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Elementary Considerations on Prime Numbers and on
the Riemann Hypothesis
Armando V.D.B. Assis
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Armando V.D.B. Assis. Elementary Considerations on Prime Numbers and on the Riemann
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Elementary Considerations on Prime Numbers and
on the Riemann Hypothesis∗
Armando V.D.B. Assis†
[email protected]
Department of Physics,
(Former) UFSC
Universidade Federal de Santa Catarina,
88040-900, Florianópolis, 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 of N, 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 function λ (n) [1] depending on the variable 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 of n,
(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
which ω (2) = 1; λ (20) = −1, since, counted with multiplicity, 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,
hence λ (p) = −1, i.e., being p ∈ N prime: ω (p) = 1.
By virtue of these considerations, one may be interested 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 matter, 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
n 6= 1 contributes nothing to ω (n), viz.:
ω (n) = ω (1 × 1 × · · · × 1 × · · · × 1 × n) ,
(3)
by the very elementary fact that we just count the quantity of prime factors of an n. Also, once a factorization
is unique, except for the order of the prime factors of a
given natural number, a given even (or odd) factorization with 2k (or 2k − 1), k ∈ N, factors turns out to be
unique under a combinatorial (combination) consideration 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 generalized sound.
Let PN = {p1 , p2 , · · · , pN } be the set of all the first N
prime numbers, so that P = P∞ is the set of the prime
numbers. One may consider 2k, k ∈ N, slots filled with
any pl element, l ∈ {1, 2, · · · , N }, from the PN set, e.g.:
p3 × p5 × p5 × p2 × p3 × p3 × p8 × · · · × p2 (4)
|
{z
}
(2k)−slots
Of course, any permutation of (4) generates the very
same configuration, generates the very same number due
to the multiplication (× between slots). One should infer 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 (choosing a different one from PN ), 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 factorization (configuration) is unique. Similarly, one may
consider 2k − 1, k ∈ N, slots filled with any pl element,
l ∈ {1, 2, · · · , N }, from the PN set, e.g.:
p3 × p5 × p5 × p3 × p3 × p8 × · · · × p2
|
{z
}
got its last slot accupied by pN , from which we turn out
to have a problem of simple q-combination of N + q − 1
elements. Hence, the quantity QN
q of unique factorizations having q prime, with multiplicity allowed, factors
taken from PN is:
QN
q = CN +q−1, q =
(5)
(2k−1)−slots
Of course, any permutation of (5) generates the very
same configuration, generates the very same number due
to the multiplication (× between slots). One should infer 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 (choosing 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 number 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, hence PN → P∞ = P, and
to consider the totality of 2k slots and to consider the
totality of 2k−1 slots, these latter totalities by completely
considering k (viz.: ∀k ∈ N).
Since a given configuration does not change with a permutation of its elements, we are dealing with a problem
of combination [2] (the order does not matter). However, elements may be repeated, hence the combinatorics
involved is neither CN,2k nor CN,2k−1 . To uniquely represent a configuration, we may define a convention. In fact,
considering an hypothetical factorization with q ∈ N factors, e.g., as represented below:
(N + q − 1) !
= |Nq | ,
(N − 1) ! q !
(10)
the totality of distinct numbers belonging to N and having got q prime, with multiplicity allowed, factors. The
totality of numbers belonging to N turns out to be:
QN
q = lim CN +q−1, q = lim
N →∞
N →∞
QN = 1 +
∞
X
= 1+
= 1+
= 1+
∞
X
q=1
∞
X
∞
X
(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 a q-combination has
lim
(11)
|Nq | = |N| = ℵ0 .
q=1
We will be interested in the infinite sum:
∞
X
λ (n) = λ (1) +
n=1
∞
X
λ (n)
n=2
= 1+
= 1+
∞
X
n=2
∞
X
k=1
λ (n)
QN
2k −
∞
X
QN
2k−1 ,
(12)
k=1
where:
QN
2k =
(q)−slots
(q)−slots
lim CN +q−1, q
N →∞
(N + q − 1) !
