding section observe that () implies that ( ) ( ) ( ) pm+n pm p m p n q m+n q m q m q n where p n q n is the nth convergent for {a m+k : k } Hence p m+n q m+n p m p n + p m q n q m p n + q m q n p p n m q + p m n p q n m q + q m n

5 5 and therefore after letting n we obtain In particular [a a n ] p m [a m+ a m+n ] + p m q m [a m+ a m+n ] + q m () [a a n ] [a m+ a m+n ] [a a n ] p m [a a n ] + p m q m [a a n ] + q m which is exactly the sort of equation for which we are looking In order to apply the preceding we must learn how to expand a number u (0 ) in a continued fraction and the key to doing so is the Gauss map T : (0 ) \ Q (0 ) \ Q given by T u u u The basic observation is that (0) u [a a n ] T m u [a m a m+n ] and a m T m u for m To prove these first note that because [a a n ] a + [a a n ] and [a a n ] (0 ) a u from which (0) is clear when m Next assume that (0) holds for m and note that because [a m a m+n ] a m + [a m+ a m++n ] one finds first that [a m+ a m++n ] T m u T m u T m u and then just as in the case when m that a m+ T u m We now know that if u admits a continued fraction expansion { [a } a n ] then the a m s are given by (0) To show that this prescription works let pn q n : n be the convergents for the {a n : n } in (0) Because we know that these convergents converge it suffices to show that p n qn is greater than or less than u depending on whether n is odd or even Since p and Here and elsewhere Q denotes the set of rational numbers

6 6 q u this is clear when n Let n and assume that the required relationship holds for n and all u (0 ) \ Q Then depending on whether n is even or odd Hence [a a n+ ] < T u or [a a n+ ] > T u [a a n+ ] a + [a a n+ ] is greater than or less than a + T u u depending on whether n is even or odd and so the required relationship holds for n + and all u (0 ) \ Q Back to squareroots: Assume that N Z + is not a square and set u N α 0 where N α 0 We know from () and (0) combined with (6) and () that () T k u u β qk + ( )k Z + p k q k α 0 + β p k q k (N α0) q k q k α 0 + β and u p k u + p k q k u + q k In particular we will know how to produce solutions to the Pell equation as well as coefficients a b c d for which () holds once we show that there is always a k for which T k u u In fact once we know that such a k exists we will know that for all m T mk u u and therefore that q mk is a solution to the Pell equation q N + ( ) mk Z + and p mk p mk q mk q mk can be taken to be the coefficients a b c d in () Referring to the preceding there are two steps in the proof that there is always a k for which T k u u The first step is to show that there is an m 0 and k such that T m+k u T m u The second step is to show that if m < n and T n u T m u then T n u T m u To carry out the first step observe that u N+α0 and therefore that T u N α N α 0 where N + N α0 α0 and α α 0 N α In particular divides N α and so T u N+α and T u N α N + α and α α N α Again divides N α and so T u where N α N + α and α α where

7 Proceeding by induction one sees that T m u N αm m where α m < N and m divides N αm Hence T m u can take only a finite number of values and so two of them must coincide To see that T n u T m u T n u T m u write T m u and T n u as N αm and N αn α m m and check that T m u T n u for l N + αl N αl α n n Hence T n u T m u + α l + α l l + N + αl m N+αm m T l u N+αn n 7 n But This shows that and so N+αl > for l and clearly N + α 0 > Therefore N + αl T l u T l u N + αl + T l u As a consequence one sees that T n u T m u T n u T m u To summarize we have now shown that if N Z + is not a square and u N then there exists a k such that T k u u Moreover if { p m qm the continued fraction expansion of u then N : m } are the convergents for β k q k N + ( )k Z + ( ) p k q k N + β k p k q k N N q k q k N + β k and u p k u + p k q k u + q k Further if k 0 min{k : T k u u} then for any l T l u u l mk 0 Indeed it is obvious that T mk 0 u u To prove the converse suppose that T l u u and write l mk 0 + r where r < k 0 Then u T r T mk 0 u T r u and so r 0 Although I have not proved it here with a little more work one can show that every solution to the Pell equations q N ± Z + corresponds to a periods of the continued fraction expansion of N N In particular if k 0 is even there are no solutions to q N Z + Some Examples () If N m + for some m then u N m u N + m a m T u u and so there are infinitely many solutions to both q N + Z + and q N Z + In addition p 0 0 p q 0 q m and so u u+m

