Wikipedia:Reference desk/Archives/Mathematics/2014 October 17

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October 17

Does a solution exist?

Firstly, not a homework question, I'm long past school age, I just dabble in mathematics recreationally. I'm playing around with something and I've narrowed it down to one problem, basically I need to find two integers $a,b$ which are not equal such that $a,b,{\frac {a}{a+b}},{\frac {b}{a+b}}$ are all positive integers. I'm beginning to suspect that no solutions exist, but can anyone help me either find a solution, or show that there are none? Organics ^talk 11:14, 17 October 2014 (UTC)

If $a$ and $b$ are positive then $a+b>a$, so ${\frac {a}{a+b}}<1$ can not be a positive integer. Similar for $b$. --CiaPan (talk) 11:21, 17 October 2014 (UTC)

(ec) Well, ${\frac {a}{a+b}}+{\frac {b}{a+b}}=1$, so it's not possible for both terms to be positive integers. If you allow non-negative integers then the terms have to be 0 and 1 in some order, so a = 0, b = any positive integer (or vice versa) will work, and these are the only solutions. AndrewWTaylor (talk) 11:26, 17 October 2014 (UTC)

That seemed too easy. If you meant $a,b,{\frac {a+b}{a}},{\frac {a+b}{b}}$ then note that (a+b)/a = 1 + b/a, and (a+b)/b = 1 + a/b. If a ≠ b then either b/a or a/b will be below 1 and not an integer. PrimeHunter (talk) 11:34, 17 October 2014 (UTC)

Is there a easy way to do this math?

PS:Nope, this is not homework, its for some musical thing I am doing.

Is there a easier way to do this math? Imagine this. Two drummers are drumming (at the same time) during one second. One at 1 bps and other at 1.5 bps. A stick will hit a drum at 2 points points of time, at 1 second, and at 2/3 of a second.

With that info, now imagine 3 drummers, one playing at 1.5 bps, the other at 1 bps, and the other at 2 bpm (the number we previously got). A stick will hit a drum 3 points of time, at half second, at 2/3 of a second and at 1 second.

With that info now imagine 4 drummers, one playing at 1 bps, the other playing at 1.5 bps, the third at 2 bps and the fourth at 3 bps (the number we previously got). A stick will hit a drum 4 points of time, at half second, at 1/3 of a second and at 2/3 of a second and at 1 second.

With that info now imagine 5 drummers, one playing at 1 bps, the other playing at 1.5 bps, the third at 2 bps and the fourth at 3 bps and another playing at 4 bps (the number we previously got). A stick will hit a drum 6 points of time, at half second, at 1/3 of a second and at 2/3 of a second and at 1 second 1/4 of a second and 3/4 of a second.

And this goes on.

Is there a easier way to do that, or some formula that continue this...?
— Preceding unsigned comment added by 201.78.201.199 (talk) 19:12, 17 October 2014 (UTC)

I'm not following this. If a drummer is playing at 1 bpm (beats per minute), why would his stick hit the drum at 1 second in? Did you mean beats per second? OldTimeNESter (talk) 20:27, 17 October 2014 (UTC)

Do what? You're not doing any math, just establishing facts. I'm not sure why you keep saying "the number we previously got". That number doesn't somehow follow from the previous scenario.--80.109.80.31 (talk) 20:30, 17 October 2014 (UTC)

The 1.5 bps drummer doesn't really fit into the pattern of 1 bps, 2 bps, 3bps etc. Apart from that, the sequence of numbers that you are describing starting 1,2,4,6 is sequence A002088 at OEIS. For more information see our article on Farey sequences. Gandalf61 (talk) 11:49, 18 October 2014 (UTC)

I already tried wolfram alpha to find the results and already saw your oeis link. The number 1 and 1.5 are the first 2 numbers used to make the thing (to ge the number 2). — Preceding unsigned comment added by 201.78.160.103 (talk) 13:06, 18 October 2014 (UTC)

So, how do 1 and 1.5 give 2? And again, what are you trying to do? Make a list of the rational numbers with denominator up to a given limit? —Tamfang (talk) 07:38, 19 October 2014 (UTC)

I think the n+1th drummer has a number of bps equal to the total number of drumbeats each second from the previous n drummers, where only a single beat is counted if two drummers hit their drum simultaneously (so the next in the sequence would be 1bps, 1.5bps, 2bps, 3bps, 4bps, 6bps). Not sure whether there is a good general form for the series, but this seems to be what the OP is getting at. MChesterMC (talk) 12:00, 20 October 2014 (UTC)

Is the following proof of Goldbach Conjecture correct?

Keywords: π(*):= Odd Prime Counting Function and Fundamental Theorem of Arithmetic (FTA) Goldbach conjecture states every positive even integer is the sum of two prime numbers. (We count one as prime in the sense of additive number theory outside of the FTA.)

