Farey Fractions. Rickard Fernström. U.U.D.M. Project Report 2017:24. Department of Mathematics Uppsala University


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1 U.U.D.M. Project Report 07:4 Frey Frctions Rickrd Fernström Exmensrete i mtemtik, 5 hp Hledre: Andres Strömergsson Exmintor: Jörgen Östensson Juni 07 Deprtment of Mthemtics Uppsl University
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3 Frey Frctions Uppsl University Rickrd Fernström June, 07
4 Introduction The Frey sequence of order n is the sequence of ll reduced frctions etween 0 with denomintor less thn or equl to n, rrnged in order of incresing size. The properties of this sequence hve een thoroughly investigted over the yers, out of intrinsic interest. The Frey sequences lso ply n importnt role in vrious more dvnced prts of numer theory. In the present tretise we give detiled development of the theory of Frey frctions, following the presenttion in Chpter 6. of the ook MNZ = I. Niven, H. S. Zuckermn, H. L. Montgomery, An Introduction to the Theory of Numers, fifth edition, John Wiley & Sons, Inc., 99, ut filling in mny more detils of the proofs. Note tht the definition of Frey sequence Frey frction which we give elow is priori different from the one given ove; however in Corollry 7 we will see tht the two definitions re in fct equivlent. Frey Frctions Frey Sequences We will ssume tht frction is the quotient of two integers, where the denomintor is positive (every rtionl numer cn e written in this wy. A reduced frction is frction where the gretest common divisor of the 3.5 numertor denomintor is. E.g. 4 is not frction, ut 7 is oth 8 frction reduced frction (even though we would normlly sy tht = 7. Also 8 0 re not frctions, since their denomintors re negtive. We will construct tle in the following wy, where the frctions in ech row of the tle re in the specified order. The st row only contins the frctions 0. If the nth row hs een constructed, then the (n + st row is constructed y copying the nth row then for ech pir of consecutive frctions tht exist in the nth row tht stisfy + n +,
5 the frction + is inserted in the (n + st row etween. For exmple, the nd row is constructed y copying the st row then inserting = etween 0, so tht the nd row is 0,,. When constructing the 3rd row, = 3 is inserted etween = 3 is inserted etween, so tht the 3rd row ecomes 0, 3,, 3,. 0 + When constructing the 4th row, + 3 = 4 is inserted etween = 3 4 is inserted etween 3, ut = is not inserted 5 etween 3 + ecuse = 3 5 is not inserted etween 3 ecuse 5 4. So the 4th row is 0, 4, 3,, 3, 3 4,. The first six rows in the tle re: Definition (Frey sequence. The sequence of frctions in the nth row in the ove tle is clled the Frey sequence of order n. Definition (Frey frction of order n. A Frey frction of order n is frction in the Frey sequence of order n. 3
6 Definition 3 (Frey frction. A Frey frction is Frey frction of some order, i.e. frction in the Frey tle. Theorem (Theorem 6. Corollry 6.3 in MNZ. If re consecutive frctions in the nth row with to the left of, then =. The frctions in the nth row re lso listed in order of their size (in strictly scending order. Proof. Bse cse (n = : 0 = the frctions in the first row re clerly listed in order of their size (in strictly scending order. re frctions Induction hypothesis: Assume for some n N >0 tht if in the nth row with to the left of, then =. Also ssume the frctions in the nth row re listed in order of their size (in strictly scending order. Induction step: We wnt to show tht the frctions in the (n + st row re listed in order of their size (in strictly scending order. We know tht the (n + st row is constructed y copying the nth row then for ech pir of consecutive frctions in the nth row, insert or 0 frctions etween those frctions we lso know y the induction hypothesis tht the frctions in the nth row re listed in order of their size (in strictly scending order. So it s sufficient to show tht for ech pir of consecutive frctions p q, r in the s nth row with p q to the left, p q < p + r q + s < r s. So let p q r e consecutive s frctions in the nth row with p to the left. Then q p q < p + r q + s p(q + s q(q + s < q(p + r q(q + s p(q + s < q(p + r rq ps > 0. But the lst inequlity holds ecuse it follows from the induction hypothesis tht rq ps =. Hence p q < p + r. Similrly it cn e shown tht q + s 4
