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


 Avice Lamb
 5 years ago
 Views:
Transcription
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
2
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
padic Egyptian Fractions
padic Egyptin Frctions Contents 1 Introduction 1 2 Trditionl Egyptin Frctions nd Greedy Algorithm 2 3 Setup 3 4 pgreedy Algorithm 5 5 pegyptin Trditionl 10 6 Conclusion 1 Introduction An Egyptin frction
More informationBases for Vector Spaces
Bses for Vector Spces 22625 A set is independent if, roughly speking, there is no redundncy in the set: You cn t uild ny vector in the set s liner comintion of the others A set spns if you cn uild everything
More informationMTH 505: Number Theory Spring 2017
MTH 505: Numer Theory Spring 207 Homework 2 Drew Armstrong The Froenius Coin Prolem. Consider the eqution x ` y c where,, c, x, y re nturl numers. We cn think of $ nd $ s two denomintions of coins nd $c
More information2.4 Linear Inequalities and Interval Notation
.4 Liner Inequlities nd Intervl Nottion We wnt to solve equtions tht hve n inequlity symol insted of n equl sign. There re four inequlity symols tht we will look t: Less thn , Less thn or
More informationIntegral points on the rational curve
Integrl points on the rtionl curve y x bx c x ;, b, c integers. Konstntine Zeltor Mthemtics University of Wisconsin  Mrinette 750 W. Byshore Street Mrinette, WI 5443453 Also: Konstntine Zeltor P.O. Box
More informationset is not closed under matrix [ multiplication, ] and does not form a group.
Prolem 2.3: Which of the following collections of 2 2 mtrices with rel entries form groups under [ mtrix ] multipliction? i) Those of the form for which c d 2 Answer: The set of such mtrices is not closed
More informationQuadratic Forms. Quadratic Forms
Qudrtic Forms Recll the Simon & Blume excerpt from n erlier lecture which sid tht the min tsk of clculus is to pproximte nonliner functions with liner functions. It s ctully more ccurte to sy tht we pproximte
More informationBridging the gap: GCSE AS Level
Bridging the gp: GCSE AS Level CONTENTS Chpter Removing rckets pge Chpter Liner equtions Chpter Simultneous equtions 8 Chpter Fctors 0 Chpter Chnge the suject of the formul Chpter 6 Solving qudrtic equtions
More information(e) if x = y + z and a divides any two of the integers x, y, or z, then a divides the remaining integer
Divisibility In this note we introduce the notion of divisibility for two integers nd b then we discuss the division lgorithm. First we give forml definition nd note some properties of the division opertion.
More informationMath 4310 Solutions to homework 1 Due 9/1/16
Mth 4310 Solutions to homework 1 Due 9/1/16 1. Use the Eucliden lgorithm to find the following gretest common divisors. () gcd(252, 180) = 36 (b) gcd(513, 187) = 1 (c) gcd(7684, 4148) = 68 252 = 180 1
More informationChapter 1: Logarithmic functions and indices
Chpter : Logrithmic functions nd indices. You cn simplify epressions y using rules of indices m n m n m n m n ( m ) n mn m m m m n m m n Emple Simplify these epressions: 5 r r c 4 4 d 6 5 e ( ) f ( ) 4
More informationUniversitaireWiskundeCompetitie. Problem 2005/4A We have k=1. Show that for every q Q satisfying 0 < q < 1, there exists a finite subset K N so that
Problemen/UWC NAW 5/7 nr juni 006 47 Problemen/UWC UniversitireWiskundeCompetitie Edition 005/4 For Session 005/4 we received submissions from Peter Vndendriessche, Vldislv Frnk, Arne Smeets, Jn vn de
More informationCoalgebra, Lecture 15: Equations for Deterministic Automata
Colger, Lecture 15: Equtions for Deterministic Automt Julin Slmnc (nd Jurrin Rot) Decemer 19, 2016 In this lecture, we will study the concept of equtions for deterministic utomt. The notes re self contined
