Is the sum of all prime numbers a prime number or a non-prime number?

It's an interesting question to pose I'd be interested to see a proposed proof.

would (the sum of all numbers) - (the sum of all non-prime numbers) = (the sum of all prime numbers) ?

A non prime number.
Proof:
[eqn] \sum_{ \text{p prime}} p \leq \sum_{n \in \mathbb{N}} n = \frac{-1}{12}
[/eqn]

Therefore the sum of all prime numbers is negative and negative numbers are not prime. Therefore, the sum of all prime numbers is not a prime number.

QED

17 + 23 (two primes) = 40 ( not a prime)
17 + 5 + 3 (three primes) = 25 (not a prime)

Meaningless question. The author of and Euclid (circa 300 BC) both knew that.

All three terms are infinite, so this is another meaningless question.

one of you is wrong

One post has a proof, the other has feelings. You can see which one is correct.

Brainlets should be gassed.

...

not an argument

What if you had the harmonic series, but replaced all the prime-numbered terms with their negative, does that converge or diverge?

you want to gas yourself?

Non-prime.
Since [math]\mathbb{P}\subset\mathbb{N}[/math], we have [math]\forall x\left(x\in\mathbb{P} \Rightarrow x\in\mathbb{N}\right)[/math], and so by contraposition, [math]\forall x\left(x\not\in\mathbb{N} \Rightarrow x\not\in\mathbb{P}\right)[/math].
Let the sum [math]s := \sum\limits_{p\in\mathbb{P}} p[/math]. Clearly, [math]s\not\in\mathbb{N}[/math], which then implies [math]s\not\in\mathbb{P}[/math].

explain

Still diverge. The primes only compensate for log log N.

[math]\mathbb{N}[/math] is the set of natural numbers, [math]\mathbb{P}[/math] is the set of all prime numbers.
[math]\mathbb{P}\subset\mathbb{N}[/math] reads 'the prime numbers are a subset of the natural numbers', which by definition means that all numbers in [math]\mathbb{P}[/math] must be in [math]\mathbb{N}[/math].
Formally, this expressed as 'for all objects x, if x is a prime, it is also a natural number', symbolically written as [math]\forall x\left(x\in\mathbb{P}\Rightarrow x\in\mathbb{N}\right)[/math].
Contraposition is the logical principle that, if P being true imples that Q is true, then Q being false requires P to be false.
Symbolically, this is written as [math] P \Rightarrow Q[/math] being equivalent to [math] \neg Q \Rightarrow \neg P [/math]
In our case, this means that that [math]x\in\mathbb{P}\Rightarrow x\in\mathbb{N} [/math] is equivalent to [math]x\not\in\mathbb{N}\Rightarrow x\not\in\mathbb{P}[/math], which we know is true for all x.
en.wikipedia.org/wiki/Contraposition
[math]s:=\sum\limits_{p\in\mathbb{P}} p[/math] is simply the sum of all prime numbers.
The sum of all prime numbers, s, is not a natural number, since if there was, there'd be a natural number larger than s, for example, s+1.
Lets pretend that such a number exists, let's call it n: [math]n\in\mathbb{N},n>s[/math]. As is known at least since Euclid, you can find an arbitrary number of prime numbers.
Therefore, we can simply find n prime numbers and add them, and we'd get a sum greater than n, since all the terms are greater than one.
Since this finite sum hasn't added all prime numbers, it is less than the full sum s, and so [math] s>n[/math].
As shown, the sum of all prime numbers is greater than any natural number, and so cannot be a natural number itself. And since it isn't a natural number, it also isn't a prime.

I hope that explanation was thorough enough for ya.

>fixing

[math]1 = 1^2[/math]
[math]1+3 = 2^2[/math]
[math]1+3+5 = 3^2[/math]
[math]1+3+5+7 = 4^2[/math]

thus

[math]1+3+...+(2n-1) = n^2[/math]

So your sum of all prime numbers, will not be prime

aren't you making the incorrect assumption that every odd number is prime?

>the sum of all prime numbers is greater than any natural number, and so cannot be a natural number itself. And since it isn't a natural number, it also isn't a prime.

then why does 1+2+3+4+5... = -1/12?

>7
>prime
fucking brainlets

You can prove this yourself using induction. It's an ancient observation that has been independently discovered for thousands of years that the sum of primes is a square.

but 7 is prime

2 is also a prime, that's the proof that every odd number is a square not primes.

Ramanujimmyjons and Hardy spent forever trying to figure out primes and partitions

what does this even mean?

what the fuck even is this thread
are we seriously asking whether or not an infinite number is even, odd, or prime?

why the paedophilia?