7 p + r q + s < r. So the frctions in the (n + st row re listed in order of their s size (in strictly scending order from tht it follows tht no frction cn pper twice in the (n + st row. Let e consecutive frctions in the (n + st row with to the left of. We wnt to show tht =. If they re lso consecutive frctions in the nth row with to the left, then it follows from the induction hypothesis tht =. So ssume they re not consecutive frctions in the nth row with to the left. We wnt to show it cn t e the cse tht oth exist in the nth row. Assume for contrdiction tht oth exist in the nth row. If they re not consecutive frctions in the nth row, then there is some frction in the nth row somewhere etween them, sy p. But ecuse of how q p rows re constructed, q must e etween in the (n + st row, which contrdicts eing consecutive frctions in the (n + st row. Assume insted for contrdiction tht re consecutive frctions in the nth row, ut with to the left. Either no frction ws dded etween them when constructing the (n + st row, in which cse will e consecutive frctions in the (n + st row, ut with to the left. This leds to contrdiction. If insted frction ws dded etween when constructing the (n + st row, then re not even consecutive frctions in the (n + st row, which is contrdiction. So it cn t e the cse tht oth exist in the nth row when they re consecutive frctions in the (n + st row with to the left they re not consecutive frctions 5
8 in the nth row with to the left. We wnt to show tht it cn t e the cse tht neither of the frctions exist in the nth row. Assume or contrdiction tht neither of the frctions exist in the nth row. Since they exist in the (n + st row, it must e the cse tht they were oth inserted etween frctions in the nth row. But ecuse t most frction is inserted etween ech pir of consecutive frctions, they must hve een inserted etween distinct pirs of frctions in the nth row. But then it s cler tht some frction must exist etween in the (n + st row, which contrdicts them eing consecutive frctions in the (n + st row. So it s impossile tht neither of the frctions exist in the nth row when they re consecutive frctions in the (n + st row with to the left. The remining cse is when one of the frctions exist in the nth row ut the other doesn t. Assume it s the left frction tht exists in the nth row (if insted it s the right frction tht exists in the nth row, then the proof cn e done nlogously. The right frction must hve een inserted in the (n + st row etween the frction in the nth row directly to the right of, sy p. Then the nth (n + st row will look something like q p Row n: q this: + p Row n + :... + q = p.... q Then = ( + p ( + q = p q =. {y induction hypothesis} 6
9 Corollry (Corollry 6. in MNZ. Every frction reduced form, i.e. gcd(, =. in the tle is in Proof. Let n N >0. We wnt to show every frction in the nth row hs gcd(, =. Let e frction in the nth row. Since there is more thn frction in the nth row, there is frction, sy p q next to in the nth row. Assume p q is to the right of (if it is to the left of, then the proof cn e done nlogously. By Theorem, p q =. Consider the diophntine eqution x + y =. It hs solution iff gcd(, =. But x = p, y = q is solution, so gcd(, =. Theorem 3. n N >0 : if row, then + n +. re consecutive frctions in the nth Proof. For n =, + = +, so it s true for n =. Induction hypothesis: Assume it s true for some n N >0. Induction step: We wnt to show it s true for n +. Let e consecutive frctions in the (n+st row with to the left. If they re lso consecutive frctions in the nth row then y the induction hypothesis, + n +. It cn t e the cse tht + = n +, ecuse if tht were the cse then + n +, so in the (n + st row frction should hve een inserted etween, ut then cn t e consecutive frctions in the (n + st row, contrdiction. So + > n +, i.e. + n +. Otherwise if they re not consecutive frctions in the nth row, exctly of the frctions pper in the nth row like in Theorem. Sy ppers in the nth row p q is the frction in the nth row directly to the right of. Then is the frction etween p q, so = + q. By the induction hypothesis, +q n+. And hence + = ++q ++q +(n+ = n+. 7
10 Theorem 4. Let n N >0 let p e frction in the nth row. Then q n q with equlity iff p q does not exist in ny previous row. Proof. We will egin y showing the first prt, i.e. frction in the nth row, then q n. n N >0 : if p q is For n = the only frctions re 0,. Induction hypothesis: Assume for some n N >0 tht for every frction p q in the nth row, q n. Induction step: Let p q tht q n +. p Cse : exists in the nth row s well. e frction in the (n + st row. We wnt to show q q n < n +. Cse : p q does not exist in the nth row. Then p q Then y induction hypothesis, must lie etween frctions tht exist in the nth row, sy. Then q = + ecuse of the rules of how rows re constructed, + n +, so q n +. Now we wnt to prove the second prt, i.e. n N >0 : if p q is frction in the nth row, then q = n iff p q does not exist in ny previous row. For n =, we see tht q = p does not exist in ny previous row, so q the theorem holds for n =. Sy n > let p q e frction in the nth row. If p exists in previous q row, then since rows re constructed y copying the previous row, p q must exist in the (n st row. This mens we re in cse of the proof of the previous prt of this theorem, so q < n, i.e. q n. If insted p doesn t exist in previous row, then we re in cse of the proof q p of the previous prt of this theorem, i.e. q lies etween the frctions tht exist in the (n st row. Then q = + n. But ecuse of 8
