THE NUMBER OF RATIONAL NUMBERS DETERMINED BY LARGE SETS OF INTEGERS

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1 HE NUMBER OF RAIONAL NUMBERS DEERMINED BY LARGE SES OF INEGERS JAVIER ILLERUELO, D.S. RAMANA, AND OLIVIER RAMARÉ ABSRA. When A and B are subsets of the integers in [, X] and [, Y ] respectively, with A αx and B βx, we show that the nuber of rational nubers expressible as a/b with (a, b in A B is (αβ +ɛ XY for any ɛ > 0, where the iplied constant depends on ɛ alone. We then construct exaples that show that this bound cannot in general be iproved to αβxy. We also resolve the natural generalisation of our proble to arbitrary subsets of the integer points in [, X] [, Y ]. Finally, we apply our results to answer a question of Sárközy concerning the differences of consecutive ters of the product sequence of a given integer sequence.. INRODUION When A and B are intervals in the integers in [, X] and [, Y ] respectively, satisfying A αx and B βy, where X, Y real nubers, α, β are real nubers in (0, ], a standard application of the Möbius inversion forula shows that the nuber of rational nubers a/b with (a, b in A B is αβxy. Our purpose is to investigate what ight be deduced when in place of intervals we consider arbitrary subsets A and B of the integers in [, X] and [, Y ] respectively with A αx and B βy. When A and B are not intervals, it ay happen that an abnorally large nuber of eleents of these sets are ultiples of certain integers, deterining which in general is not easy. Nevertheless, since the sets under consideration are large, popular heuristics suggest that a non-trivial conclusion should still be accessible. What is pleasing is that we in fact have the following theore, which is our principal conclusion. In the stateent of this theore and thereafter we write A/B to denote the subset of Q consisting of all rational nubers expressible as a/b with (a, b in A B for any A and B subsets of the integers. HEOREM.. Let α and β be real nubers in (0, ] and X and Y real nubers. When A and B are subsets of the integers in [, X] and [, Y ] respectively, with A αx and B βy we have A/B (αβ +ɛ XY for any ɛ > 0, where the iplied constant depends on ɛ alone Matheatics Subject lassification : B05 Keywords : rational nubers, large subsets, gaps, product sequence.

2 2 JAVIER ILLERUELO, D.S. RAMANA, AND OLIVIER RAMARÉ Deferring the detailed proof of heore. to Section 2, let us suarize our arguent with the aid of the following notation. For any integer d, A and B subsets of the integers, we write M(A, B, d to denote the subset of A B consisting of all (a, b in A B with gcd(a, b = d. We show in Proposition 2. that for A and B as in heore. we have sup d M(A, B, d 8 (αβ2 XY. Starting fro this initial bound we then obtain sup d M(A, B, d (αβ +ɛ XY by a bootstrapping arguent. heore. follows iediately fro this last bound, since for any integer d we have a/b a /b for any two points (a, b and (a, b of M(A, B, d, and therefore A/B sup d M(A, B, d. We suppleent heore. with the following result which shows that the bound provided by heore. cannot be replaced with A/B αβxy. his bound, as we have already rearked, holds when A and B are intervals. HEOREM.2. For any ɛ > 0, there exists α > 0 such that for all sufficiently large X there exists a subset A of the integers in [, X] satisfying A αx and A/A < ɛα 2 X 2. We prove heore.2 in Section 3. Our ethod depends on the observation that for any ɛ > 0 and any set of prie nubers P with P sufficiently large, we have S(P/S(P ɛ S(P 2, where S(P is the set of squarefree integers d fored fro the pries in the subsets of P containing about half the pries in P. By eans of this observation we deduce that, for a suitable P, the set of ultiples of the eleents of S(P in [, X], eets the conditions of heore.2. he questions answered by the above theores ay be viewed as particular cases of a ore general proble naely, for X and Y real nubers and γ in (0, ], given a subset of the integer points in [, X] [, Y ] satisfying γxy, to deterine in ters of γ, X and Y an optial lower bound for Frac(, the nuber of rational nubers a/b with (a, b in. Plainly, the above theores take up the special case when is of the for A B, that is, when is equal to the product of its projections onto the co-ordinate axes. It turns out, however, that aforeentioned general proble is soewhat easily resolved. In effect, the ethod of Proposition 2. generalizes without additional effort to give the bound Frac( 8 γ2 XY and, interestingly, this bound is in fact optial upto the constant. More precisely, we have the following theore. 8 HEOREM.3. For any γ in (0, ] and all sufficiently large X and Y there exists a subset of the integer points in [, X] [, Y ] satisfying γ γ2 XY and Frac( XY. 8 2 We prove heore.3 at the end of Section 3 by explicitly describing sets that satisfy the conditions of this theore. Such sets are in general far fro being of the for A B, which is only natural on account of heore.. Indeed, our bootstrapping

