O The Prie Nubers I Itervals arxiv:1706.01009v1 [ath.nt] 4 Ju 2017 Kyle D. Balliet A Thesis Preseted to the Faculty of the Departet of Matheatics West Chester Uiversity West Chester, Pesylvaia I Partial Fulfillet of the Requireets For the Degree of Master of Arts 2015 Copyright by Kyle D. Balliet, 2015. All rights reserved.
Abstract Bertrad s postulate establishes that for all positive itegers > 1 there exists a prie uber betwee ad 2. We cosider a geeralizatio of this theore as: for itegers k 2 is there a prie uber betwee k ad (k + 1)? We use eleetary ethods of bioial coefficiets ad the Chebyshev fuctios to establish the cases for 2 k 8. We the ove to a aalytic uber theory approach to show that there is a prie uber i the iterval (k,(k + 1)) for at least k ad 2 k 519. We the cosider Legedre s cojecture o the existece of a prie uber betwee 2 ad (+1) 2 for all itegers 1. To this ed, we show that there is always a prie uber betwee 2 ad (+1) 2.000001 for all 1. Furtherore, we ote that there exists a prie uber i the iterval [ 2,(+1) 2+ε ] for ay ε > 0 ad sufficietly large. We also cosider the questio of how ay prie ubers there are betwee ad k for positive itegers k ad for each of our results ad i the geeral case. Furtherore, we show that the uber of prie ubers i the iterval (,k) is icreasig ad that there are at least k 1 prie ubers i (,k) for k 2. Fially, we copare our results to the prie uber theore ad obtai explicit lower bouds for the uber of prie ubers i each of our results.
Ackowledgeets I would like to express y appreciatio to everyoe who allowed this thesis to coe to fruitio. This icludes ot oly the professors of West Chester Uiversity, but also to ayoe who dared to questio. Thak you to Li Ta for his ecourageet, questios, ad coets which helped shape this research. I a also grateful to the staff of Bloosburg Uiversity. Perhaps ost otably Willia Calhou whose kowledge of uber theory ad correspodece would later ispire the work withi these pages. Most iportatly for last. Charlee who would allow e to droe o ad o, ad o soe ore, about Matheatics. Sorry.
Mia eso restis alferitaj, eio alia.
Cotets List of Notatio................................. List of Tables.................................. i ii 1 Itroductio 1 2 Prie Nubers I Itervals 5 2.1 Pries i the Iterval [4, 5]...................... 6 2.2 Pries i the Iterval [8, 9]...................... 16 2.3 Pries i the Iterval [519, 520]................... 39 2.4 Pries betwee 2 ad (+1) 2+ε.................... 43 3 The Nuber of Prie Nubers 47 3.1 Betwee ad 5............................. 48 3.2 Betwee ad 9............................. 49 3.3 Betwee ad 520........................... 51 3.4 Betwee ad k............................. 52 3.5 Betwee Successive Powers........................ 54 4 The Prie Nuber Theore 57 5 Suary ad Coclusios 62 Bibliography 63
List of Notatio N The set of positive itegers: {1,2,3,...} Z The set of all itegers: {..., 2, 1,0,1,2,...} logx ϑ(x) ψ(x) The logarith of x with base e (that is, the atural logarith) The first Chebyshev fuctio: logp The secod Chebyshev fuctio: p x logp = p k x k N! The factorial of :! = 1 2 3 ( 1) ϑ(x 1/ ) =1 π() exp(x) A\B The uber of prie ubers less tha or equal to Equivalet otatio for e x The relative copleet of B i A. That is, A\B = {x A x B} u() A upper boud approxiatio for!:! < u() = 2π +1 2 e + 1 12 l() A lower boud approxiatio for!:! > l() = 2π +1 2 e + 1 12+1 [x] The floor of x. That is, [x] = ax{ Z x} {x} { x y} B k (,) The sawtooth fuctio of x. That is, {x} = x [x] Bioial coefficiet with floored ters. That is, { } ( x y = δ(y,x) [x] [y]), where δ(y,x) = 1 if {x} {y} ad δ(y,x) = [x y]+1 if {x} < {y} } The divisio of bioial coefficiets: { (k+1) k }/{ (k+1) 2 k 2 i
List of Tables 4.1 Copariso of Theore 4.0.1 with PNT................. 59 4.2 Copariso of Theore 4.0.3 with PNT................. 60 ii
Chapter 1 Itroductio Bertrad s postulate is stated as: For > 1 there is a prie uber betwee ad 2. This, relatively siply stated, cojecture was stated by Joseph Bertrad i 1845 ad was ultiately solved by Pafuty Chebyshev i 1850, see [3]. Later, i 1919, it was show by Sriivasa Raauja i [13] usig properties of the gaa fuctio. Fially, Paul Erdős [6, 8] i 1932 showed the theore usig eleetary properties, ad approxiatios of bioial coefficiets ad the Chebyshev fuctios: ϑ(x) = p x logp, ψ(x) = p k x k N logp. I 2006, Mohaed El Bachraoui [1] showed that for > 1 there is a prie uber betwee 2 ad 3. This siilar result to Bertrad s postulate provided a saller iterval for a prie uber to exist. As a exaple, Bertrad s postulate guaratees 1
p (10, 20) whereas Bachraoui guaratees p (10, 15). Moreover, Bachraoui also questioed if there was a prie uber betwee k ad (k +1) for all k 2. Fially, i 2011 Ady Loo [11] shorteed these itervals to p (3,4) for > 1. A. Loo wet o to prove that as approaches ifiity the uber of prie ubers betwee 3 ad 4 also teds to ifiity - a result which is iplied by the prie uber theore. A coo feature exists i each of these eleetary proofs, aily that each iproveet requires checkig ore cases by had/coputer tha the previous result. Ufortuately, this is a coditio of the ethod of proof used ad the ature of the proble, as we shall see i Chapter 2. The priary idea with these proofs is to cosider the bioial coefficiet ( ) (k+1) k ad subdivide the categories of pries as sall, ediu, ad large as: T 1 = p (k+1) p β(p), T 2 = (k+1)<p k p β(p), ad T 3 = p k+1 p (k+1) such that ( ) (k +1) = T 1 T 2 T 3, k where β(p) is the largest power of p, say α, such that p α divides ( ) (k+1) k. It should also be oted that we ay readily calculate β(p) by β(p) = ([ ] [ ] [ (k +1) k t=1 p t p t p t ]), a fact show by P. Erdős i [8, p. 24]. 2
