So that
people won’t mistakenly deduce that I’m some cool kat magician, err . . .
mathematician, I proclaim that we humans do not invent or create anything,
rather, we discover. These discoveries are most aptly described by the mythic
literature as boons from the gods and goddesses. This proof is just such a
boon.
My own
discoveries are granted by the infinitely brilliant and feminine presence I
identify as the Muse. Awhile back I went to the art supply store to purchase
some paint and in the same shopping complex was a Half Price Bookstore. I went
into the bookstore hoping to find a copy of Ben Goertzel’s, The Hidden Pattern. I wasn’t lucky enough to
find a copy of The Hidden Pattern but, while browsing the math and
science section, I came across a textbook, Mathematical Ideas. When I
saw the book Mathematical Ideas I received the “ping” in my mind which
lets me know the Muse is permeating state space. I pulled the book and scanned
through it.
Mathematical
Ideas is a textbook, 11th
edition, for liberal arts students. It gives a concise sketch of all the major
mathematical concepts, such as number theory, set theory, group theory,
functions, logic, etc.. I found the book interesting but when I encountered the
Goldbach Conjecture the “pinging” just went crazy. I took due notice and put
the book back on the shelf, knowing I could find a more in depth textbook
online.
I purchased
two books from the store: The Equation That Couldn’t Be Solved, by Mario
Livio; The Quark and the Jaguar, by Murray Gell-Mann. The Goldbach
Conjecture was mentioned three different times in The Quark and the Jaguar.
I
conducted an online search and found the website of George Cain, Department of Mathematics, Georgia Institute of
Technology, Atlanta, Georgia, which is another synchronicity in and of itself.
On Dr. Cain’s website I found Proofs and Concepts: The Fundamentals
of Abstract Mathematics, a most extraordinary
book.
Proofs
and Concepts: The Fundamentals of Abstract Mathematics, by Dave and Joy Morris, Department of Mathematics and
Computer Science, University of Lethbridge, Lethbridge, Alberta, in Canada, is
extremely user friendly; one concept flows into the next with unfettered ease.
Furthermore, the exercises are designed such that one not only learns the rote
of the language, but they force one to take the same journey the mathematical
forefathers took; hence, one learns the why and whereof of the language as
well. I finally have the core perspective of mathematics and I’m left with
nothing but admiration, respect, and enthusiasm for the discipline. Proofs
and Concepts: The Fundamentals of Abstract Mathematics is a masterwork, a
boon from the gods and goddesses. And this little textbook led me straightaway
to the following Proof which I hereby submit, with modifications, as a poem:
A Simple Proof To Close The Binary Goldbach Conjecture
By Wesley W. Hansen Copyright Creative Commons Attribution/NoDeriv
"Then there's the poetic truth, and it's different than the factual truth
but has a better and more meaningful place, maybe, than any facts you think you
could know. Knowing something on an intuitive or imaginative level is maybe a
more true kind of knowledge than thinking you have some sort of fact."
- Elisa Ambrogio in an interview
by Rin Kelley, L. A. Record, Issue #117, with permission.
Abstract. We define the set of positive even
integers, the set of prime numbers, and the Cartesian product on the set of
prime numbers. We then define a set composed of the sums of all ordered pairs
in the Cartesian product on the set of primes. Finally, we demonstrate the
existence of a bijection between this set of sums and the set of positive even
integers and conclude by demonstrating the existence of an identity between the
domain and codomain of this bijection, thus closing the binary Goldbach
Conjecture.
1.
Introduction. We wish to prove the “strong” or
“binary” Goldbach Conjecture as reformulated by Leonhard Euler:
“All positive even integers greater
than or equal to four can be expressed as the sum of two primes.”
By implication, this proof
generalizes to the weak conjectures.
From elementary Number Theory:
The set of all
positive integers greater than zero is equal to the set of all natural numbers.
