Bertrad s Postulate Lola Thompso Ross Program July 3, 2009 Lola Thompso (Ross Program Bertrad s Postulate July 3, 2009 1 / 33
Bertrad s Postulate I ve said it oce ad I ll say it agai: There s always a prime betwee ad. -Joseph Bertrad, cojectured i 1845 Lola Thompso (Ross Program Bertrad s Postulate July 3, 2009 2 / 33
Bertrad s Postulate Bertrad did ot prove his postulate. He verified the statemet (by had for all positive itegers up to 6,000,000. I 1850, Chebyshev proved Bertrad s Postulate. For this reaso, it is also referred to as Chebyshev s Theorem. Lola Thompso (Ross Program Bertrad s Postulate July 3, 2009 3 / 33
Bertrad s Postulate May other proofs have bee foud i the time sice Chebyshev first proved this theorem. We will follow a proof due to Ramauja ad Erdös. Lola Thompso (Ross Program Bertrad s Postulate July 3, 2009 4 / 33
Outlie 1 Fu With Biomial Coefficiets 2 A Few New Arithmetic Fuctios 3 A Upper Boud for ψ(x 4 Provig Bertrad s Postulate The Setup: ABC = ( A Lower Boud for ( A Upper Boud for C A Upper Boud for B Puttig Everythig Together 5 Geeralizatios 6 Some Neat Applicatios Lola Thompso (Ross Program Bertrad s Postulate July 3, 2009 5 / 33
Fu With Biomial Coefficiets Defiitio! = # of ways of orderig elemets Defiitio ( k = # of ways of choosig k elemets from a set cotaiig objects, without worryig about order. Lola Thompso (Ross Program Bertrad s Postulate July 3, 2009 6 / 33
Fu With Biomial Coefficiets There is a curious relatioship betwee ( k ad Pascal s Triagle. Lola Thompso (Ross Program Bertrad s Postulate July 3, 2009 7 / 33
Fu With Biomial Coefficiets There is a curious relatioship betwee ( k ad Pascal s Triagle. If we write dow Pascal s Triagle, the first few rows are: 1 1 1 1 2 1 1 3 3 1 Lola Thompso (Ross Program Bertrad s Postulate July 3, 2009 7 / 33
Fu With Biomial Coefficiets There is a curious relatioship betwee ( k ad Pascal s Triagle. If we write dow Pascal s Triagle, the first few rows are: 1 1 1 1 2 1 1 3 3 1 We ca obtai the umbers i the ext row by addig adjacet pairs of umbers from the previous row: 1 3 3 1 1 4 6 4 1 Lola Thompso (Ross Program Bertrad s Postulate July 3, 2009 7 / 33
Fu With Biomial Coefficiets There is a curious relatioship betwee ( k ad Pascal s Triagle. If we write dow Pascal s Triagle, the first few rows are: 1 1 1 1 2 1 1 3 3 1 We ca obtai the umbers i the ext row by addig adjacet pairs of umbers from the previous row: 1 3 3 1 1 4 6 4 1 The elemets i the 4 th row are precisely ( 4 0 through ( 4 4. Lola Thompso (Ross Program Bertrad s Postulate July 3, 2009 7 / 33
Fu With Biomial Coefficiets Aother iterestig observatio that you might make is that, at least i the examples above, the sum of all of the elemets i a row is a power of 2. Lola Thompso (Ross Program Bertrad s Postulate July 3, 2009 8 / 33
Fu With Biomial Coefficiets Aother iterestig observatio that you might make is that, at least i the examples above, the sum of all of the elemets i a row is a power of 2. For example, 1 + 2 + 1 = 2 2 ad 1 + 3 + 3 + 1 = 2 3. Lola Thompso (Ross Program Bertrad s Postulate July 3, 2009 8 / 33
Fu With Biomial Coefficiets Aother iterestig observatio that you might make is that, at least i the examples above, the sum of all of the elemets i a row is a power of 2. For example, 1 + 2 + 1 = 2 2 ad 1 + 3 + 3 + 1 = 2 3. ( Exercise: k = 2. k=0 Lola Thompso (Ross Program Bertrad s Postulate July 3, 2009 8 / 33
