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The headline seems inaccurate to me. The proof in the article states that there is a 100% chance of a random elliptic curve being either rank 0 (50% chance) or
by claar 8y ago
The headline seems inaccurate to me.
The proof in the article states that there is a 100% chance of a random elliptic curve being either rank 0 (50% chance) or rank 1 (50% chance).
However, the article also informs us that there are an infinite number of elliptic curves with rank 2 or more. For example, the elliptic curve `y2+y=x3+x2−2x` is of rank 2.
So is it truly accurate to say that elliptic curves come only in two types?
- baq 8y agoThis is explained in the article, too.
- gus_massa 8y agoI think that the correct mathematical terminology is that almost all the elliptic curves [of this family] come in only too types. (I'm not sure that this is official.) For a similar example see https://en.wikipedia.org/wiki/Almost_everywhere https://en.wikipedia.org/wiki/Almost_everywhere For example, we know that almost all the numbers are normal, but we have a prof that any particular number is normal (but we have some good candidates). Moreover, we know plenty of numbers that are not normal. https://en.wikipedia.org/wiki/Normal_number https://en.wikipedia.org/wiki/Normal_number
- deleted 8y ago[deleted]
- thaumasiotes 8y ago> we [don't] have a pro[o]f that any particular number is normal (but we have some good candidates) https://en.wikipedia.org/wiki/Champernowne_constant https://en.wikipedia.org/wiki/Champernowne_constant ?
- Someone 8y agoEach Ck is normal in base k, but that is a much weaker claim than the claim without limiting it to a base. That, we don’t know (”It is an open problem whether Ck is normal in bases b ≠ k”) Each Chaitin constant (https://en.wikipedia.org/wiki/Chaitin%27s_constant https://en.wikipedia.org/wiki/Chaitin%27s_constant) is normal in that stronger sense, but we don’t know, and can’t know, much more about them than that they are normal numbers between zero and one.
- gus_massa 8y agoI totally forgot the "don't"! Too late for editing :(. Thanks for noticing it.
- ducttapecrown 8y agoI'm an undergrad taking some measure theory right now. This is probably not exactly the mathematical formalism they are using, but it could be similar. If I asked you to pick a random real number from 0 to 1, what do you think the probability is that the real number is rational? A natural way to answer this question is to try to generalize the way we say the interval from 0 to 1 is length 1 to more kinds of sets. Measure theory does exactly this, but we find out that the measure of the rational numbers is 0! This means that the probability of picking a rational number is 0. But that's clearly impossible you say, because there are an infinite number of rational numbers from 0 to 1! But in a precise mathematical way, the probability is 0. Now a funny think I said was "pick a random real number". Since the computable numbers are also a measure 0 subset of the real numbers, it's literally impossible to randomly pick a real number with a computer...
- darawk 8y agoYes, but that doesn't mean that there are only irrational numbers. There are rational numbers, they just have zero measure. Similarly, these curves come in more than two types - those other types just have zero measure in the probability space of the curves.
- aportnoy 8y agoNot only that, we should expect to always get an irrational number, but always get rationals from a finite subset of Q. I recently asked a closely related question on Math.SE: https://math.stackexchange.com/questions/2952087/sampling-from-a-continuous-distribution/ https://math.stackexchange.com/questions/2952087/sampling-fr...
- cix_pkez 8y agoHey, non-mathematician computer science type here. If I follow correctly, the issue with randomly picking any real number in that interval is that irrational numbers would require infinite computational steps to resolve. So the probability is really 0 that you'll get an irrational. If you have a finite number of computations, you're guaranteed to resolve to a rational, while if you have an infinite number of computations, you never resolve to anything. Is that a decent lay interpretation?
- throwaway080383 8y agoIt's not accurate. Looking at the arxiv link (https://arxiv.org/abs/1702.02325 https://arxiv.org/abs/1702.02325), the idea is you fix some elliptic curve E, and you look at its "twists" E^d, where the parameter d is a nonzero integer. If you only look at twists where |d| < N, you can ask "What proportion of these are rank zero, rank one, rank two, ...?" The Theorem in the paper is that as you let N go off to infinity, these proportions tend to 1/2, 1/2, 0, 0, 0, ... respectively. Here's the thing: this does not correspond to a measure on the set of rational elliptic curves, and indeed, there is no reasonable way to define a uniform probability measure on a countable set. Consequently, statements like "half of all elliptic curves..." are kind of misleading and meaningless.