11 Theorem 3, + n. Since q n q n we must hve tht q = n. Lemm 5. Let n N >0 let x N e such tht 0 x n + gcd(x, n + =. Let e the gretest frction in the nth row tht is less x thn thn let e the smllest frction in the nth row tht is greter n + x n +. Then + = x + = n +. Proof. It cn t e the cse tht x n + would contrdict Theorem 4 (n + n. So then frctions in the nth row with to the left < is frction in the nth row, since tht x n + <. re consecutive We see tht < x x (n + > 0 x (n +. n + Consider the diophntine eqution u v = k, where k := x (n + u v re the unknowns. It hs prticulr solution u, v = k, k since k k = k( = k. { y Theorem } We hve tht gcd(, = ecuse of Corollry, so it hs the generl solution u, v = k + e, k + e, e Z. But we lso know it hs prticulr solution u, v = x, n +, so m Z : x = k + m n + = k + m. Let m Z e such tht k + m = n + ( k + m = x. ( 9
12 We wnt to show tht k = m =. Since 0 x n + gcd(x, n + =, it must e the cse tht x > 0. We hve tht x n + < n + > x (n + x > 0 (n + x (k + m (k + m {y ( (} ( m m. { y Theorem } This shows tht m. We lso know tht k. If we ssume for contrdiction tht k > or m >, then + < m + k = n +, {y (} ut tht s impossile, since y Theorem 3, + n +. This shows tht k m. Since lso m k, it follows tht k = m =. Eqution ( now ecomes + = n + eqution ( ecomes + = x. Theorem 6 (Theorem 6.5 in MNZ. If 0 x y, gcd(x, y =, then the frction x ppers in the yth ll lter rows. y Proof. It is cler tht if x y ppers in the yth row then it lso ppers in ll lter rows, so it is enough to show tht x y ppers in the yth row. If y = then either x = 0 or x =. Both 0 pper in the st row, so the theorem holds for y =. We wnt to show the theorem holds when y, i.e. when y = n + for some n N >0. Let n N >0, 0 x n + gcd(x, n + =. If x = 0 or x = n + then gcd(x, n + = n + >, so it must e the cse tht 0 < x < n +. 0
13 x We wnt to show tht n + ppers in the (n + st row. Let e the x gretest frction in the nth row tht is less thn let e the n + x smllest frction in the nth row tht is greter thn. By Lemm 5 we n + know tht + = x + = n +. From the first few lines of the proof x of Lemm 5 we know tht n + does not exist in the nth row, so must e consecutive frctions in the nth row. Becuse + = x + n + x how rows re constructed, we know tht must e inserted in the n + (n + st row etween. Corollry 7 (Corollry 6.6 in MNZ. The nth row consists exctly of ll reduced frctions such tht 0 0 < n. The frctions re listed in order of their size. Proof. Let n N >0 let e reduced frction such tht 0 0 < n. We wnt to show tht ppers in the nth row. Clerly 0 n. By Theorem 6 we hve tht ppers in the th ll lter rows, so ppers in the nth row. So every reduced frction such tht 0 0 < n exist in the nth row, they re listed in order of their size y Theorem. We wnt to show tht the nth row doesn t contin ny other frctions thn these. Let p e frction in the q nth row. The frction p must e reduced y Corollry. By Theorem 4 q we hve tht q n q > 0 y the definition of frctions. So 0 < q n. Becuse rows re constructed y copying the previous row then inserting frctions etween the frctions the fct tht rows re ordered y size, we cn never get row with frction tht is strictly greter thn less thn 0. So 0 p q. or strictly
14 Theorem 8. If re Frey frctions with = ( thus <, then re consecutive Frey frctions of some order. Proof. Let e Frey frctions with = (3 ( thus <. Let c d e Frey frction such tht < c d <. We wnt to show tht d > mx(,. From < c d c d < { c d > 0 d c > 0. it follows tht Let k := c d l := d c. Then k l re positive integers { c d = k (4 d c = l. Consider the diophntine eqution We know tht (k (k = k( x y = k. (5 = k. {y (3} Hence x, y = k, k is prticulr solution to (5. Becuse gcd(, = (this follows from Corollry, the generl solution to (5 is x, y = k + m, k + m = k, + m,, m Z. Consider the diophntine eqution Since (l + (l = l( x + y = l. (6 = l {y (3}