N →∞ (N − 1) ! q !
q=1
it may be rearranged respecting the order of the indexes:
x1 × x3 × x4 × x6 × x7 × x10 × x14 × · · · × xN +q−1
|
{z
}
QN
q
q=1
(q)−slots
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.:
(9)
and the quantity QN
q of unique factorizations having q
prime factors turns out to be:
pN × p5 × p1 × p2 × p3 × p3 × p8 × · · · × p2 , (6)
|
{z
}
p1 × p2 × p2 × p3 × p3 × p5 × p8 × · · · × pN , (7)
|
{z
}
(N + q − 1) !
,
(N − 1) ! q !
=
lim CN +2k−1, 2k
N →∞
lim
N →∞
(N + 2k − 1) !
,
(N − 1) ! (2k) !
(13)
and:
QN
2k−1 =
=
lim CN +(2k−1)−1, 2k−1
N →∞
lim
N →∞
(N + 2k − 2) !
,
(N − 1) ! (2k − 1) !
(14)
k ∈ N. Considering the partial difference:
N
DkN = QN
2k − Q2k−1 ,
which, by virtue of the Eq. (9), leads to:
(15)
3
(N + 2k − 1) !
(N + 2k − 2) !
−
(N − 1) ! (2k) ! (N − 1) ! (2k − 1) !
(N + 2k − 1) ! − 2k(N + 2k − 2) !
(N + 2k − 1) (N + 2k − 2) ! − 2k(N + 2k − 2) !
=
=
(N − 1) ! (2k) !
(N − 1) ! (2k) !
(N + 2k − 1 − 2k) (N + 2k − 2) !
(N − 1) (N + 2k − 2) !
(N − 1) (N + 2k − 2) !
=
=
=
(N − 1) ! (2k) !
(N − 1) ! (2k) !
(N − 1) (N − 2) ! (2k) !
DkN = CN +2k−1, 2k − CN +(2k−1)−1, 2k−1 =
2k factors
z
}|
{
(N + 2k − 2) !
[(N − 2) + 2k] !
[(N − 2) + 2k] [(N − 2) + 2k − 1] × · · · × [(N − 2) + 1] (N − 2) !
=
=
=
(N − 2) ! (2k) !
(N − 2) ! (2k) !
(N − 2) ! (2k) !
2k factors
z
}|
{
[(N − 2) + 2k] [(N − 2) + 2k − 1] × · · · × [(N − 2) + 1]
=
(2k) !
2k
1 Y
[(N − 2) + j]
=
(2k) ! j=1
> dN
k ,
(16)
where:
(N − 2)
(2k) !
dN
k =
2k
.
(17)
Hence, from the Eqs. (15), (16) and (17):
∞
X
DkN =
k=1
=
>
∞
X
QN
2k −
∞
X
QN
2k−1
k=1
k=1
∞
X
2k
Y
k=1
∞
X
k=1
n→1+
1
[(N − 2) + j]
(2k) ! j=1
dN
k =
∞
X
k=1
sentially the set N − {1}, the quantity of possible factorizations exhaustively grows [a given finite set of factorizations will have a factorization, say f , representing
a greatest factorized number in this set and being oneto-one to an an = f for this set, with the remaining
factorizations (< an )] as n → ∞. Hence, as discussed in
the appendix [6]:
(N − 2)
(2k) !
∞
X
k=1
lim QN
2k +
N →∞
∞
X
k=1
lim QN
2k−1 ,
N →∞
(20)
viz.:
2k
= cosh (N − 2) − 1.
n→1+
an = 1 + (n − 1)1 = n ⇔ an = n,
(19)
with an ∈ N − {1}, shows that the number of possible
factorizations grows as n grows. Hence, since this is es-
N
QN
2k + Q2k−1 = ∞.
(21)
k=1
(18)
Now, consider the set N − {1}, hence N − {1} exhausts
any prime factorization. This is so, since a given prime
factorization represents one and only one element in N −
{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 is N − {1}. The arithmetic progression:
∞
X
By virtue of Eq. (21):
"
n
1
2 +
→ 1+
∞
X
# 12 +
N
QN
2k + Q2k−1
= ∞.