For any natural number n, n is even iff n = 2m, where m is any positive integer greater than
zero, and n is odd iff n = 2m + 1,
where m is any positive integer
greater than or equal to zero. Since, in the expression, 2m + 1, m is any positive integer greater than
or equal to zero, we can safely conclude that 2m + 1 defines a positive integer
greater than zero. From this it follows that the sum of any two odd natural
numbers yields an even natural number since, (2m + 1) + (2m + 1) = 4m + 2 =
2(2m + 1).
The only even
number in the set of prime numbers is two. If we eliminate two from the set of
prime numbers, we guarantee that the sum of any two of the remaining prime
numbers will yield an even natural number; however, the positive even integer,
four, can only be expressed as the sum, 2 + 2 = 4. We express this identity
here allowing us to eliminate two from the set of prime numbers utilized in the
main body of our work.
We conclude our
introduction by stating an obvious fact of specific interest to the problem at
hand: the set of all even integers greater than or equal to four is equal to
the set of all even natural numbers greater than or equal to four.
Notation. We use the following
notation:
N+ │ the
set of all natural numbers
Acknowlegments. I wish to express
sincere gratitude to the Drs. Dave and Joy Morris, Department of Mathematics
and Computer Science, University of Lethbridge, Lethbridge, Alberta, Canada,
for their unsurpassed free-access textbook, Proofs and Concepts: The Fundamentals
of Abstract Mathematics; to Dr. Paul Dawkins, Department of Mathematics,
Lamar University, Beaumont, Texas, for his equally unsurpassed Calculus
Notes; to Dr. Duane Kouba, Department of Mathematics, University of
California, Davis, California, and, by extension, the entire Mathematics
Department at UC-Davis for exquisite problem sets and their internet Calculus
Page; to Dr. George Cain, Department of Mathematics, Georgia Institute of
Technology, Atlanta, Georgia, for his website, Online Mathematical Textbooks;
to Dr. William Tiller, Professor Emeritus, Stanford University, Stanford,
California, and Dr. Ben Goertzel, Research Professor, Xiamen University,
Xiamen, China, for invaluable inspiration.
2. Definitions. We define our
mathematical entities using standard terminology:
Definition 2.1. We define the set of
even natural numbers greater than or equal to six:
E = {2n │ n є N+,
n ≥ 3}
Definition 2.2. We define the set of
prime numbers greater than two:
P = {p │ p is
prime, p > 2}
Definition 2.3. We define a set on the
Cartesian product of Definition 2.2:
C = {(a,b) │ (a,b)
є P X P}
Definition 2.4. We define a set whose
elements are defined on the elements of Definition 2.3:
D = {d │ d = a + b
for every element, (a,b), of C}
Definition 2.5. We define an ordered
sequence on the elements of Definition 2.1 (reference Calculus II, Sequence and
Series, in [DAW] and Chapter 16 in [MM]), where en < en+1:
{e1, e2, e3,
… , en, … }
Definition 2.6. We define a predicate representing the
partial sums on Definition 2.5 (reference Calculus II, Sequence and Series, in
[DAW] and Chapter 16 in [MM]):
P(n): e1
+ e2 + e3 + … + en + … + 2(n + 2) = n2
+ 5n
Definition 2.7. We define an ordered
sequence on the elements of Definition 2.4 (reference Calculus II, Sequence and
Series, in [DAW] and Chapter 16 in [MM]), where dn < dn+1
and, by Definitions 2.2, 2.3, and 2.4, d1 = 6, d2 = 8, d3
= 10, d4 = 12, d5 = 14,
d6 = 16, d7 = 18, … , d20 = 44, …:
{d1, d2,
d3, … , dn, … }
Definition 2.8. We define a predicate on
Definition 2.7:
P(n): dn
= dn – 1 + 2 = 2(n + 2)
3. Arguments. We demonstrate our
arguments using standard terminology and theorems which form the foundation of
abstract mathematics. Specifically, from the foundation of abstract
mathematics, we’re given that the set of natural numbers, N+, is
countably infinite and that any subset of N+ is either finite or
countably infinite (reference Chapter 15 in [MM]).
