PRIME NUMBERS YANKI LEKILI
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1 PRIME NUMBERS YANKI LEKILI We denote by N the set of natural numbers: 1,2,..., These are constructed using Peano axioms. We will not get into the philosophical questions related to this and simply assume the usual properties of natural numbers: There is an addition and multiplication law on numbers. These satisfy the commutative, associative and distributive laws. There is an order on N so that either a < b or b < a for distinct natural numbers. Furthermore, every non-empty set in N has a smallest element, i.e. the order on N is a well-ordering. Finally, we shall appeal to the principle of mathematical induction. We write Z for all integers: {..., 2, 1, 0, 1, 2,...} and Z 0 for non-negative integers. 1. Divisibility Definition 1. An integer a is divisible by b if there is a third integer c such that a = bc We write b a if b divides a and b a if b does not divide a. The relation is reflexive, a a; transitive, b a and c b implies c a, but not symmetric, if b a then it is not usually the case that a b. In fact, if b and a are positive integers and b a, then we have b a. Definition 2. A positive integer p is said to be a prime number (or simply a prime) if p > 1 and p has no positive divisors except 1 and p. The first few prime numbers are: 2, 3, 5, 7, 11, 13, 17, 19, 23,... The primes are the building blocks of numbers. The following theorem makes this precise: Theorem 3. Every positive integer except 1 is a product of primes. Proof. Let n N be a number. Either n is prime, when there is nothing to prove or n has divisors between 1 and n. Let S be the set of divisors of n greater than 1. Then this set has a smallest element m. We claim that m is a prime. Otherwise, there would be natural number l with 1 < l < m such that l m but since m n, by transitivity, we have that l n. We obtained an element of S, namely l, that is smaller than the smallest element of S. This is a contradiction. Hence, m must have been a prime. Therefore, n is either prime or divisible by a prime less than n, say p 1 in which case, n = p 1 n 1, 1 < n 1 < n 1
2 2 YANKI LEKILI Here n 1 is either a prime, in which case the proof is completed or it is divisible by a prime p 2, in which case, we have n = p 1 n 1 = p 1 p 2 n 2, 1 < n 2 < n 1 < n Repeating the argument, we obtain a sequence of decreasing numbers n, n 1,..., n k 1,... The sequence stops when n k is prime for some k, and then we have: n = p 1 p 2... p k Note that p i s in the above proof do not have to be distinct, we can group them together and write: n = p e 1 1 p e p es s to get the prime factorisation of the integer n. For example, we have: 666 = We will see later that the factors p e i i are unique apart from rearrangement of factors. But, first we need to develop our understanding of division a little more. Lemma 4. (Division Algorithm) Given a Z and b > 0, there exists a unique q, r Z such that a = qb + r and 0 r < b. Proof. Consider the arithmetic progression..., a 3b, a 2b, a b, a, a + b, a + 2b, a + 3b... extending indefinitely in both directions. Let S be the set of nonnegative elements in this list. S is non-empty: Either a is nonnegative then a S, or if a is negative,then a ab = a(1 b) 0 hence a ab S. Now, S is non-empty, hence has a smallest element r. Thus, by definition, we have r = a qb for some q and r 0. Also r < b because otherwise r b = a (q + 1)b would be an element of S that is smaller than r. Next, to prove uniqueness, let a = q 1 b + r 1 = q 2 b + r 2 satisfying the same conditions. Then (q 1 q 2 )b = r 2 r 1. Taking absolute values, we get q 1 q 2 b = r 2 r 1, hence b r 2 r 1, but 0 r 1, r 2 < b hence r 2 r 1 < b. Hence, it must be that r 2 r 1 = 0 and q 2 q 1 = 0. In other words, r 1 = r 2 and q 1 = q 2. Definition 5. Let a, b N. The greatest common divisor of a and b is the greatest number d N such that d a and d b. We write (a, b) (or gcd(a, b)) for the greatest common divisor of a and b. The numbers a and b are said to be coprime (or relatively prime) if (a, b) = 1. For example, by listing all divisors of 12 and all divisors of 8, one can easily compute (12, 8) = 4 but soon we will do this in a much more efficient way. Theorem 6. If d = (a, b) then there exists integers x 0 and y 0 such that d = (a, b) = ax 0 + by 0