Fu With Biomial Coefficiets Oe particular type of biomial coefficiet will be of iterest to us as we prove Bertrad s Postulate. That coefficiet is (, the coefficiet i the ceter of the th row of Pascal s Triagle. Lola Thompso (Ross Program Bertrad s Postulate July 3, 2009 9 / 33
Fu With Biomial Coefficiets Oe particular type of biomial coefficiet will be of iterest to us as we prove Bertrad s Postulate. That coefficiet is (, the coefficiet i the ceter of the th row of Pascal s Triagle. Which primes divide (? Let s look at some examples. Lola Thompso (Ross Program Bertrad s Postulate July 3, 2009 9 / 33
Fu With Biomial Coefficiets Oe particular type of biomial coefficiet will be of iterest to us as we prove Bertrad s Postulate. That coefficiet is (, the coefficiet i the ceter of the th row of Pascal s Triagle. Which primes divide (? Let s look at some examples. ( 4 ( 2 = 6. Which primes divide 6? What about 6 ( 3? 8 ( 4? 10 5? Ay cojectures about whe a prime has to divide (? Lola Thompso (Ross Program Bertrad s Postulate July 3, 2009 9 / 33
Fu With Biomial Coefficiets Oe particular type of biomial coefficiet will be of iterest to us as we prove Bertrad s Postulate. That coefficiet is (, the coefficiet i the ceter of the th row of Pascal s Triagle. Which primes divide (? Let s look at some examples. ( 4 ( 2 = 6. Which primes divide 6? What about 6 ( 3? 8 ( 4? 10 5? Ay cojectures about whe a prime has to divide (? Exercise: p ( if there exists a positive iteger j with < p j Lola Thompso (Ross Program Bertrad s Postulate July 3, 2009 9 / 33
Fu With Biomial Coefficiets Assumig that those exercises are true, we ca prove two results: Lemma (1 p p:<p j (. Lola Thompso (Ross Program Bertrad s Postulate July 3, 2009 10 / 33
Fu With Biomial Coefficiets Assumig that those exercises are true, we ca prove two results: Lemma (1 p p:<p j (. Proof We kow that each prime p with < p j divides.sice the primes are distict, they are pairwise relatively ( prime.thus, their product must divide (. Lola Thompso (Ross Program Bertrad s Postulate July 3, 2009 10 / 33
Fu With Biomial Coefficiets Assumig that those exercises are true, we ca prove two results: Lemma (1 p p:<p j (. Proof We kow that each prime p with < p j divides.sice the primes are distict, they are pairwise relatively ( prime.thus, their product must divide (. Lemma (2 ( 4 for all 0. Lola Thompso (Ross Program Bertrad s Postulate July 3, 2009 10 / 33
Fu With Biomial Coefficiets Assumig that those exercises are true, we ca prove two results: Lemma (1 p p:<p j (. Proof We kow that each prime p with < p j divides.sice the primes are distict, they are pairwise relatively ( prime.thus, their product must divide (. Lemma (2 ( 4 for all 0. Proof Look at the th row of Pascal s Triagle: ( is i the ceter, 2 = sum of all terms i the row. Lola Thompso (Ross Program Bertrad s Postulate July 3, 2009 10 / 33
A Few New Arithmetic Fuctios We re already familiar with σ, τ ad ϕ. Let s defie some ew arithmetic fuctios: Defiitio { log p, = p k, k > 0 Λ( = 0, otherwise Lola Thompso (Ross Program Bertrad s Postulate July 3, 2009 11 / 33
A Few New Arithmetic Fuctios We re already familiar with σ, τ ad ϕ. Let s defie some ew arithmetic fuctios: Defiitio { log p, = p k, k > 0 Λ( = 0, otherwise Examples: Λ(2 = log 2, Λ(4 = log 2, Λ(6 = 0. Lola Thompso (Ross Program Bertrad s Postulate July 3, 2009 11 / 33