We the boud T 1 ad T 2 fro above by easily coputable approxiatios. Siilarly, we boud ( ) (k+1) k fro below ad show that ( ) (k +1) 1 T 3 = > 1. k T 1 T 2 Sice the T 3 > 1, there is at least oe prie uber betwee k ad (k +1). We will use ethods siilar to these i Sectio 2.1 to show that there is a prie uber betwee 4 ad 5 for all itegers > 2. However, i Sectio 2.2 we will ot oly apply these ethods, but we will eed to defie a divisio of bioial coefficiets i order to efficietly boud T 2. We will defie B k (,) as B k (,) = { (k+1) k } / { (k+1) } 2 k 2 ad establish both upper ad lower bouds for this divisio. I doig so we will be able to exted our proof techique to the case whe k = 8. We the ove fro our eleetary approach i favor of a aalytic uber theory approach that was utilized by S. Raauja i [13] ad J. Nagura i [12], showig that ( ) (k +1) ϑ ϑ() > 0 k provided that ad k are itegers such that 1 k 519 ad k. We will the show that this theore yields a corollary that there is at least oe prie uber betwee 519 ad 520 for all > 15. We coclude Chapter 2 by showig that there exists a prie uber p such that 2 < p < (+1) 2.000001. This result is close to Legedre s cojecture, which asserts 3
that there exists a prie uber betwee 2 ad (+1) 2. Moreover, we ote that there exists a prie uber p such that 2 < p < (+1) 2+ε for ay ε > 0 ad soe sufficietly large. I Chapter 3 we will tur our attetio to deteriig the uber of prie ubers betwee ad k based upo our previous results. We will also show that there are at least k 1 prie ubers betwee ad k for ay k 2 ad k. I doig so, we will also ote that the uber of prie ubers betwee ad k teds to ifiity by showig that ( ( k ϑ(k) ϑ() > k 1 3.965 log 2 (k) + 1 )) log 2 > 0, where log 2 x = (logx) 2. Lastly, we show that for every positive iteger there exists a positive iteger L such that for all L there are at least prie ubers i each of our itervals. We the copare this to the prie uber theore which states that π(x) x logx. 4
Chapter 2 Prie Nubers I Itervals I this chapter we will show that there is always a prie uber i certai itervals for soe positive iteger. The eleetary eas eployed i Sectios 2.1 ad 2.2 are essetially the sae as those used by P. Erdős [6], M. El Bachraoui [1], ad A. Loo [11]. The aalytic eas eployed i Sectio 2.3 are the ethods used by S. Raauja i [13] ad J. Nagura i [12]. We will extesively utilize the followig bouds of the factorial (gaa) fuctio as provided by H. Robbis [14] i the ext two sectios ad so we preset the as a lea. Lea 2.0.1. Let l() = 2π +1 2 e + 1 12+1 ad u() = 2π + 1 2 e + 1 12, the l() <! < u() for 1. Moreover, we will also use the fact that 1 δ(r,s) s as show i the followig lea. 5
Lea 2.0.2. Let r ad s be real ubers satisfyig s > r 1 ad { } ( s r = δ(r,s) [s] [r]), the 1 δ(r,s) s. Proof. Let z > 0 ad let [z] be the greatest iteger less tha or equal to z. Defie {z} = z [z]. Let r ad s be real ubers satisfyig s > r 1. Observe that the uber of itegers i the iterval (s r,s] is [s] [s r], which is [r] if {s} {r} ad [r]+1 if {s} < {r}. Let N be the set of all atural ubers ad defie { } s = r k ( ) [s] k = δ(r,s) [r] k (s r,s] N k (0,r] N where δ(r,s) = 1 if {s} {r} ad δ(r,s) = [s r]+1 if {s} < {r}. I either case, 1 δ(r,s) s. Lea 2.0.3. The followig are true: 1. Let c 1 be a fixed costat. For x 1, u(x+c) 12 2 l(c)l(x) is icreasig. 2. Let c be a fixed positive costat ad defie h 2 (x) = u(c) l(x)l(c x). The h 2(x) > 0 whe 1 2 x c 2, h 2 (x) = 0 whe x = c 2, ad h 2 (x) < 0 whe c 2 < x c 1 2. Proof. See Leas 1.4 ad 1.5 of [11]. 2.1 Pries i the Iterval [4, 5] I this sectio we will show that there is always a prie uber i the iterval (4,5) for ay positive iteger > 2. I order to do so, we will begi by showig two iequalities which will be vital to the ai proof which follows. 6
For our ai proof we will cosider the bioial coefficiet ( 5 4) ad subdivide the prie ubers coposig ( 5 4) ito three products based o their size. That is, we will categorize the prie ubers as T 1 = p β(p), T 2 = p β(p), ad T 3 = p 5 5<p 4 p 4+1 p 5 so that ( ) 5 = T 1 T 2 T 3. 4 We will the show that T 3 = ( ) 5 1 4 T 1 T 2 is greater tha 1 ad therefore there is at least oe prie uber i T 3. That is, there is at least oe prie uber i the iterval (4, 5). Lea 2.1.1. The followig iequalities hold: 1. For 6818, e 1 /(60 + 1) 1/48 1/12 0.999986. 2. For 1, e 1 /30 1/(24 + 1) 1/(6 + 1) 1. 3. For 1, e 1 /20 1/(16 + 1) 1/(4 + 1) 1. 4. For 6815, 4+3 3 < 4.002202. Proof (1). The followig iequalities are equivalet for 6818: e 1 /(60 + 1) 1/48 1/12 0.999986 1 1 1 log0.999986 60+1 48 12 252+5 2880 2 +48 log0.999986 0.000014. 7
Now the left-had side is decreasig i ad so it suffices to verify the case whe = 6818. Whe = 6818, we obtai 1718141 < 0.000013 < log0.999986. 133877484384 Proof (2). The followig iequalities are equivalet for 1: e 1 /30 1/(24 + 1) 1/(6 + 1) 1 1 30 1 24+1 1 6+1 0 1 756 2 +30. Which clearly holds for all 1. Proof (3). The followig iequalities are equivalet for 1: e 1 /20 1/(16 + 1) 1/(4 + 1) 1 1 1 1 0 20 16+1 4+1 1 336 2 +20. Which clearly holds for all 1. Proof (4). Theiequalityfollowsdirectlybyotigthat 4+3 3 < 4.002202isequivalet to 0 < 0.002202 15.006606 which clearly holds for all 6815. Lea 2.1.2. For all 6818 the followig iequality holds: 0.054886 2 2 3 2 ( ) 3125 6 2.51012 5 > (5) log(5). 256 8
Proof. The followig are equivalet for 6818: 0.054886 2 2 3 2 ( ) 3125 6 2.51012 5 > (5) log(5) 256 log0.054886 2 log2 3 2 log+ 6 log3125 6 log256 > 2.51012 5 6 [log3125 3log2 log256] > 2.51012 5+ 3 2 log log0.054886 1 2.51012 5 [5log5 11log2] > + 3 ( ) log log0.054886. 6 2 Now the right-had side is decreasig i ad so it suffices to verify the case whe = 6818. Whe = 6818, we obtai 1 6 2.51012 5 [5log5 11log2] > 0.0742 > + 3 6818 2 ( ) log6818 6818 log0.054886. 6818 We ow proceed with the proof of our ai theore for this sectio; that is, there is always a prie uber betwee 4 ad 5 for all itegers > 2. Theore 2.1.3. For ay positive iteger > 2 there is a prie uber betwee 4 ad 5. Proof. It ca be easily verified that for = 3,4,...,6817 there is always a prie uber betwee 4 ad 5. Now let 6818 ad cosider: ( ) 5 4 = (4+1)(4+2) (5). 1 2 The product of pries betwee 4 ad 5, if there are ay, divides ( 5 4). 9