8 8 () If N m + then u N m Thus N + m N m u a m T u T u N + m a m T u u Hence there are infinitely many solutions to q N + Z but none to q N Z + Furthermore p p m q m and q m + Thus β m + and u u + m mu + m + () If m and N m + then u N m (i) If m n then N + m N m u a n T u T u N + m a m T u u Thus there are infinitely many solutions to q N + Z + but none to q N Z Furthermore p p m q n q m + and so β m + and (ii) If m n + then u u + m nu + m + N + m N (m ) u a n T u N + m N T u a T u m m N + N (m ) T u m a T u m N + m N m T u a n T T N + m a 5 m T 5 u u Thus both q N + Z + and q N Z + have infinitely many solutions Furthermore p m p 5 (m + ) q m + q 5 m(m + ) and so β 5 m(m +) and u mu + (m + ) (m + )u + m(m + )

9 () If N m + m then u N m and N + m N m u m a T u m T u N + m a m T u u Thus q N + Z + has infinitely many solutions but q N Z + has none Furthermore p p m q q m and so β m + and u u + m u + m (5) If N m + with m then u N m (i) If m n then N + m N m u a n T u T u N + m a m T u u Thus q N + Z + has infinitely many solutions and q N Z + has none Furthermore p p m q n q mn + and so β mn + and u u + m nu + mn (ii) If m {n + n + } the outcome depends on properties of n N ( + ) + then For example if + u a T u + T u a T u T u a T u + T u a T u T u a 5 T 5 u T 5 u + a 6 8 T 6 u u

10 0 Thus there are infinitely many solutions to q + Z + and none to q N Z + Furthermore p 5 p 6 7 q 5 ; q 6 6 and so β 6 70 and If N 5 ( + ) + then u u + 7 u u a T u T u a T u T u a T u T u a T u T u a 5 T 5 u T 5 u 5 7 a 6 T 6 u u Thus there are infinitely many solutions to 5q + Z + and none to q N Z + Furthermore p 5 p 6 70 q 5 0 q 6 7 and so β 6 6 and If N 8 ( + ) + then u u u u a T u T u a T u T u a T u T u a 0 T u u Thus there infinitely many solutions to q N + Z + and none to q N Z + Furthermore p 7 p 7 q q 7 and so β 7 and u 7u + 7 u + 7

11 If N 67 ( + ) + then u a 5 T u T u a T u T u a T u T u a T u T u a 5 7 T 5 u T 5 u a 6 T 6 u T 6 u a 7 T 7 u T 7 u a 8 T 8 u T 8 u a 5 T u T u a 0 6 T 0 u 8 Thus there are infinitely many solutions to 67q + Z + and none to q N Z + Furthermore p 06 p 0 70 q 567 q and so β 0 88 and u 06u u Exercise Show that if u (0 ) is a rational number then there exists a unique n(u) Z + such that T m u 0 for m n(u) and T m u 0 for m < n(u) In addition show that T n(u) u Z+ \{} and that u [a a n(u) ] where a m T u for m n(u) Conclude that there is a m one-to-one correspondence between the set of rational numbers u (0 ) and the set of continued fractions [a a n ] where n Z + and a n Finally given an example which shows that uniqueness is lost when one drops the condition that a n

CHAPTER 1 REAL NUMBERS KEY POINTS

CHAPTER 1 REAL NUMBERS KEY POINTS CHAPTER 1 REAL NUMBERS 1. Euclid s division lemma : KEY POINTS For given positive integers a and b there exist unique whole numbers q and r satisfying the relation a = bq + r, 0 r < b. 2. Euclid s division