Bertrand s Postulate. Theorem (Bertrand s Postulate): For every positive integer n, there is a prime p satisfying n < p 2n.

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1 Bertrad s Postulate Our goal is to prove the followig Theorem Bertrad s Postulate: For every positive iteger, there is a prime p satisfyig < p We remark that Bertrad s Postulate is true by ispectio for,, 3 ad 4, so from ow o we may assume that 5 To prove Bertrad s Postulate we will derive a upper boud o the itegers for which the desired property does ot hold The it will simply be a matter of verifyig Bertrad s Postulate up to that poit This approach will revolve aroud the biomial coefficiet!/! Lemma : If there is o prime p with < p, the ay prime factor of is o greater tha Proof: We easily have!! +! Clearly ay prime factor of this quatity is o greater tha the largest factor i the umerator, Lemma : For ay positive iteger m ad ay prime p, m! [ ] m p r, where [ ] deotes the greatest iteger fuctio p r are Proof: Fix a expoet r for the momet The positive itegers o larger tha m that are multiples of ad those that are multiples of p r+ are p r, p r,, p r+, p r+,, [ ] m p r p r [ ] m p r+ p r+ Thus there are precisely [m/p r ] [m/p r+ ] positive itegers m with r We ow have m m! m r r [m/p r ] [m/p r+ ] r r[m/p r ] r[m/p r+ ] r[m/p r ] [m/p r r ] [ ] m p r

2 Lemma 3: Suppose 5 ad p is a prime divisor of If p, the p /3 Proof: Suppose p is a prime divisor of ad p If p > /3, the p > 4 /9 > sice 5 Hece by Lemma,! [ ] p r [ ] p ad! [ ] p r [ ] p Now sice p > /3 we have 3/ > /p, so that!!! ordp! [ ] [ ] p p 0 I other words, p is ot a factor of the biomial coefficiet We will ow cosider the arithmetic fuctio θx p x Lemma 4: Assume 5 is such that there are o primes p satisfyig < p The θ/3 + Proof: As i the proof of Lemma 3, for all primes p we have [ ] [ ] p r p r p r [ ] Note that this order is ecessarily at most if p > Now by Lemma ay prime p dividig must

3 satisfy p, ad by Lemma 3 such a prime ecessarily satisfies p /3 Therefore p p /3 θ/ > > p θ/3 + [ ] We wat to use the iequality i Lemma 4 to deduce a upper boud for Towards that ed we eed a large lower boud for This will lead to a sharper upper boud for the theta fuctio as well Lemma 5: For all positive itegers, Proof: Cosider the product Note that < < + P : i i i!! i i + i < for all i Thus > + P This gives the upper boud i the lemma O the other had, i < for all i, so that > i i i 4P, i i i i i i 3

4 yieldig the other iequality Lemma 6: For all positive itegers, θ < Proof: By Lemma 5 Sice!! we have via Lemma! whece <p < +!!!,! [ /p r ] [ /p r ] [/p r ] θ θ, θ θ < +! We ow proceed by iductio The lemma is true by ispectio for ad, so suppose that m > ad the lemma is true for all itegers < m If m is odd, the m for some iteger sice m > By the above iequality ad the iductio hypothesis, θm θ < θ + + < sice m If m is eve, the m for some iteger with m > ad m is ot a prime Thus θm θm ad by what we have already show, θm θm < m < m 4

5 Propositio: Suppose 5 ad there is o prime p with p < The < Proof: By Lemmas 4, 5 ad 6 < θ/3 + < Corollary: If < 4 3 +, the < 500 I particular, Bertrad s Postulate is true for all 500 Proof: We will assume the iequality holds with 500 ad obtai a cotradictio First we rearrage the iequality to get /3 < + / + /3 < + + Cosider the fuctio fx x/x Sice f x x x, fx is decreasig for x e Set c 500/ 500 The x c x for all x 500 I particular, sice 500 by hypothesis we have with the help of a calculator so that /3 < + 8 c < 0 + < , Now we cosider the fuctio gx 099x 6 x Sice g x 099 8x /, we ca see that g is icreasig for all x 8/ I particular, gx g500 > for all x 500 This fiishes our proof Proof of Bertrad s Postulate: The primes 503, 57, 3, 67, 4, 3, 3 ad 7 show that Bertrad s Postulate is true for all 5 500, sice ay such umber satisfies < p for at least oe of these primes p Combiig this with the Corollary to the Propositio ad our iitial observatio about 4 completes the proof 5