Followig the otatio used i [1, 6, 11], we let T 1 = p β(p), T 2 = p β(p), ad T 3 = p 5 5<p 4 p 4+1 p 5 such that ( ) 5 = T 1 T 2 T 3. 4 The prie decopositio of ( 5 4) iplies that the powers i T2 are less tha 2; see [8, p. 24] for the prie decopositio of ( j). I additio, the prie decopositio of ( 5 4), see [8, p. 24], yields the upper boud for T1 : T 1 < (5) π( 5). But π(x) 1.25506x, see [15]. So we obtai logx T 1 < (5) π( 5) (5) 2.51012 5 log(5). Now let A = { } { 5/2 2 ad B = 5/3 4/3} ad observe the followig for a prie uber p i T 2 : If 5 2 < p 4, the < p 4 < 5 < 2p. Hece β(p) = 0. Clearly p divides A. 2<p 5 2 10
If 5 3 < p 2, the < p < 2p 4 < 5 < 3p. Hece β(p) = 0. Clearly 4 3 <p 5 3 p divides B. If 5 4 < p 4 3, the < p < 3p 4 < 5 < 4p. Hece β(p) = 0. If < p 5, the 4 Hece p divides A. 2 < p < 2 < 2p 5 2 < 3p. <p 5 4 If 5 6 < p, the p < 2p < 4p 4 < 5p 5 < 6p. Hece β(p) = 0. If 2 3 < p 5 6, the Hece 2 3 <p 5 6 p divides B. 3 < p < 4 3 < 2p 5 3 < 3p. 11
If 5 8 < p 2 3, the p < < 2p < 6p 4 < 7p < 5 < 8p. Hece β(p) = 0. If 2 < p 5 8, the Hece 2 <p 5 8 p divides A. If 5 11 < p 2, the 2 < p < 3p < 2 < 4p 5 2 < 5p. p < 2p < 3p < 8p 4 < 9p < 10p 5 < 11p. Hece β(p) = 0. If 4 9 < p 5 11, the Hece 4 9 <p 5 11 p divides B. If 5 12 < p 4 9, the 3 < p < 2p < 4 3 < 3p < 5 3 < 4p. p < 2p < < 3p < 9p 4 < 10p < 11p < 5 < 12p. Hece β(p) = 0. 12
If 3 < p 5 12, the Hece 3 <p 5 12 p divides B. If 5 16 < p 3, the 3 < p < 3p < 4 3 < 4p 5 3 < 5p. p < 2p < 3p < 4p < 12p 4 < 13p < 14p < 15p 5 < 16p. Hece β(p) = 0. If 2 7 < p 5 16, the Hece 2 7 <p 5 16 p divides A. If 5 18 < p 2 7, the p < 2 < 2p < 6p < 2 < 7p < 8p 5 2 < 9p. p < 2p < 3p < < 4p < 14p 4 < 15p < 16p < 17p < 5 < 18p. Hece β(p) = 0. If 4 < p 5 18, the Hece 4 <p 5 18 p divides A. p < 2 < 2p < 7p < 2 < 8p < 9p 5 2 < 10p. 13
By the fact that p xp < 4 x, see [8, p. 167], we obtai 5<p 4 p < 4 4 = 2 2. Now, to suarize, we obtai T 2 = 5<p 4 p β(p) < 2 2 AB. By Leas 2.0.1 ad 2.1.1, we obtai ( ) 5 = (5)! 4 (4)!(!) > l(5) u(4)u() ( 5 3125 = 8π 256 > 0.446024 1 2 ) e 1 60+1 1 48 1 12 ( 3125 256 ). Siilarly, by Leas 2.0.1, 2.0.2, 2.0.3, ad 2.1.1, we obtai { 5 } ( A = 2 [ 5 = δ(2, 5/2) ] ) 2 2 2 5 ( [ 5 ] ) 2 5 2 2 2 u([ 5]) 2 l(2)l([ 5] 2) 2 5 2 u( 5) 2 l(2)l( ) 2 = 5 ( 5 3125 4 π 256 < 1.576958 1 2 ) 2 e 1 30 1 24+1 1 6+1 ( ) 3125 256 2. 14
Lastly, we obtai the upper boud approxiatio for B as: { 5 } 3 B = 4 5 3 3 = 5 3 5 3 ( [ 5 = δ(4/3, 5/3) ( [ 5 ] ) 3 [ 4] 3 [ 4 3 ]+1 [ 5 3 ] [4 3 ] 4 +1 3 5 1 4 3 3 3 ] [ 4 3 ] ( [ 5 ] ) 3 [ 4]+1 3 5 3 4+3 3 u( 5) 3 l( 4 3 )l() 3 ( ) ( 125 4+3 3125 = 1 2 24π 3 256 ( ) < 5.153158 1 3125 3 2. 256 ) u([ 5 3 ]) l([ 4 3 ]+1)l([5 3 ] ([4]+1)) 3 ) 3 e 1 20 1 16+1 1 4+1 Thus we obtai ( ) 5 1 T 3 = 4 T 2 T ( ) 1 5 1 > 4 2 2AB > 2 2 ( > 0.054886 2 2 3 2 1 0.446024 1 2 )( 1.576958 1 2( 3125 256 (5) 2.51012 5 log(5) ( 3125 ) 256 ) 2 ( ) 3125 256 6 (5) 2.51012 5 log(5) 5.153158 1 2( 3125 256 1 > 1, ) ) 3 1 (5) 2.51012 5 log(5) where the last iequality follows by Lea 2.1.2. Cosequetly the product T 3 of prie ubers betwee 4 ad 5 is greater tha 1 ad therefore the existece of such ubers is prove. 15
With the proof of the previous theore coplete we ay also show that there is always a prie uber betwee ad (5+15)/4 for all positive itegers > 2 as i the followig theore. Theore 2.1.4. For ay positive iteger > 2 there exists a prie uber p satisfyig < p < 5(+3) 4. Proof. Whe = 3, we obtai 3 < 5 < 7 < 15. Let 4. By the divisio algorith 2 4 (+r) for soe r {0,1,2,3} ad by Theore 2.1.3 there exists a prie uber ( ) ( ) ( ) p such that p +r, 5(+r). Sice +r, 5(+r) is cotaied i, 5(+3) for 4 4 4 ( ) all 0 r 3 ad > 2, p as desired., 5(+3) 4 2.2 Pries i the Iterval [8, 9] If we attept to cotiue usig the process fro the previous sectio, the we would ru ito a issue. For exaple, cosider tryig to prove that there is always a prie uber betwee 5 ad 6 for > 1. We cosider the bioial ( 6 5) ad let T 1 = p 6 p β(p), T 2 = 6<p 5 p β(p), ad T 3 = p. 5+1 p 6 As i Sectio 2.1, we approxiate T 1 easily eough as T 1 < (6) π( 6), however whe we attept to approxiate T 2 the issue presets itself. To see this, let be a atural uber ad cosider a prie uber p satisfyig 6 < p 5 6. Now if, < 5, the < +1 +1 hece p does ot divide ( 6 5). 16 < p p 5 < 6 < (+1)p ad
Therefore, T 2 = 6<p 5 p βp = 6<p p βp <p 6 5 p βp 5 4 <p 3 2 p βp 5 3 <p 2 p βp 5 2 <p 3 p βp. Certaily 5 4 <p 3 2 p p, 5 2 <p 3 5 3 <p 2 p, ad <p 6 5 p divide ( ) ( 3 5/2, 2 ) ( 5/3,ad 6/5 ), respectively. The we have T 3 = ( ) 6 1 > 5 T 1 T 2 ( 6 ) 5 ( 3 )( 2 )( 6/5 5/2 5/3 ) p β(p) 6<p 1 T 1. However, 1 > ( 6 ) 5 ( 3 )( 2 )( 6/5 5/2 5/3 ) for all > 40 ad thus the ethod is icoclusive. Therefore we eed to fid a sharper approxiatio for each of the products of pries which copose T 2. We will do so by defiig B k (,) to be a divisio of bioial coefficiets as B k (,) = { (k+1) k } / { (k+1) } 2. k 2 We will the boud this approxiatio fro both above ad below. The reaso for acquirig a lower boud is due to the fact, as i our exaple, just because 5 2 <p 3p divides B 5 (,2) does ot ea that B 5 (,2) is a upper boud for the product of 17
pries. Therefore we also eed to show that B 5 (,2) 5 2 <p 3 1 p 1 ad thus B 5 (,2) is a upper boud for the product of pries betwee 5 2 ad 3. While our proof i this sectio is ot etirely eleetary due to the ethods utilized i Leas 2.2.4 ad 2.2.5, the results of this sectio provide isight ad proof techiques which ay be used to show other cases via eleetary eas. Also, the proof of Lea 2.2.4 provides a shift ito the proof techiques of Sectios 2.3 ad 2.4. Our first priority is to show that if k,, ad are atural ubers with 2 ad k 2, the ay prie uber satisfyig k < p (k+1) B k (,) as i the ext lea. divides Lea 2.2.1. For k,, N with 2 ad k 2. If p is a prie uber such that k < p (k+1), the p divides B k (,) = { (k+1) k } / { (k+1) } 2. k 2 Proof. Let k,, N with 2 ad k 2. Let p be a prie uber such that k < p (k+1) ad observe that k (k +1) < 2 2 < k < p (k +1). Hece p divides B k (,) as desired. 18