3 HE NUMBER OF RAIONAL NUMBERS DEERMINED BY LARGE SES OF INEGERS 3 arguent for heore. depends crucially on the fact that this theore is, fro the ore general point view, about sets which are of the for A B. We conclude this note with Section 4 where we apply heore. to obtain a nearoptial answer to the following question of Sárközy. When A, B are sequences of integers, let A.B be the sequence whose ters are the integers of the for ab, for soe a A, b B. hen Sárközy [4] asks if it is true that for any α > 0 and A such that the lower asyptotic density d(a > α there is a c(α such that there are infinitely any pairs of consecutive ters of A.A the difference between which is bounded by c(α. Berczi [] responded to the aforeentioned question of Sárközy by showing that the iniu of the differences between consecutive ters of A.A is α 4, where α = d(a. Sandor [3] subsequently iproved this by showing that this iniu is in fact α 3, with α now the upper asyptotic density d(a of A. illeruelo and Le [2] obtained the sae bound when α is the upper Banach density of A and showed that this is the best possible bound for this density. he following result iproves upon and generalizes Sandor s conclusion. HEOREM.4. Let α and β be real nubers in (0, ] and let ɛ be > 0. When A and B are infinite sequences of integers with upper asyptotic densities α and β respectively, there are infinitely any pairs of consecutive ters of the product sequence A.B the difference between which is (αβ +ɛ, where the iplied constant depends on ɛ alone. When A and B are the sequences of ultiples of the integers h and k respectively, the difference between any two consecutive ters of the sequence A.B is hk. Since we have d(a = and d(b =, we see that the conclusion of heore.4 is optial up h k to a factor (αβ ɛ. hroughout this note, X, Y shall denote real nubers and α, β, γ real nubers in (0, ]. Also, the letter p shall denote a prie nuber. When I and J are subsets of a given set, I \ J shall denote the set of eleents of I that are not in J. In addition to the notation introduced so far, we shall write A d to denote the subset of a set of integers A consisting of all ultiples of d in A for any integer d. Finally, if B = {b} with b, we siply write A/b in place of A/B, by an abuse of notation. 2. PROOF OF HE BOUND Let A and B be finite subsets of the integers. hen the faily of subsets M(A, B, d of A B, with d varying over the integers, is a partition of A B. onsequently, we have ( A B = d M(A, B, d.

4 4 JAVIER ILLERUELO, D.S. RAMANA, AND OLIVIER RAMARÉ When A and B are contained in [, X] and [, Y ] respectively, we have A d X/d and B d Y/d, for any d. Since M(A, B, d is contained in A d B d, we then obtain M(A, B, d A d B d XY d 2, for all d. PROPOSIION 2.. When A and B are subsets of the integers in the intervals [, X] and [, Y ] respectively, with A αx and B βy, we have sup d M(A, B, d (αβ2 XY 8. PROOF. We adapt an arguent fro [2]. Fro ( we have for any integer that (2 A B = M(A, B, d + M(A, B, d d <d d M(A, B, d + XY, where the last inequality follows fro <d M(A, B, d <d A B αβxy we conclude fro (2 that XY d 2 XY. Since (3 sup M(A, B, d d d M(A, B, d ( αβ XY for any integer. Since 2 > αβ, the interval [ 2, 4 ] contains an integer. he αβ αβ proposition now follows on setting in (3 to be any such integer. DEFINIION 2. We call a real nuber an adissible exponent if there exists a real nuber > 0 such that for any α, β real nubers in (0, ], any X, Y real nubers and any subsets A and B of the integers in [, X] and [, Y ] with A αx and B βy, we have sup d M(A, B, d (αβ XY. We call a satisfying these conditions a constant associated to the adissible exponent. Proposition 2. says that = 2 is an adissible exponent. Proposition 2.2 will allow us to conclude that every > is an adissible exponent. he following lea prepares us for an application of Hölder s inequality within the proof of Proposition 2.2. For any integer n let τ(n denote, as usual, the nuber of integers that divide n. When D is an integer we write τ D (n to denote the nuber of divisors d of n satisfying the condition p d = p D for any prie nuber p. LEMMA 2.. When q is an integer 0 there is a real nuber c(q > 0 such that for all real nubers X and integers D we have (4 n X τ D (n q c(qdx, PROOF. In effect, we have