15 it follows tht x, y = l, l is prticulr solution to (6. We hve tht gcd(, = ecuse of Corollry, so the generl solution to (6 is x, {( y = ( l } + n, l + n = l, + n,, n Z. We know tht, is sis for the vector spce R, since ( det = = ( = {y (3} 0. ( ( u u But then every vector R cn e written in unique wy s = ( ( v v c +c, where c, c R. From (4 it follows tht x, y = c, d is prticulr solution( to oth (5 ( (6. ( By looking t the generl solution c of (5 we see tht = k d + m for some m Z. By looking ( ( ( c t the generl solution of (6 we see tht = l + n d for some ( c n Z. By compring these wys of writing using the fct tht {( ( } d, is sis for R we conclude tht l = m k = n. Thus ( ( ( c = k d + l. But then d = k + l + > mx(, =: m. In the ove equtions we used the fct tht k l re positive integers. By Corollry 7 it follows tht c does not exist in the mth row, since d > m. d But c d ws n ritrry Frey frction lying strictly etween. So no Frey frction lying strictly etween exist in the mth row, ut oth exist in the mth row. So must e consecutive Frey frctions in the mth row. 3
16 Corollry 9. Let e Frey frctions such tht <. Then = iff re consecutive Frey frctions of some order. Proof. = : Assume =. Then it follows from Theorem 8 tht re consecutive Frey frctions of some order. = : Assume re consecutive Frey frctions of some order. Then it follows from Theorem tht =. Theorem 0 (Theorem 6.4 in MNZ. If re consecutive frctions in the nth row (with to the left, then mong ll frctions p such tht q < p q <, p q = + is the unique frction with smllest denomintor. + Proof. Let n N >0 let e consecutive frctions in the nth row, with to the left. There re possiilities. Cse : + n +. Then + n + y Theorem 3, so + = n +. We hve tht + will e inserted in the (n + st row etween +. Cse : + > n +. Then for ll integers k such tht n k < + : will e consecutive frctions in the kth row + will e + inserted etween in the ( + th row. This is ecuse the kth row is constructed y copying the (k st row inserting frctions etween the frctions. If re consecutive frctions in the (k st row + ws not inserted etween them in the kth row (i.e. + k, then + will e consecutive frctions in the kth row s well. If re consecutive frctions in the (k st row then frction is only inserted etween them in row k if + k. 4
17 In oth cses we get tht, + + re 3 consecutive frctions in the ( + th row tht re consecutive frctions in the ( + st row. Let x y e reduced frction such tht < x y <. Since re frctions in the nth row, it follows from Corollry 7 tht 0 0. Since < x y <, we hve tht 0 x. If we ssume for y contrdiction tht y < +, then y Theorem 6, x y ppers in the (+ st row. But tht s impossile since we know from Theorem tht frctions re listed in order of their size there is no frction strictly etween in the (+ st row. So y +, i.e. reduced frction tht lies strictly etween cn t hve denomintor smller thn +. Suppose x y hs miniml denomintor, i.e. y = +. By Theorem 6, x exists in the y ( + th row. But since, + re 3 consecutive frctions in the + (+ th row the (+ th row is ordered y size y Theorem, the only frction in the ( + th row tht lies strictly etween x y must e the frction + +. is + +. So If you hve frction p q tht lies strictly etween tht is not reduced, then it cn t e the cse tht q +, since then you could simplify p q into reduced frction where the denomintor is strictly less thn +, ut tht s impossile. Sy you hve frction r s tht lies strictly etween. Then s +, since if r s is not reduced, then s > + if r s is reduced, then s +. Also since + exists in the ( + th row it must e reduced frction + ecuse of Corollry. So the smllest denomintor r s cn hve is +, 5
18 in tht cse r must e reduced frction. So if you hve frction tht s lies strictly etween with smllest denomintor, then it must e reduced frction with denomintor + (since + is the smllest denomintor. But the only reduced frction with denomintor + tht lies strictly etween is +. So mong ll frctions tht lie strictly etween +, + is the unique frction with smllest denomintor. + Proposition (Prolem 6.. in MNZ. Let n e positive integer such tht n > let e the Frey frctions immeditely to the left the right of respectively in the Frey sequence of order n. Then = n = +, i.e. is the gretest odd integer n. It is lso true tht + =. Proof. By Corollry 7 we know tht exists in the Frey sequence of order n since n. We will prove the Proposition y induction. Bse cse (n = : Then we see from the Frey tle on pge 3 tht is 0 is. We hve tht =, is the gretest odd integer 0 + =. Induction hypothesis: Assume for some n tht if re the Frey frctions immeditely to the left the right of the frction respectively in the Frey sequence of order n, then = is the gretest odd integer n + =. Induction step: Let e the Frey frctions immeditely to the left respectively in the Frey sequence of order n. the right of the frction Let c d c d e the Frey frctions immeditely to the left the right of the frction respectively in the Frey sequence of order n +. We wnt to 6