We ow tur our attetio to showig a upper approxiatio for B k (,). To accoplish this we will utilize Lea 2.0.1 to approxiate the factorial fuctio. Lea 2.2.2. For all k,, N with 2 ad k 2, B k (,) < ( ) (k +1)(k+)(k++2)(+2) (k +1) k+1 2 e E 4 2 3 ( ) ( (k +1)(k+)(k++2)(+2) (k +1) k+1 4 2 3 ( ) k k k k ) 2 e 12029 111150 where E = (2k2 +5k +2) 12k(k +1) 12k+ 12+ 6(k +1)+. Proof. By Leas 2.0.1, 2.0.2, ad 2.0.3, we obtai { (k+1) } ( ) 2 [(k +1)/] = δ(k/,(k +1)/) k [k/] 2 ( ) (k +1) [(k +1)/] [k/] [ (k +1) k ] ( ) = +1 [(k +1)/] [ ] (k+1) [ ] k [k/]+1 ( ) k (k +1) +1 u (k+1) (k+1) k 1 l ( ( k ) l ) ( ) (k +1) (k +1)(k+) (k +1) k+1 = e Eu, 2πk k k where E u = 12(k +1) 12k+ 12+. 19
Siilarly, we obtai the lower boud approxiatio for { } (k+1)/2 k/2 as: { (k+1) } ( ) 2 [(k +1)/2] = δ(k/2,(k +1)/2) k [k/2] 2 ( ) [(k +1)/2] 1 [k/2] = > = ([ ] (k+1) +1 2 1 )([ (k+1) 2 1 ( )( (k+1) +1 (k+1) k 2 2 4 2 ((k +1)+2)(+2) 4 2 (k++2)(+2) ([ (k+1) 2 ] ) +1! ] [ ] ) [ k k ] ] 2 2! ([ (k+1) [ k 2 2 ([ ] ) l (k+1) +1 2 ) +1 u ([ ]) ([ ] k 2 2 u (k+1) [ k 2 2 ( ) l (k+1) 2 u ( ) ( k 2 u ) 2 ( ) (k +1) (k +1) k+1 2 e E l, πk k k ] 1 )! ] 1 ) where E l = 6(k+1)+ 6k 6 = 6(k+1)+ (k +1). 6k Therefore B k (,) < ( ) (k +1)(k+)(k++2)(+2) 4 (k +1) k+1 2 e E, 2 3 ( ) k k where E = E u E l = ( ) 2k 2 +5k +2 12k(k +1) 1 12k +/ 1 12+/ 1. 6(k+1)+/ 20
Sice k 2, clearly k 2 + 3k > 1. I other words, 2k 2 + 5k + 2 > k 2 + 2k + 3. Dividig both sides by 12k(k +1) ad rewritig the right-had side, we obtai 2k 2 +5k +2 12k(k +1) > 1 12 + 1 12k + 1 6(k +1). Moreover, sice 2, 0 < 1 ad hece 2k 2 +5k +2 12k(k +1) > 1 12 + 1 12k + 1 6(k +1) 1 > 12+/ + 1 12k+/ + 1 6(k +1)+/ = 12+ + 12k+ + 6(k +1)+. Multiplyig both sides by, we obtai (2k 2 +5k +2) 12k(k +1) > 12+ + 12k+ + 6(k +1)+. Hece E is decreasig i. Furtherore, sice, [, ) ad hece E is axiu whe =. Whe =, we obtai E = 2k2 +5k +2 12k(k +1) 1 12k +1 1 13 1 6(k+1)+1 = 1008k4 +2268k 3 +2684k 2 +1463k +182 11232k 4 +25272k 3 +15132k 2 +1092k = 1008+2268/k+2684/k2 +1463/k 3 +182/k 4 11232+25272/k+15132/k 2 +1092/k 3 7 78 21
Now sice E is decreasig i k ad k 2 we have that E is axiu whe k = 2. Hece 7 E 12029. Thus we obtai 78 111150 B k (,) < ( ) (k +1)(k+)(k++2)(+2) (k +1) k+1 2 e E 4 2 3 ( ) ( (k +1)(k+)(k++2)(+2) (k +1) k+1 4 2 3 ( ) k k k k ) 2 e 12029 111150. We ust show that B k (,) is a upper boud for the product of prie ubers betwee k ad (k+1). Oe way to accoplish this task is to boud B k(,) fro below, boud the product of pries fro above, ad deterie their iequality. We will do so for k = 8 ad therefore show that B 8 (,) is a upper boud for the product of pries betwee 8 ad 9. Lea 2.2.3. For all k,, N with 2 ad k 2, ( 2 3 ( 2) (k +1) k+1 B k (,) > (k +1)(k++)(+)(k+2) k k ( 2 3 ( 2) (k +1) k+1 (k +1)(k++)(+)(k+2) k k ) 2 e F ) 2 e 5 84 where Proof. Let ad F = 12(k+1)+ + 6k+ + 6+ (k2 +4k +1) 12k(k+1). F l = F u = (k +1) 12(k +1)+ 12k 6(k +1) 6k+ 6+, 22
the by Leas 2.0.1, 2.0.2, ad 2.0.3, we obtai { (k+1) } ( ) [(k +1)/] = δ(k/,(k +1)/) k [k/] ( ) [(k +1)/] 1 [k/] = > = ([ ] (k+1) +1 ( (k+1) 1 )([ (k+1) ] [ ] ) k ([ (k+1) ] [ ] k ]! ([ (k+1) 1 l )( ) +1 (k+1) k +1 u ([ k ( ) l (k+1) u ( ( k ) u ) ( (k +1) (k +1) k+1 2πk k k 2 (k++)(+) 2 (k++)(+) ([ (k+1) ] ]) u ([ (k+1) ) +1! [ ] ) k 1! ) +1 ] ) e F l. [ k ] 1 ) Siilarly, we obtai the upper boud approxiatio of { } (k+1)/2 k/2 as: { (k+1) } ( ) 2 [(k +1)/2] = δ(k/2,(k +1)/2) k [k/2] 2 ( ) (k +1) [(k +1)/2] 2 [k/2] [ (k +1) k ( ) = +1 [(k +1)/2] 2 ] ] [k/2]+1 = (k +1) 2 [ (k+1) 2 k +1 2 (k+1) 2 (k +1) 2 k+2 2 (k +1)(k+2) 2( 2) [ k 2 1 k l ([ k 2 2 ( ) (k+1) u ([ ]) (k+1) 2 ] ] +1 ) l ([ (k+1) 2 u 2 l ( ) ( k 2 l ) 2 ( (k +1) (k +1) k+1 πk k k ) 2 e Fu. [ ] ) k 2 1 23
Thus B k (,) > ( 2 3 ( 2) (k +1)(k++)(+)(k+2) (k +1) k+1 k k ) 2 e F where F = F l F u = 12(k +1)+ + 6k+ + 6+ (k2 +4k +1) 12k(k +1) = ( ) 12(k+1)+/ + 6k +/ + 6+/ k2 +4k +1. 12k(k +1) Sice k 2, clearly 360k 4 +810k 3 +809k 2 +338k > 91. Equivaletly, 864k 4 +3456k 3 +3924k 2 +1332k > 504k 4 +2646k 3 +3115k 2 +994k+91 ad so 72k 2 +216k +111 504k 2 +630k +91 > k2 +4k +1 12k 2 +12k. Factorig both sides, we obtai 1 12k +13 + 1 6k +1 + 1 7 > k2 +4k +1 12k(k +1). Moreover, sice 2, 0 < 1, we have k 2 +4k +1 12k(k +1) < 1 12k +13 + 1 6k +1 + 1 7 1 12k +12+/ + 1 6k +/ + 1 6+/ 24
Multiplyig both sides by, we obtai 12(k +1)+ + 6k+ + 6+ > (k2 +4k +1) 12k(k +1). Thus F is icreasig i ad is iiu whe =. Whe =, we obtai 1 F = 12k +13 + 1 6k +1 + 1 7 k2 +4k +1 12k(k+1) = 360k4 +810k 3 +809k 2 +338k 91 6048k 4 +13608k 3 +8652k 2 +1092k. 16061 Now sice F is decreasig i k ad k 2, F 5. Thus we obtai 242424 84 ( 2 3 ( 2) (k +1) k+1 B k (,) > (k +1)(k++)(+)(k+2) k k ( 2 3 ( 2) (k +1) k+1 (k +1)(k++)(+)(k+2) k k ) 2 e F ) 2 e 5 84. Lea 2.2.4. For, N with 3 7 ad 10437, 8 <p 9 p < e 1.129918( ). Proof. Observe the well-kow idetity that 0.985x < ϑ(x) < 1.001102x where the left-had iequality holds for x 11927, the right-had side for x 1, ad where ϑ(x) = p xlogp is the first Chebyshev fuctio, see [16]. That is, e 0.985x < p x p < e 1.001102x 25
ad hece 8 <p 9 p < e 1.001102(9 ) 0.985(8 )=e1.129918( ). Now sice 8 11927 11927 ad 3 7,. Thus 10437 > 11927 7 8 8 assures that the iequality holds for 3 7. Now we ay show that B 8 (,) is a upper boud for the product of prie ubers betwee 8 ad 9 as was desired. Lea 2.2.5. For, N with 3 7 ad 10437, B 8 (,) > p. 8 <p 9 Proof. Let, N such that 3 7 ad 10437. Clearly the iequality ( ) 1 log(9 9 /8 8 ) 1.129918 > log2 2 2 + 3log3 + log + log(9+) + log(+) + log(4+) holds for = 10437 ad = 3,4,5,6, ad 7. Moreover, the right-had side of the iequality is decreasig i ad hece the iequality also holds for all 10437. Addig 3log + log( 2) + 5 84 to the left-had side, we obtai ( ) 1 log(9 9 /8 8 ) 1.129918 + 3log + log( 2) + 5 2 84 > log2 2 + 3log3 + log + log(9+) 26 + log(+) + log(4+).
