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You have to be very careful with these "plausibility arguments" that physicists like to make. It may well happen, for example, that there are counterexamples to

• hn_throwaway_99 10y ago
  I'm not a mathematician, nor a physicist, but I don't understand this statement: similar to, for example, the zero probability that a random real number in some interval is a rational number. Is it just a consequence of rational numbers being countable and real numbers not being countable? Seems like an odd way to define probability in this instance.
  • alextheparrot 10y ago
    It comes from how we define infinity/the number line. Any number divided by infinity is 0, that's easy. But between any two points there is an infinite number of rational numbers and between any two rational numbers you can derive an infinite number of irrational numbers. I know this isn't kosher, but it is pretty much infinity/(infinity ^ infinity). That leads us back to the probability of picking a rational number as 1/infinity, because for every rational number there are infinity irrational numbers that we also could have chosen. David Foster Wallace had a really good book called Everything and More where he explored the history of infinity. Might be of interest to you.
    • SomeStupidPoint 10y ago
      That's a really bad argument, since between any two irrational numbers, there's an infinite number of rational numbers. Ed: For an explicit construction of such numbers -- Suppose we have a<b, two irrational numbers. Let b-a = d, the difference between them. Then there exists N, an integer, such that 10^(-N) < d. Let b1 be b truncated at the (N+1)th digit past the decimal point. Then b1 is rational (finitely many digits) and b-b1 < d (and b1 < b), so b1 is in (a,b). You can then add back one digit of b at a time, to get b2, b3, etc which form an infinite sequence of rational numbers converging to b from below. (We get infinite unique numbers from the fact b is irrational. Not all the b_n need be distinct.)
    • posterboy 10y ago
      > Any number divided by infinity is 0, that's easy That's a dirty hack for the sake of argument, but it isn't really true. The convergence limit of division with infinitely increasing divisors is zero. Corollary division by zero isn't infinity, it's simply not defined.
    • thanatropism 10y ago
      I wish there was a third alternative to upvote/downvote -- "sounds like a good chap acting in good faith, but knows less than nothing of this subject".
  • fahadkhan 10y ago
    I think the point being made is that whilst the probability of picking one at random is "zero" the rational numbers exists.
  • n4r9 10y ago
    Countability does have to do with it. Measure theory [0] allows us to extend the classical notions of discrete probability to continuous domains. [0] https://en.wikipedia.org/wiki/Measure_(mathematics) https://en.wikipedia.org/wiki/Measure_(mathematics) https://en.wikipedia.org/wiki/Probability_measure https://en.wikipedia.org/wiki/Probability_measure
  • jordigh 10y ago
    It is a consequence of countability, but I like to think about it this way: a real number is, roughly, an infinite sequence of "random" digits. What is the probability that your infinite sequence of random digits will randomly settle into an infinite repeating pattern? That's obviously, zero, right? No way that a random process will produce an infinitely repeating pattern. But just because it's "probabilistically impossible" to pick a rational number doesn't mean that they don't exist.
    • hn_throwaway_99 10y ago
      Thanks, this was the explanation that made sense to me.
  • thaumasiotes 10y ago
    It is a consequence of the rationals being countable, in that all countable sets necessarily have this property. The intuition (well, one way of looking at it) is that if you have a line of length 40, and a subset of that line which has total length 2, then a randomly-chosen point on the line has a 5% chance (2 in 40) of being part of that subset. It's not important that the subset be connected; the intervals (3,4) and (28,29) have length 1 each, length 2 in total. We want to think in terms of intervals because they're very easy to measure: the interval (low,high) is (high-low) amount of stuff. It turns out that there are so few rational numbers that you can define a series of intervals with the following properties: - every rational number is contained within one or more of the intervals - the (infinite) sum of the lengths of each interval is arbitrarily small So for any value, you can capture the entire set of rationals in a set of intervals whose total length is less than that value. This means the "amount of stuff" in the set of rational numbers cannot be bounded away from 0; removing the rationals from a line of probability-stuff leaves no less probability-stuff than was there before.
• tgb 10y ago
  What specifically is your first paragraph attacking? I just read their paper and have no idea what you're going after. They present a concrete strategy for a rigorous proof, but with gaps left. They also check that the suggested strategy has the properties we expect given what they want it to do. They do not make a plausibility argument like you suggest, except in so far as that they argue that their proof strategy might plausibly work. The distinction, I think, is very significant; the end result here would be a rigorous proof.
  • deleted 10y ago
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  • jordigh 10y ago
    Oh, I'm probably wrong and just arguing from my prejudices. I obviously haven't tried reading their paper. What is their argument for why the strategy would work?
    • tgb 10y ago
      Uh, I'm out of my field so I'm not sure I can summarize it better than the article did, but I can try. They prove that the Riemann hypothesis is equivalent to a property of a specific operator, however they can't prove that property (that's it's self-adjoint under some inner product) since they don't know what the appropriate domain of the operator is. But the operator does do several things: first, it's classical version is what was expected. Secondly, they suggest an approach towards getting an inner product that would the operator self-adjoint. Lastly, the authors conclude "The fact that iH [the operator in question] is PT symmetric, with a broken PT symmetry, offers a fresh and optimistic outlook" which I don't understand well enough to add to.