5 HE NUMBER OF RAIONAL NUMBERS DEERMINED BY LARGE SES OF INEGERS 5 (5 n X τ D (n q X(log 2D 2q (2 q! DX, where the iplied constants are absolute. Plainly, the second inequality results fro the eleentary inequality (log t n n! t for t. We now prove the first inequality in (5. Let us write D for the set of integers satisfying the condition p = p D. For any integer n, let k(n be the largest of the divisors of n lying in D. We then have that (6 n X τ D (n q = X, D τ( q n X, k(n= X D τ( q, where we have used the upper bound X/ for the nuber of integers n in [, X] with k(n =. Let us write S(q for any integer q 0 to denote the last su in (6. Since Merten s forula gives p D ( e γ, with γ here being Euler s constant, we p log D have (7 S(0 = D = p D ( + p + p = p D ( p log 2D, where the iplied constant is absolute. On noting that every divisor of an integer in D is again in D and using τ(dk τ(dτ(k, valid for any integers d and k, we obtain (8 τ( q D = τ( q = D d τ(dk q (d,k D D dk ( 2 τ(d q. d d D In other words, S(q S(q 2, for any q. An induction on q then shows that for any integer q 0 we have S(q S(0 2q (log D 2q, where the iplied constant is absolute. On cobining this bound with (6 we obtain the first inequality in (5. PROPOSIION 2.2. If > is an adissible exponent then so is 3(+/q 2 2 integer q. for every PROOF. Let q be a given integer and, for the sake of conciseness, let us write to denote 3(+/q 2 2, which is > since >. When is a constant associated to, let us set to be the unique real nuber > 0 satisfying (9 8 = ( 2( 8 (4c(q q(,

6 6 JAVIER ILLERUELO, D.S. RAMANA, AND OLIVIER RAMARÉ where c(q is the iplied constant in (4 of Lea 2.. It is easily seen fro (9 that by replacing with a saller constant associated to if necessary we ay assue that 4. We shall show that is an adissible exponent with a constant associated to. hus let α, β be real nubers in (0, ] and X, Y real nubers. Also, let A and B be any subsets of the integers in [, X] and [, Y ] satisfying A αx and B βy. We shall show that (0 sup M(A, B, d (αβ XY. d Replacing α and β with α α and β β such that α A 2α and β B 2β if necessary, we reduce to the case when A 2αX and B 2βY. Let us first dispose of the possibility that an abnorally large nuber of the integers in A and B are ultiples of a given integer. hus let α d = A d /X and β d = B d /Y, for any integer d. Suppose that there exists an integer d such that ( ( α d β d (αβ 2 d 2. hen A d and B d are both non-epty and therefore X and Y are both d. Further, the sets A d /d and B d /d are subsets of the integers in [, X/d] and [, Y/d]. Since is an adissible exponent, a constant associated to, and we have A d /d = (dα d X/d, B d /d = (dβ d X/d, there exists an integer d such that (2 M(A d /d, B d /d, d (d 2 α d β d XY d 2 (αβ XY, where the last inequality follows fro (. Since M(A d /d, B d /d, d does not exceed M(A, B, dd, we obtain (0 fro (2. We ay therefore verify (0 assuing that for every integer d we have ( (3 α d β d < (αβ 2 d 2. With the aid of (3 we shall in fact obtain a ore precise conclusion than (0. Let us set K = (αβ 8 and L = + [K]. We shall show that (4 L d L M(A, B, d (αβ XY, so that we have M(A, B, d (αβ XY for soe integer d L, which of course iplies (0. Note that since L is roughly about (αβ, (4 is what one ight expect fro (.