A Few New Arithmetic Fuctios We re already familiar with σ, τ ad ϕ. Let s defie some ew arithmetic fuctios: Defiitio { log p, = p k, k > 0 Λ( = 0, otherwise Examples: Λ(2 = log 2, Λ(4 = log 2, Λ(6 = 0. Defiitio ψ(x = x Λ( Lola Thompso (Ross Program Bertrad s Postulate July 3, 2009 11 / 33
A Few New Arithmetic Fuctios We re already familiar with σ, τ ad ϕ. Let s defie some ew arithmetic fuctios: Defiitio { log p, = p k, k > 0 Λ( = 0, otherwise Examples: Λ(2 = log 2, Λ(4 = log 2, Λ(6 = 0. Defiitio ψ(x = x Λ( Example: ψ(6 = 0 + log 2 + log 3 + log 2 + log 5 + 0 = log(2 2 3 5 = log(60. Lola Thompso (Ross Program Bertrad s Postulate July 3, 2009 11 / 33
A Upper Boud for ψ(x Lemma ψ(x x log 4 For ow, let s preted that x Z. This will allow us to use the Well-Orderig Priciple. Base Case: If = 1, the ψ(1 = 1 Λ( = Λ(1 = 0. (This is certaily 1 log 4 Lola Thompso (Ross Program Bertrad s Postulate July 3, 2009 12 / 33
Proof of Upper Boud for ψ(x Let S = {x Z + ψ(x > x log 4}. Assume that S is o-empty. The, by Well-Orderig Priciple, S has a least elemet. Call it l. Case 1: l = 2k ψ(2k ψ(k Lola Thompso (Ross Program Bertrad s Postulate July 3, 2009 13 / 33
Proof of Upper Boud for ψ(x Let S = {x Z + ψ(x > x log 4}. Assume that S is o-empty. The, by Well-Orderig Priciple, S has a least elemet. Call it l. Case 1: l = 2k ψ(2k ψ(k = 2k Λ( k Λ( (from defiitio of ψ Lola Thompso (Ross Program Bertrad s Postulate July 3, 2009 13 / 33
Proof of Upper Boud for ψ(x Let S = {x Z + ψ(x > x log 4}. Assume that S is o-empty. The, by Well-Orderig Priciple, S has a least elemet. Call it l. Case 1: l = 2k ψ(2k ψ(k = Λ( Λ( (from defiitio of ψ 2k k = Λ( (cacelig terms k< 2k Lola Thompso (Ross Program Bertrad s Postulate July 3, 2009 13 / 33
Proof of Upper Boud for ψ(x Let S = {x Z + ψ(x > x log 4}. Assume that S is o-empty. The, by Well-Orderig Priciple, S has a least elemet. Call it l. Case 1: l = 2k ψ(2k ψ(k = Λ( Λ( (from defiitio of ψ 2k k = Λ( (cacelig terms k< 2k log ( 2k k (sice Λ( = p if = p j ad 0 otherwise, so this lie follows from Lemma (1 after takig log of both sides Lola Thompso (Ross Program Bertrad s Postulate July 3, 2009 13 / 33
Proof of Upper Boud for ψ(x From the previous slide: ψ(2k ψ(k log ( 2k k. Lola Thompso (Ross Program Bertrad s Postulate July 3, 2009 14 / 33
Proof of Upper Boud for ψ(x From the previous slide: ψ(2k ψ(k log ( 2k k. From Lemma (2, ( 2k k 4 k, i.e. ψ(2k ψ(k log(4 k. Lola Thompso (Ross Program Bertrad s Postulate July 3, 2009 14 / 33
Proof of Upper Boud for ψ(x From the previous slide: ψ(2k ψ(k log ( 2k k. From Lemma (2, ( 2k k 4 k, i.e. ψ(2k ψ(k log(4 k. From high school math: log(4 k = k log 4. Lola Thompso (Ross Program Bertrad s Postulate July 3, 2009 14 / 33
Proof of Upper Boud for ψ(x From the previous slide: ψ(2k ψ(k log ( 2k k. From Lemma (2, ( 2k k 4 k, i.e. ψ(2k ψ(k log(4 k. From high school math: log(4 k = k log 4. So ψ(2k ψ(k + k log 4 Lola Thompso (Ross Program Bertrad s Postulate July 3, 2009 14 / 33
Proof of Upper Boud for ψ(x From the previous slide: ψ(2k ψ(k log ( 2k k. From Lemma (2, ( 2k k 4 k, i.e. ψ(2k ψ(k log(4 k. From high school math: log(4 k = k log 4. So ψ(2k ψ(k + k log 4 k log 4+k log 4 (sice l = 2k was the smallest elemet i S, so k / S ψ(k k log 4 Lola Thompso (Ross Program Bertrad s Postulate July 3, 2009 14 / 33
Proof of Upper Boud for ψ(x From the previous slide: ψ(2k ψ(k log ( 2k k. From Lemma (2, ( 2k k 4 k, i.e. ψ(2k ψ(k log(4 k. From high school math: log(4 k = k log 4. So ψ(2k ψ(k + k log 4 k log 4+k log 4 (sice l = 2k was the smallest elemet i S, so k / S ψ(k k log 4 ψ(2k 2k log 4, ie. l caot be eve. Lola Thompso (Ross Program Bertrad s Postulate July 3, 2009 14 / 33