      • bermanoid 10y ago
        Taking this down to an even more basic level that doesn't require much knowledge of this stuff - the basic strategy of this program is to find a quantum system that they can prove (rigorously) has energy levels that map directly to the Riemann zeta zeros (well, the non-trivial ones) under a simple shift/rotation in the complex plane. Since energy levels of these systems are always real (and in a straight line on the complex plane), this would establish that the zeros are all in a straight line as well, namely the one where Re(z) = 1/2. The tough bit is finding that system. These researchers seem to think they've made headway; I'm withholding judgment for now, since by my reading the whole motivation for reaching for a physical system (real eigenvalues) is exactly what they still need to establish about their proposed system. They might have just succeeded in recasting the Riemann hypothesis in a different but no more tractable form, we'll have to see.
• enugu 10y ago
  There is no dichotomy here. Mathematicians also use heuristics to figure out what is probably true. This is a very important part of mathematical activity - generating good, not yet worked out ideas based on rough insights. What the physicists are doing here is also how the Fermat's Last Theorem was approached - a conjectural analogy to a different field(elliptic curves). The only difference is that they are more careful about separating heuristics from proofs. If you read physics textbooks, they are less interested in being formally precise. Mathematical activity is not all about formal reasoning, though that is the output.
• tzs 10y ago
  > I don't understand the particulars of the spectrum of this operator that they have constructed, but I have heard others describe this approach as a simple reformulation of one hard problem in terms of another equally hard problem. I don't know if it is something inherent in the RH, or just because it has received so much attention, but it seems to have a propensity for having equivalent formations in areas that do not have any obvious connection to it. A couple examples: • RH is equivalent to the assertion that for integers n >= 3, |log lcm(1,2,...,n) - n| < sqrt(n) log^2(n). • For integers n >= 1, let s(n) = sum of divisors of n. E.g., s(6) = 1 + 2 + 3 + 6 = 12. Let H(n) = 1 + 1/2 + 1/3 + ... + 1/n. Then RH is equivalent to the assertion that s(n) <= H(n) + exp(H(n)) log(H(n)), with equality only at n = 1. More at: http://mathoverflow.net/questions/39944/collection-of-equivalent-forms-of-riemann-hypothesis http://mathoverflow.net/questions/39944/collection-of-equiva...
• Houshalter 10y ago
  There is a theory that many famous unproved conjectures are true but unprovable with the traditional methods you hold dear. Of course Godel showed there are true unprovable theorems. But those are weird paradoxical self referencing things. That can sort of be proved if you use a different system of axioms. But it's possible there are unprovable true theorems that aren't like that. Freeman Dyson created an example of one. He conjectured that no power of 5, it's base 10 representation reversed, is a power of 2. This is almost certainly true. There is no particular correlation with powers of numbers and their base 10 representation. They are basically random except for the last digit. There is no pattern to it. And because the base 10 representations get longer and longer, it gets increasingly unlikely there exists a counterexample. You can calculate the probability and it's very low. The thing is that statement is very likely to be unprovable. If it's true, it's not true for any particular reason. Nothing at all says a counterexample can't exist. It just happens to be increasingly unlikely to happen. A mathematical coincidence. You can prove some famous conjectures using probabilistic arguments. For instance Goldbach's conjecture that every even number is the sum of 2 primes. Because primes are distributed basically randomly, as numbers get bigger, it gets more and more likely there are two primes that add to get that number. There are an awful lot of prime numbers after all. But again, there's not any reason to suspect there's a mathematical reason a counterexample can't exist. It's just unlikely. This is really hard to formalize. Obviously prime numbers are deterministic, not random. Obviously base 10 representations aren't really random. It'd be more precise to say that they aren't correlated with the particular properties we are interested in. But even that is pretty hard to pin down precisely. Mathematicians strongly disrespect anything that isn't a rigorous proof. But if you do that you might miss out on a lot of interesting things. And there is absolutely no guarantee that the universe will always give us that!
• adyavanapalli 10y ago
  I think you're mistaken in thinking that the arguments Bender et al are making are physical. Operator theory, Hilbert spaces, Lebegue measures, etc. are pure mathematical tools used to construct and derive insight from Quantum Mechanics. The tools themselves are grounded in the usual sense of mathematical rigor. The inspiration to attack this conjecture, the Riemann hypothesis, framed as a quantum system might come from physical intuition, but everything else is rigor.
  • jordigh 10y ago
    I wasn't thinking that they're physical, just that physicists are used to playing fast and loose with approximations because it's a technique that usually works in physics. Or to not really care about what a Dirac distribution really is. Or that defining a tensor (field) as something that transforms like a tensor is a satisfying definition. I am skeptical of the methods that they're used to. But it doesn't look like they're doing that, so I'm probably wrong.