and I am sorry for the self-serving post, but I recently formulated a proof on Goldbach's Conjecture, and I was wondering why it was wrong.

Let n be an even natural number greater than two, $n \in$ $N^+$ and $n > 2$; let $k \in$ $N^+$. $\forall n$, $n = 2k$.

Claim: $\forall n$ $\geq$ 4, $\forall k \geq 2$, $\exists x, y$ that are both prime such that $n = x + y$ or $2k = x + y$.

Part 1: Induction (The definition of n was only included because that is the universal accepted version of Goldbach's Conjecture, but the proof will use the $2k = x + y$ version more frequently)

Base Case: $k = 2$

$$2(2) = x + y$$

$$4 = 2 + 2 \checkmark$$

Inductive Step: For the purposes of induction, assume $1, 2, 3, ... , k$ to be true. Prove $k + 1$ to be true. Let $L = k + 1$; this implies $L \in$$N^+$ by our inductive hypothesis being true.

$$2L = a + b $$ where $a$ and $b$ are some other numbers

To prove this, let $A$ be the set of prime numbers less than $2L$. Formally, let $p_n$ denote the $nth$ prime, $A = ${$p_n|\forall p_n \leq 2L$} = ${2, 3, 5, ..., p_{n-1}, p_n}$ and let $a_n$ denote the $nth$ term in the set where $a_n$ = $p_n$ Now, there are composite solutions to this equations, but to find a solution where both $a$ and $b$ are prime let $a$ take on values from $A$. If we let $a$, with no loss of generality from a to b, take on values from $A$, we can solving the equation to get

$$2L-p_n = b$$

Let's define another set $B$ where each element $b_n = 2*L- p_n$. B is formally defined as $B = \{2*L- p_n\mid\forall p_n\leq 2L\} = \{2L-2, 2L-3, 2L-5,..., 2L-p_{n-1}, 2L-p_n\}$. Since we can assume that $a$ is prime because $a$ comes from a set of prime numbers, we must prove that $\exists$ $b_n$ that is prime.

Part 2 (Lemma): Contradiction

For the purposes of contradiction, assume that all members of $B$ are composite, $\forall$$b_n$, $b_n$ is composite.

Since $A$ is a set of prime numbers and $B$ is now a set of composite numbers, $A$ and $B$ are disjoint, meaning A $\cap$ B = $\emptyset$. Another definition of disjoint is that for all index $i \in A$, and for all index $j \in B$, $$a_i\neq b_j$$ This implies that this inequality must hold when $i = j$, the indices are at the same place in the set. We will use $i$ with no loss of generality. $$a_i\neq b_i$$ This means our formulas must also hold. $$p_i\neq 2L-p_i$$ $$2p_i\neq 2L$$ $$p_i\neq L$$ $$L\neq p_i$$ This implies that $L$ cannot be prime itself. We have reached an inconsistency. From our inductive hypothesis, that is true, we proved that $k \in$ $N^+$, and it can be trivially proven that $L \in$ $N^+$ as well.

If we let $C$ be the set of Composite Numbers, the conclusion derived from our contradictory hypothesis claims that $L \in C$. This implies that $C = N^+$, and this is trivially false

Q.E.D. There must be at least one element in $B$ that is prime.

Since all the elements in $A$ are prime and at least one element in $B$ is prime, let $b$ take on the an element in $B$ that is prime, and $a$ take on the element in $A$ that resolves our inductive step equation, meaning it has to be prime. We have proved that there exists at least one prime $a$ and $b$ that solves the inductive equation $$2L = a + b$$ Which is what we needed to show. Q.E.D. $\forall n$ $\geq$ 4, $\forall k \geq$ 2, $\exists x, y$ that are both prime such that $n = x + y$ or $2k = x + y$.

Please tell me what I am getting wrong, thank you and sorry for taking up your time.

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    $\begingroup$ In order for $\{$ to render, you should do \{ and likewise for $\}$. Additionally, if you are going to use $\mid$ in your set notation, you should use \mid not the ASCII character. \mid automatically puts a space before and after it to keep the symbols decluttered. $\endgroup$ – Cameron Williams Mar 23 '14 at 21:44
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    $\begingroup$ As a first comment, I don't understand what your proof of Lemma 2 shows. In particular, how is "L cannot be a prime" a contradiction? $\endgroup$ – Johannes Kloos Mar 23 '14 at 21:48
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    $\begingroup$ As far as I can tell, all you have managed to prove is that if $L=p_i$ for some prime $p_i$, then assuming that $2L$ can't be expressed as a sum of primes gives you a contradiction. $\endgroup$ – Joshua Pepper Mar 23 '14 at 22:11

What you proved is that if $2k - p$ is composed for every prime $p < 2k$, then $k$ cannot be prime. This is obvious because if $k$ was prime, letting $p = k$ we have $2k - p = 2k - k = k$ which is prime, contraddicting the hypothesis.

But how is this supposed to prove the conjecture?

If we let $C$ be the set of Composite Numbers, the conclusion derived from our contradictory hypothesis claims that $L \in C$. This implies that $C = \mathbb N_+$, and this is trivially false

This is the mistake: the conclusion is actually that if $2L$ is not the sum of two prime numbers, then $L$ is in $C$. But this doesn't hold for every $L \in \mathbb N_+$; in fact, there are obviously some $L$ for which $2L$ is the sum of two primes.

  • $\begingroup$ So, I cannot say since the contradictory hypothesis produces an mismatched domain, Composite numbers, than what was assumed was right, Natural numbers, still does not provide a direct contradiction to my lemma so it can't be used. Is that right? $\endgroup$ – user137451 Mar 23 '14 at 22:26
  • $\begingroup$ @user137451 I've edited my answer $\endgroup$ – dani_s Mar 23 '14 at 22:37

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