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Elementary Considerations on Prime Numbers and on the Riemann Hypothesis
Armando V.D.B. Assis
To cite this version:
Armando V.D.B. Assis. Elementary Considerations on Prime Numbers and on the Riemann Hypoth-
esis . 2015. �hal-01220071�
on the Riemann Hypothesis
∗Armando V.D.B. Assis† armando.assis@outlook.com.br
Department of Physics, (Former) UFSC
Universidade Federal de Santa Catarina, 88040-900, Florian´opolis, SC, Brazil
(Date: October 23, 2015)
The Riemann Hypothesis states the non trivial zeros of a mathematical function, the Riemann zeta function, are, all of them, points pertaining to a vertical line in the Argand-Gauss plane.
Technically, this implies there exists an intrinsical order, a balance, within the structure of natural numbers, related to the building blocks, these the prime numbers. We prove the natural numbers are free to exist, i.e., the Riemann Hypothesis is false, which is the main purpose of this paper:
to provide an elementary disproof of the Riemann Hypothesis. This version contains an appendix that should suffice for explanation on the cardinality ofN,ℵ0, which is important here, emphasizing denumerable means countable [countably infinity], from which the scope of combinatorics is justified.
Keywords: Riemann hypothesis, prime number, number theory, disproof
COMBINATORICS ON PARTIAL SUM OF THE LIOUVILLE FUNCTION UP TO A PRIME
SUPERIOR QUOTA, LIMITING AND SUBSIDIARY COMPLEMENTS
The Liouville functionλ(n) [1] depending on the vari- able n ∈ N = {1,2,3,4,· · · } (in this paper, 0 ∈/ N) is given by:
λ(n) = (−1)ω(n), (1)
where:
ω(n) = Number of prime factors ofn, (2) being these prime factors not necessarily distinct, counted with multiplicity. Hence, the image of λ(n) is {1,−1}. E.g.: λ(1) = 1, since 1 has not got any prime factors (ω(1) = 0); λ(2) =−1, since 2 has got just one prime factor (itself, since 2 is a prime number), from whichω(2) = 1;λ(20) =−1, since, counted with multi- plicity, 20 has got 3 prime factors (20 = 2×2×5), from which ω(20) = 3. Of course, all the numbers n ∈ N, except the number 1, will have prime factors, being such quantity of prime factors, exactly ω(n), either odd or even. A prime number has got itself as its prime factor, henceλ(p) =−1, i.e., beingp∈Nprime: ω(p) = 1.
By virtue of these considerations, one may be inter- ested in a generic factorization with an even number of prime factors and also be interested in a factorization with an odd number of prime numbers. The presence
∗Mathematics Subject Classification (2010)Primary 97K20, 11Axx; Secondary 26D15, 03D60, 11B25
of 1 as a factor in a given factorization does not mat- ter, as well as for any quantity of the number 1 as a factor, except for the unique case of the factorization of the number 1, this latter being trivial and unique, once a consideration of 1 factor(s) in a given natural number n6= 1 contributes nothing toω(n), viz.:
ω(n) =ω(1×1× · · · ×1× · · · ×1×n), (3) by the very elementary fact that we just count the quan- tity of prime factors of ann. Also, once a factorization is unique, except for the order of the prime factors of a given natural number, a given even (or odd) factoriza- tion with 2k(or 2k−1), k∈N, factors turns out to be unique under a combinatorial (combination) considera- tion with repetition with 2k (or 2k−1) elements. As a matter of fact and clarification, here, we should start to put these assertions under a more mathematically gener- alized sound.