...

Nice proof user!

Mind if I borrow this proof for my math class?

I can't tell if this sarcasm or not

superposition.

all prime numbers are odd

odd + odd = even.
even + odd = odd.

thus, after each successive prime the sum switches between even and odd. if we assume that there are an infinite number of primes, then we can know that it can be neither, but a superposition of both.

1 + 3 + 5 + 7 + 11 = ?

The street shitter already told us.

~25

>includes 1 as prime
>doesn't include 2 as prime

Well you must find if "infinity is prime number".

It would odd because 2 is a prime number.

there are no prime numbers you can multiply to get another prime. they would be visible by howver many primes you multiplied by.

3*5*7 = 15*7 = 105 = not prime because it is divisible by 15 and 7. You should be able to do this for any prime number.

>divergent sum
>brainlets misinterpreting -1/12
but, adding primes ends up, 3+5 8 even, + 7 15 odd, + 11 26 even, etc etc, so half of the time it's even and not prime (two doesn't affect jack shit). We can therefore conclude the sum of all prime numbers is half prime, just as 1-1+1-1+1...=1/2

>all prime numbers are odd
JUST

impressive

it's hard to assign a value to that infinite sum, but here's some related ideas

mathworld.wolfram.com/PrimeZetaFunction.html

Schroedinger's cat.

how the fuck is something half prime

The sum of all the primes diverges.

I actually understood all of that. Is this proof intro set theory stuff? I have electives and I was thinking about doing a course involving set theory.

because the sum of all natural numbers isn't greater than any natural number.

does that apply to primes too?

save it, it's all yours my friend

It's easy to confuse yourself with this shit but it's quite simple.

All that the ramanujan summation stuff, cutoff and zeta regularization does, is look at the smoothed curve at x = 0.
What sums usually do is look at the value as x->inf.

It's just a unique value you can assign to a sum, really they have many such values.

en.wikipedia.org/wiki/File:Sum1234Summary.svg

[math] \displaystyle
\zeta \neq \Sigma
[/math]

>(2n-1)

Here is a relevant computation. It turns out the sum diverges since log of zeta at negative even integers is undefined.
Damn, , beat me to it.

The hand-wavyness of these arguments is disgusting. You guys should stick to philosophy.

>The hand-wavyness of these arguments is disgusting. You guys should stick to philosophy.
>literally the whole thread
how will Veeky Forums ever recover

This is the only one that makes sense

Please somebody help me I cannot find a single thing wrong with this argument please I need help

this is very basic set theory yes
he could've just wrote that the sum of all the primes is infinite, and infinity isn't a number therefore it can't be a natural number and therefore can't be a prime number, and therefore is non-prime.

anything not a number is also not prime, meaning any non number is also a non prime.
be careful though, "non prime" and "non prime number" indicate two different things. non prime numbers are a subset of non primes.

can you explain why their arguments are wrong and your argument is correct?

There are an infinite number of primes

In 3 years time, will you still always be linking to this image even though nobody knows where it came from or how to compute that interpolation curve? Will you keep ignorig that a picture without the generating math isn't a valid explanation?

yes? and?

They cannot be summed. Seems obvious.

i don't really get it either because if you look at that graph, it hits the sum value at the half marks rather than at the unit marks

and if you actually had that function, it'd just be an offset of f(x)=x(x+1)/2, in fact it's just f(x-0.5), which has a y intercept of -1/8, not -1/12

>the illuminati has taken over wikipedia
get out your tin foil hats, we're under attack

>how to compute
terrytao.wordpress.com/2010/04/10/the-euler-maclaurin-formula-bernoulli-numbers-the-zeta-function-and-real-variable-analytic-continuation/

b-but 1/3 isn't 0.3...

here we go....

We'll memed

But that's wrong you fucking brainlet.
[eqn]\prod_{p \ \text{prime}} \frac{1}{1-p} = \sum_{n \in \mathbb{N}} n = -\frac{1}{12}[/eqn][eqn]\prod_{p \ \text{prime}} (p-1) = 12[/eqn][eqn]\sum_{p \ \text{prime}} \text{ln}(p-1) = \text{ln}(12)[/eqn][eqn]\sum_{p \ \text{prime}} p \geq \sum_{p \ \text{prime}} \text{ln}(p-1) = \text{ln}(12) \approx 2.485[/eqn]
as ln(x-1) ≤ x for all x > 1.

...

Fine, not counting 2.

Sum of all naturals converges, and since primes form a subsequence of naturals it's sum is strictly smaller than that of all naturals, therefore it converges

whoa whoa whoa
slow down