Multiplyig both sides by ad rearragig ters, we obtai 3log+log( 2) 3log3 log log(9+) log(+) log2 2 log(4+)+ 2 log(99 /8 8 )+ 5 84 > 1.129918(/). Takig both sides to the base e, we obtai 3 ( 2) 9 2(9+)(+)(4+) ( 9 9 8 8 ) 2 e 5 84 > e 1.129918( ). Now by Leas 2.2.3 ad 2.2.4, we obtai B 8 (,) > 3 ( 2) 9 2(9+)(+)(4+) ( 9 9 8 8 ) 2 e 5 84 > e 1.129918( ) > 8 <p 9 p as desired. We ow eed to boud the bioial coefficiet ( ) (k+1) k fro below, ad Bk (,) ad ( ) 9/2 4 fro above for later use i our ai result as i the ext three leas. Lea 2.2.6. For a positive iteger 28327, ( ) ( ) 9 9 > 0.4231409 1/2 9. 8 8 8 Proof. By Lea 2.0.1, we obtai ( ) (k +1) l((k +1)) > k u(k)u() ( ) k +1 (k +1) k+1 = e 2πk k k 1 12(k+1)+1 k+1 12k. 27
Now whe k = 8, we obtai ( ) ( 9 9 9 9 > 8 16π 8 8 ) e 1 108+1 3 32. 1 Furtherore, sice 3 < 0 ad 1 3 is icreasig i, e 1 108+1 3 32 108+1 32 108+1 32 is icreasig i. Moreover, sice 28327, e Thus ( ) 9 9 > 8 16π 9 > 16π ( ) 9 9 e 8 8 ( 9 9 8 8 1 108+1 3 1 108+1 3 32 ) e 8271487 2773160725088 ( ) 9 > 0.4231409 1/2 9. 8 8 2773160725088. 32 e 8271487 Lea 2.2.7. The followig iequalities hold for 93: 1. B 8 (,3) < 0.065661 4 ( 9 9 ) 6 8 8 ( ) 2. B 8 (,5) < 0.014183 4 9 9 10. 8 8 ( ) 3. B 8 (,7) < 0.005169 4 9 9 14. 8 8. Proof. By Lea 2.2.2, we have B 8 (,) < 9(8+)(9+2)(+2) 4 2 3 ( ) ( 9 9 8 8 ) 2 e 12029 111150. Now cosider (8+)(9+2)(+2) 28 < 4,
equivaletly 72 4 +169 3 +52 2 2 +4 3 < 5 4. Dividig both sides by 5 ad rearragig ters, we obtai 72 + 169 + 522 + 43 + 2 3 4 < 1. Now the left-had side is decreasig i ad it suffices to verify the case whe = 93 for the respective values of. Whe = 93, we obtai 799450 923521 < 1, 69365294 74014306 < 1, ad 74805201 74805201 < 1 whe = 3, = 5, ad = 7, respectively. Thus (8+)(9+2)(+2) < 4 ad the results follow directly. Lea 2.2.8. For a positive iteger, Proof. By Lea 2.0.1, we obtai { } 9/2 9 4 2 9 2 { } 9/2 < 2.692861 ( 9 9 4 8 8 ([ 9 2 ) 2. ]) < 9 4 2 u ([ ]) 9 2 l(4)l ([ ] ) 9 2 4 u ( ) 9 2 l(4)l ( ) = 9 ( 9 9 9 2 8π 8 8 2 29 ) 2 e 1 54 1 6+1 1 48+1
Now sice 1 54 1 6+1 1 48+1 < 0 ad 1 54 1 6+1 1 48+1 e 1 54 1 6+1 1 48+1 is icreasig i. Hece e 1 54 1 6+1 1 48+1 1 ad so ( ) 9/2 < 9 4 2 ( 9 9 9 8π 8 8 ) 2 < 2.692861 ( 9 9 8 8 is icreasig i, ) 2. Lea 2.2.9. For all 56833, ( 9 9 8 8 )17 105 > e 1.001102( 9 19) 13 (9) 7.53036 log(9). Proof. The followig iequalities are equivalet for all 56833: 17 105 log 17 105 log ( 9 9 8 8 )17 105 > e 1.001102( 9 19) 13 (9) 7.53036 log(9) ( ) ( ) 9 9 9 1.001102 > 13log+7.53036 8 8 19 ( ) ( ) 9 9 9 1.001102 > 13log + 7.53036. 8 8 19 Now the right-had side is decreasig i, so it suffices to verify the case whe = 56833. Whe = 56833, we obtai 17 105 log ( ) 9 9 8 8 > 0.0340918 > 13log56833 56833 + 7.53036 56833. Theore 2.2.10. For ay positive iteger > 4 there is a prie uber betwee 8 ad 9. 30
Proof. It ca be easily verified that for = 5,6,...,56832 there is always a prie betwee 8 ad 9. Now let 56833 ad cosider: ( ) 9 8 = (8+1)(8+2) (9). 1 2 The product of pries betwee 8 ad9, if there are ay, divides ( 9 8). Followig the otatio used i [1, 6, 11], we let T 1 = p 9 p β(p), T 2 = 9<p 8 p β(p), ad T 3 = p 8+1 p 9 such that ( ) 9 = T 1 T 2 T 3. 8 The prie decopositio of ( 9 8) iplies that the powers i T2 are less tha 2, see [8, p. 24] for the prie decopositio of ( j). I additio, the prie decopositio of ( 9 8) yields the upper boud for T1 : T 1 < (9) π( 9). See [8, p. 24]. But π(x) 1.25506x, see [15]. Thus logx T 1 < (9) π( 9) (9) 2.51012 9 log(9). By Lea 2.2.1, we kow that if, N, 2, ad p is a prie uber such that 8 < p 9, the p B 8(,). So let satisfy 7 3, let A = { } 9/2 4, ad observe the followig for a prie uber p i T 2 : 31
If 9 2 < p 8, the < p 8 < 9 < 2p. Hece β(p) = 0. If 4 < p 9, the p divides A. 2 Hece p divides A. 9 2 <p 4 If 3 < p 4, the < p < 2p 8 < 9 < 3p. Hece β(p) = 0. If 8 < p 3, the p divides B 3 8(,3). Hece p divides B 8 (,3). 8 3 <p 3 If 9 4 < p 8 3, the < p < 3p 8 < 9 < 4p. Hece β(p) = 0. If 2 < p 9 4, the 2 < 2 < p 9 4 < 4 < 2p 9 2 < 3p. Hece p divides A. 2<p 9 4 If 9 5 < p 2, the < p < 4p 8 < 9 < 5p. Hece β(p) = 0. 32
If 8 < p 9, the p divides B 5 5 8(,5). Hece p divides B 8 (,5). 8 5 <p 9 5 If 3 2 < p 8 5, the < p < 5p 8 < 9 < 6p. Hece β(p) = 0. If 4 3 < p 3 2, the 2 < 4 3 < p 3 2 < 2p < 4 < 3p 9 2 < 4p. Hece 4 3 <p 3 2 p divides A. If 9 7 < p 4 3, the < p < 6p 8 < 9 < 7p. Hece β(p) = 0. If 8 < p 9, the p divides B 7 7 8(,7). Hece p divides B 8 (,7). 8 7 <p 9 7 If 9 8 < p 8 7, the < p < 7p 8 < 9 < 8p. Hece β(p) = 0. 33
If < p 9 8, the Hece p divides A. 2 < < p < 3p < 4 < 4p 9 2 < 5p. If 9 10 <p 9 8 < p, the p < 2p < 8p 8 < 9p 9 < 10p. Hece β(p) = 0. If 4 5 < p 9 10, the Hece 4 5 <p 9 10 p divides A. If 3 4 < p 4 5, the 2 < p < 4p < 4 < 5p 9 2 < 6p. p < < 2p < 10p 8 < 11p < 9 < 12p. Hece β(p) = 0. If 2 3 < p 3 4, the 2 < p < 5p < 4 < 6p 9 2 < 7p. Hece 2 3 <p 3 4 p divides A. 34
If 9 14 < p 2 3, the p < < 2p < 12p 8 < 13p < 9 < 14p. Hece β(p) = 0. If 4 7 < p 9 14, the Hece 4 7 <p 9 14 p divides A. If 9 16 < p 4 7, the 2 < p < 6p < 4 < 7p 9 2 < 8p. p < < 2p < 14p 8 < 15p < 9 < 16p. Hece β(p) = 0. If 2 < p 9 16, the Hece 2 <p 9 16 p divides A. If 9 19 < p 2, the 2 < p < 7p < 4 < 8p 9 2 < 9p. p < 2p < 3p < 16p 8 < 17p < 18p 9 < 19p. Hece β(p) = 0. 35
By the fact that p xp < e 1.001102x as show i [16], we obtai p 9<p 9 19 p 9 19 p < e 1.001102(9 19 ). By Leas 2.2.7 ad 2.2.8, we obtai { } 9/2 T 2 < e 1.001102(9 19 ) B 8 (,3)B 8 (,5)B 8 (,7) 4 ( ) < 0.000013e 1.001102(9 19 ) 25 9 9 2 + 6 + 10 + 14 2 = 0.000013e 1.001102(9 19 ) 25 2 8 8 ( 9 9 8 8 )88 105. By Lea 2.2.6, ( ) ( ) 9 9 > 0.4231409 1/2 9. 8 8 8 Thus we obtai T 3 = ( ) 9 1 8 T 1 T 2 9 0.000013 e 1.001102( > 0.4231409 19 ) 13 ( 9 9 ( )17 9 > e 1.001102(9 19 ) 13 9 105 (9) 2.51012 9 8 8 log(9) > 1, 8 8 )17 105 (9) 2.51012 9 log(9) where the last iequality follows by Lea 2.2.9. Cosequetly the product T 3 of prie ubers betwee 8 ad 9 is greater tha 1 ad therefore the existece of such ubers is prove. 36
With the proof of the previous theore coplete we ay also show that there is always a prie uber betwee ad 9+63 8 for ay positive iteger as i the followig theore. Theore 2.2.11. For ay positive iteger there is a prie uber betwee ad 9+63 8. Proof. The cases whe {1,2,3,4} ay be verified directly. Now let 5 be a positive iteger. Bythedivisio algorith8 (+r)forsoe r {0,1,2,3,4,5,6,7}. By Theore 2.2.10 there exists a prie uber p such that p ( +r, 9(+r) ). 8 Sice ( desired. +r, 9(+r) 8 ) is cotaied i (, 9+63 8 ) ( ) for all r ad, p, 9+63 8 as Sice we, i essece, skipped the cases whe k = 5,6, ad 7 i this chapter, we will show that they readily follow as corollaries of Theore 2.2.10. Corollary 2.2.12. For ay positive iteger > 1 there is a prie uber betwee 5 ad 6. Proof. The cases for 2 61 ay be verified directly. Let 62. By the divisio algorith = 8k +j for soe k N ad j {0,1,...,7}. Observe that 40k+5j 40k +8j < 45k+9j 48k +6j ad as a cosequece (8(5k+j),9(5k +j)) (5(8k +j),6(8k+j)) = (5,6). By Theore 2.2.10, p (8(5k +j),9(5k+j)) ad therefore p (5,6). 37
Corollary 2.2.13. For ay positive iteger > 4 there is a prie uber betwee 6 ad 7. Proof. The cases for 5 62 ay be verified directly. Let 63. By the divisio algorith = 8k +j for soe k N ad j {0,1,...,7}. Observe that 48k+6j 48k +8j < 54k+9j 56k +7j ad as a cosequece (8(6k+j),9(6k +j)) (6(8k +j),7(8k+j)) = (6,7). By Theore 2.2.10, p (8(6k + j),9(6k + j)) ad therefore p (6,7) as desired. Corollary 2.2.14. For ay positive iteger > 2 there is a prie uber betwee 7 ad 8. Proof. The cases for 3 63 ay be verified directly. Let 64. By the divisio algorith = 8k +j for soe k N ad j {0,1,...,7}. Observe that 56k+7j 56k +8j < 63k+9j 64k +8j ad as a cosequece (8(7k+j),9(7k +j)) (7(8k +j),8(8k+j)) = (7,8). By Theore 2.2.10, p (8(7k + j),9(7k + j)) ad therefore p (7,8) as desired. 38