7 HE NUMBER OF RAIONAL NUMBERS DEERMINED BY LARGE SES OF INEGERS 7 Let D be an integer in [ 2, 4 ]. hus in particular D >. When L D we obtain (4 αβ αβ even without (3. In effect, we then have K and hence that L < 2K or, what is the sae thing, that L < (αβ fro which (4 follows on noting that for any integer 4 D, and in particular for = L, we have fro (3 that (5 d M(A, B, d ( αβ XY ( αβ D XY αβxy 2. Suppose now that L < D. Let us first verify that for any integer such that < D we have the following inequality on account of (3. (6 M(A, B, d ( ( 2 (αβ 2 (XY 2 A d 2 ( B d 2. Indeed, for any integer d satisfying < d D we have that (7 A d B d = (α d Xβ d Y 2 Ad 2 Bd 2 ( (αβ 2 (XY 2 Ad 2 Bd 2, where the last inequality follows fro (3 on noting that d for d satisfying < d D, since. On cobining the bound M(A, B, d A d B d with (7 and an application of the auchy-schwarz inequality we obtain (6. We now estiate the sus on the right hand side of (6. An application of Hölder s inequality gives (8 A d = ( τ D (n A q n A n A, d n n X τ D (n q q. Fro Lea 2. we have the upper bound c(qdx for the last su in (8. Since A 2αX and D 4, we deduce fro (8 that αβ (9 A d (2α 2 q β q (4c(q q X. Arguing siilarly, we obtain the bound (20 B d (2β 2 q α q (4c(q q Y. With these estiates we conclude fro (6 that for any integer satisfying < D we have

8 8 JAVIER ILLERUELO, D.S. RAMANA, AND OLIVIER RAMARÉ (2 ( 2 M(A, B, d 2 (αβ q (4c(q q XY, We now reveal that our choices for and were ade so that K satisfies the relation ( 2 (22 2 (αβ q K (4c(q q = αβ 4, as ay be confired by a odest calculation using the expressions defining and in ters of and. We see that αβ L<d D M(A, B, d XY using (2 for = L together with (22 and 4 noting that K < L. Since (5 applied with = D gives us d D M(A, B, d αβ XY, we conclude that when L < D we have 2 (23 L d L M(A, B, d αβ 4L XY. If L = we obtain (4 fro (23 on noting that αβ 4 (αβ, since 4 and. When < L < D we have K and hence L < (αβ so that (4 results fro (23 in 4 this final case as well. OROLLARY 2.. Every > is an adissible exponent. PROOF. Let q be any integer 4 and let { n (q} n the sequence of real nubers deterined by the relations (q = 2 and (24 n+ (q = ( 3 n (q + 2 q 2 n (q for n. hen each n (q is an adissible exponent by Propositions 2. and 2.2. It is easily verified that the sequence n (q is decreasing and has a liit (q given by the relation (25 (q = + 3 4q q + 9 4q 2. Plainly, any > (q is an adissible exponent. he corollary now follows on taking q arbitrarily large in (25. heore. follows fro the above corollary and the definition of adissible exponents on recalling that A/B sup d M(A, B, d. 3. OUNEREXAMPLES

9 HE NUMBER OF RAIONAL NUMBERS DEERMINED BY LARGE SES OF INEGERS 9 Let us first prove heore.2. o this end, given an integer let P denote any set of 2 prie nubers and, for any subset I of P, let d(i = p I p. If S(P denotes the set of all d(i with I =, we have the following lea. LEMMA 3.. For any ɛ > 0, we have S(P/S(P ɛ S(P 2 for all sufficiently large. PROOF. Plainly, we have S(P = ( 2. Let Q be the set of ordered pairs of disjoint subsets of P. hen, for any I and J subsets of P, we have ( d(i d(j = d(i \ J d(j \ I, and since I \ J and J \ I are disjoint, (I \ J, J \ I is in Q. hus S(P/S(P Q. Let us associate any (U, V in Q to the ap fro P to the three eleent set {, 2, 3} that takes U to, V to 2 and the copleent of U V in P to 3. It is easily seen that this association in fact gives a bijection fro Q onto the set of aps fro P to {, 2, 3} and hence that Q = 3 2. In suary, we deduce that ( 2 (2 S(P/S(P Q = 3 2 = ( S(P 2 ( S(P 2, where we have used the inequality ( he lea follows fro (2 on noting 2+ that (2 + ( as +. PROOF OF HEOREM.2. Given an integer, it is easily deduced fro the prie nuber theore that the interval [, + /] contains at least 2 prie nubers when is sufficiently large. For such a, let P be a subset of 2 prie nubers in [, + /]. If A(P is the sequence of integers that are divisible by at least one of the integers d(i in S(P then a siple application of the principle of inclusion and exclusion iplies that A(P has an asyptotic density α(p that is given by the relation (3 α(p = r ( 2 ( r+ i <i 2...<i r ( 2 d(i i I i2... I ir, where I, I 2,..., I ( 2 are the subsets of cardinality in P. For any i we have d(i i ( + < e. onsequently, for the ter r = in (3 we obtain (4 i ( 2 d(i i ( 2 e.

