2
$\begingroup$

My professor introduced the Goldbach Conjecture to me a couple months ago, and I've been intrigued by it ever since. The Goldbach Conjecture states that every even number greater than 4 can be written as the sum of two primes.

If we look at the number $2m$, where we let $m$ be the number containing every prime greater than $2$, then $2m$ = $2\cdot3\cdot5\cdot7\cdot11\cdot13\cdot17\cdots$

Then every number less than $2m -1$ will be contained in $2m$, so no matter what number $x$ you subtract from $2m$, you get

$2m - x = x\cdot\left(\frac{2m}{x}-1\right)$ which is composite, so this even number can't be written as sum of two primes.

I'm assuming this is not a valid counter-example, but why not?

$\endgroup$
  • 6
    $\begingroup$ Lol, nice try.${}{}{}$ $\endgroup$ – Jorge Fernández Hidalgo Jul 26 '16 at 18:27
  • 1
    $\begingroup$ Can you clarify about the construction of $m$? It's not clear if an infinite number of primes is involved or not. In the former case, $m$ is not a real number because the product diverges, as there are infinitely many primes (the first proof of this fact, credited to Euclid, is really slick). $\endgroup$ – rubik Jul 26 '16 at 18:29
10
$\begingroup$

$m$ doesn't exist, since there are infinitely many primes.


In a little more detail: the product of infinitely many natural numbers is not, in general, a natural number. Similarly, the sum of infinitely many natural numbers is not, in general, a natural number: for example, $$1+1+1+1+...$$ is not a natural number. The Goldbach conjecture is a statement about natural numbers, so looking to infinite products for a counterexample isn't going to work.

There are contexts where we can make sense of an infinite expression like the above (see e.g. Why does $1+2+3+\cdots = -\frac{1}{12}$?), but it's something that takes serious work to make precise - and often results in surprising properties, or the failure of properties we usually take for granted.


Addressing rubik's comments: what if we only use finitely many primes in the construction of $m$?

Well, then the whole shebang breaks down! Suppose $$m=3\cdot 5\cdot . . . \cdot p_k$$ is the product of the first $(k-1)$-many primes bigger than $2$. Then their may be a prime which is $<2m-1$, but which is not a factor of $2m$ - namely, $p_{k+1}$! So we can't conclude that $2m$ is a counterexample to Goldbach.

$\endgroup$
  • $\begingroup$ $m=0{}{}{}{}{}$ $\endgroup$ – Jorge Fernández Hidalgo Jul 26 '16 at 18:27
  • $\begingroup$ @CarryonSmiling Um, what? The OP has defined $m$ as $3\cdot 5\cdot 7\cdot . . .$ - how on earth is that $0$? $\endgroup$ – Noah Schweber Jul 26 '16 at 18:28
  • 1
    $\begingroup$ well, maybe with the OP's first definition $m=0$ works.(an integer containing all primes $>2$) $\endgroup$ – Jorge Fernández Hidalgo Jul 26 '16 at 18:29
  • $\begingroup$ @rubik I disagree, I don't see that in the question at all. Their argument crucially relies on the assumption that $m$ is divisible by every prime; until they weigh in otherwise, I think my answer addresses their question. $\endgroup$ – Noah Schweber Jul 26 '16 at 18:35
  • $\begingroup$ @NoahSchweber Thank you for your answer, it was very clear and I understand now why it's not valid. $\endgroup$ – Ash Jul 26 '16 at 18:40
2
$\begingroup$

There are infinitely many primes, so $m$ doesn't exist.

Pretend like $m$ could exist i.e. there are finitely many primes. Then clearly $m+1$ has a prime factor, say $q$. Since $m$ is the product of all primes, $q$ must be a prime factor of $m$ as well. Thus, $q$ must be a prime factor of $m+1-m=1$ as well, yet $1$ has no prime factors. Therefore, $m$ cannot exist, so there must be infinitely many primes.

$\endgroup$
0
$\begingroup$

Apart from the concern about the nature of $ m $ as mentioned by others, your claim that every number less than $2m - 1$ is contained in $2m $ is not true. For example consider number 4; then for $ x=4$ you do not conclude anything.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.