2.3 Pries i the Iterval [519, 520] Wewillowoveawayfrotheeleetaryethodsuseditheprevioustwosectios ad ove towards a aalytic uber theory approach to establish a iproved result. Our proof cofors to S. Raauja s [13] proof of Bertrad s postulate. It also cofors with J. Nagura s [12] proof of pries i the iterval to 6 5. The basis of our proof is to approxiate the first Chebyshev fuctio ϑ usig the secod Chebyshev fuctio ψ. That is, the fuctios: ψ(x) = =1ϑ(x 1/ ) ad ϑ(x) = p x p prie logp. If we are able to show that ϑ( 520 ) ϑ() > 0, the by takig both sides to the 519 base e, we obtai: p 520 519 p prie p > p p prie However, the the product of pries betwee ad 520 519 is greater tha 1, ad so there ust be at least oe prie uber betwee ad 520 519. I 1976, L. Schoefeld [16] showed that for all x e 19, the upper ad lower bouds of ψ(x) are give by the iequality 0.99903839x < ψ(x) < 1.00096161x. I his paper he achieved these approxiatios by usig aalytic ethods to show that ψ(x) x < 0.00096161x fro which the double iequality follows. We will use this approxiatio i the followig theore. p. Theore 2.3.1. For 31409, there exists at least oe prie uber p such that < p < 520 519. 39
Proof. I order to prove ϑ( 520 ) ϑ() > 0 for the values of as sall as possible, 519 let us use ad ψ(x) ψ(x 1/2 ) ψ(x 1/3 ) ψ(x 1/503 ) = ψ(x) p 503 ϑ(x) ψ(x) ψ(x 1/2 ) ψ(x 1/3 ) ψ(x 1/509 ) = ψ(x) ψ(x 1/p ) ϑ(x) (2.1) p 509 ψ(x 1/p ). (2.2) Note that we chose 503 ad 509 as the suatio upper liit i the suatios of equatios 2.1 ad 2.2, respectively. These choices are the secod largest prie ad the largest prie less tha 519, respectively. ϑ Thus we obtai ( ) 520 ϑ() ψ 519 ( ) 520 ψ 519 p 509 ( (520 ) ) 1/p ψ()+ ψ( 1/p ). 519 p 503 By usig the approxiatio for ψ(x), 0.99903839x < ψ(x) < 1.00096161x, we obtai ϑ ( ) ( 520 520 ϑ() > 0.99903839 519 519 + ) 1/p p 503 ( 1.00096161 + ( ) ) 1/p 520 519 p 509 which is positive for e 19. 40
However, we ay verify the cases for 31409 e 19 usig a progra such as Matheatica ad our theore is thus proved. Fro the previous theore we ay prove a corollary that ϑ( (k+1) ) ϑ() > 0 for k all k ad such that 31409 ad 519 k 1. We will also apply this corollary to show that there is a prie uber betwee 519 ad 520 for all 15. This theore also shows that there is always a prie betwee k ad (k+1) for all k ad 519 k 2 which is a sigificat iproveet i the uber of cases for k that we were able to show i Sectios 2.1 ad 2.2. Corollary 2.3.2. For k, N with 31409 ad 519 k 1, ( ) (k +1) ϑ ϑ() > 0. k Proof. The iequality follows directly by otig that for all 31409, ϑ(2) ϑ(3/2) ϑ(4/3)... ϑ(519/518) ϑ(520/519). Hece ϑ(2) ϑ() ϑ(3/2) ϑ()... ϑ(520/519) ϑ() > 0 where the last iequality follows by Theore 2.3.1. Theore 2.3.3. For 15 there is a prie uber betwee 519 ad 520. Proof. The cases for 15 31408 ay be verified directly. Now let 31409 ad cosider ϑ( 520 ) ϑ() > 0 as show i Corollary 2.3.2. Takig both sides of 519 41
the iequality to the base e, we obtai e ϑ(520 519 ) ϑ() = <p 520 519 p > 1. Therefore there exists at least oe prie uber betwee ad 520 519. Allowig = 519 for soe N, we deduce that there exists at least oe prie uber betwee 519 ad 520 as desired. Theore 2.3.4. For k, N with k ad 519 k 2, there is at least oe prie uber betwee k ad (k +1). Proof. For ay choice of k such that 2 k 519, the cases for k 31408 ay be verified directly. By Theore 2.3.1, for all 31409, π( 520 ) π() 1. Now, 519 π(2) π(3/2) π(4/3)... π(520/519) π()+1. Therefore, π(2) π() π(3/2) π()... π(520/519) π() 1. (2.3) Now by lettig =, = 2,..., = 519 for soe N i the respective iequalities above, we obtai π(2) π() 1, π(3) π(2) 1,..., π(520) π(519) 1 as desired. Fro equatio (2.3) i the previous theore we obtai a direct corollary. Corollary 2.3.5. For k, N with k ad 519 k 2, there is at least oe prie uber betwee ad (k+1) k. 42
2.4 Pries betwee 2 ad (+1) 2+ε If we allow k = i our previous questio of a prie uber p satisfyig k < p < (k +1), the we obtai a sharper iequality tha Legedre s cojecture. Legedre s cojecture states that for every positive iteger there exists a prie uber betwee 2 ad (+1) 2. That is, there is always a prie uber betwee ay two successive perfect squares. This questio is oe of the faous Ladau probles proposed by Edud Ladau i 1912 at the Iteratioal Cogress of Matheaticias, see [9]. While this questio reais ope to date, soe progress has bee ade for sufficietly large. Perhaps ost otably, Che Jigru [4] has show that there exists a uber p satisfyig 2 < p < (+1) 2 such that p is either a prie uber or a seiprie; where a seiprie is a product of two prie ubers, ot ecessarily distict. Furtherore, there is always a prie uber betwee θ ad for θ = 23/42 ad θ = 0.525, see [9, p. 415], [10], ad [2]. While we offer o proof of Legedre s cojecture, we do show that for all positive itegers there exists a prie uber p such that 2 < p < (+1) 2+ε for ε = 0.00011516865557559264 ad ε = 0.000001. Theore 2.4.1. For a positive iteger, there exists a prie uber betwee 2 ad (+1) 2+ε where ε = 0.00011516865557559264. Proof. The cases for = 1,2,...,4407 ay be verified directly. By Theore 5.2 of [5] we have for all x 3594641: ϑ(x) x < 0.2x log 2 x. 43
Let 4408 so that (+1) 2+ε > 2 > 3594641. Now ϑ((+1) 2+ε ) ϑ( 2 ) > ( ) (+1) 2+ε 0.2(+1)2+ε log 2 (+1) 2+ε ( 2 + 0.22 log 2 2 ). Cosider the followig equivalet iequalities for 4408: (+1) 2+ε > 2 + 0.2(+1)2+ε 0.22 log 2 2+ε + (+1) log 2 2, 1 > ( ) 2 ( ) 2 1 0.2 +1 (+1) ε + (2+ε) 2 log 2 (+1) + 0.05 +1 (+1) ε log 2. Now the right-had side is decreasig i ad so it suffices to verify the case whe = 4408. Whe = 4408, we obtai 1 > ( ) 2 4408 1 4409 4409 ε + 0.2 (2+ε) 2 log 2 4409 + ( ) 2 4408 0.05 4409 4409 ε log 2 4408. Therefore ( ) (+1) 2+ε 0.2(+1)2+ε log 2 (+1) 2+ε ) ( 2 + 0.22 log 2 > 0 2 ad hece ϑ((+1) 2+ε ) ϑ( 2 ) > 0 as desired. Although siilar results ay be show for ay ε > 0, we shall see that as ε 0 the uber of base cases for which ust be verified directly icreases. I the ext theore we show that ε = 0.000001 is sufficiet, however this icreases the uber of base cases which ust be verified by 26,010,188. As etioed previously, this icrease i the uber of base cases to verify is expected due to the techiques used withi the proofs of this chapter. 44
Theore 2.4.2. For a positive iteger, there exists a prie uber betwee 2 ad (+1) 2.000001. Proof. The cases for = 1,2,...,26014595 ay be verified directly. By Theore 5.2 of [5] we have for all x 7713133853, ϑ(x) x < 0.01x log 2 x. Let 26014596 so that (+1) 2.000001 > 2 > 7713133853. Now ϑ((+1) 2.000001 ) ϑ( 2 ) > ( (+1) 2.000001 0.2(+1)2.000001 log 2 (+1) 2.000001 ) ) ( 2 + 0.22 log 2. 2 Cosider the followig equivalet iequalities for 26014596: (+1) 2.000001 > 2 + 0.01(+1)2.000001 0.012 log 2 2.000001 + (+1) log 2 2, 1 > ( ) 2 1 +1 (+1) 0.000001 + 0.01 2.000001 2 log 2 (+1) ( + +1 ) 2 0.0025 (+1) 0.000001 log 2. Now the right-had side is decreasig i ad so it suffices to verify the case whe = 26014596. Whe = 26014596, we obtai 1 > ( ) 2 26014596 1 26014597 26014597 0.000001 + 0.01 2.000001 2 log 2 26014597 ( 26014596 + 26014597 45 ) 2 0.05 26014597 0.000001 log 2 26014596.