(22)
k=1
Considering the partial sum:
N
N
sN
k = Q2k + Q2k−1 ,
which, by virtue of the Eq. (9), leads to:
(23)
4
(N + 2k − 1) !
(N + 2k − 2) !
+
(N − 1) ! (2k) ! (N − 1) ! (2k − 1) !
(N + 2k − 1) ! + 2k(N + 2k − 2) !
(N + 2k − 1) (N + 2k − 2) ! + 2k(N + 2k − 2) !
=
=
(N − 1) ! (2k) !
(N − 1) ! (2k) !
(N + 2k − 1 + 2k) (N + 2k − 2) !
(N + 4k − 1) (N + 2k − 2) !
(N + 4k − 1) [(N − 1) + (2k − 1)] !
=
=
=
(N − 1) ! (2k) !
(N − 1) ! (2k) !
(N − 1) ! (2k) !
sN
k = CN +2k−1, 2k + CN +(2k−1)−1, 2k−1 =
2k−1 factors
z
}|
{
(N + 4k − 1) [(N − 1) + (2k − 1)] {(N − 1) + [(2k − 1) − 1]} × · · · × [(N − 1) + 1] (N − 1) !
=
(N − 1) ! (2k) !
2k−1 factors
z
}|
{
(N + 4k − 1) [(N − 1) + (2k − 1)] {(N − 1) + [(2k − 1) − 1]} × · · · × [(N − 1) + 1]
=
(2k) !
=
2k−1
N + 4k − 1 Y
N + 4k − 1 + 1
2k−1
[(N − 1) + j] <
[(N − 1) + 2k − 1]
(2k) !
(2k)
!
j=1
<
N + 4k
N + 4k
2k−1
2k−1
[(N − 1) + 2k − 1 + 2k]
<
(N + 4k − 2 + 2)
(2k) !
(2k) !
∴
sN
k
< SkN ,
(24)
where:
2k
SkN =
(N + 4k)
N + 4k
2k−1
(N + 4k)
=
(2k) !
(2k) !
,
(25)
remembering: k ∈ N. Furthermore:
4α0 k = [4α0 k] + {4α0 k} ,
SkN < ΞN
k ,
(26)
where:
2k
ΞN
k
(N + N/α)
=
(2k) !
=
1
1+
α
Suppose the Eq. (30) is correct. Hence, there exists, by
virtue of the Eq. (30), an α0 ∈ R∗+ such that the successive values of N to exhaustively cover the set of prime
numbers never exceed 4α0 k. Putting such 4α0 k, with the
obeyer number of the Eq. (30), a fortiori:
2k
N 2k
,
(2k) !
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 demand 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,
with [4α0 k] and {4α0 k} being, respectively, the integer
and the fractionary parts of 4α0 k, one is led to a superior
quota:
(27)
∀ fixed α ∈ R∗+ . The Eq. (26) follows from the proposition:
(30)
one turns out to be, by hypothesis, considering an implied superior quota for the existence of prime numbers.
(31)
N = [4α0 k] ,
(32)
since N ∈ N. By virtue of this superior quota for the
quantity of prime numbers, there exist only Q
finitely many
N
primes p1 < p2 < · · · < pN . Hence, let η = f =1 pf > 2,
and consider η − 1 ∈ N. Since η − 1 is a product of
primes, it turns out to have a common prime divisor pf
with η, implying this common prime divisor pf divides
η − (η − 1) = 1: an absurd! Henceforth, Eq. (30) is an
absurd, the proposition given by the Eq. (28) is correct
∀ fixed k ∈ N and, by virtue of the Eqs. (23), (24), (26)
and (27), one turns out to be led to:
2k
1
1
N
N
N
N
sk = Q2k + Q2k−1 < Ξk =
1+
N
.
(2k) !