19 show tht d = d tht d is the gretest odd integer n+ c+c = d. Cse : + = + = n + ( = y induction hypothesis. Then when constructing the (n + st row the frction + + is inserted etween lso the frction + + is inserted etween. Then it must e the cse tht c + c is the frction is the frction +, so then d + d + d = + d = +. Also from the induction hypothesis we see tht re odd, since + = n + we hve tht n = +, i.e. n is even. From the induction hypothesis we know tht is the gretest odd integer n. Since d = + we need to show tht + is the gretest odd integer n +, ut this is ovious, since we know tht n is even. Since d = + d = + = y induction hypothesis, it follows tht d = d, so d = d d is the gretest odd integer n +. We lso see tht c + c = {c = + c = + } = + {y induction hypothesis} = d. Cse : + = + > n + ( = y induction hypothesis. Then no frction will e inserted etween or etween constructing the (n + st row. So c d is the frction. We know tht n + < + n +, { n y Theorem 4} when c is the frction d so it follows tht + = n +, i.e. = n. But is odd y the induction hypothesis, so n must e odd. We know from the induction hypothesis tht is the gretest odd integer n d =. We need to show tht is the gretest odd integer n + But tht is ovious. 7
20 Since = y induction hypothesis, = d = d it follows tht d = d. So d = d d is the gretest odd integer n +. We lso see tht c + c = + { = c = c } = {y induction hypothesis} = d. There is no cse where + = + < n + ecuse of Theorem 3. Theorem (Prolem 6..9 in MNZ. For ech Frey frction ( ( let C denote the circle in the plne of rdius ( center, (. These circles re clled the Ford circles. The interior of Ford circle contins ( ( no point of ny other Ford circle two Ford circles C C re tngent if only if order. re consecutive Frey frctions of some Proof. First ssume tht re distinct Frey frctions such tht ( < (. We wnt to show tht neither of the Ford circles C C ( contin point tht is n interior point of the other Ford circle. C hs rdius center r := (, ( =, r. 8
21 ( C hs rdius center s := ( (, ( ( =, s. ( ( Let d e the distnce etween the center of C the center of C. Then ( d = + (s r = ( + s + r rs. (7 ( ( ( In order to show tht the Ford circles C C don t contin ny interior points of the other Ford circle, it is sufficient to show tht d r + s, i.e. d r + s + rs. Using (7 we see tht we must show tht ( + s + r rs r + s + rs ( ( ( 4rs = ( (. (8 It cn t e the cse tht = 0, since then = 0 = =, {divide oth sides y } which contrdicts the Frey frctions eing distinct. But then is nonzero integer, hence its squre must e greter thn or equl to, i.e. (8 ( ( is true. So the Ford circles C C don t contin ny interior 9
22 points of the other Ford circle d r + s. ( ( We wnt to show tht C C re tngent iff consecutive Frey frctions of some order. If re the sme Frey ( ( frction, then the Ford circles C C re the sme re therefore not tngent re not consecutive Frey frctions of some order, so the Theorem holds in tht cse. So ssume re distinct. Without loss of generlity ssume tht < like efore. We know ( ( tht C C re tngent if only if d = r + s or d = r s. Since r s re positve, it follows tht r s mx(r, s < r + s. But efore we showed tht d r + s, so it cn t e the cse tht d = r s. ( ( Therefore C C re tngent if only if d = r + s. But d = r + s iff d = r + s + rs. Using (7 we see tht d = r + s + rs ( ( ( ( = 4rs = ( ( = = ±. + s + r rs = r + s + rs But it cn esily e shown tht < iff > 0, since we ssumed tht <, it must e the cse tht > 0. Therefore = ± =. re From Corollry 9 it follows tht = re consecutive 0
23 ( Frey frctions of some order. So C C if re consecutive Frey frctions of some order. ( re tngent if only Remrk. We will now use n lterntive method to show tht if ( ( re consecutive Frey frctions of some order, then C C re tngent. When using this method we will find explicitly the coordintes for the intersection of the Ford circles. Proof. Let e consecutive Frey frctions of order n, with <. ( ( We wnt to show tht C C re tngent. The eqution for ( C is ( x ( + y = (9 4 4 ( the eqution for C is ( (x + y = ( 4(. (0 4 ( Then the xcoordinte of center of C is less thn (i.e. to the left of the ( xcoordinte of the center of C. Let L e the line tht goes through
24 ( the center of C the center of C ( ( ( / ( ( ( = / ( ( ( ( = / ( = (. ( The eqution for L is y ( = ( (x ( L intersects C. It s slope is {y Theorem }. We wnt to see where, so we plug the eqution for L into (0 get ( (x ( + (x = 4( ( 4 ( ( ( + x = 4( 4 ( 4 ( 4 ( + ( ( + (( (x = 4 ( 4( 4 ( + ( (x = 4( 4 (x = ( ( + ( x = ± ( ( + ( x = ± ( + (. We only cre out the left intersection, i.e. where x = ( + (.