Proof of Goldbach Conjecture:

Suppose there exists a positive even integer, e > 4, that is not the sum of two odd prime numbers or 1. e ≠ p + q

over S = {all odd prime numbers less than e} and where k = card(S) = π(e). Therefore, e ≠ p + q over S, (p,q є S) , implies the following system of equations over S, 1 = e - n1 * q1, 3 = e - n2 * q2, ..., pk = e - nk * qk, according to the Fundamental Theorem of Arithmetic where 1 < qj ≤ (nj * qj)^.5 ≤ nj for 1 ≤ j ≤ k where pj, qj є S and nj is a positive integer. Note: If qj = 1, then nj є S, or nj is an odd prime less than e.

Therefore, 2. Probability(e ≠ p + q over S) = ∏(j=1 to k→∞)[Probability (qj ≠ 1 | e - pj = nj * qj over S) * Probability (e - pj = nj * qj over S)]

= ∏(j=1 to k→∞)[(π((nj * qj)^.5) - 1)/π((nj * qj)^.5)] → 0 (This is an increasingly fast convergence for this almost everywhere monotonic non-increasing expression. This implies that the expected value of e ≠ p + q over S is practically zero, or E[e ≠ p + q over S] = e * Probability(e ≠ p + q over S) ≈ 0 for all e ≥ 100.)

Note: Probability (e - pj = nj * qj over S) = 1 for 1 ≤ j ≤ k.

In addition, empirical evidence has confirmed the validity of the conjecture for all positive even integers up to at least an order of 10^18. Therefore, we conclude the conjecture is true. Euler was right! Thank God! Praise God! --David Cole, primesdor@gmail.com — Preceding unsigned comment added by Primesdegold (talk • contribs) 20:04, 17 October 2014 (UTC)

You have a couple of misconceptions about the Goldbach Conjecture: first, we do not consider 1 to be prime for this purpose; second, we do not require the primes to be distinct. Those aren't the real problems with your argument, though.

As far as I can tell, you're putting a uniform distribution on the integers less than $e$ and then assuming that $m$ and $e-m$ being prime are independent events. There's no reason to make that assumption.

You can find the same thinking in our Goldbach's Conjecture article under heuristic justification.--80.109.80.31 (talk) 21:01, 17 October 2014 (UTC)

(edit conflict) This is not a proof. You make arguments similar to Goldbach's conjecture#Heuristic justification. Such arguments can only show that if we make certain unproven assumptions that numbers behave sufficiently "randomly" and independently, then the chance of a large counter example is extremely small. We don't know whether the assumptions are correct and even if they are in som sense, each large untested number would still have a tiny chance of being a counter example. That chance would quickly approach 0 for e > 4×10¹⁸ (the current search limit), but it would never actually be 0. PrimeHunter (talk) 21:15, 17 October 2014 (UTC)

First, as PrimeHunter notes, a "proof" using heuristic arguments is not considered a proof. Goldbach's Conjecture is thought by most mathematicians to be almost certainly true, but there is the annoying possibility that it may be true but unprovable, an example of Gödel's incompleteness theorems. (If so, its unprovability is itself unprovable. It isn't like the Halting Problem, which was proved to be undecidable.) The same possibility was advanced with regards to Fermat's Last Conjecture until its proof by Wiles. The challenge is to find a formally valid proof of Goldbach's Conjecture, which may require branches of mathematics not in existence in Euler's time, as is the case with Wiles's proof of Fermat's Last Conjecture. Robert McClenon (talk) 15:29, 20 October 2014 (UTC)

Any proof is probably beyond the scope of "classical" number theory. Otherwise Gauss would have found it if Euler didn't find it. Robert McClenon (talk) 15:45, 20 October 2014 (UTC)

Please don't add wording to your proposed proof after their have been responses. It causes confusion because the replies may not appear to address the updated version of the proof. However, it doesn't change the original objections. A "proof" using heuristic arguments is not a rigorous proof, and only a rigorous proof is acceptable. Robert McClenon (talk) 18:19, 20 October 2014 (UTC)

Is the following proof of Riemann Hypothesis correct?

Riemann Hypothesis states that the real part of all non-trivial zeros of the Riemann zeta function, or ζ(s) = Σ(k=1 to ∞) 1/k^s = 0, equals one-half. For the non-trivial zero, s, a complex number, we have s = a + bi where Re(s)= a = 1/2.

Proof of Riemann Hypothesis

Fact I: The real part of all non-trivial zeros of the Riemann zeta function are located in the critical strip, [0, 1], according to a Riemann Theorem.

Fact II: There are infinitely many non-trivial zeros of the Riemann zeta function whose real part equals one-half according to a Hardy Theorem.