LetPN ={p1, p2,· · ·, pN}be the set of all the firstN prime numbers, so thatP =P∞ is the set of the prime numbers. One may consider 2k, k∈ N, slots filled with anypl element,l∈ {1,2,· · ·, N}, from thePN set, e.g.:
p3 ×p5 ×p5 × p2 × p3 ×p3 ×p8 × · · · × p2
| {z }
(2k)−slots
(4)
Of course, any permutation of (4) generates the very same configuration, generates the very same number due to the multiplication (×between slots). One should in- fer that repetition is allowed, viz., the elements in (4) do not need to be different. However, for an instantaneously fixed order of (4), a change in a given slot element (choos- ing a different one fromPN), with the elements within the remaining slots not changed, would lead to a new configuration for (4). A permutation of this latter new
2 configuration does not change it. A given configuration
represents a unique natural number with an even number of prime factors, since, in spite of permutation, its fac- torization (configuration) is unique. Similarly, one may consider 2k−1,k∈ N, slots filled with any pl element, l∈ {1,2,· · ·, N}, from thePN set, e.g.:
p3 ×p5 ×p5 × p3 × p3 ×p8 × · · · × p2
| {z }
(2k−1)−slots
(5)
Of course, any permutation of (5) generates the very same configuration, generates the very same number due to the multiplication (×between slots). One should in- fer that repetition is allowed, viz., the elements in (5) do not need to be different. However, for an instantaneously fixed order of (5), a change in a given slot element (choos- ing a different one from PN), with the elements within the remaining slots not changed, would lead to a new configuration for (5). A permutation of this latter new configuration does not change it. A given configuration represents a unique natural number with an odd num- ber of prime factors, since, in spite of permutation, its factorization (configuration) is unique.
To consider the totality of numbers in N, one needs to impose N → ∞, N ∈ N, hencePN → P∞ =P, and to consider the totality of 2k slots and to consider the totality of 2k−1 slots, these latter totalities by completely consideringk(viz.: ∀k∈N).
Since a given configuration does not change with a per- mutation of its elements, we are dealing with a problem of combination [2] (the order does not matter). How- ever, elements may be repeated, hence the combinatorics involved is neitherCN,2knorCN,2k−1. To uniquely repre- sent a configuration, we may define a convention. In fact, considering an hypothetical factorization withq∈Nfac- tors, e.g., as represented below:
pN × p5 × p1 ×p2 ×p3 × p3 × p8 × · · · × p2
| {z }
(q)−slots
, (6)
it may be rearranged respecting the order of the indexes:
p1 ×p2 × p2 × p3 × p3 ×p5 × p8 × · · · × pN
| {z }
(q)−slots
, (7)
which is the very same factorization preserving its same elements, now written in increasing order of indexes, which, to uniquely be represented, may have each index increased by an amount exactly equal to the quantity of previous elements (which is a unique characteristic per element), also changing the label to x, viz.:
x1 ×x3 ×x4 ×x6×x7× x10 × x14 × · · · ×xN+q−1
| {z }
(q)−slots
(8) With this convention, each q-combination turns out to have got all of its elements with different indexes, with the maximum index occurring when aq-combination has
got its last slot accupied bypN, from which we turn out to have a problem of simpleq-combination of N+q−1 elements. Hence, the quantity QNq of unique factoriza- tions having q prime, with multiplicity allowed, factors taken fromPN is:
QNq =CN+q−1, q =(N+q−1) !
(N−1) !q! , (9) and the quantity QNq of unique factorizations having q prime factors turns out to be:
QNq = lim
N→∞CN+q−1, q= lim
N→∞
(N+q−1) !
(N−1) !q! =|Nq|, (10) the totality of distinct numbers belonging toNand hav- ing gotq prime, with multiplicity allowed, factors. The totality of numbers belonging toNturns out to be:
QN = 1 +
∞
X
q=1
QNq
= 1 +
∞
X
q=1
Nlim→∞CN+q−1, q
= 1 +
∞
X
q=1
lim
N→∞
(N+q−1) !
(N−1) !q! (11)
= 1 +
∞
X
q=1
|Nq|=|N|=ℵ0. We will be interested in the infinite sum:
∞
X
n=1
λ(n) = λ(1) +
∞
X
n=2
λ(n)
= 1 +
∞
X
n=2
λ(n)
= 1 +
∞
X
k=1
QN2k−
∞
X
k=1
QN2k−1, (12) where:
QN2k = lim
N→∞CN+2k−1,2k
= lim
N→∞
(N+ 2k−1) !
(N−1) ! (2k) !, (13) and:
QN2k−1 = lim
N→∞CN+(2k−1)−1,2k−1
= lim
N→∞
(N+ 2k−2) !
(N−1) ! (2k−1) !, (14) k∈N. Considering the partial difference:
DNk =QN2k−QN2k−1, (15) which, by virtue of the Eq. (9), leads to:
DNk = CN+2k−1,2k−CN+(2k−1)−1,2k−1= (N+ 2k−1) !
