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.
Best Answer
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?
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.