Fact III: The sum of the complex conjugate pairs of non-trivial zeros, s = a + bi and s' = c + di where ζ(s) = Σ(k=1 to ∞) 1/k^s = 0 and ζ(s') = Σ(k=1 to ∞) 1/k^s' = 0, of the Riemann zeta function equals one according to the Fundamental Theorem of Arithmetic and the Harmonic Series (H):(Note: Euler and others have proven that there exists an infinite set of primes in H. And that the divergence of H is a key reason for that result.)

H = Σ(k=1 to ∞) 1/k = Σ(k=1 to ∞) (1/k^s)(1/k^s') = Σ(k=1 to ∞) 1/k^(s+s') = ∝. Therefore, according to Facts I, II, and III, we have the following properties for all non-trivial zeros of the Riemann zeta function:

s + s' = 1 which implies a + c = 1 and b + d = 0 such that 0 ≤ a ≤ 1 and 0 ≤ c ≤ 1.

And of course, k^s = k^(a + bi) implies k^(s-bi) = k^a, and k^s' = k^(c + di) implies k^(s-di) = k^c.

Fact IV: For all k > 1, k is a positive integer, there exists a prime number, p, so that p|k such that p = k or p ≤ k^(1/2).

Therefore, according to Facts I, II, III, and IV, we have:

k^(1/2) ≤ k^a ≤ k, k^(1/2) ≤ k^c ≤ k, and a + c = 1.

Hence, k^a = k^c = k^(1/2) which implies a = c = 1/2. Riemann Hypothesis is true! Riemann was right! Thank God! Praise God!

Note the Importance of the Harmonic Series (H) with regards to prime numbers:

H = Σ(k=1 to ∞) 1/k = 2Σ(k=1 to ∞) [1/(2k) = 1/(q - p)] = ∝ where q > p, and p,q are odd prime numbers. H1 = 2( 1/(5-3) + 1/(11-7) + ...) = ∝; There is a infinite set, P1, of primes, p, generated from H1, and there is a infinite set, Q1, of primes, q, generated from H1. H2 = 2( 1/(7-5) + 1/(17-13) + ...) = ∝; There is a infinite set, Q2, of primes, q, generated from H2. H3 =2( 1/(13-11) + 1/(23-19) + ...) = ∝; There is a infinite set, Q3, of primes, q, generated from H3. ... H∞ = 2(...) = ∝; There is a infinite set (Q∞) of primes, q, generated from H∞. Therefore, the infinite set of all odd primes or ℙ\{2} = P1 ∪ Q1 ∪ Q2 ∪ Q3 ∪ ... ∪ Q∞. (Note: If we accept one as prime in the sense of additive number theory outside of FTA, this inclusion of 1 as a prime will change H1 through H∞, slightly. For example, H1 = 2( 1/(3-1) + 1/(11-7) + ...) = ∝, and H2 = 2( 1/(5-3) + 1/(17-13) + ...) = ∝, ...,H∞ = 2(...) = ∝ ; Thank God! Praise God!

Fact/Proposition: There are infinitely many more positive integers than there are prime numbers, or prime numbers have a zero density relative to the positive integers, and prime numbers generate the positive even integers efficiently so that gaps between two consecutive prime numbers increase without bound. Thank God! Praise God!

Fact/Proposition: π(e = mg = 1 + p2n) = 2π(g = 1 + pn)= 2n ∈2ℕ where π():=Prime Counting Function, p2n, pn ≥ 3 ∈ ℙ\{1,2}, and 2 < m ∈ ℚ ≤ 3. As g → ∞, m → 2. Thank God! Praise God! --David Cole, primesdor@gmail.com

 — Preceding unsigned comment added by Primesdegold (talk • contribs) 22:29, 17 October 2014 (UTC) 

My guess is that it is about as right as your proof of Goldbach's conjecture above. Bubba73 ^{You talkin' to me?} 00:54, 18 October 2014 (UTC)

I don't see what implies your 2nd to last line, I don't believe that anything does. If you wouldn't mind elaborating on your reasoning, we could through things a bit more thoroughly and see what is going on; at the moment, it just appears as jump. I'd love to discuss more if you're willing. Though, I do want to point out, while it can be rewarding to ponder over, and workout, our own notions about such problems, the odds that conjectures such as Goldbach and Riemann being solved in a few lines of elementary mathematics is, essentially, nil. If you would like to learn more about such questions, I can definitely recommend a few books and articles that would be of use. Best of luck - there is something captivating about these problems, to be sure:-) Phoenixia1177 (talk) 03:59, 18 October 2014 (UTC)

Fact III is wrong. By itself this would trivially imply the Riemann hypothesis if true. I don't see what the harmonic series has to do with it, but to me it seems that you have assumed what you wanted to prove here. Sławomir Biały (talk) 12:21, 18 October 2014 (UTC)

Despite recent incomprehensible attempts to clarify the role of the harmonic series in proving "fact" III, the claim that s+s'=1 follows from the divergence of the harmonic series remains a non-sequitur. Sławomir Biały (talk) 14:00, 19 October 2014 (UTC)