(N−1) ! (2k) !− (N+ 2k−2) ! (N−1) ! (2k−1) !
= (N+ 2k−1) !−2k(N+ 2k−2) !
(N−1) ! (2k) ! = (N+ 2k−1) (N+ 2k−2) !−2k(N+ 2k−2) ! (N−1) ! (2k) !
= (N+ 2k−1−2k) (N+ 2k−2) !
(N−1) ! (2k) ! = (N−1) (N+ 2k−2) !
(N−1) ! (2k) ! = (N−1) (N+ 2k−2) ! (N−1) (N−2) ! (2k) !
= (N+ 2k−2) !
(N−2) ! (2k) ! =[(N−2) + 2k] ! (N−2) ! (2k) ! =
2kfactors
z }| {
[(N−2) + 2k] [(N−2) + 2k−1]× · · · ×[(N−2) + 1] (N−2) ! (N−2) ! (2k) !
=
2kfactors
z }| {
[(N−2) + 2k] [(N−2) + 2k−1]× · · · ×[(N−2) + 1]
(2k) !
= 1
(2k) !
2k
Y
j=1
[(N−2) +j]
> dNk, (16)
where:
dNk =(N−2)2k
(2k) ! . (17)
Hence, from the Eqs. (15), (16) and (17):
∞
X
k=1
DNk =
∞
X
k=1
QN2k−
∞
X
k=1
QN2k−1
=
∞
X
k=1
1 (2k) !
2k
Y
j=1
[(N−2) +j]
>
∞
X
k=1
dNk =
∞
X
k=1
(N−2)2k
(2k) ! = cosh (N−2)−1.
(18) Now, consider the setN− {1}, henceN− {1}exhausts any prime factorization. This is so, since a given prime factorization represents one and only one element inN− {1} and a given element of N− {1} has got one and only one prime factorization. Hence, all the factorizations exhaust N− {1}. N− {1} is the entire set of possible prime factorizations and the entire set of possible prime factorizations isN− {1}. The arithmetic progression:
an = 1 + (n−1)1 =n⇔an=n, (19) with an ∈ N− {1}, shows that the number of possible factorizations grows as ngrows. Hence, since this is es-
sentially the set N− {1}, the quantity of possible fac- torizations exhaustively grows [a given finite set of fac- torizations will have a factorization, sayf, representing a greatest factorized number in this set and being one- to-one to an an = f for this set, with the remaining factorizations (< an)] as n→ ∞. Hence, as discussed in the appendix [6]:
n→1 +
∞
X
k=1
lim
N→∞QN2k+
∞
X
k=1
N→∞lim QN2k−1, (20) viz.:
n→1 +
∞
X
k=1
QN2k+QN2k−1
=∞. (21)
By virtue of Eq. (21):
n12+→
"
1 +
∞
X
k=1
QN2k+QN2k−1
#12+
=∞. (22)
Considering the partial sum:
sNk =QN2k+QN2k−1, (23) which, by virtue of the Eq. (9), leads to:
4
sNk = CN+2k−1,2k+CN+(2k−1)−1,2k−1= (N+ 2k−1) !
(N−1) ! (2k) !+ (N+ 2k−2) ! (N−1) ! (2k−1) !
= (N+ 2k−1) ! + 2k(N+ 2k−2) !
(N−1) ! (2k) ! = (N+ 2k−1) (N+ 2k−2) ! + 2k(N+ 2k−2) ! (N−1) ! (2k) !
= (N+ 2k−1 + 2k) (N+ 2k−2) !
(N−1) ! (2k) ! = (N+ 4k−1) (N+ 2k−2) !
(N−1) ! (2k) ! =(N+ 4k−1) [(N−1) + (2k−1)] ! (N−1) ! (2k) !
= (N+ 4k−1)
2k−1factors
z }| {
[(N−1) + (2k−1)]{(N−1) + [(2k−1)−1]} × · · · ×[(N−1) + 1] (N−1) ! (N−1) ! (2k) !
= (N+ 4k−1)
2k−1factors
z }| {
[(N−1) + (2k−1)]{(N−1) + [(2k−1)−1]} × · · · ×[(N−1) + 1]
(2k) !
= N+ 4k−1 (2k) !