Proof of Upper Boud for ψ(x Case 2: l = 2k + 1. The argumet for the case where l is odd is similar to the case where l is eve. The result is the same, i.e. it shows that l caot be odd. Sice the least elemet of S is either eve or odd the S must empty. Remark: The iequality ψ(x x log 4 holds for ALL x 1 (ot just itegers. Why? Lola Thompso (Ross Program Bertrad s Postulate July 3, 2009 15 / 33
Proof of Upper Boud for ψ(x Case 2: l = 2k + 1. The argumet for the case where l is odd is similar to the case where l is eve. The result is the same, i.e. it shows that l caot be odd. Sice the least elemet of S is either eve or odd the S must empty. Remark: The iequality ψ(x x log 4 holds for ALL x 1 (ot just itegers. Why? ψ(x = ψ( x x log4 x log 4. Lola Thompso (Ross Program Bertrad s Postulate July 3, 2009 15 / 33
Provig Bertrad s Postulate We will use what we have leared about ( ad ψ(x i order to prove our mai result: Theorem (Bertrad s Postulate For every Z +, there exists a prime p such that < p. Lola Thompso (Ross Program Bertrad s Postulate July 3, 2009 16 / 33
The Setup : ABC = ( Let A = <p p. Lola Thompso (Ross Program Bertrad s Postulate July 3, 2009 17 / 33
The Setup : ABC = ( Let A = <p p. From Lemma (1, we kow that A (. So, there exists m Z st. A m = (. Lola Thompso (Ross Program Bertrad s Postulate July 3, 2009 17 / 33
The Setup : ABC = ( Let A = <p p. From Lemma (1, we kow that A (. So, there exists m Z st. A m = (. By the Uique Factorizatio Theorem, we ca factor m ito a product of primes (uiquely. Lola Thompso (Ross Program Bertrad s Postulate July 3, 2009 17 / 33
The Setup : ABC = ( Let A = <p p. From Lemma (1, we kow that A (. So, there exists m Z st. A m = (. By the Uique Factorizatio Theorem, we ca factor m ito a product of primes (uiquely. Let B = cotributio to ( from primes p (, ]. Let C = cotributio to ( from primes p. Lola Thompso (Ross Program Bertrad s Postulate July 3, 2009 17 / 33
The Setup : ABC = ( Let A = <p p. From Lemma (1, we kow that A (. So, there exists m Z st. A m = (. By the Uique Factorizatio Theorem, we ca factor m ito a product of primes (uiquely. Let B = cotributio to ( from primes p (, ]. Let C = cotributio to ( from primes p. The ABC = (. Lola Thompso (Ross Program Bertrad s Postulate July 3, 2009 17 / 33
The Setup : ABC = ( Goal: We wat to show that BC < (. Why does this imply that Bertrad s Postulate holds? If BC < ( the A 1. Lola Thompso (Ross Program Bertrad s Postulate July 3, 2009 18 / 33
The Setup : ABC = ( Goal: We wat to show that BC < (. Why does this imply that Bertrad s Postulate holds? If BC < ( the A 1. But A = p p, so there must be a prime dividig A. Lola Thompso (Ross Program Bertrad s Postulate July 3, 2009 18 / 33
The Setup : ABC = ( Goal: We wat to show that BC < (. Why does this imply that Bertrad s Postulate holds? If BC < ( the A 1. But A = p p, so there must be a prime dividig A. I other words, there must be a prime betwee ad dividig (. Lola Thompso (Ross Program Bertrad s Postulate July 3, 2009 18 / 33
The Setup : ABC = ( Goal: We wat to show that BC < (. Why does this imply that Bertrad s Postulate holds? If BC < ( the A 1. But A = p p, so there must be a prime dividig A. I other words, there must be a prime betwee ad dividig (. This proves Bertrad s Postulate because it proves the existece of a prime betwee ad. Lola Thompso (Ross Program Bertrad s Postulate July 3, 2009 18 / 33