Therefore (+1) 2.000001 0.2(+1)2.000001 log 2 (+1) 2.000001 2 0.22 log 2 2 > 0 ad hece ϑ((+1) 2.000001 ) ϑ( 2 ) > 0 as desired. 46
Chapter 3 The Nuber of Prie Nubers I this chapter we cosider the questio: How ay prie ubers are there betwee ad k? For istace, Bertrad s postulate states that there is at least oe prie uber betwee ad 2 for all > 1. Moreover, by M. El Bachraoui [1], we kow there exists a prie uber betwee 2 ad 3 for all > 1. Therefore, there are at least two prie ubers betwee ad 3 for > 1. Furtherore, if we take ito accout A. Loo s [11] result that there is a prie uber betwee 3 ad 4 for > 1, the there are three prie ubers betwee ad 4 for all > 1. I the followig sectios we will exted these ethods to iprove upo the uber of pries betwee ad k usig our previous results. 47
3.1 Betwee ad 5 I the followig theore we exted the previous facts by showig that there are at least four prie ubers betwee ad 5. I order to accoplish this task we will use Theore 2.1.4; that is, there is a prie uber betwee ad (5+15)/4 for all > 2. Theore 3.1.1. For ay positive iteger > 2, there are at least four prie ubers betwee ad 5. Proof. The cases whe = 3,4,...,14 ay be verified directly. Now let 15 ad by Theore 2.1.4 we kow there exists prie ubers p 1, p 2, ad p 3 such that < p 1 < 5+15 4, 2 < p 2 < 10+15 4, 3 < p 3 < 15+15 4, ad by Theore 2.1.3 there exist a prie uber p 4 such that 4 < p 4 < 5. Hece < p 1 < 5+15 4 < 2 < p 2 < 10+15 4 < 3 < p 3 < 15+15 4 < p 4 < 5. 4 I the ext theore we iprove o the uber of prie ubers betwee ad 5 to show that there are at least seve prie ubers betwee ad 5 for > 5. Theore 3.1.2. For all > 5 there are at least seve prie ubers betwee ad 5. Proof. The cases whe = 6,7,...,244 ay be verified directly. Now let f() = 5+15 4 for 245 ad let f () = f(f 1 ()). By Theore 2.1.4 there exists a prie uber betwee ad f(). Furtherore, there exists a prie 48
uber betwee f 1 () ad f () for all N\{1}. I geeral, ( f () = 1 5 +3 4 ) 1 5 k 4 k. k=0 Cosider ( ) f () = 1 1 5 +3 5 k 4 k 5. 4 k=0 Solvig for, we obtai 1 3 5 k 4 k. 5 4 5 k=0 However, 5 4 5 is positive oly for 7. So let = 7, the for 245 > 3 5 4 7 5 7 6 5 7 k 4 k k=0 there are at least seve prie ubers betwee ad 5. 3.2 Betwee ad 9 Siilar to the previous sectio, we ay show that there are at least 8 prie ubers betwee ad 9 for all positive itegers > 2. Theore 3.2.1. For ay positive iteger > 2 there are at least eight prie ubers betwee ad 9. 49
Proof. The cases for = 3,4,...,63 ay be verified directly. Now let 64 ad observe that by Theore 2.2.11 there exists prie ubers p 1,p 2,...,p 8 such that < p 1 < 9+63 8 < p 4 < 36+63 8 < p 7 < 63+63 8 < 2 < p 2 < 18+63 8 < 5 < p 5 < 45+63 8 < 8 < p 8 < 9 < 3 < p 3 < 27+63 8 < 6 < p 6 < 54+63 8 < 4 < 7 where 8 < p 8 < 9 by Theore 2.2.10. Therefore there are at least eight prie ubers betwee ad 9 for ay positive iteger > 2. I the ext theore we iprove o the uber of prie ubers betwee ad 9 to show that there are at least eightee prie ubers betwee ad 9 for > 8. Theore 3.2.2. For all > 8 there are at least eightee prie ubers betwee ad 9. Proof. The cases whe = 9,10,...,691 ay be verified directly. Now let f() = 9+63 8 for 692 ad let f () = f(f 1 ()). By Theore 2.2.11 there exists a prie uber betwee ad f(). Furtherore, there exists a prie uber betwee f 1 () ad f () for all N\{1}. I geeral, ( f () = 1 9 +7 8 ) 1 9 k 8 k. k=0 Cosider ( ) f () = 1 1 9 +7 9 k 8 k 9. 8 k=0 50
Solvig for, we obtai 1 7 9 k 8 k. 9 8 9 However, 9 8 9 is positive oly for 18. So let = 18, the for k=0 692 > 1 9 8 18 9 18 17 k=0 9 18 k 8 k there are at least eightee prie ubers betwee ad 9. 3.3 Betwee ad 520 Siilarly to the previous two sectios, usig Theore 2.3.4, we ay establish that there are at least 519 prie ubers betwee ad 520 for all positive itegers > 7. Theore 3.3.1. For > 7 there are at least 519 prie ubers betwee ad 520. Proof. The cases for = 8,9,...,31408 ay be verified directly. Now let 31409. By Theore 2.3.4 we kow there exists pries p 1, p 2,..., ad p 519 such that p 1 (,2), p 2 (2,3),..., ad p 519 (519,520). Thus p 1,p 2,...,p 519 (,520) as desired. I the ext theore we iprove o the uber of prie ubers betwee ad 520 to show that there are at least 3248 prie ubers betwee ad 520 for > 58. 51
Theore 3.3.2. For > 58 there are at least 3248 prie ubers betwee ad 520. Proof. The cases whe = 59, 60,..., 31408 ay be verified directly. Now let f() = 520 519 for 31409 ad let f () = f(f 1 ()). By Theore 2.3.1 there exists a prie uber betwee ad f(). Furtherore, there exists a prie uber betwee f 1 () ad f () for all N\{1}. Cosider: f 3248 () = 29744 5 3248 13 3248 3 3248 173 3248 < 519.14 < 520. Which holds for all > 0 ad our proof is coplete. 3.4 Betwee ad k We will ow geeralize our previous results by showig that for ay k 2 there are at least k 1 prie ubers i the iterval (,k). I order to obtai our result we will use the followig iequality as show by P. Dusart i Theore 5.2 of [5]: ϑ(x) x < 3.965x log 2 x, where log 2 x = (logx) 2. Theore 3.4.1. For all k 2, there are at least k 1 prie ubers betwee ad k. Proof. The cases for k ad 7 k 2 follow directly by [1], [11], Theore 2.1.3, Corollary 2.2.12, Corollary 2.2.13, Corollary 2.2.14, ad Theore 2.2.10. 52
Let k 8. By Theore 5.2 of [5, p. 4], for x 2, we obtai ϑ(x) x < 3.965x log 2 x. Now ϑ(k) ϑ() > ( k 3.965k ) ( log 2 + 3.965 ) k log 2. (3.1) Moreover, observe that the followig iequalities are equivalet for all k 8: k 3.965k log 2 k 3.965 log 2 > (k 1)logk, k 1 > (k 1)logk + 3.965k log 2 k + 3.965 log 2. Now the right-had side is decreasig i, so it suffices to verify the case whe = k. Whe = k, we obtai the followig equivalet iequalities: k 1 > (k 1)logk2 k + 3.965k log 2 k 2 + 3.965 log 2 k, 1 > 2logk k + 3.965k 4(k 1)log 2 k + 3.965 (k 1)log 2 k. Now the right-had side is decreasig i k, so it suffices to verify the case whe k = 8. Whe k = 8, we obtai 1 > 2log8 8 + 3.965 8 28log 2 8 + 3.965 7log 2 8. 53