More informationTorsion in Groups of Integral Triangles
Advnces in Pure Mthemtics, 01,, 11610 http://dxdoiorg/1046/pm011015 Pulished Online Jnury 01 (http://wwwscirporg/journl/pm) Torsion in Groups of Integrl Tringles Will Murry Deprtment of Mthemtics nd Sttistics,
More informationARITHMETIC OPERATIONS. The real numbers have the following properties: a b c ab ac
REVIEW OF ALGEBRA Here we review the bsic rules nd procedures of lgebr tht you need to know in order to be successful in clculus. ARITHMETIC OPERATIONS The rel numbers hve the following properties: b b
More informationMath 61CM  Solutions to homework 9
Mth 61CM  Solutions to homework 9 Cédric De Groote November 30 th, 2018 Problem 1: Recll tht the left limit of function f t point c is defined s follows: lim f(x) = l x c if for ny > 0 there exists δ
More informationThe Regulated and Riemann Integrals
Chpter 1 The Regulted nd Riemnn Integrls 1.1 Introduction We will consider severl different pproches to defining the definite integrl f(x) dx of function f(x). These definitions will ll ssign the sme vlue
More informationMinimal DFA. minimal DFA for L starting from any other
Miniml DFA Among the mny DFAs ccepting the sme regulr lnguge L, there is exctly one (up to renming of sttes) which hs the smllest possile numer of sttes. Moreover, it is possile to otin tht miniml DFA
More informationChapter 1: Fundamentals
Chpter 1: Fundmentls 1.1 Rel Numbers Types of Rel Numbers: Nturl Numbers: {1, 2, 3,...}; These re the counting numbers. Integers: {... 3, 2, 1, 0, 1, 2, 3,...}; These re ll the nturl numbers, their negtives,
More informationLecture 3 ( ) (translated and slightly adapted from lecture notes by Martin Klazar)
Lecture 3 (5.3.2018) (trnslted nd slightly dpted from lecture notes by Mrtin Klzr) Riemnn integrl Now we define precisely the concept of the re, in prticulr, the re of figure U(, b, f) under the grph of
More informationQUADRATIC EQUATIONS OBJECTIVE PROBLEMS
QUADRATIC EQUATIONS OBJECTIVE PROBLEMS +. The solution of the eqution will e (), () 0,, 5, 5. The roots of the given eqution ( p q) ( q r) ( r p) 0 + + re p q r p (), r p p q, q r p q (), (d), q r p q.
More informationLecture 3: Equivalence Relations
Mthcmp Crsh Course Instructor: Pdric Brtlett Lecture 3: Equivlence Reltions Week 1 Mthcmp 2014 In our lst three tlks of this clss, we shift the focus of our tlks from proof techniques to proof concepts
More information378 Relations Solutions for Chapter 16. Section 16.1 Exercises. 3. Let A = {0,1,2,3,4,5}. Write out the relation R that expresses on A.
378 Reltions 16.7 Solutions for Chpter 16 Section 16.1 Exercises 1. Let A = {0,1,2,3,4,5}. Write out the reltion R tht expresses > on A. Then illustrte it with digrm. 2 1 R = { (5,4),(5,3),(5,2),(5,1),(5,0),(4,3),(4,2),(4,1),
More information8. Complex Numbers. We can combine the real numbers with this new imaginary number to form the complex numbers.
8. Complex Numers The rel numer system is dequte for solving mny mthemticl prolems. But it is necessry to extend the rel numer system to solve numer of importnt prolems. Complex numers do not chnge the
More informationDefinite Integrals. The area under a curve can be approximated by adding up the areas of rectangles = 1 1 +
Definite Integrls 5 The re under curve cn e pproximted y dding up the res of rectngles. Exmple. Approximte the re under y = from x = to x = using equl suintervls nd + x evluting the function t the lefthnd
More informationSuppose we want to find the area under the parabola and above the x axis, between the lines x = 2 and x = 2.