25 If we plug this xvlue into the eqution for L solve for y we get So x = ( C y = ( + ( ( + ( = ( ( ( ( + ( + ( = ( ( + ( ( ( ( + ( = + (. ( + ( y = + ( is the left intersection of L (. Plugging. We wnt to show tht this point lso elongs to C 3
26 the xvlue yvlue into the left h side of (9 gives ( ( + ( ( + + ( ( = ( + ( ( + ( + ( + ( ( + ( ( ( = ( + {y Theorem } ( + ( ( + ( ( + ( = ( + ( ( + ( ( ( + ( + ( ( ( 3 ( ( ( 4 = + ( ( + ( ( ( + ( = 4 ( ( 4 ( 6 + ( ( 4 ( + ( = ( 4 ( + ( 4 4 ( 4 ( + ( = 4, 4 ( which mens tht ( + (, elongs to C + ( ( ( L, so it must e the cse tht C C (, C re tngent. ( Theorem 3 (Prolem 6.. in MNZ. The numer of Frey frctions of order n stisfying the inequlities 0 n is + φ(j their sum is exctly hlf this vlue. Proof. We wnt to prove the first prt of the theorem. Let n e positive integer. Every Frey frction of order n must stisfy the inequlities 4 j=
27 0 y Corollry 7, so the Frey frctions of order n stisfying the inequlities 0 re precisely the Frey frctions of order n. By Corollry 7 the Frey frctions of order n re precisely the reduced frctions such tht 0 0 < n. For integers j such tht j n, define A(j := the numer of reduced frctions j such tht 0 j. Also let F (n := the numer of Frey frctions of order n. n Then clerly F (n = A(j. It is cler tht A(j is the numer of Frey j= frctions of order n with denomintor j. We hve tht A( =, ecuse 0 re the only reduced frctions with denomintor tht lie etween 0. The definition of A(j cn e rewritten y multiplying the lst inequlities y j using the definition of reduced frction to otin A(j = the numer of frctions j such tht gcd(, j = 0 j. Different vlues for give different frctions, so we cn rewrite A(j s j A(j = the numer of integers such tht gcd(, j = 0 j. For j we hve tht gcd(0, j = j, so for j we cn rewrite A(j s So A(j = the numer of integers such tht gcd(, j = j F (n = n A(j = A( + j= n A(j = + j= n φ(j j= = + φ( + = + n φ(j j= n φ(j. j= = φ(j. 5
28 We wnt to prove the second prt of the theorem. Let S(n e ( the sum of the Frey frctions of order n. We wnt to show tht S(n = n + φ(j. S(n = is Frey frction of order n If is Frey frction of order n, then = is lso Frey frction of order n. This is ecuse if is Frey frction of order n then y Corollry 7 we hve tht 0, so 0 if gcd(, =, then gcd(, = hence y Corollry 7, is lso Frey frction of order n. If 3, then it cn t e the cse tht = if is Frey frction of order n, since tht would imply =, i.e., ut lso = >, so tht is common divisor of tht is strictly greter thn, which mens tht gcd(,. So S(n = + is Frey frction of order n. j= is Frey frction of order n > ( The terms in the right sum cn e pired up s,. The terms in ech pir dd up to. If we ssume n then the only Frey frctions of order n with denomintor re 0, n, so there re + φ(j 3 j= ( n terms eing dded in the right sum, so there re φ(j pirs eing j= ( n summed in the second sum. Thus the right sum equls φ(j. The left sum equls 3, so ( n ( S(n = 3 + φ(j = + j= 6 j= n φ(j. j=.
29 ( If n =, then S(n = 0 + = = + for ll n. φ(j, so the theorem holds j= 3 Rtionl Approximtions Theorem 4 (Theorem 6.7 in MNZ. Let c e Frey frctions of d order n such tht no other Frey frction of order n lies etween them. Then + c + d = ( + d (n + c d + c + d = d( + d d(n +. Proof. For the first formul we hve + c + d = ( + d ( + c ( + d d c = ( + d { } = y Theorem ( + d. {since + d n + y Theorem 3} (n + The second formul is otined in similr wy. Theorem 5 (Theorem 6.8 in MNZ. Let n N >0 x R. Then there is rtionl numer such tht 0 < n x (n +. 7
30 Proof. Let n N >0 x R. Let k e the unique integer such tht 0 x + k <. There re cses. Cse : x+k hs the sme vlue s some Frey frction of order n. Let p q e the Frey frction of order n tht x+k simplifies to. Let the rtionl numer e p k q. Since p is Frey frction of order n, q n y Theorem 4. q q So is rtionl numer such tht 0 < n x = x p k q q = x + k p k q k q q q = p q p q = 0 (n +. Cse : x + k does not hve the sme vlue s some Frey frction of order n. It cn t e the cse tht x + k = 0, since 0 is Frey frction of order n. So 0 < x + k <. Let p e the gretest Frey frction of order n tht is q smller thn x + k let r s e the smllest Frey frction of order n tht is greter thn x + k. There is no Frey frction of order n tht lies etween p q r s. Since p q r re Frey frctions of order n, it follows from Theorem s 4 tht q n s n. There re cses now. p + r Cse. : q + s x + k. Let the rtionl numer e r k s. We hve s 8