10 0 JAVIER ILLERUELO, D.S. RAMANA, AND OLIVIER RAMARÉ When r 2, we have that d(i i I i2... I ir, for any distinct indices i, i 2..., i r, has at least k + prie factors in P and hence is +. It follows fro (3 and these bounds that we have ( 2 ( 2 (5 α(p e 2(2 + 3 when is sufficiently large. In particular, on recalling that S(P = ( 2, we obtain that for any integer, we have (6 α(p S(P 3 for all sufficiently large and P any set of 2 prie nubers in [, + /]. Finally, for P as above and any X, let us set A = A(P [, X]. Since α(p is the asyptotic density of A(P, we have fro (6 that A S(P X, for all large enough 4 X and. learly, each integer in A is of the for d(in, for soe d(i in S(P and an integer n, which ust necessarily be X, since A is in [, X] and d(i. onsequently, we have we have A/A S(P/S(P X 2, for all large enough X and 2. On coparing A and A/A by eans of Lea 3., we see that A eets the conditions of heore.2 when, and X are all sufficiently large. PROOF OF HEOREM.3. he nuber of priitive integer points, that is, integer points with coprie co-ordinates, in [, γx] [, γy ] is 6 π 2 γ 2 XY as X, Y. hus for any γ in (0, ] and all sufficiently large X and Y, there is a subset S of the priitive integer points in [, γx] [, γy ] satisfying γ2 γ2 XY S XY. Let us take for the 4 2 union of the sets d.s with d varying over the interval [, ], where each d.s is the set γ of (da, db with (a, b varying over S. hen is contained in [, X] [, Y ]. Moreover, the sets d.s are disjoint but Frac(d.S = Frac(S, for each d, and Frac(S = S. We therefore have = [ ] S γ γ2 XY but Frac( = Frac(S = S XY. γ GAPS IN PRODU SEQUENES We now deduce heore.4 fro heore.. Let A and B be sequences with upper asyptotic densities α and β. hen there exist infinitely any real nubers X and Y such that A ( X, X] αx and B ( Y, Y ] βy. For such X and Y let us apply heore. to the sets A = A ( X, X] and B = B ( Y, Y ]. We then have that 2 2 A/B (αβ +ɛ XY, where the iplied constant depends on ɛ alone. Since A/B is a subset of the interval [ X, 2X ], which is of length X, we deduce that there are distinct 2Y Y Y a/b and a /b in A/B such that ( 0 < a b a b X/Y (αβ +ɛ XY = (αβ +ɛ Y 2.

11 HE NUMBER OF RAIONAL NUMBERS DEERMINED BY LARGE SES OF INEGERS Since bb Y 2, it follows fro ( that difference between the distinct ters ba and b a of the product sequence A.B is (αβ +ɛ. Since there are infinitely any distinct X and Y satisfying the required conditions, there are infinitely any such pairs of ters in the product sequence A.B. Acknowledgents : We arrived at heore.3 in response to a question of Professor Adrian Ubis, who we gladly thank. We also wish to thank the RM, Barcelona and HRI, Allahabad for opportunities that supported discussions on the probles addressed here. J. illeruelo was supported by Grant G08-UAM/ESP-3906 and MM of the MYI, Spain during the course of preparation of this note. D.S. Raana is with the Harish-handra Research Institute which is a constituent institution of the Hoi Bhabha National Institute, Mubai, India. Olivier Raaré is supported by the NRS, France. REFERENES [] G. Berczi. On the distribution of products of ebers of a sequence with positive density, Per. Math. Hung., 44(2002: [2] J. illeruelo and.h. Le, Gaps in product sequences. Israel Journal of Matheatics, to appear. [3]. Sandor, On the inial gaps between products of ebers of a sequence with positive density. Ann. Univ. Sci. Budapest Eötvös Sect. Math. 28(2005, 3-7. [4] A. Sarközy, Unsolved probles in nuber theory, Per. Math. Hung., 42(200, INSIUO DE IENIAS MAEMÁIAS (SI-UAM-U3M-UM AND DEPARAMENO DE MAEMÁIAS, UNIVERSIDAD AUÓNOMA DE MADRID, MADRID-28049, SPAIN E-ail address: franciscojavier.cilleruelo@ua.es HARISH-HANDRA RESEARH INSIUE, JHUNSI, ALLAHABAD -2 09, INDIA. E-ail address: suri@hri.res.in LABORAOIRE PAUL PAINLEVÉ, UNIVERSIÉ LILLE, VILLENEUVE D ASQ EDEX, FRANE E-ail address: raare@ath.univ-lille.fr