Mth 43 Section 6. Section 6.: Definite Integrl Suppose we wnt to find the re of region tht is not so nicely shped. For exmple, consider the function shown elow. The re elow the curve nd ove the x xis cnnot
More informationMath 154B Elementary Algebra2 nd Half Spring 2015
Mth 154B Elementry Alger nd Hlf Spring 015 Study Guide for Exm 4, Chpter 9 Exm 4 is scheduled for Thursdy, April rd. You my use " x 5" note crd (oth sides) nd scientific clcultor. You re expected to know
More informationSUMMER KNOWHOW STUDY AND LEARNING CENTRE
SUMMER KNOWHOW STUDY AND LEARNING CENTRE Indices & Logrithms 2 Contents Indices.2 Frctionl Indices.4 Logrithms 6 Exponentil equtions. Simplifying Surds 13 Opertions on Surds..16 Scientific Nottion..18
More informationThe area under the graph of f and above the xaxis between a and b is denoted by. f(x) dx. π O
1 Section 5. The Definite Integrl Suppose tht function f is continuous nd positive over n intervl [, ]. y = f(x) x The re under the grph of f nd ove the xxis etween nd is denoted y f(x) dx nd clled the
More informationTheoretical foundations of Gaussian quadrature
Theoreticl foundtions of Gussin qudrture 1 Inner product vector spce Definition 1. A vector spce (or liner spce) is set V = {u, v, w,...} in which the following two opertions re defined: (A) Addition of
More informationPolynomials and Division Theory
Higher Checklist (Unit ) Higher Checklist (Unit ) Polynomils nd Division Theory Skill Achieved? Know tht polynomil (expression) is of the form: n x + n x n + n x n + + n x + x + 0 where the i R re the
More informationMATH 573 FINAL EXAM. May 30, 2007
MATH 573 FINAL EXAM My 30, 007 NAME: Solutions 1. This exm is due Wednesdy, June 6 efore the 1:30 pm. After 1:30 pm I will NOT ccept the exm.. This exm hs 1 pges including this cover. There re 10 prolems.
More informationSurface maps into free groups
Surfce mps into free groups lden Wlker Novemer 10, 2014 Free groups wedge X of two circles: Set F = π 1 (X ) =,. We write cpitl letters for inverse, so = 1. e.g. () 1 = Commuttors Let x nd y e loops. The
More informationAQA Further Pure 1. Complex Numbers. Section 1: Introduction to Complex Numbers. The number system
Complex Numbers Section 1: Introduction to Complex Numbers Notes nd Exmples These notes contin subsections on The number system Adding nd subtrcting complex numbers Multiplying complex numbers Complex
More informationMORE FUNCTION GRAPHING; OPTIMIZATION. (Last edited October 28, 2013 at 11:09pm.)
MORE FUNCTION GRAPHING; OPTIMIZATION FRI, OCT 25, 203 (Lst edited October 28, 203 t :09pm.) Exercise. Let n be n rbitrry positive integer. Give n exmple of function with exctly n verticl symptotes. Give
More informationLecture 3. In this lecture, we will discuss algorithms for solving systems of linear equations.
Lecture 3 3 Solving liner equtions In this lecture we will discuss lgorithms for solving systems of liner equtions Multiplictive identity Let us restrict ourselves to considering squre mtrices since one
More informationImproper Integrals. The First Fundamental Theorem of Calculus, as we ve discussed in class, goes as follows:
Improper Integrls The First Fundmentl Theorem of Clculus, s we ve discussed in clss, goes s follows: If f is continuous on the intervl [, ] nd F is function for which F t = ft, then ftdt = F F. An integrl
More information1 PYTHAGORAS THEOREM 1. Given a right angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.