- throwaway080383 8y agoTo back up my last claim, let me prove that 100% of positive integers are even in the same spirit: Fix any odd positive integer x, and consider its "twists" x{d}, which for positive d I define to be x*2^d. Every integer is of the form x{d} for a unique choice of x and d. Now, for fixed N, if I consider all "twists" for which d<N, the proportion which are even is (N-1)/N. Thus, as N tends to infinity, the proportion tends to 1.
- rocqua 8y agoHow does a twist x{d} being even equate to the integer d being even?
- throwaway080383 8y agoThe twist x{d} is even as soon as d>0.
- rocqua 8y agoYes, but that doesn't mean the integer d is even.
- 8y ago
- nabla9 8y agoProbability theory is based on measure theory. You can think measure as abstraction of the size of a set. As others point out in different ways, probability 1 and zero-probability don't mean exactly what you think they mean. Simply put: it’s possible to have a non-empty set with zero "size".
- Sharlin 8y agoYes, it’s inaccurate. The rigorous way to say it would be “Almost all ecliptic curves are one of two types” but to a layman not familiar with the technical definition of almost all it may not seem particularly impressive.
- throwaway080383 8y agoEven the phrase almost all is a bit misleading here, since there is no canonical choice of measure on the set of elliptic curves in question.
- monster_group 8y agoThe article also explains why it is OK to say that elliptic curves come only in two types. Read the two paragraphs starting from "You may be wondering how it’s possible...". It seems like the curves of rank 0 and 1 are infinite and curves of rank greater than 1 are also infinite. But infinities are strange. The infinity of curves with rank 0 and 1 is much much greater than those of rank > 1 so essentially the latter can be ignored. That's what I understood.
- vladislav 8y agoThese statements are consistent modulo zero measure (probability) sets. For instance: there are infinitely many rational numbers, but the probability of picking a rational number at random from the interval [0,1] is zero.
- hota_mazi 8y agoYes, because of the way infinities work. Even though the quantities of ranks 0/1/2+ are infinities (countable or not), there are vastly more 0/1 than any other, so that mathematically, it's accurate to say that the odds of finding 2+ are zero. But it's not really zero. Infinity is confusing :)
- max_likelihood 8y agoI agree it's confusing. If you consider a random variable X whose value is the rank of a randomly chosen elliptic curve. Then I believe this article is saying X converges in probability [1] to 50%/50%. Similarly related is the concept of "Almost Surely" [2]. Wikipedia includes a great example with throwing a dart. [1] : https://en.wikipedia.org/wiki/Convergence_of_random_variables#Convergence_in_probability https://en.wikipedia.org/wiki/Convergence_of_random_variable... [2] : https://en.wikipedia.org/wiki/Almost_surely#Throwing_a_dart https://en.wikipedia.org/wiki/Almost_surely#Throwing_a_dart
- bsder 8y agoInfinity has odd properties and there are multiple infinities. See: The Aleph and Beth infinity series. For example, the natural numbers are infinite. They are aleph-null. The ordinal numbers are also infinite. They are aleph-one. Even thought both are infinite, aleph-one has more elements than aleph-null. This is the realm of set theory and started with Georg Cantor.
- ars 8y agoIt's essentially aleph 0 / aleph 1. Aleph 1 is so much larger than aleph 0 that this equation "equals" zero. The real problem here is that this equation doesn't really make any sense: You can't divide infinity and expect answers that make sense in ordinary language. So you can have an infinite number of curves of rank > 1, and yet there is 0% chance you'll come across one by accident (since you'll need to sample an infinite number of rank <= 1 curves in order to do that - but of course, you can't, there's an infinite number of them).
- vostok 8y agoThey're not really dividing aleph_0 by aleph_1. The important thing is that it has measure 0, but you can easily have a set with cardinality aleph_1 and measure 0. There are also no ironic quotes around "equals". The measure really is 0.