Lemma
3.1. The set E from Definition 2.1 is
countably infinite.
Proof. Given that N+ is
countably infinite, we can list the elements of N+ in an infinite
sequence, a1, a2, a3, …, where, for any an
there exists an an+1. If we shift the index of this infinite sequence
by any finite amount we still have an infinite sequence in that for any an
there exists an an+1, hence, the set of all natural numbers greater
than or equal to three is countably infinite. It then follows, by Definition
2.1, that there exists a bijection between the set of all natural numbers
greater than or equal to three and the set E defined by, f(n) = 2n, hence, E is
countably infinite, as desired. □
Lemma
3.2. The set P from Definition 2.2 is
countably infinite.
Proof. Given that the set of prime
numbers is a subset of N+, we know that P is either finite or
countably infinite. Since we can list the prime numbers and, hence, the
elements of P, in an infinite sequence such that for any an there exists an an+1, the set P is
countably infinite, as desired. □
Lemma
3.3. The set C from Definition 2.3 is
countably infinite.
Proof. Given that we can list the
elements of set P, from Definition 2.2, in an infinite sequence, a1,
a2, a3, …, then, by the definition of Cartesian product
(reference Chapter 6, Section C, in [MM]), we can also list the elements of set
C in an infinite sequence, (a,b)1, (a,b)2, (a,b)3,
…, where for any (a,b)n there exists an (a,b)n+1
(reference Theorem 15.43(2) in [MM]), hence, C is countably infinite, as
desired. □
Lemma
3.4 The set D from Definition 2.4 is
countably infinite.
Proof. As defined, for any element, d, of D, there exists an element, (a,b), of C, such that f(a,b) = a + b =
d, hence, there exists a surjection, f : C → D, and, since f : C → D is surjective, f(C) =
D. Therefore, given that the image of a countably infinite set is countably
infinite (reference Theorem 15.43(3) in [MM]), D is countably infinite, as
desired. □
Lemma
3.5. The elements of set E, defined by
Definition 2.1 and Definition 2.5, generate a series, ∑en, which
diverges to infinity with an equation for partial sums, n2 + 5n.
Proof. We induct on n using Definition 2.6, where, by Definition 2.1, e1 =
6.
Base case. For n = 1, we have:
6 + 8 + 10 + 12 + …
+ 2(n + 2) = 12 + 5(1) = 6
Since an identity exists, P(1) is
true.
Assume P(k – 1) is
true, then:
6 + 8 + 10 + 12 + …
+ 2((k – 1) + 2) = k2 + 3k – 4
Hence:
(6 + 8 + 10 + 12 +
… + 2((k – 1) + 2)) + 2(k + 2) = k2 + 5k
k2 + 3k
– 4 + 2k + 4 = k2 + 5k
k2 + 5k
= k2 + 5k
Since an identity exists, P(k) is
true; therefore, by the Principle of Mathematical Induction, Definition 2.5
generates a series, ∑en, with an equation for partial sums
(reference Calculus II, Sequence and Series, in [DAW]), n2 + 5n, as
desired.
Let {Sn}
be the sequence of partial sums on the series ∑en, then, from
calculus (reference Calculus II, Sequence and Series, in [DAW]), it’s given
that ∑en diverges to infinity iff {Sn} diverges to
infinity. It is plain to see that n2 + 5n goes to infinity as n goes to infinity, therefore, ∑en diverges
to infinity, as desired. □
Lemma
3.6. For any dn in the
sequence {dn}, defined by Definition 2.7,
dn = 2(n + 2).
Proof. We induct on n using Definition 2.8 with strong induction and multiple base
cases:
Base case. For n = 1, we have:
6 = 2(1 + 2) = 2(3)
= 6
Since an identity exists, P(1) is
true.