1 PYTHAGORAS THEOREM 1 1 Pythgors Theorem In this setion we will present geometri proof of the fmous theorem of Pythgors. Given right ngled tringle, the squre of the hypotenuse is equl to the sum of the
More informationSection 6.1 Definite Integral
Section 6.1 Definite Integrl Suppose we wnt to find the re of region tht is not so nicely shped. For exmple, consider the function shown elow. The re elow the curve nd ove the x xis cnnot e determined
More informationConvert the NFA into DFA
Convert the NF into F For ech NF we cn find F ccepting the sme lnguge. The numer of sttes of the F could e exponentil in the numer of sttes of the NF, ut in prctice this worst cse occurs rrely. lgorithm:
More informationI1 = I2 I1 = I2 + I3 I1 + I2 = I3 + I4 I 3
2 The Prllel Circuit Electric Circuits: Figure 2 elow show ttery nd multiple resistors rrnged in prllel. Ech resistor receives portion of the current from the ttery sed on its resistnce. The split is
More informationMatrix Algebra. Matrix Addition, Scalar Multiplication and Transposition. Linear Algebra I 24
Mtrix lger Mtrix ddition, Sclr Multipliction nd rnsposition Mtrix lger Section.. Mtrix ddition, Sclr Multipliction nd rnsposition rectngulr rry of numers is clled mtrix ( the plurl is mtrices ) nd the
More informationList all of the possible rational roots of each equation. Then find all solutions (both real and imaginary) of the equation. 1.
Mth Anlysis CP WS 4.X Section 4.4.4 Review Complete ech question without the use of grphing clcultor.. Compre the mening of the words: roots, zeros nd fctors.. Determine whether  is root of 0. Show
More informationI do slope intercept form With my shades on MartinGay, Developmental Mathematics
AATA Dte: 1//1 SWBAT simplify rdicls. Do Now: ACT Prep HW Requests: Pg 49 #1745 odds Continue Vocb sheet In Clss: Complete Skills Prctice WS HW: Complete Worksheets For Wednesdy stmped pges Bring stmped
More informationRegular Language. Nonregular Languages The Pumping Lemma. The pumping lemma. Regular Language. The pumping lemma. Infinitely long words 3/17/15
Regulr Lnguge Nonregulr Lnguges The Pumping Lemm Models of Comput=on Chpter 10 Recll, tht ny lnguge tht cn e descried y regulr expression is clled regulr lnguge In this lecture we will prove tht not ll
More informationLinear Inequalities. Work Sheet 1
Work Sheet 1 Liner Inequlities RentHep, cr rentl compny,chrges $ 15 per week plus $ 0.0 per mile to rent one of their crs. Suppose you re limited y how much money you cn spend for the week : You cn spend
More informationAn Alternative Approach to Estimating the Bounds of the Denominators of Egyptian Fractions
Leonrdo Journl of Sciences ISSN 580 Issue, JnuryJune 0 p. 0 An Alterntive Approch to Estimting the Bounds of the Denomintors of Egyptin Frctions School of Humn Life Sciences, University of Tsmni, Locked
More informationCS 330 Formal Methods and Models
CS 330 Forml Methods nd Models Dn Richrds, George Mson University, Spring 2017 Quiz Solutions Quiz 1, Propositionl Logic Dte: Ferury 2 1. Prove ((( p q) q) p) is tutology () (3pts) y truth tle. p q p q
More informationUSA Mathematical Talent Search Round 1 Solutions Year 21 Academic Year
1/1/21. Fill in the circles in the picture t right with the digits 18, one digit in ech circle with no digit repeted, so tht no two circles tht re connected by line segment contin consecutive digits.
More informationCIRCULAR COLOURING THE PLANE
CIRCULAR COLOURING THE PLANE MATT DEVOS, JAVAD EBRAHIMI, MOHAMMAD GHEBLEH, LUIS GODDYN, BOJAN MOHAR, AND REZA NASERASR Astrct. The unit distnce grph R is the grph with vertex set R 2 in which two vertices
More informationMathematics. Area under Curve.