3 PRIME NUMBERS 3 Another way to state this fundamental result is that the greatest common divisor of a and b is a Z-linear combination of a and b. Proof. Consider the set S of all natural numbers of the form ax + by with x and y in Z. The set is non-empty, for example it contains a and b. Hence, S has a smallest element m. So m is a natural number of the form m = ax 0 + by 0 for some integers x 0 and y 0. Every common divisor of a and b divides m, hence in particular d m. To conclude that d = m, we shall show that m a and m b. Using the division algorithm, write a = qm + r for 0 r < m. Let x = (1 qx 0 ) and y = qy 0. Then, we have ax + by = a qax 0 qby 0 = a qm = r Hence, by the minimality property of m it follows that r = 0. This shows that m a, similarly we show that m b. Note that the integers x 0, y 0 are not uniquely determined. Indeed, given one solution (x 0, y 0 ) to d = ax 0 + by 0, we can obtain infinitely many other solutions as: d = a(x 0 + nb) + b(y 0 na) for n Z. The previous theorem gives a characterization of the greatest common divisor of a and b. Namely, it is least positive integer value of ax + by where x and y ranges over all integers. But how to compute this value? (and the integers x 0, y 0? ) We shall use Euclid s algorithm. The crucial observation is the following lemma: Lemma 7. If a = qb + r then (a, b) = (b, r). Proof. If d is a common divisor of a and b, then d a qb = r, hence d is a common divisor of b and r. Conversely, if d is a common divisor of b and r, then d qb + r = a, hence d is a common divisor of a and b. Therefore, the set of common divisors of a and b agree with the set of common divisors of b and r, hence the greatest common divisors are the same. Now, given a, b N, Euclid s algorithm works as follows to determine (a, b). Without loss of generality, suppose b < a, then we apply division algorithm to write a = q 1 b+r 1 with 0 r 1 < b. If r 1 0, then we apply division algorithm to b and r 1 to write b = q 2 r 1 + r 2 with 0 r 2 < r 1. We repeat this until we find a remainder which is zero. (This must happen at some finite step, since b > r 1 > r Thus, we have a system of equations: a = q 1 b + r 1, 0 < r 1 < b b = q 2 r 1 + r 2, 0 < r 2 < r 1 r 1 = q 3 r 2 + r 3, 0 < r 3 < r 2. r n 3 = q n 1 r n 2 + r n 1, r n 2 = q n r n 1 + r n, r n 1 = q n r n < r n 1 < r n 2 0 < r n < r n 1
4 4 YANKI LEKILI We apply the above lemma repeatedly to deduce Thus, we proved: (a, b) = (b, r 1 ) = (r 1, r 2 ) =... = (r n 2, r n 1 ) = (r n 1, r n ) = (r n, 0) = r n Theorem 8. The last non-zero remainder r n of Euclid s algorithm is the greatest common divisor of a and b. Here is an example. Let us compute (187, 35). We have 187 = = = = 11.1 Thus, we see that (187, 35) = 1. Thus, we should be able to find integers x 0 and y 0 such that 187x y 0 = 1 Euclid s algorithm also gives a way to do this. Namely, we have 1 = = 12 1.( ) 1 = (35 2.( )) = Indeed, one can use Euclid s algorithm to give another proof of Theorem 6. We will now return back to factorisations of natural numbers into primes. We start with the following: Theorem 9. (Euclid s first theorem) Let p be a prime number and let a 1, a 2 N. If p a 1 a 2 then p a 1 or p a 2. More generally, if a prime p divides a product a 1 a 2... a n, then p divides a i for some i. Proof. The case of a product with n factors follows easily from the case with two factors. Suppose p is a primes and p a 1 a 2. If p a 1 then (a 1, p) = 1 and therefore, by Theorem 6, there are an x 0 and a y 0 for which Multiplying this by a 2 gives x 0 a 1 + y 0 p = 1 x 0 a 1 a 2 + y 0 pa 2 = a 2 Now, p x 0 a 1 a 2 and p y 0 pa 2, hence p a 2. Let s use this to prove a theorem due to Pythagoras. Theorem is irrational.
