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Is this a proof for Goldbach's Theorem? I'm assuming not, but where does it go wrong?

The theorem states that any even number greater than 2 can be represented as the sum of two numbers.

Let $p_1$ and $p_2$ be two prime numbers.

$n = p_1 + p_2; \quad (n>2)$

If $n$ is even, then $n = p_1 + p_2; \quad (n>2)$ (This is the representation of the above statement. This is what we are trying to prove.)

$n = 2k$ for some integer $k>1$. (Because n is even)

For all $k>1, 2k = p_1 + p_2$ (Statement with 2k replacing n)

For all $k>1, 2k-p_2 = p_1$ (Subtract p2 from both sides)

For all $k>1, 2k \equiv p_2 \pmod{p_1}$ (Definition of congruency from above statement)

For all $k>1, 2k \bmod p_1 = p_2 \bmod p_1$ (Restating congruency)

$p_2 \not \equiv 0 \pmod{p_1}$ or else $p_2$ would not be prime (it would be a multiple of $p_1$)

For all $k>1, 2k \not \equiv 0 \pmod{p_1}$

For all $k>1, \frac{2k}{p_1} > 0$

Contrapositive: If $\frac{2k}{p_1} = 0, k\le 1$

Only true if $2k = 0$, so $k=0$ which means $k \le 1$ so contrapositive is true therefore Goldbach’s theorem proven!

EDIT: Proof is wrong from the beginning by assuming statement is true. This proves the converse, not the contrapositive.

Luke1195
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  • Your "proof" is too difficult to parse and also it's too difficult to see where you are heading with the proof. – Stefan4024 Oct 14 '16 at 16:37
  • Okay thank you for answering! I'll try to clean it up a bit. – Luke1195 Oct 14 '16 at 16:38
  • Why the downvotes when he clearly asks where the mistake is? – Marko Karbevski Oct 14 '16 at 16:59
  • @Luke1195: First of all in para 2 of your question you need to correct the statement of the Goldbach conjecture by stating that any even number > 2 can be represented as a sum of two **prime** numbers. –  Oct 14 '16 at 17:15

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You have shown that if $n=2k=p_1+p_2$ for some primes $p_1$ and $p_2$, then $k>1$. However, this is not the contrapositive of Goldbach's conjecture, it is the converse! The contrapositive of Goldbach's conjecture: if $n>2$ cannot be written as the sum of two primes, then $n$ is not even.

kccu
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