Mthemtics Are under Curve www.testprepkrt.com Tle of Content 1. Introduction.. Procedure of Curve Sketching. 3. Sketching of Some common Curves. 4. Are of Bounded Regions. 5. Sign convention for finding
More informationQUADRATIC EQUATION EXERCISE  01 CHECK YOUR GRASP
QUADRATIC EQUATION EXERCISE  0 CHECK YOUR GRASP. Sine sum of oeffiients 0. Hint : It's one root is nd other root is 8 nd 5 5. tn other root 9. q 4p 0 q p q p, q 4 p,,, 4 Hene 7 vlues of (p, q) 7 equtions
More informationGeometric Sequences. Geometric Sequence a sequence whose consecutive terms have a common ratio.
Geometric Sequences Geometric Sequence sequence whose consecutive terms hve common rtio. Geometric Sequence A sequence is geometric if the rtios of consecutive terms re the sme. 2 3 4... 2 3 The number
More informationThings to Memorize: A Partial List. January 27, 2017
Things to Memorize: A Prtil List Jnury 27, 2017 Chpter 2 Vectors  Bsic Fcts A vector hs mgnitude (lso clled size/length/norm) nd direction. It does not hve fixed position, so the sme vector cn e moved
More informationVectors , (0,0). 5. A vector is commonly denoted by putting an arrow above its symbol, as in the picture above. Here are some 3dimensional vectors:
Vectors 1232018 I ll look t vectors from n lgeric point of view nd geometric point of view. Algericlly, vector is n ordered list of (usully) rel numers. Here re some 2dimensionl vectors: (2, 3), ( )
More informationalong the vector 5 a) Find the plane s coordinate after 1 hour. b) Find the plane s coordinate after 2 hours. c) Find the plane s coordinate
L8 VECTOR EQUATIONS OF LINES HL Mth  Sntowski Vector eqution of line 1 A plne strts journey t the point (4,1) moves ech hour long the vector. ) Find the plne s coordinte fter 1 hour. b) Find the plne
More informationThe graphs of Rational Functions
Lecture 4 5A: The its of Rtionl Functions s x nd s x + The grphs of Rtionl Functions The grphs of rtionl functions hve severl differences compred to power functions. One of the differences is the behvior
More informationMATH 101A: ALGEBRA I PART B: RINGS AND MODULES 35
MATH 101A: ALGEBRA I PART B: RINGS AND MODULES 35 9. Modules over PID This week we re proving the fundmentl theorem for finitely generted modules over PID, nmely tht they re ll direct sums of cyclic modules.
More informationLecture 3: Curves in Calculus. Table of contents
Mth 348 Fll 7 Lecture 3: Curves in Clculus Disclimer. As we hve textook, this lecture note is for guidnce nd supplement only. It should not e relied on when prepring for exms. In this lecture we set up
More informationW. We shall do so one by one, starting with I 1, and we shall do it greedily, trying
Vitli covers 1 Definition. A Vitli cover of set E R is set V of closed intervls with positive length so tht, for every δ > 0 nd every x E, there is some I V with λ(i ) < δ nd x I. 2 Lemm (Vitli covering)
More information1B40 Practical Skills
B40 Prcticl Skills Comining uncertinties from severl quntities error propgtion We usully encounter situtions where the result of n experiment is given in terms of two (or more) quntities. We then need
More informationAssignment 1 Automata, Languages, and Computability. 1 Finite State Automata and Regular Languages
Deprtment of Computer Science, Austrlin Ntionl University COMP2600 Forml Methods for Softwre Engineering Semester 2, 206 Assignment Automt, Lnguges, nd Computility Smple Solutions Finite Stte Automt nd
More informationLecture 1. Functional series. Pointwise and uniform convergence.