α
(33)
Back to the interest carried from the Eqs. (20), (21)
and (22), now, we consider the consequence to the sum:
∞
X
k=1
sN
k =
∞
X
k=1
QN
2k +
∞
X
k=1
QN
2k−1 ,
(34)
5
i.e.:
∞
X
sN
k =
k=1
<
∞
X
k=1
∞
X
QN
2k +
ΞN
k
k=1
=
∞
X
k=1
∞
X
k=1
QN
2k−1
1
(2k) !
1
1+
α
From the reasonings we have carried throughout the
march that has been led from that Eq. (12), accom-
Pn
lim
n→∞
l=1 λ (l)
1
n 2 +
= lim
1+
n→∞
>
l=2
n
lim N →∞
Pn
2k
N
= cosh
1
1+
α
N −1
(35)
plished from its previous combinatorics, to the Eq. (35),
one reaches, a fortiori, the following implied consequence:
λ (l)
1
2 +
cosh (N − 2)
1 + [cosh (N − 2) − 1]
12 + = Nlim
12 +
→∞
1
1
1 + cosh 1 +
N −1
cosh 1 +
N
α
α
∴
Pn
lim
n→∞
l=1 λ (l)
1
n 2 +
e(N −2) 1 + e−2(N −2)
e(N −2) + e(2−N )
2
2
> √ lim = √ lim 1
1
2 N →∞ e(1+1/α)N + e−(1+1/α)N 2 +
2 N →∞ e(1+1/α)N 1 + e−2(1+1/α)N 2 +
∴
Pn
lim
n→∞
l=1 λ (l)
1
n 2 +
>
eN 1 + e−2(N −2)
2
√
lim 1
e2 2 N →∞ e(1+1/α)N 1 + e−2(1+1/α)N 2 +
∴
Pn
lim
n→∞
l=1 λ (l)
1
n 2 +
>
2
√
lim
e2 2 N →∞
1 + e−2(N −2)
N [1−(+1/2)(1+1/α)]
.
12 + e
−2(1+1/α)N
1+e
Since the Riemann Hypothesis [3] is true if and only if
[4, 5]:
(36)
Supposing the Riemann Hypothesis is true, the condition:
1 − ( + 1/2) (1 + 1/α) < 0
(38)
Pn
lim
n→∞
l=1 λ (l)
1
n 2 +
= 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.
holds ∀ fixed > 0. Hence:
( + 1/2) (1 + 1/α) > 1,
(39)
∀ fixed > 0. But this latter condition is an absurd,
since, for = 1/8 and α = 2, e.g.:
( + 1/2) (1 + 1/α)|(α,)=(2,1/8) =
15
< 1,
16
(40)
implying the Riemann hypothesis turns out to be false.
6
APPENDIX
The Eq. (20) must be correct, otherwise the Fundamental Theorem of Arithmetic turns out to be contradicted, as one grasps. This theorem asserts the factorization 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] factorizations, with the degeneracy of a given factorization removed from the convention we have adopted in this paper, 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 domain N:
torizations is [denumerable, hence countable] too. Counting, 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 mathematical field, Human Beings and primes. A.V.D.B.A is
deeply grateful to his Family.
f (n) = 1 = f −1 (n) , for n = 1;
f (n) = Product of the prime factors of n
= n
= f −1 (n) , for n ∈ N and n 6= 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 cardinality of N (remember our definition of N, at the beginning 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 sequence (1, 2, 3, 4, · · · , n, · · · ) tends to the infinity that
is symbolized by the cardinal ℵ0 . 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.
Since N is denumerable, hence countable, the set of fac-
†
Electronic address: [email protected]
[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 Mathematical Problems with Elementary Solutions, Volume 1, Combinatorial Analysis and Probability Theory (Dover Publications, Inc., New York, 1987), american edition ed.
[3] B. Riemann, Ueber die Anzahl der Primzahlen unter einer
gegebenen Grösse (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, P
Tokyo, 2007).
[5] E. Landau, Neuer Beweis der Gleichung ∞
k=1 µ (k) /k =
0 (Friedrich-Wilhelms, Berlin, 1899).
[6] See, also, the Eq. (11) and its subsequent Eq. which emphasizes the cardinal context of ℵ0 .