The Setup : ABC = ( I order to show that BC < (, we will fid upper bouds for B ad C ad we will fid a lower boud for (. Lola Thompso (Ross Program Bertrad s Postulate July 3, 2009 19 / 33
A Lower Boud for ( Lemma ( 4 Proof I a earlier exercise, we showed that j=0 ( j = 4. Lola Thompso (Ross Program Bertrad s Postulate July 3, 2009 20 / 33
A Lower Boud for ( Lemma ( 4 Proof I a earlier exercise, we showed that j=0 ( j = 4. Sice the two ed terms i a row of Pascal s triagle are both 1, the 1 = 2 +. j=0 ( j j=1 ( j Lola Thompso (Ross Program Bertrad s Postulate July 3, 2009 20 / 33
A Lower Boud for ( Lemma ( 4 Proof I a earlier exercise, we showed that j=0 ( j = 4. Sice the two ed terms i a row of Pascal s triagle are both 1, the 1 = 2 +. j=0 ( j j=1 ( j We kow that the middle term, (, is the largest term i the sum. Lola Thompso (Ross Program Bertrad s Postulate July 3, 2009 20 / 33
A Lower Boud for ( Lemma ( 4 Proof I a earlier exercise, we showed that j=0 ( j = 4. Sice the two ed terms i a row of Pascal s triagle are both 1, the 1 = 2 +. j=0 ( j j=1 ( j We kow that the middle term, (, is the largest term i the sum. 1 ( Thus, 2 + ( j ( sice we are summig terms. j=1 Lola Thompso (Ross Program Bertrad s Postulate July 3, 2009 20 / 33
Aother Pair of Useful Exercises The followig exercises will also be useful i boudig BC: Exercise Prove or disprove ad salvage if possible: If x R, the 2x 2 x = 0 or 1. Exercise (a What power of the prime p appears i the prime factorizatio of!? (b What power of p appears i the factorizatio of ( k? Lola Thompso (Ross Program Bertrad s Postulate July 3, 2009 21 / 33
A Upper Boud for C Lemma C ( 1 Proof Let k be the highest power of p dividig (. Have you solved both of the exercises o the previous slide? Oce you have, you will see that k 1. j:p j Lola Thompso (Ross Program Bertrad s Postulate July 3, 2009 22 / 33
A Upper Boud for C Lemma C ( 1 Proof Let k be the highest power of p dividig (. Have you solved both of the exercises o the previous slide? Oce you have, you will see that k 1. j:p j The sum o the right couts the umber of j that satisfy p j. Lola Thompso (Ross Program Bertrad s Postulate July 3, 2009 22 / 33
A Upper Boud for C Lemma C ( 1 Proof Let k be the highest power of p dividig (. Have you solved both of the exercises o the previous slide? Oce you have, you will see that k 1. j:p j The sum o the right couts the umber of j that satisfy p j. I order to determie that umber, we eed to solve the equatio p x. But this is the same as solvig x log p log. Lola Thompso (Ross Program Bertrad s Postulate July 3, 2009 22 / 33
A Upper Boud for C Lemma C ( 1 Proof Let k be the highest power of p dividig (. Have you solved both of the exercises o the previous slide? Oce you have, you will see that k 1. j:p j The sum o the right couts the umber of j that satisfy p j. I order to determie that umber, we eed to solve the equatio p x. But this is the same as solvig x log p log. Thus, x log log p. Lola Thompso (Ross Program Bertrad s Postulate July 3, 2009 22 / 33
A Upper Boud for C Sice the j that satisfy p j must be itegers, the log 1 = log p. j:p j Recall that C = p p ( p. Suppose that a prime p is ot icluded i C, i.e. p > ad p. log The log p = 1. So all of the primes p such that p k ( for k > 1 must occur i C. Lola Thompso (Ross Program Bertrad s Postulate July 3, 2009 23 / 33