1 Introduction. Lecture 1. Functionl series. Pointwise nd uniform convergence. In this course we study mongst other things Fourier series. The Fourier series for periodic function f(x) with period 2π is
More informationReview of Gaussian Quadrature method
Review of Gussin Qudrture method Nsser M. Asi Spring 006 compiled on Sundy Decemer 1, 017 t 09:1 PM 1 The prolem To find numericl vlue for the integrl of rel vlued function of rel vrile over specific rnge
More information1 ELEMENTARY ALGEBRA and GEOMETRY READINESS DIAGNOSTIC TEST PRACTICE
ELEMENTARY ALGEBRA nd GEOMETRY READINESS DIAGNOSTIC TEST PRACTICE Directions: Study the exmples, work the prolems, then check your nswers t the end of ech topic. If you don t get the nswer given, check
More informationRudin s Principles of Mathematical Analysis: Solutions to Selected Exercises. Sam Blinstein UCLA Department of Mathematics
Rudin s Principles of Mthemticl Anlysis: Solutions to Selected Exercises Sm Blinstein UCLA Deprtment of Mthemtics Mrch 29, 2008 Contents Chpter : The Rel nd Complex Number Systems 2 Chpter 2: Bsic Topology
More informationHomework 4. 0 ε 0. (00) ε 0 ε 0 (00) (11) CS 341: Foundations of Computer Science II Prof. Marvin Nakayama
CS 341: Foundtions of Computer Science II Prof. Mrvin Nkym Homework 4 1. UsetheproceduredescriedinLemm1.55toconverttheregulrexpression(((00) (11)) 01) into n NFA. Answer: 0 0 1 1 00 0 0 11 1 1 01 0 1 (00)
More informationIdentify graphs of linear inequalities on a number line.
COMPETENCY 1.0 KNOWLEDGE OF ALGEBRA SKILL 1.1 Identify grphs of liner inequlities on number line.  When grphing firstdegree eqution, solve for the vrible. The grph of this solution will be single point
More informationSeptember 13 Homework Solutions
College of Engineering nd Computer Science Mechnicl Engineering Deprtment Mechnicl Engineering 5A Seminr in Engineering Anlysis Fll Ticket: 5966 Instructor: Lrry Cretto Septemer Homework Solutions. Are
More informationChapter 9 Definite Integrals
Chpter 9 Definite Integrls In the previous chpter we found how to tke n ntiderivtive nd investigted the indefinite integrl. In this chpter the connection etween ntiderivtives nd definite integrls is estlished
More informationA study of Pythagoras Theorem
CHAPTER 19 A study of Pythgors Theorem Reson is immortl, ll else mortl. Pythgors, Diogenes Lertius (Lives of Eminent Philosophers) Pythgors Theorem is proly the estknown mthemticl theorem. Even most nonmthemticins
More information7. Indefinite Integrals
7. Indefinite Integrls These lecture notes present my interprettion of Ruth Lwrence s lecture notes (in Herew) 7. Prolem sttement By the fundmentl theorem of clculus, to clculte n integrl we need to find
More information2 b. , a. area is S= 2π xds. Again, understand where these formulas came from (pages ).
AP Clculus BC Review Chpter 8 Prt nd Chpter 9 Things to Know nd Be Ale to Do Know everything from the first prt of Chpter 8 Given n integrnd figure out how to ntidifferentite it using ny of the following
More informationInfinite Geometric Series
Infinite Geometric Series Finite Geometric Series ( finite SUM) Let 0 < r < 1, nd let n be positive integer. Consider the finite sum It turns out there is simple lgebric expression tht is equivlent to
More informationBefore we can begin Ch. 3 on Radicals, we need to be familiar with perfect squares, cubes, etc. Try and do as many as you can without a calculator!!!