Base case. For n = 2, we have:
8 = 6 + 2 = 2(2 +
2) = 2(4) = 8
Since an identity exists, P(2) is
true.
Base case. For n = 3, we have:
10 = 8 + 2 = 2(3 +
2) = 2(5) = 10
Since an identity exists, P(3) is
true.
Base case. For n = 4, we have:
12 = 10 + 2 = 2(4 +
2) = 2(6) = 12
Since an identity exists, P(4) is
true.
Base case. For n = 5, we have:
14 = 12 + 2 = 2(5 +
2) = 2(7) = 14
Since an identity exists, P(5) is
true.
Assume P(k – 5),
P(k – 4), P(k – 3), P(k – 2), and P(k – 1) are all true, then:
P(k – 5) = 2(k – 5
+ 2) = 2k – 6
P(k – 4) = 2k – 6 +
2 = 2k – 4 = 2(k – 4 + 2) = 2(k – 2)
P(k – 3) = 2k – 4 +
2 = 2k – 2 = 2(k – 3 + 2) = 2(k – 1)
P(k – 2) = 2k – 2 +
2 = 2k = 2(k – 2 + 2) = 2k
P(k – 1) = 2k + 2 = 2(k – 1 + 2) = 2(k + 1)
P(k) = 2k + 2 + 2 =
2k + 4 = 2(k + 2)
Since an identity exists, P(k) is
true; therefore, by the Principle of Mathematical Induction using strong induction
with multiple base cases, for any dn in the sequence {dn},
defined by Definition 2.7, dn = 2(n + 2), as desired. □
4. Proof of the conjecture. Our proof rests on
demonstrating that set E, from Definition 2.1, is equal to set D, from
Definition 2.4. Since, by Lemma 3.1 and Lemma 3.4, set E and set D are
countably infinite, we know there exists a bijection between them (reference
Chapter 12 and Chapter 15, Section E, in [MM]).
Assume the
bijection, f: D → E, is defined by f(d) = d, then the elements of D, defined by
Definitions 2.2, 2.3, 2.4, and 2.7, generate a series which diverges to
infinity with an equation for partial sums, n2 + 5n. We define a
predicate on Definition 2.7 where, by Lemma 3.6, dn = 2(n + 2):
P(n): 6 + 8 + 10 +
12 + … + 2(n + 2) = n2 + 5n
We now induct on n, making a note that, due to the
distributed search for a counterexample, there existed, as of July 14, 2008, 6
x 1017 base cases (reference [WMW]).
Base case. For n = 1, we have:
6 + 8 + 10 + 12 + …
+ 2(n + 2) = 12 + 5(1) = 6
Since an identity exists, P(1) is
true.
Assume P(k – 1) is
true, then:
6 + 8 + 10 + 12 + …
+ 2((k – 1) + 2) = k2 + 3k – 4
Hence:
(6 + 8 + 10 + 12 +
… + 2((k – 1) + 2)) + 2(k + 2) = k2 + 5k
k2 + 3k
– 4 + 2k + 4 = k2 + 5k
k2 + 5k
= k2 + 5k
Since an identity
exists, P(k) is true; therefore, Definition 2.7 generates a series, ∑dn,
with an equation for partial sums (reference Calculus II, Sequence and Series,
in [DAW]), n2 + 5n, as desired.
Let {Tn}
be the sequence of partial sums on the series ∑dn, then, from
calculus (reference Calculus II, Sequence and Series, in [DAW]), it’s given
that ∑dn diverges to infinity iff {Tn} diverges to
infinity. It is plain to see that n2 + 5n goes to infinity as n goes to infinity, therefore, ∑dn diverges
to infinity, as desired. □
As the forgoing
well demonstrates, the bijection, f: D → E, is defined by f(d) = d, hence, the
domain D equals the codomain E proving the binary Goldbach Conjecture, as
desired. □
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REFERENCES
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