5 PRIME NUMBERS 5 Proof. If 2 is rational, we can write it as 2 = a for integers a, b such that (a, b) = 1. b Then, a and b satisfy the equation: a 2 = 2b 2 Hence, b a 2. Therefore, p a 2 for any prime factor p of b. It follows from Theorem 9 that p a. But, this is contrary to the assumption that (a, b) = 1. Hence b = 1, and this also is clearly false. We now come to one of the main tools of elementary number theory: Theorem 11. (Fundamental theorem of arithmetic) Every positive integer a > 1 has a factorisation into prime factors as a = p e 1 1 p e p es s, and apart from rearrangement of factors, this factorisation is unique. Proof. We have already seen the existence of a factorisation in Theorem 3. Now, we show uniqueness. Suppose that a = p 1... p k = q 1... q j are two prime factorisations of a. Then, by Theorem 9, p 1 q i1 for some i 1. Since q i1 is a prime, this implies that p 1 = q i1. We can then divide out p 1 and q i1 from both sides to get two prime factorisations of a/p 1 = p 2... p k = q 1... q i1 1q i q j. We can then match p 2 with q i2 for some i 2 by the same argument. Continuing this way, we get that for all s, p s = q is for some i s. After cancelling out all of p 1, p 2,... p k, the remaining product must equal 1. Hence, there are no remaining factors on the right hand side either. Hence k = j and the matching p 1 = q i1, p 2 = q i2,... p k = q ik shows that the two factorisations differ by only a rearrangement of factors. It is now clear why 1 should not be counted as a prime. If it were, then Theorem 11 would be false, since we could insert any number of unit factors. If we know the prime factorisation of positive integer a then we can immediately write down all positive divisors: if a = p e 1 1 p e p e k k then b a if and only if b has a prime factori- with 0 f i e i for all i. This observation gives the sation of the form b = p f 1 1 p f p f k k following lemma: Lemma 12. If m, n N are coprime then every natural number d with d mn can be written uniquely as d = d 1 d 2 where d 1, d 2 N and d 1 m and d 2 n. Proof. Since m and n are coprime, they don t have any prime factors in common. So, m = p k p kr r and n = p k r+1 r+1... p k r+s r+s where all the p i are distinct. If d mn then d = p l p l r+s r+s with 0 l i k i for 1 i r +s. Let d 1 = p l p lr r and d 2 = p l r+1 r+1... p l r+s r+s. Then, obviously d = d 1 d 2, d 1 m and d r n. Conversely, if d = d 1d 2, d 1 m and d 2 n then we must have d 1 = p l p l r r and d 2 = p l r+1 r+1... p l r+s r+s. From d = d 1d 2 it follows that l i = l i for 1 i r + s and therefore d 1 = d 1 and d 2 = d 2. This shows uniqueness. We will also need the following lemma. Lemma 13. If p 1, p 2... p r be distinct prime numbers and let n be any integer. If p i n for all i then p 1 p 2... p r n.
6 6 YANKI LEKILI Proof. For every r 1, we must show the following statement: If p 1, p 2..., p r are distinct primes, n is any integer and p i n for i = 1, 2,..., r, then p 1 p 2... p r n. To do this, we use induction on r. The case r = 1 is obviously true. Now assume r 2 and that we have shown the statement for r 1. Let p 1..., p r be distinct primes and n an integer such that p i n for all i. By induction hypothesis, we get that p 1 p 2... p r 1 n. So, we can write n = p 1 p 2... p r 1 m for some m Z. Now, since p r n and p r p i for 1 i r 1, it follows, by Theorem 9, that p r m. Hence, it follows that p 1 p 2... p r p 1 p 2... p r 1 m = n as desired. Exercise: Give an alternative proof of Lemma 13 using FTA. Finally, let us discuss factorials and their prime factorisations. natural number N, we have the notation: Recall that given a N! := N i = (N 1).N i=1 This number is equal to the number of permutations of a set of N elements. Let us observe that if p N!, then p must divide one of the numbers 1, 2,..., N and therefore p N. On the other hand, every prime number p N is a prime factor of N!. So, to find a prime factorisation of N!, we need to determine the exponent of each prime p N which divides N!. Let us write this first as: N! = p N p ep where the product is over all p N and e p are non-negative integers. To state the next result, we find convenient to introduce the following notation: Definition 14. For any real number x R, one signifies by [x] the largest integer x, that is, the unique integer such that x 1 < [x] x. This function is called the integral part of x. Lemma 15. N! = p N pep where e p = m=1 [ N p m ] Note that the sum m=1 [ N p m ] has only finitely many non-zero terms because if p m > N, then [ N p m ] = 0. Proof. Consider a prime number p N. We must count how often p appears in the product N! = 1.2..N. Clearly, [ N ] of the factors 1, 2,..., N are multiples of p; [ N ] p p 2 factors are multiples of p 2 etc. Hence, in e p = [ N ] + [ N ] +... we have counted once the p p 2 number of factors which are divisible by p but not p 2 (as part of [ N ]), we have counted p twice the number of factors which are divisible by p 2 but not p 3 (as part of [ N ] + [ N ]) etc. p p 2 This completes the proof.