Thus ϑ(k) ϑ() > ( k 3.965k ) ( log 2 + 3.965 ) k log 2 > (k 1)logk for k 8. That is, ϑ(k) ϑ() = <p k log(p) > (k 1)logk. However, logk logp for ay p satisfyig < p k ad hece for the above iequality to be true there ust be at least k 1 prie ubers betwee ad k. Equatio (3.1) i the previous theore provides us with a better lower approxiatio for ϑ(k) ϑ() as i the followig corollary. Corollary 3.4.2. For all k 8, ( ( k ϑ(k) ϑ() > k 1 3.965 log 2 k + 1 )) log 2. 3.5 Betwee Successive Powers If we allow k = i Theore 3.4.1, the there are at least 1 prie ubers betwee ad 2 for all 64. However, we ay geeralize our ethod of proof to show that there are at least d (logd)/2 ay prie ubers betwee d ad d+1 as i the ext theore. Theore 3.5.1. For 8 ad d 1, there are at least d (logd)/2 prie ubers betwee d ad d+1. 54
Proof. By Theore 5.2 of [5, p. 4], for x 2, we obtai ϑ(x) x < 3.965x log 2 x. Now ) ) ϑ( d+1 ) ϑ( d ) > ( d+1 3.965d+1 log 2 ( d + 3.965d d+1 log 2. d d 1: Moreover, observe that the followig iequalities are equivalet for all 8 ad d+1 > d (logd)/2 log d+1 + d + 3.965d+1 3.965d log 2 + d+1 log 2, d 1 > logd+1 + 1 1+(logd)/2 + 3.965 3.965 log 2 + d+1 log 2 d. Now the right-had side is decreasig i both ad d, so it suffices to verify the case whe = 8 ad d = 1. Whe = 8 ad d = 1, we obtai 1 > 3log2 4 + 1 8 + 3.965 36log 2 2 + 3.965 72log 2 2. Thus ) ) ϑ( d+1 ) ϑ( d ) > ( d+1 3.965d+1 log 2 ( d + 3.965d d+1 log 2 > d (logd)/2 log d+1 d for 8 ad d 1. 55
That is, ϑ( d+1 ) ϑ( d ) = d <p d+1 logp > d (logd)/2 log d+1. However, log d+1 logp for ay p satisfyig d < p d+1 ad hece for the above iequality to be true there ust be at least d (logd)/2 prie ubers betwee d ad d+1. 56
Chapter 4 The Prie Nuber Theore The prie uber theore describes the asyptotic distributio of the prie ubers aog the positive itegers. That is to say, the prie uber theore asserts that the prie ubers becoe rarer as they becoe larger. This is foralized by li x π(x) x/logx = 1 which is kow as the asyptotic law of distributio of prie ubers. Equivaletly, π(x) x logx. (4.1) The first truly eleetary proofs of the prie uber theore appeared withi [7, 17] ad was also cause to the Erdős Selberg priority dispute, see [18]. Now by equatio (4.1) for k a positive iteger we would expect the uber of prie ubers i the iterval (k,(k +1)) to icrease as rus through the positive itegers. Siply applyig the prie uber theore estiates that we could 57
expect, very roughly, (k,(k +1)). (k+1) log(k+1) k logk logk ay prie ubers i the iterval We will establish explicit lower bouds for the uber of prie ubers i each of our itervals i the previous two chapters. I each case we first show a lower approxiatio for the uber of prie ubers i a particular iterval ad the show that our results coicide with the prie uber theore. To establish a explicit boud for the uber of prie ubers i the iterval (4,5) we ay boud each prie i the iterval fro above by 5 as follows. Theore 4.0.1. For > 2, the uber of prie ubers i the iterval (4,5) is at least [ 0.054886 log 5 2 2 3 2 ( ) 3125 256 6 (5) 2.51012 5 log(5) ]. Proof. I Theore 2.1.3 we approxiated the product of prie ubers betwee 4 ad 5 fro below by 0.054886 2 2 3 2 ( ) 3125 6 (5) 2.51012 5 log(5). 256 Boudig each of the prie ubers betwee 4 ad5froabove by 5, we obtai: [ 0.054886 log 5 2 2 3 2 > ( ) 3125 256 6 (5) 2.51012 5 log(5) = log0.054886 log2+ 2 6 log+log5 ( ) 1 3125 log 1 log2 2.51012 5 6 256 2 > log ] log 3125 256 2.51012 5+ 3 2 log5 + 2log ( 0.035214 2.8063995 ) 168033 100000. 3 2 3 log5+log0.054886 2 3 log+log5 2 58
2.8063995 Now observe that li = 0. Moreover li log = +. Coparig the previous result with the weak versio of the prie uber theore, (k+1) log(k+) k logk with k = 4, we obtai the followig table. Result Weak PNT π(5) π(4) 10 4 11.7 846.4 930 10 5 399.9 7093.3 7678 10 6 4200.5 61023.5 65367 10 7 38725.5 535330.3 567480 10 8 348808.9 4767502.3 5019541 Table 4.1: Copariso of Theore 4.0.1 with PNT. By Theore 4.0.1 we have the followig theore. Theore 4.0.2. As, the uber of prie ubers i the iterval [4,5] goes to ifiity. That is, for every positive iteger, there exists a positive iteger L such that for all L, there are at least prie ubers i the iterval [4,5]. To establish a explicit lower boud for the uber of prie ubers i the iterval (8,9) we ay boud each prie i the iterval fro above by 9 as i the ext theore. Theore 4.0.3. For > 4, the uber of prie ubers i the iterval (8,9) is at least [ ( log 9 32549.3e 1.001102(9 19) 13 9 9 8 8 )17 105 (9) 2.51012 9 log(9) ]. Proof. I Theore 2.2.10 we approxiated the product of prie ubers betwee 8 ad 9 fro below by 32549.3e 1.001102(9 19) 13 59 ( 9 9 8 8 )17 105 (9) 2.51012 9 log(9).
Boudig each of the prie ubers betwee 8 ad9froabove by 9, we obtai: log 9 [ ( 32549.3e 1.001102(9 19) 13 9 9 8 8 )17 105 (9) 2.51012 9 log(9) > ( ( ) ( ) 17 9 9 9 2log 105 log 1.001102 + log32549.3 13log 7.53036 ) 8 8 19 > ( 0.0170459 13log 3.76518 ). log ( 13log Now observe that li + 3.76518 ) = 0. Moreover li log = +. Coparig the previous result with the weak versio of the prie uber theore, (k+1) log(k+) k logk with k = 8, we obtai the followig table. Result Weak PNT π(9) π(8) 10 4-55.8 803.4 876 10 5 71.9 6788.2 7323 10 6 1732.4 58748.2 62712 10 7 17669.9 517719.7 547572 10 8 163011.7 4627221.6 4863036 Table 4.2: Copariso of Theore 4.0.3 with PNT. ] By Theore 4.0.3 we have the followig theore. Theore 4.0.4. As, the uber of prie ubers i the iterval [8,9] goes to ifiity. That is, for every positive iteger, there exists a positive iteger L such that for all L, there are at least prie ubers i the iterval [8,9]. Siilar results ay be show about the uber of prie ubers betwee ad k. That is, the uber of prie ubers betwee ad k teds to ifiity as teds to ifiity. 60