Journal of Science and Arts
Year 15, No. 3(32), pp. 211-214, 2015
ORIGINAL PAPER
PARTITIONS OF THE SET OF POSITIVE INTEGERS
IN TRANSLATED SETS
VASILE POP1
_________________________________________________
Manuscript received: 10.07.2015; Accepted paper: 02.09.2015;
Published online: 30.09.2015.
Abstract. If F = { Ai | i Î I } is a family of subsets Ai Ì N , i Î I , we say that F is a
translated family if there is a subset B Ì N and a sequence {ai | i Î I } Ì Z such that
B = ai + Ai , i Î I . In this paper we study the existence of partitions of the set N in translated
sets.
Keywords: Translated families, partitions, arithmetic progression.
1. INTRODUCTION
Many of actual problems regarding partition and in particular partitions of N and Z are
from Ramsey Theory: a collection of results which, given a finite partition (coloring) of some
structure guarantee the existence of certain monochromatic configuration or substructures.
Van der Waerden's Theorem [2] is the result which has determined the research in Ramsey
Theory of Integers [1].
In this paper we study a different type of problems: the possibility of partition of N in
translated sets, problems studied in [3] and [4].
2. INFINITE PARTITIONS IN TRANSLATED SETS
The problems which we discuss in this section is whether there are partitions of the set
N in an infinity of translated sets, that is there are subsets B Ì N and sequences
{an | n Î N} Ì N such that the sets An = an + B , n Î N form a partition of N.
First we notice that any two translated sets have the same cardinality, so the sets An ,
n Î N are all finite or all infinite. In the case of finite sets we can construct infinite partition
of any cardinal k Î N :
Example 2.1. We take B = {0,1,..., k - 1} = [1, k - 1] and an = n × k , n Î N .
We have: A0 = 0 + B = B , A1 = k + B = [k ,2k - 1] , A2 = 2k + B = [2k ,3k - 1] ,...,
An = nk + B = [ nk , (n + 1) k - 1] ,..., and the sets A0 , A1 , A2 ,..., An ,... form a partition of N.
It is obvious that using this idea it is not possible to construct partitions in infinite sets
but the answers for the case of infinite sets is affirmative.
1
Technical University of Cluj-Napoca, Department of Mathematics, 400114 Cluj-Napoca, Romania.
E-mail: [email protected].
ISSN: 1844 – 9581
Mathematics Section
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212
Vasile Pop
Theorem 2.1. There are partitions of N into an infinite number of infinite translated
sets.
¥
Proof: We look for a partition of the form N =
UA
n
, in which all the sets An are
n =0
infinite and can be obtained by the translation of the set B = A0 , the set from the partition
which contain 0.
Let T = {t0 , t1 , t 2 ,..., tn ,...} be the set of translations and let An = tn + B , n Î N . The
condition that the sets An , n Î N form a partition of N is about showing that any number
n Î N can be written uniquely as n = t + b , with t Î T and b Î B .
We define the set B as the set of natural numbers which in base 2 are of the form:
x = xn xn -1...x2 x1x0 and all digits with odd index are 0: 0 = x1 = x3 = x5 = ...
We define the set T as the set of numbers x which in base 2 have all the even index
digits equal to 0. Note that N = T + B and that any positive integer n can be written in a
unique form: n = t + b , t Î T and b Î B .
3. FINITE PARTITION OF N IN TRANSLATED SETS
The most famous partition of N in n subsets is the partition in equivalence classes
modulo n : N = C0 È C1 È ... È Cn -1 , where C0 = n × N , C1 = 1 + n × N ,..., Cn -1 = n - 1 + n × N ,
such that the sets C1, C2 ,..., Cn -1 can be obtained by the translation of the set C0 and we have
the relations:
n + C0 = n - 1 + C1 = n - 2 + C2 = ... = 1 + Cn -1 = n × N* .
We propose to characterize the n -tuples (a1, a2 ,..., an ) for which there are partition of
the set N : N = A1 È A2 È ... È An such that a1 + A1 = a2 + A2 = ... = an + An .
We obtain a characterization of arithmetic progression with a prime number of terms.
Theorem 3.1. Let n ³ 3 be a prime number and let a1 < a2 < ... < an be integers.
Then a1 , a2 ,..., an is an arithmetic progression if and only if there exists a partition of
N with classes A1, A2 ,..., An such that a1 + A1 = a2 + A2 = ... = an + An .
Proof: If a1, a2 ,..., an is an arithmetic progression with step r then the partition
( A, A - r , A - 2r ,..., A - ( n - 1)r ) ,
with
A=
U{knr + (n - 1)r + i | i Î 0, r - 1}
fulfils
the
k ³0
required condition.