7 PRIME NUMBERS 7 Note that it follows easily from above that N e p [ p 1 ] for the sum m=1 [ N p m ] < m=1 N p m = N( 1 p + 1 p ) = N p 1. As an example, let us find the largest integer k such that 7 k 50!. We can compute this as: k = [ 50 7 ] + [50 7 ] = = Computational problems. Lemma 16. A positive integer n is composite if and only if n has a prime divisor p n Proof. If n is composite then n = ab with 1 < a < n and 1 < b < n. We can assume that a b. Let p be a prime factor of a. Then p is also a prime factor of n and p a = a.a a.b = n. A primality test is an algorithm that determines whether an integer n > 1 is a prime or composite. The above lemma gives the following test. For every prime p n test whether n is divisible by p or not. We know that if p n for some p n then n is composite, otherwise n is prime. The sieve of Eratosthenes is the method of computing list of primes up to a number n by using this algorithm. Write down all the integers n and cross out multiples of 2, then cross out multiples of 3 and continue until we cross out multiples of all primes p n. Then the remaining numbers are prime. This method is useful for small numbers but, of course, it is clear that it is not very efficient for large numbers. In 2002, Agrawal, Kayal and Saxena developed the first polynomial time primality test (known as the AKS primality test). Polynomial time means that there exists constants C, k such that for every integer n > 1 the algorithm needs at most C.(log n) k many steps to decide whether n is prime or not. Note that AKS test determines whether n is prime or not without finding a prime factor. 2. Basic distribution results Recall that fundamental theorem arithmetic leads us to think that prime numbers are building blocks of all numbers. We begin with the famous result of Euclid that says that there are infinitely many primes: Theorem 17. (Euclid s second theorem) The number of primes is infinite. We will give two proofs of this result. Here is Euclid s own proof: Proof. Let 2, 3, 5,..., p be the list of primes up to p, and let q = p + 1 Then q is not divisible by any of the numbers 2, 3, 5,..., p. It is therefore either prime, or divisible by a prime between p and q. In either case, there is a prime greater than p, which proves the theorem.
8 8 YANKI LEKILI To study the distribution of prime numbers among all natural numbers, we give the following definition: Definition 18. For a real number x R, we let π(x) = the number of primes that are not greater than x For example, π(7) = 4, π(10) = 4, π(π) = 2, π(1) = 0. Note that if we know the function π then we also know all the prime numbers: an integer n is prime if and only if π(n) > π(n 1). Obviously, π(x) x for all x 0. Euclid s second theorem implies that π(x) as x. Can we show that π(x) is given by a formula in terms of more familiar functions? Let us try to deduce a bit more from Euclid s argument. Let p n denote the n th prime. So p 1 = 2, p 2 = 3,... etc.. Let us define q = p 1.p 2.p 3... p n + 1 Then, since p i < p n for i = 1,..., n 1, we deduce that for n > 1, and so that q < p n n + 1 p n+1 < p n n + 1 as either q = p n+1 or there is a prime bigger than p 1,..., p n that divides q. In fact, we can do a bit better. Suppose that p n < 2 for n = 1, 2,... N, then Euclid s argument shows that p N+1 p 1 p 2... p N + 1 < N + 1 < 2 2N+1 Since, p 1 = 2 < 2 2 = 4, by induction we conclude that p n < 2, for all n. Therefore, we have that π(2 ) n. Now, given any positive real number x, we can sandwich it as e en 1 < x e en, for some n since the series e em for m = 1, 2,..., is monotonically increasing. In other words, we have: n 1 < log log x n Let us suppose n 3, so that e n 1 > 2 n and so e en 1 > 2, then we see π(x) π(e en 1 ) π(2 ) n log log x Note that the only place where we used the assumption n 3 was in deducing π(e en 1 ) n. This can easily be checked directly for n = 1 and n = 2 as well. Hence, we proved that π(x) log log x for all x > e. We also note that log log x 0 for 1 < x e, hence we get that: Theorem 19. π(x) log log x, for x > 1.