Nme: Algebr II Honors PreChpter Homework Before we cn begin Ch on Rdicls, we need to be fmilir with perfect squres, cubes, etc Try nd do s mny s you cn without clcultor!!! n The nth root of n n Be ble
More informationHamiltonian Cycle in Complete Multipartite Graphs
Annls of Pure nd Applied Mthemtics Vol 13, No 2, 2017, 223228 ISSN: 2279087X (P), 22790888(online) Pulished on 18 April 2017 wwwreserchmthsciorg DOI: http://dxdoiorg/1022457/pmv13n28 Annls of Hmiltonin
More informationConvex Sets and Functions
B Convex Sets nd Functions Definition B1 Let L, +, ) be rel liner spce nd let C be subset of L The set C is convex if, for ll x,y C nd ll [, 1], we hve 1 )x+y C In other words, every point on the line
More information13.3 CLASSICAL STRAIGHTEDGE AND COMPASS CONSTRUCTIONS
33 CLASSICAL STRAIGHTEDGE AND COMPASS CONSTRUCTIONS As simple ppliction of the results we hve obtined on lgebric extensions, nd in prticulr on the multiplictivity of extension degrees, we cn nswer (in
More informationImproper Integrals. Introduction. Type 1: Improper Integrals on Infinite Intervals. When we defined the definite integral.
Improper Integrls Introduction When we defined the definite integrl f d we ssumed tht f ws continuous on [, ] where [, ] ws finite, closed intervl There re t lest two wys this definition cn fil to e stisfied:
More informationThomas Whitham Sixth Form
Thoms Whithm Sith Form Pure Mthemtics Unit C Alger Trigonometry Geometry Clculus Vectors Trigonometry Compound ngle formule sin sin cos cos Pge A B sin Acos B cos Asin B A B sin Acos B cos Asin B A B cos
More informationAPPENDIX. Precalculus Review D.1. Real Numbers and the Real Number Line
APPENDIX D Preclculus Review APPENDIX D.1 Rel Numers n the Rel Numer Line Rel Numers n the Rel Numer Line Orer n Inequlities Asolute Vlue n Distnce Rel Numers n the Rel Numer Line Rel numers cn e represente
More informationHow do we solve these things, especially when they get complicated? How do we know when a system has a solution, and when is it unique?
XII. LINEAR ALGEBRA: SOLVING SYSTEMS OF EQUATIONS Tody we re going to tlk out solving systems of liner equtions. These re prolems tht give couple of equtions with couple of unknowns, like: 6= x + x 7=
More informationQuadratic Residues. Chapter Quadratic residues
Chter 8 Qudrtic Residues 8. Qudrtic residues Let n>be given ositive integer, nd gcd, n. We sy tht Z n is qudrtic residue mod n if the congruence x mod n is solvble. Otherwise, is clled qudrtic nonresidue
More informationMath 554 Integration
Mth 554 Integrtion Hndout #9 4/12/96 Defn. A collection of n + 1 distinct points of the intervl [, b] P := {x 0 = < x 1 < < x i 1 < x i < < b =: x n } is clled prtition of the intervl. In this cse, we
More informationLecture 2: January 27
CS 684: Algorithmic Gme Theory Spring 217 Lecturer: Év Trdos Lecture 2: Jnury 27 Scrie: Alert Julius Liu 2.1 Logistics Scrie notes must e sumitted within 24 hours of the corresponding lecture for full
More information0.1 THE REAL NUMBER LINE AND ORDER
6000_000.qd //0 :6 AM Pge 00 CHAPTER 0 A Preclculus Review 0. THE REAL NUMBER LINE AND ORDER Represent, clssify, nd order rel numers. Use inequlities to represent sets of rel numers. Solve inequlities.