For the converse, denote ri = an - an - i and Bi = An - i for i Î 0, n - 1 , so Bi = B0 + ri
for i ³ 1 . We will call segment of length k of the set Bi every set {a, a + 1,..., a + k - 1} Ì Bi
such that a -1 Ï Bi and a + k Ï Bi .
We will firstly prove that each Bi is an union of segments of the same lengths r = r1 .
From x Î B0 Þ x + r Î B1 we get x Î B0 Þ x + r Ï B0 , therefore all the segments of
B0 must have lengths less that r + 1 . If B0 contains segments of lengths less than r , let S be
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213
the first of these. Then S + r Ì B1 and between S and S + r there exists a segment S ¢ of
some Bi , i ¹ 0 . But, in this case the segment S ¢ - ri would be in B0 , would have length less
that r and would be before S , which contradicts the way S was chosen.
This proves that all the segments of B0 have length r and, from Bi = B0 + ri , this is
also true for every Bi .
We will now prove that the first segment of Bi is Si = {ir , ir + 1,..., ir + r - 1} for every
i Î 0, n - 1 . Since x Î Bi for i ³ 1 Þ x ³ ri ³ r it follows that B0 must contain the segment
S0 = {0,1,..., r - 1} and therefore B1 contains the segment S1 = {r, r + 1,...,2r - 1} .
Suppose now that there exists k < n ( k ³ 2) such that S0 Î B0 , S1 Î B1 ,..., Sk -1 and
S k Ï Bk . Then S k must be a segment of some Bi (since all the segments have the same
length), i must be less than k and Bi must be B0 (because the second segment of every Bi ,
i ³ 1 must come after the second segment of B0 ).
This leads to Sk +1 Î B0 , Sk + 2 Î B2 ,..., S2k -1 Î Bk -1 and, repeating the above
judgement if necessary, the first segment of Bk must be of the form S lk , l ³ 2 . This leads to
rk = lk , therefore S (l +1) k = S k + rk Î Bk . The segment Slk +1 cannot be in B0 (it would lead to
S( l +1) k Î Bk -1 ) or in any of the Bi¢s , i Î 0, k - 1 (the set {S0 , S1 ,..., Slk +1} would contain more
segments from Bi than segments from B0 ) therefore Slk +1 Î Bk +1 .
In the same way Slk + 2 Î Bk + 2 ,..., Slk + k -1 Î B2 k -1 and the sequence of segments from
( Bk , Bk +1,..., B2k -1 ) will repeat itself a number of times before the appearance of a segment
from a new set (which might be B0 or B2 k ).
We notice that a judgement as above shows that each time when a segment from a
new set Bsk appears, then he must be followed immediately by segments from the sets
Bsk +1, Bsk + 2 ,..., Bsk + k -1 , so the number n of Bi¢s must be a multiple of k , 1 < k < n , a
contradiction with the premises.
Thus Si Î Bi for every i Î 0, n - 1 , therefore ri = ir for every i Î 1, n - 1 and
a1, a2 ,..., an is an arithmetic progression.
Remark 3.1. If the number of terms of progression is not a prime number one of the
implication of Theorem 3.1 is not valid. More precisely, there are partitions of the set N in
n = p × q translated subsets ( p ³ 2 , q ³ 2 ) but the set T of translations does not form an
arithmetic progression, as we shall see in the next example.
éiù
Example 3.1. B0 = {2nk | k Î N} È {2nk + p | k Î N} and ri = i + p ê ú for i Î 1, n - 1
ë pû
which corresponds to the periodic sequence of segments of length 1 obtained by repeating the
block [ B0 B1...B p -1B0 ...B p -1 ] [ B p B p +1...B2 p -1 B p ...B2 p -1 ] ... [ Bn - p Bn - p +1...Bn -1Bn - p ...Bn -1 ] .
For p = q = 2 , n = 4 , we obtain the partition:
A0 = {8k ,8k + 2 | k Î N} , A1 = {8k + 1,8k + 3 | k Î N} ,
A2 = {8k + 4,8k + 6 | k Î N} , A3 = {8k + 5,8k + 7 | k Î N}
ISSN: 1844 – 9581
Mathematics Section
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a0 = 5 , a1 = 4 , a2 = 1 , a3 = 0 , which is not an arithmetic progression, but:
a0 + A0 = a1 + A1 = a2 + A2 = a3 + A3 .
REFERENCES
[1] Landman, B.M., Robertson, A., Ramsey Theory on the Integers, A.M.S. 2004.
[2] van der Waerden, B.L., Bewis einer Baudetschen Vermutung, Nieuw Arch. Wiskunde,
15, 212, 1927.
[3] Pop, V., Teleuca, M., Probleme de combinatorică elementară, Ed. Matrix Rom, 2012.
[4] Pop, V., Romanian Mathematical Competitition 1998, S.S.M.R. (Problem 12, pp. 30, 4243).
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