9 PRIME NUMBERS 9 However, log log x is a rather weak bound. For example, for x = 10 9, it gives π(x) 3, whereas the value of π(x) is over 50 million. Figure 1 shows the graph of π(x) for 0 x 100. Figure 1. Graph of π(x) for 0 x 100 To draw this for yourselves, go to Mathematica and type Plot[PrimePi[x],{x,0,100}]. Around 1800, several mathematicians conjectured approximations for π(x) as x. Legendre suggested in 1798 that π(x) is approximately equal to the function x. Figure log x 2 shows how π(x) and x compares for x 400. log x It is close but does not quite match. Of course, we are able to see this much easily today with the help of computers. A better approximation to π(x) is provided by the logarithmic integral, defined by: Li(x) := x 2 dt log t Here is the Mathematica code for playing with these functions: p1 = P lot[p rimep i[x], {x, 2, }, ImageSize Large] p2 = P lot[x/log[x], {x, 2, }, P lotstyle {Green}, ImageSize Large] p3 = P lot[logintegral[x], {x, 2, }, P lotstyle {Red}, ImageSize Large] Show[p1, p2, p3] We will return to this interesting subject of asymptotic approximation of π(x) later in the next section.
10 10 YANKI LEKILI Figure 2. Graph of π(x) vs. x/ log x for 2 x 400 Let s discuss a second proof of Euclid s theorem. The proof uses a special sequence of rather fast growing numbers called Fermat s numbers. They are defined by: so that F n = F 1 = 5, F 2 = 17, F 3 = 257, F 4 = It can be checked easily that F 1, F 2, F 3, F 4 are prime numbers. Fermat conjectured that all F n are primes but in fact this was disproved by Euler in 1732 when he showed that F 5 = = = An easy proof was given by Coxeter. Indeed, 641 = = hence divides each of = 2 28 ( ) and (since x 4 1 = (x + 1)(x 1)(x 2 + 1)). So, it divides their difference. In fact, there is no known prime number of the form F n for n > 4. People are still searching for them (why?). Here is a website for it: However, we have a more theoretical use of Fermat numbers in mind. Namely, let us prove the following: Theorem 20. (Goldbach) No two Fermat numbers have a common divisor greater than 1.
11 PRIME NUMBERS 11 Proof. Suppose that F n and F n+k where k > 0 are two Fermat numbers and that m F n and m F n+k. Letting x = 2 we can write: F n+k 2 F n = x2k 1 x + 1 = 1 x2k x 2k 2 + x 2k = 2+k So, we see that F n F n+k 2, but now since m F n+k and m F n+k 2, we deduce that m 2. So m = 1 or 2 but it can t be 2 since all the Fermat numbers are odd. Hence, the theorem follows. Corollary 21. There are infinitely many primes. Proof. Each F i is either prime or have an odd prime divisor which does not divide any other. So, there are at least as many primes as there are Fermat numbers. There are infinitely many Fermat numbers by definition. This proof also gives p n+1 F n = 2 + 1, which is slightly better than what we have seen before (but not much). Can one find a simple function f : N N such that for each natural number n, f(n) is a prime number? Clearly, Fermat s sequence could not do this job. There is no known satisfactory answer to this. Of course, one could define f(n) = p n, the n th prime number but this is by no means a simple function. There is a remarkable polynomial function given by f(n) = n 2 n + 41 It turns out this polynomial takes prime values for all n with 0 n 40, but obviously f(41) is composite, since 41 f(41). The following theorem says that polynomial functions with integer coefficients are no good for answering the above question. Theorem 22. No polynomial f(n) with integral coefficients, not a constant, can be a prime for all n, or for all sufficiently large n. Proof. Consider a polynomial given by f(x) = a 0 x k + a 1 x k a k We may assume that the leading coefficient a 0 > 0 so that f(n) as n since otherwise f(x) will become negative when x is big. Thus, f(x) > 1 for all x sufficiently large, so say f(x) > 1 for all x > N. Let x 0 > N be such a number, and let us take Then, for all integers r Z, we see that y = f(x 0 ) > 1. f(ry + x 0 ) = a 0 (ry + x 0 ) k +... is divisible y by the binomial expansion theorem. Now, f(ry +x 0 ) as r. Hence, there are infinitely many composite values of f(n).