A Upper Boud for C Recallig that C = C p p log p log p log p log p p p ( p, we have Lola Thompso (Ross Program Bertrad s Postulate July 3, 2009 24 / 33
A Upper Boud for C Recallig that C = C p p = p log p log p log p log p p p ( p, we have e log (log rule: p α = e α log p Lola Thompso (Ross Program Bertrad s Postulate July 3, 2009 24 / 33
A Upper Boud for C Recallig that C = C p p = p = p log p log p log p log p p p ( p, we have e log (log rule: p α = e α log p Lola Thompso (Ross Program Bertrad s Postulate July 3, 2009 24 / 33
A Upper Boud for C But = ( π(, where π( = # of primes. p Lola Thompso (Ross Program Bertrad s Postulate July 3, 2009 25 / 33
A Upper Boud for C But = ( π(, where π( = # of primes. p Sice 1 is ot prime, the π( 1. Lola Thompso (Ross Program Bertrad s Postulate July 3, 2009 25 / 33
A Upper Boud for C But = ( π(, where π( = # of primes. p Sice 1 is ot prime, the π( 1. Thus, we see that C ( 1. Lola Thompso (Ross Program Bertrad s Postulate July 3, 2009 25 / 33
A Upper Boud for B Lemma B 4 2 3 Proof Recall that B = p. <p p ( Lola Thompso (Ross Program Bertrad s Postulate July 3, 2009 26 / 33
A Upper Boud for B Lemma B 4 2 3 Proof Recall that B = <p p ( For all > 4.5, < 2 3 (square both sides, the divide both sides by p. Lola Thompso (Ross Program Bertrad s Postulate July 3, 2009 26 / 33
A Upper Boud for B Lemma B 4 2 3 Proof Recall that B = <p p ( For all > 4.5, < 2 3 (square both sides, the divide both sides by We will separate B ito two products: p. <p 2 3 p ad p. 2 3 <p Lola Thompso (Ross Program Bertrad s Postulate July 3, 2009 26 / 33
A Upper Boud for B Let p ( 2 3, ]. Lola Thompso (Ross Program Bertrad s Postulate July 3, 2009 27 / 33
A Upper Boud for B Let p ( 2 3, ]. We will show that k = 0 whe p is i this rage, where k is the highest power of p dividig (. Lola Thompso (Ross Program Bertrad s Postulate July 3, 2009 27 / 33
A Upper Boud for B Let p ( 2 3, ]. We will show that k = 0 whe p is i this rage, where k is the highest power of p dividig (. Sice p ( 2 3, ], we have 1 p < 3 2. Thus p = 1 + r, with 0 r < 1 2. Lola Thompso (Ross Program Bertrad s Postulate July 3, 2009 27 / 33
A Upper Boud for B Let p ( 2 3, ]. We will show that k = 0 whe p is i this rage, where k is the highest power of p dividig (. Sice p ( 2 3, ], we have 1 p < 3 2. Thus p = 1 + r, with 0 r < 1 2. Usig this fact with the formula for k that you foud i the problem set yields k = 0. So, the product p does t cotribute ay 2 3 <p primes to B. Lola Thompso (Ross Program Bertrad s Postulate July 3, 2009 27 / 33
A Upper Boud for B Let p ( 2 3, ]. We will show that k = 0 whe p is i this rage, where k is the highest power of p dividig (. Sice p ( 2 3, ], we have 1 p < 3 2. Thus p = 1 + r, with 0 r < 1 2. Usig this fact with the formula for k that you foud i the problem set yields k = 0. So, the product p does t cotribute ay 2 3 <p primes to B. B = p 2 p 4 3 (sice ψ(x x log 4. <p p 2 3 p ( Lola Thompso (Ross Program Bertrad s Postulate July 3, 2009 27 / 33
Puttig Everythig Together What we have show: ( = ABC ( 4 C ( 1 B 4 2 3 So, A = ( /(BC Lola Thompso (Ross Program Bertrad s Postulate July 3, 2009 28 / 33
Puttig Everythig Together What we have show: ( = ABC ( 4 C ( 1 B 4 2 3 So, A = ( /(BC 4 ( 1 4 2 3 Lola Thompso (Ross Program Bertrad s Postulate July 3, 2009 28 / 33
Puttig Everythig Together What we have show: ( = ABC ( 4 C ( 1 B 4 2 3 So, A = ( /(BC 4 ( 1 4 2 3 = 4 1 3 ( Lola Thompso (Ross Program Bertrad s Postulate July 3, 2009 28 / 33