2k−1
Y
j=1
[(N−1) +j]<N+ 4k−1 + 1
(2k) ! [(N−1) + 2k−1]2k−1
< N+ 4k
(2k) ! [(N−1) + 2k−1 + 2k]2k−1<N+ 4k
(2k) ! (N+ 4k−2 + 2)2k−1
∴
sNk < SkN, (24)
where:
SkN =N+ 4k
(2k) ! (N+ 4k)2k−1= (N+ 4k)2k
(2k) ! , (25) remembering: k∈N. Furthermore:
SNk <ΞNk, (26) where:
ΞNk = (N+N/α)2k (2k) ! =
1 + 1
α
2k N2k
(2k) !, (27)
∀ fixedα∈R∗+. The Eq. (26) follows from the proposi- tion:
4αk < N, ∀α∈R∗+. (28) Since N is to exhaustively cover the set P∞ = P, this latter being the entire set of prime numbers, viz., in:
PN ={p1, p2, p3,· · · , pN}={2,3,5,· · · , pN}, (29) N has been taken, defined, to exhaustively cover on de- mand the set of prime numbers, as we have defined from the beginning of this paper. Now, to conversely suppose the condition stated by the Eq. (28), i.e.:
∃α∈R∗+|4αk≥N, (30) one turns out to be, by hypothesis, considering an im- plied superior quota for the existence of prime numbers.
Suppose the Eq. (30) is correct. Hence, there exists, by virtue of the Eq. (30), an α0 ∈ R∗+ such that the suc- cessive values ofN to exhaustively cover the set of prime numbers never exceed 4α0k. Putting such 4α0k, with the obeyer number of the Eq. (30),a fortiori:
4α0k= [4α0k] +{4α0k}, (31) with [4α0k] and {4α0k} being, respectively, the integer and the fractionary parts of 4α0k, one is led to a superior quota:
N = [4α0k], (32)
since N ∈ N. By virtue of this superior quota for the quantity of prime numbers, there exist only finitely many primesp1< p2<· · ·< pN. Hence, letη=QN
f=1pf >2, and consider η −1 ∈ N. Since η−1 is a product of primes, it turns out to have a common prime divisorpf
with η, implying this common prime divisorpf divides η−(η−1) = 1: an absurd! Henceforth, Eq. (30) is an absurd, the proposition given by the Eq. (28) is correct
∀fixedk∈Nand, by virtue of the Eqs. (23), (24), (26) and (27), one turns out to be led to:
sNk =QN2k+QN2k−1<ΞNk = 1 (2k) !
1 + 1
α
N 2k
. (33) Back to the interest carried from the Eqs. (20), (21) and (22), now, we consider the consequence to the sum:
∞
X
k=1
sNk =
∞
X
k=1
QN2k+
∞
X
k=1
QN2k−1, (34)
i.e.:
∞
X
k=1
sNk =
∞
X
k=1
QN2k+
∞
X
k=1
QN2k−1
<
∞
X
k=1
ΞNk =
∞
X
k=1
1 (2k) !
1 + 1
α
N 2k
= cosh
1 + 1 α
N
−1 (35)
From the reasonings we have carried throughout the march that has been led from that Eq. (12), accom-
plished from its previous combinatorics, to the Eq. (35), one reaches,a fortiori, the following implied consequence:
n→∞lim Pn
l=1λ(l)
n12+ = lim
n→∞
1 +Pn l=2λ(l) n12+
> lim
N→∞
1 + [cosh (N−2)−1]
1 + cosh
1 + 1 α
N
−1 12+
= lim
N→∞
cosh (N−2)
cosh
1 + 1 α
N
12+
∴
n→∞lim Pn
l=1λ(l) n12+ > 2
√2 lim
N→∞
e(N−2)+e(2−N)
e(1+1/α)N +e−(1+1/α)N12+ = 2
√2 lim
N→∞
e(N−2) 1 +e−2(N−2) e(1+1/α)N 1 +e−2(1+1/α)N12+
∴
n→∞lim Pn
l=1λ(l)
n12+ > 2 e2√
2 lim
N→∞
eN 1 +e−2(N−2) e(1+1/α)N 1 +e−2(1+1/α)N12+
∴
n→∞lim Pn
l=1λ(l)
n12+ > 2 e2√
2 lim
N→∞
1 +e−2(N−2)
1 +e−2(1+1/α)N12+eN[1−(+1/2)(1+1/α)]. (36)
Since the Riemann Hypothesis [3] is trueif and only if [4, 5]:
n→∞lim Pn
l=1λ(l)
n12+ = 0, ∀ fixed >0, (37) it follows that the convergence of the right-hand side of the Eq. (36), ∀ fixed >0, is a necessary condition for the validity of the Riemann Hypothesis. It follows, then, that:
• If there exists some >0 such that the right-hand side of the Eq. (36) diverges, then, the Riemann Hypothesis turns out to be false.