12 12 YANKI LEKILI What about simple functions f : N N such that f(n) is prime for infinitely many n? If f(n) is of the form f(n) = an + b for some a, b with a > 0 and (a, b) = 1 then there is a nice answer to this. Let s first practice in a few examples. Theorem 23. There are infinitely many primes of the from 4n + 3. Proof. Let p 1,..., p n be the first n primes, then consider: q = p n 1 Then q is of the form 4n + 3, and is not divisible by any of the primes p 1,..., p n, hence it should have prime divisors p > p n. Furthermore, it cannot be that all the prime divisors of q are of the form 4n + 1 since the product of such numbers is of the same form, hence there must be at least one prime divisor of q of the form 4n + 3 and greater than p n. By letting n, we can construct infinitely many primes of this form. Here is a similar result: Theorem 24. There are infinitely many primes of the from 6n + 5. Proof. The proof is similar. We define q by q = p n 1 and observe that any prime number, except 2 or 3 is of the form 6n + 1 or 6n + 5 (why?), and the product of two numbers of the form 6n + 1 is again of the same form. All these theorems are particular cases of a famous theorem of Dirichlet: Theorem 25. (Dirichlet 1837) If a > 0 and b are integers such that (a, b) = 1, then there are infinitely many primes of the form an + b. The proof of this theorem uses analytical methods too difficult to discuss here. We shall not cover its proof in this course. That deals with the linear functions. What about quadratic polynomials? The question becomes much harder and we don t even know whether the following conjecture is true or not. Conjecture 26. There are infinitely many primes of the form n Chebyshev s theorem We now return back to our study of the prime distribution function π(x). We will give a proof of a theorem due to Chebyshev (also spelled Tchebychef): Theorem 27. There exists constants c 1, c 2 > 0 such that x c 1 log x < π(x) < c x 2 log x
13 PRIME NUMBERS 13 The closer the constants c j are to 1, the more technical the proof becomes. Here we will show that c 1 = log 2 and c 2 2 = 6 log(2). The proof will use the following elementary facts: (1) 2 ( ) n 2 (2) ( ) n is not divisible by any p >. (3) ( ) n is divisible by all primes n < p. The first one follows from (1 + 1) = ( m=0 m), the second and third follows from our formula for the prime factorizations of factorials ()! and n!. Proof. (Proof of Chebyshev s theorem) Upper bound: Any p with n < p divides ( ) n so by Lemma 13, the product p divides ( n ). Therefore, we have n π() π(n) n<p n<p p ( ) 2 n Taking the log gives n π() π(n) 2 log(2) log n Using induction, we now easily see that π(2 k ) 3 2k k In fact, this is checked directly for k 5; k > 5, we argue by induction: π(2 k+1 ) π(2 k ) + 2k+1 3.2k k k + 2.2k k 5.2k k 3.2k+1 k + 1 Next, since the function f(x) = x is monotonically increasing for x e, we have that log x if 4 2 k < x 2 k+1, then Since π(x) 6 log 2 x log x bound for all x. Lower bound π(x) π(2 k+1 2 k ) 6. k log 2 2k log 2 6 log 2 x k log x for x 4 as well, we have now established the claimed upper Put N = ( ) n, let vp (N) denote the highest power of p that divides N. By the formula from Lemma 15, we now that v p (N) = m 1 Now, we use the following lemma: ([ ] p m [ ]) n 2 p m
14 14 YANKI LEKILI Lemma 28. For all x R, we have [2x] 2[x] {0, 1}. Proof. Let us write x = [x] + {x} where {x} is called the fractional part of x. Now, if {x} < 1/2, then 2x = [2x] + {2x}, hence [2x] 2[x] = 0. Otherwise, if {x} 1/2, then we get [2x] 2[x] = 1. [ ] [ ] If p m log > or equivalently m >, then we have that n 2 = 0. Thus, we log p p m p m find that [ ] log v p (N) log p Now, log ( ) n p log 2 log log ( ), because 2 n ( ) n [ ] log log p, because N = p vp(n), and v p (N) log p p p This yields the lower bound We claim that this implies that [ ] log log p log = π() log log p p π() log 2 log 1 π(x) log 2 2 x, for all x 2 log x [ ] log log p This inequality can be checked directly for x 16, hence it suffices to prove it for x > 16. Pick an integer n with 16 < x + 2. Then, hence, as required. π(x) π() log 2 log log n + 1 log = n 1 log 7 4 log 2 > 1 log 2 1 (n + 1) log 2 log() (n + 1) log 2 log( + 2) log 2 x 2 log x We will next prove Bertrand s postulate. It was conjectured by Bertrand in 1845 and proved by Chebyshev in Theorem 29. For every integer n 1, there is a prime p satisfying n < p.