31 tht is rtionl numer such tht 0 < n x = x r k s x s = r + k s = r s (x + k r s p + r { q + s since p + r q + s x + k < r } s { } y Theorem 4 s(n + = (n +. p + r Cse. : q + s > x + k. Let the rtionl numer e p k q. We hve q tht is rtionl numer such tht 0 < n x = x p k q q = x + k p q p + r q + s p { q since p q < x + k < p + r } q + s = p q p + r q + s { } y Theorem 4 q(n + = (n +. Theorem 6 (Theorem 6.9 in MNZ. If ξ R \ Q, then there re infinitely mny distinct rtionl numers such tht ξ <. 9
32 Proof. Let ξ R \ Q. For ech n N >0, let { ξ A n := Q 0 < n }. (n + Let X := {A n n N >0 } By Theorem 5, ech A n is nonempty, so tht X is set of nonempty sets. By the xiom of choice, there exists choice function f : X X such tht A X : f(a A. So let f : X X e function such tht A X : f(a A. For ech n N >0, let n e the rtionl numer f(a n. Then ξ n n (n + n n < n {since n + > n } for ll n N >0. We wnt to show there re infinitely mny distinct n. Assume for contrdiction there re only finitely mny distinct n n. Then there re only finitely n mny distinct vlues for ξ n. Let d := min ξ n. Clerly d 0 d 0, since d = 0 would imply tht ξ = k n n N >0 k n for some k N >0, ut tht contrdicts ξ eing irrtionl. So d > 0. Then for ll n N >0 we hve tht d ξ n n n (n + n +. {since n } But if we let n e sufficiently lrge (n d will work, then n + < n d. But then d < d, contrdiction. Lemm 7 (Lemm 6.0 in MNZ. If x y re positive integers then not oth of the inequlities xy ( 5 x + y x(x + y ( 5 x + (x + y cn hold. 30
33 Proof. We will rewrite the inequlities. We see tht xy ( 5 x + y xy y + x 5 (xy { 5xy y + x multiply oth sides y } 5(xy x(x + y ( 5 x + (x + y x(x + y (x + y + x 5 (x(x + y 5x(x + y (x + y + x. {multiply oth sides y 5(x(x + y } Assume for contrdiction tht oth 5xy y + x ( 5x(x + y (x + y + x ( re true. By dding the inequlities ( ( we get 5(x + xy 3x + xy + y 3x + xy + y 5(x + xy (3 5x ( 5 xy + y 0 (5 5 + x 4( 5 xy + 4y 0 {multiply y } (y ( 5 x 0 y ( 5 x = 0 5 = y + x, x ut tht contrdicts 5 eing irrtionl. 3 {since squres re nonnegtive} { solve for } 5
34 Lemm 8. If c d re consecutive Frey frctions of order n with to the left, then c d n. Proof. The frctions 0 n, n, n,, n n, n n (3 re Frey frctions of order n if they re simplified y Corollry 7 (the simplified frctions lie etween 0 the denomintor of ech simplified frction lies etween n. It cn t e the cse tht < j n c d > j n for some j n in the sequence (3, since then j would e Frey frction of order n n (if simplified tht lies strictly etween c, which would contrdict d c d eing consecutive Frey frctions of order n. So c must oth d lie etween consecutive elements in (3, which mens c d n, since the distnce etween consecutive elements of (3 is lwys n. Theorem 9 (Theorem 6. in MNZ. Given ny ξ R \ Q, there exist infinitely mny different rtionl numers h such tht k ξ h k <. 5k Proof. Let µ e the unique integer such tht 0 ξ + µ <. It cn t e the cse tht ξ + µ = 0, since tht would imply tht ξ is rtionl, which is not the cse. So 0 < ξ + µ <. Let λ := ξ + µ. Then 0 < λ <. Let n e positive integer. Let n e the gretest Frey frction of order n tht is less thn λ let c n d n thn λ. Then n n n e the smllest Frey frction of order n tht is greter c n d n re consecutive frctions in the nth row. We wnt 3