More informationQuadratic reciprocity
Qudrtic recirocity Frncisc Bozgn Los Angeles Mth Circle Octoer 8, 01 1 Qudrtic Recirocity nd Legendre Symol In the eginning of this lecture, we recll some sic knowledge out modulr rithmetic: Definition
More informationLet S be a numerical semigroup generated by a generalized arithmetic sequence,
Abstrct We give closed form for the ctenry degree of ny element in numericl monoid generted by generlized rithmetic sequence in embedding dimension three. While it is known in generl tht the lrgest nd
More informationNumber systems: the Real Number System
Numer systems: the Rel Numer System syllusref eferenceence Core topic: Rel nd complex numer systems In this ch chpter A Clssifiction of numers B Recurring decimls C Rel nd complex numers D Surds: suset
More information2. VECTORS AND MATRICES IN 3 DIMENSIONS
2 VECTORS AND MATRICES IN 3 DIMENSIONS 21 Extending the Theory of 2dimensionl Vectors x A point in 3dimensionl spce cn e represented y column vector of the form y z zxis yxis z x y xxis Most of the
More information5. (±±) Λ = fw j w is string of even lengthg [ 00 = f11,00g 7. (11 [ 00)± Λ = fw j w egins with either 11 or 00g 8. (0 [ ffl)1 Λ = 01 Λ [ 1 Λ 9.
Regulr Expressions, Pumping Lemm, Right Liner Grmmrs Ling 106 Mrch 25, 2002 1 Regulr Expressions A regulr expression descries or genertes lnguge: it is kind of shorthnd for listing the memers of lnguge.
More informationAdvanced Algebra & Trigonometry Midterm Review Packet
Nme Dte Advnced Alger & Trigonometry Midterm Review Pcket The Advnced Alger & Trigonometry midterm em will test your generl knowledge of the mteril we hve covered since the eginning of the school yer.
More informationHarvard University Computer Science 121 Midterm October 23, 2012
Hrvrd University Computer Science 121 Midterm Octoer 23, 2012 This is closedook exmintion. You my use ny result from lecture, Sipser, prolem sets, or section, s long s you quote it clerly. The lphet is
More informationPolynomial Approximations for the Natural Logarithm and Arctangent Functions. Math 230
Polynomil Approimtions for the Nturl Logrithm nd Arctngent Functions Mth 23 You recll from first semester clculus how one cn use the derivtive to find n eqution for the tngent line to function t given
More informationChapters Five Notes SN AA U1C5
Chpters Five Notes SN AA U1C5 Nme Period Section 5: Fctoring Qudrtic Epressions When you took lger, you lerned tht the first thing involved in fctoring is to mke sure to fctor out ny numers or vriles
More informationJim Lambers MAT 169 Fall Semester Lecture 4 Notes
Jim Lmbers MAT 169 Fll Semester 200910 Lecture 4 Notes These notes correspond to Section 8.2 in the text. Series Wht is Series? An infinte series, usully referred to simply s series, is n sum of ll of
More informationChapter 6 Techniques of Integration
MA Techniques of Integrtion Asst.Prof.Dr.Suprnee Liswdi Chpter 6 Techniques of Integrtion Recll: Some importnt integrls tht we hve lernt so fr. Tle of Integrls n+ n d = + C n + e d = e + C ( n ) d = ln
More informationEach term is formed by adding a constant to the previous term. Geometric progression
Chpter 4 Mthemticl Progressions PROGRESSION AND SEQUENCE Sequence A sequence is succession of numbers ech of which is formed ccording to definite lw tht is the sme throughout the sequence. Arithmetic Progression
More information7.8 Improper Integrals
7.8 7.8 Improper Integrls The Completeness Axiom of the Rel Numers Roughly speking, the rel numers re clled complete ecuse they hve no holes. The completeness of the rel numers hs numer of importnt consequences.
More information1 Sets Functions and Relations Mathematical Induction Equivalence of Sets and Countability The Real Numbers...
Contents 1 Sets 1 1.1 Functions nd Reltions....................... 3 1.2 Mthemticl Induction....................... 5 1.3 Equivlence of Sets nd Countbility................ 6 1.4 The Rel Numbers..........................
More information