Supposing the Riemann Hypothesis is true, the condition:
1−(+ 1/2) (1 + 1/α)<0 (38) holds∀fixed >0. Hence:
(+ 1/2) (1 + 1/α)>1, (39)
∀ fixed > 0. But this latter condition is an absurd, since, for= 1/8 andα= 2, e.g.:
(+ 1/2) (1 + 1/α)|(α,)=(2,1/8)=15
16 <1, (40) implying the Riemann hypothesis turns out to be false.
6 APPENDIX
The Eq. (20) must be correct, otherwise the Funda- mental Theorem of Arithmetic turns out to be contra- dicted, as one grasps. This theorem asserts the factoriza- tion is unique, except for the order of the prime factors being multiplied to generate a natural number. Hence, the∞, the upper sum limit, is a symbol for the natural numerical limit of ℵ0. From this, follows the bijection between the natural numbers and their [unique] factor- izations, with the degeneracy of a given factorization re- moved from the convention we have adopted in this pa- per, completely exhausting the members of both the sets, N and [their unique] factorizations, one-to-one. In fact, one can define the following function with domainN:
f(n) = 1 =f−1(n), forn= 1;
f(n) = Product of the prime factors ofn
= n
= f−1(n), forn∈Nandn6= 1. (41) which is obviously invertible, with image N, having got, hence, a one-to-one correspondence, a bijection, where f−1(n) is the inverse function of f(n). The cardinal- ity of N (remember our definition of N, at the begin- ning of this paper, the positive integers) is ℵ0. By Eq.
(41), it is crystalline the set of factorizations is the very same N, hence with the same cardinality, ℵ0. The se- quence (1,2,3,4,· · · , n,· · ·) tends to the infinity that is symbolized by the cardinalℵ0. Hence, the cardinality of the factorizations must be ℵ0, otherwise there would exist some number without factorization and vice-versa, contradicting the Fundamental Theorem of Arithmetic.
SinceNis denumerable, hence countable, the set of fac-
torizations is [denumerable, hence countable] too. Count- ing, n must tend to as given by the Eq. (20), since n tends to the infinity that is symbolized by the cardinal ℵ0, asseverating, as well as the infinity quantity of all the factorizations, Eq. (20).
ACKNOWLEDGMENTS
A.V.D.B.A is grateful to the Riemann Hypothesis.
A.V.D.B.A is grateful to the Giants within the mathe- matical field, Human Beings and primes. A.V.D.B.A is deeply grateful to his Family.
† Electronic address: armando.assis@outlook.com.br [1] T. M. Apostol, Introduction to analytic number theory
(Springer-Verlag, New York, Heidelberg, Berlin, 1976), undergraduate texts in mathematics ed.
[2] A. M. Yaglom and I. M. Yaglom,Challenging Mathemati- cal Problems with Elementary Solutions, Volume 1, Com- binatorial Analysis and Probability Theory (Dover Publi- cations, Inc., New York, 1987), american edition ed.
[3] B. Riemann,Ueber die Anzahl der Primzahlen unter einer gegebenen Gr¨osse(Monatsberichte der Berliner Akademie, 1859).
[4] P. Borwein, S. Choi, B. Rooney, and A. Weirathmueller, The Riemann Hypothesis: A Resource for the Afficionado and Virtuoso Alike (Springer, Berlin, Heidelberg, New York, Hong Kong, London, Milan, Paris, Tokyo, 2007).
[5] E. Landau,Neuer Beweis der GleichungP∞
k=1µ(k)/k= 0 (Friedrich-Wilhelms, Berlin, 1899).
[6] See, also, the Eq. (11) and its subsequent Eq. which em- phasizes the cardinal context ofℵ0.