35 to show tht t lest one of the numers n, c n n + c n will work s h n d n n + d n k in the theorem (when ξ = λ. Assume for contrdiction this is not the cse. Then λ n, (4 n 5 n Cse : n + c n n + d n < λ. Then (6 ecomes c n λ (5 d n 5d n λ n + c n n + d n 5(n + d n. (6 λ n + c n n + d n 5(n + d n. By dding the inequlities (4 (5 we get c n n ( + d n n 5 n d n nc n n d n ( + n d n 5 n d n ( + n d n 5 n d n { y Theorem } (7 y dding the inequlities (5 (6 we get c n n + c n ( + d n n + d n 5 d n ( n + d n c n( n + d n d n ( n + c n d n ( n + d n nc n n d n d n ( n + d n ( + 5 d n d n ( n + d n 5 ( d n + ( + 5 d n ( n + d n ( n + d n. ( n + d n { y Theorem } (8 33
36 But not oth the inequlities (7 (8 cn e true y Lemm 7 (if we let x = d n y = n in the lemm, so we get contrdiction. Cse : n + c n n + d n > λ. Then (6 ecomes n + c n n + d n λ 5(n + d n. Just like in cse we cn dd the inequlities (4 (5 to get (7. By dding the inequlities (4 (6 we get n + c n n ( + n + d n n 5 n ( n + d n n( n + c n n ( n + d n ( + n ( n + d n 5 n nc n n d n n ( n + d n ( + 5 n ( n + d n n ( n + d n ( +. 5 n ( n + d n ( n + d n { y Theorem } (9 But not oth inequlities (7 (9 cn e true y Lemm 7 (if we let x = n y = d n in the lemm, so we get contrdiction. Thus one of the numers n n, c n d n in the the n + c n n + d n orem (when ξ = λ. So for ech n N >0, let h n c n d n n + c n n + d n such tht k n λ h n <, 5k n k n will work s h k e one of the frctions n n, where n is the gretest Frey frction of order n tht is less thn λ c n n d n is the smllest Frey frction of order n tht is greter thn λ. We wnt to show tht for every ɛ > 0 there exists Frey frction h n k n (s 34
37 k n defined ove such tht λ h n < ɛ. Let ɛ > 0 let n e positive integer such tht n < ɛ (i.e. n > ɛ. We know tht n n c n d n re consecutive frctions of order n, so we know y Lemm 8 tht c n n < ɛ. Clerly d n n n λ n n < ɛ λ c n d n < ɛ. Also we know tht n < n + c n < c n, so n n + d n d n λ h n < ɛ, since h n is one of the frctions n, c n n + c n. k n n d n n + d n k n We wnt to show tht there re infinitely mny h n with distinct vlues. k n Assume for contrdiction tht there re only finitely mny h n. Then let { } k n hn A := k n n N >0 ɛ := min hn λ h n kn A k n. But s we ve shown there is Frey frction h n such tht k n λ h n < ɛ, which contrdicts ɛ eing miniml. So there re infinitely mny h n k n with distinct vlues. For every n N >0 : k n λ h n k n < 5k n λ µ h n µ k n k n < 5k n ξ h n µ k n k n < 5k n there re infinitely mny h n µ k n k n infinitely mny h n with distinct vlues. This is ecuse h n µ k n k n k n the function g : R R, x x µ is ijective. with distinct vlues since there re = h n k n µ 35
38 Theorem 0 (Theorem 6. in MNZ. The constnt 5 in Theorem 9 is the est possile, i.e. Theorem 9 does not hold if 5 is replced y ny lrger vlue. Proof. It s enough to find one ξ such tht 5 cnnot e replced y ny lrger vlue in Theorem 9. Let ξ := + 5. We see tht ( (x ξ x ( 5 = x + ( 5 x 5 = x x. So if we let h k e integers with k > 0, then h k ξ h k ξ + 5 (0 ( ( h = k ξ h k 5 = h k h k = k h hk k. ( We know tht (0 cn t e zero since tht would imply ξ = h k or h k = ξ 5, ut neither ξ nor ξ 5 re rtionl. We know tht h hk k is nonnegtive integer. It cn t e the cse tht h hk k = 0, since then eqution ( is zero hence eqution (0 is zero. Since h hk k it must e the cse tht h hk k ecuse eqution ( k k is equl to eqution (0 it follows tht h k ξ h k ξ + 5 k. ( Let m e positive rel numer ssume we hve n infinite sequence of rtionl numers h j, k j > 0 such tht k j h j ξ k j < mk j. (3 36
39 If we multiply y k j on oth sides of (3 we get From this we cn see tht h j k j ξ < mk j. k j ξ mk j < h j < k j ξ + mk j, so tht for ech rtionl numer h j k j in the sequence there cn only e finitely mny other frctions in the sequence with the sme denomintor, ut different numertor. It follows tht lim j k j =. We see tht k j h j ξ h j k j ξ + 5 k j {y (} < h j mkj ξ + 5 k j {y (3} ( h j ξ mkj k j + 5 {y inequlity} < ( + 5. {y (3} mkj mkj Multiplying y mk j we get Therefore ( m lim j mkj m < mk j + 5 = { } lim k j = j So if m > 5 then there is no infinite sequence of rtionl numers h j, k j k j > 0 such tht h j ξ k j < mk j 37
40 for every h j k j in the sequence. 38
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