15 PRIME NUMBERS 15 Chebyshev introduced an auxiliary function, the θ-function. It is defined by θ(x) = p x log p for real numbers x (summation over all prime numbers p x). For example, θ(10) = log 2 + log 3 + log 5 + log 7 Chebyshev proved upper and lower estimates for the function θ, and then deduced upper and lower estimates for the function π. We have the following upper estimate for θ. Lemma 30. If x > 0, then θ(x) < log(4) x. Proof. The statement is clearly true for 0 < x < 1, so we can assume that x 1. Since θ(x) = θ([x]), it is enough to prove the statement θ(n) < log(4) n for n N. For this, we use induction on n. The cases n = 1 and n = 2 are obviously true. Now, assume that n 3 and that θ(m) < log(4) m for m < n. We must distinguish the cases n even and n odd. If n is even then θ(n) = θ(n 1) < log(4) (n 1) < log(4) n, as required. If n is odd, let n = 2m + 1 for m 1. We will show below that θ(2m + 1) θ(m + 1) < log(4) m. It then follows that θ(n) = θ(2m+1) θ(m+1)+θ(m+1) < log(4) m+log(4) (m+1) = log(4)(2m+1) = log(4) n as required. So, it remains to show that θ(2m + 1) θ(m) < log(4) m for every m 1. Consider M = ( ) 2m+1 m = (2m+1)2m (m+2). If p is a prime number with m+2 p 2m+1, m! then p divides M (because p divides the numerator but not the denominator). Hence, by Lemma 13, the product p m+2 p 2m+1 divides M, in particular, is less than or equal to M. On the other hand, M < 2 2m because ( ) ( ) 2m + 1 2m + 1 2M = + < (1 + 1) 2m+1 m m + 1 It follows that θ(2m+1) θ(m+1) = m+2 p 2m+1 log p = log ( m+2 p 2m+1 p ) log M < log 2 2m = m log 4 Proof. (Proof of Bertrand s postulate) Recall first that any prime number p that divides N = ( n ) has to satisfy p and if there is any prime n < p then it divides N.
16 16 YANKI LEKILI Next, let us observe that for n 3 if < p n, then p does not divide N = ( ) 3 n. Indeed, < 3p p 2, hence 2 < 3. Thus, p [ ] [ ] n v p (N) = 2 = 2 2 = 0 p p Now, we prove Bertrand s postulate by contradiction. Suppose that there is an integer n and there is no prime in the interval (n, ]. By the discussion above, this implies that there is no prime p > 2 n that divides N. 3 Next, consider primes p N, such that v p (N) > 1. They satisfy p 2 p vp(n), hence we must have p for such primes. The number of such primes is clearly bounded by. Now, we have ( ) 2 p vp(n) p () p n Taking logs, we get: v p(n)>1 (log(2)) log() log()+log( v p(n)=1 v p(n)=1 v p(n)=1 )p log()+θ( 3 ) log()+log(4)( 3 ) Reorganizing, we get log(2) 3(1 + ) log() x Now, since is monotonically increasing for x > 3, this inequality cannot hold log x for large n. In fact, it is false for n 512 and we arrive at a contradiction if n 512. For n < 512, Bertrand s postulate is proved by looking at the sequence of primes 7, 13, 23, 43, 83, 163, 317, 631. Corollary 31. Let p n denote the n-th prime number. Then p n 2 n. Proof. By Bertrand s postulate we know that each of the intervals (1, 2], (2, 4], (4, 8], etc. contains at least one prime number. Hence the interval (1, 2 n ] contains at least n prime numbers. Thus the n-th prime number must be contained in this interval, i.e. p n 2 n.
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