Puttig Everythig Together Remember that i order for Bertrad s Postulate to hold, we eed to show that A > 1. Thus, we eed to determie whe 4 1 3 > (. Lola Thompso (Ross Program Bertrad s Postulate July 3, 2009 29 / 33
Puttig Everythig Together Remember that i order for Bertrad s Postulate to hold, we eed to show that A > 1. Thus, we eed to determie whe 4 1 3 > (. Usig high school math, we ca show that 4 1 3 > ( holds whe > 450. Lola Thompso (Ross Program Bertrad s Postulate July 3, 2009 29 / 33
Puttig Everythig Together Remember that i order for Bertrad s Postulate to hold, we eed to show that A > 1. Thus, we eed to determie whe 4 1 3 > (. Usig high school math, we ca show that 4 1 3 > ( holds whe > 450. Bertrad s Postulate holds for all > 450. Lola Thompso (Ross Program Bertrad s Postulate July 3, 2009 29 / 33
Fiishig Up I order to coclude that Bertrad s Postulate is true for all Z +, we just eed to check values of 450. Remember that Bertrad checked all up to 6,000,000, so if you believe him the we re doe! If ot, cosider the list of primes 2, 3, 5, 7, 13, 23, 43, 83, 163, 317, 631, where each is less tha twice the precedig. This proves Bertrad s Postulate for all < 631, sice ay such ca be squeezed betwee two umbers o the list. Lola Thompso (Ross Program Bertrad s Postulate July 3, 2009 30 / 33
Geeralizatios Oe way that a mathematicia fids ew problems to solve is by lookig at a result that has already bee prove ad askig Does this hold i a more geeral settig? Of course, more geeral ca mea may differet thigs. For example, we showed that ( 4 for all 0. Perhaps we could have show a similar result for ay iteger. Aother geeralizatio would be to try to boud ( k, k Z +. Ca you thik of a more geeral statemet of Bertrad s Postulate? Lola Thompso (Ross Program Bertrad s Postulate July 3, 2009 31 / 33
Geeralizatios Here are a few well-kow geeralizatios of Bertrad s Postulate: Theorem (Sylvester The product of k cosecutive itegers greater tha k is divisible by a prime greater tha k. Lola Thompso (Ross Program Bertrad s Postulate July 3, 2009 32 / 33
Geeralizatios Here are a few well-kow geeralizatios of Bertrad s Postulate: Theorem (Sylvester The product of k cosecutive itegers greater tha k is divisible by a prime greater tha k. Theorem (Erdös For ay positive iteger k, there is a atural umber N such that for all > N, there are at least k primes betwee ad. Lola Thompso (Ross Program Bertrad s Postulate July 3, 2009 32 / 33
Geeralizatios Here are a few well-kow geeralizatios of Bertrad s Postulate: Theorem (Sylvester The product of k cosecutive itegers greater tha k is divisible by a prime greater tha k. Theorem (Erdös For ay positive iteger k, there is a atural umber N such that for all > N, there are at least k primes betwee ad. Cojecture (Legedre For every > 1, there is a prime p such that 2 < p < ( + 1 2. Lola Thompso (Ross Program Bertrad s Postulate July 3, 2009 32 / 33
Some Neat Applicatios Usig Bertrad s Postulate, we ca also prove may other iterestig results, icludig: Every iteger > 6 ca be writte as a sum of distict primes. Lola Thompso (Ross Program Bertrad s Postulate July 3, 2009 33 / 33
Some Neat Applicatios Usig Bertrad s Postulate, we ca also prove may other iterestig results, icludig: Every iteger > 6 ca be writte as a sum of distict primes. N N, there exists a eve iteger k > 0 for which there are at least N prime pairs p, p + k. Lola Thompso (Ross Program Bertrad s Postulate July 3, 2009 33 / 33