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December 6

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Extending Shoenfield's absoluteness theorem about the Analytical Hierachy

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Are there axioms that can be added to ZF to get a system, call it ZF+, that will be weaker than ZFC, but satisfy a stronger version of the absoluteness theorem, with regards to removing choice from proofs, in the sense indicated in the title? More generally, since we can do the same for ZF + ◊, and anything else that holds in the L, can we just as easily replace choice with something else; or say something about how these middle systems' non-ZF axioms relate to how far up the AH the result goes? Final question: something like the above, but systems between ZF and ZF + V=L; and, in that case, would the degree up the hierarchy be related to "how close" the system was to V = L? (I am having a slightly stupid day and feel like this question may not make perfect sense, please excuse me if I am saying something confused).Phoenixia1177 (talk) 08:04, 6 December 2014 (UTC)[reply]

Philosophical Objections to Choice and V = L

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For those that reject/dislike choice on philosophical grounds, is there anything compelling about choice holding in L? I realize that L isn't "constructive", but for a certain type of dislike of choice, it seems that this would be persuasive on some level. Then again, I don't, to be perfectly honest, really "get" objections to choice that fall under this nature - intellectually, yes; but philosophically, no.Phoenixia1177 (talk) 08:20, 6 December 2014 (UTC)[reply]

why can't pi be an integer?

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I think it would be quite convenient for pi to be an integer (like 3). Is there any reason that this obviously would not have been coincidentally possible? 212.96.61.236 (talk) 14:29, 6 December 2014 (UTC)[reply]

Your question has either of two different meanings. First, you may be asking whether it is possible that pi, contrary to modern mathematical thinking, actually is an integer, such as 3. The answer is no. The value of pi has been known since ancient times not to be an integer. Archimedes spent much of his career improving the bounds on pi, for instance. In modern times it has been calculated to at least a million decimal places. Pi is not an integer, because it is approximately 3.1415927.... Second, you may be asking why pi isn't an integer, which would have been convenient. The answer is basically: That isn't the way it is. I would guess that there might be other inconveniences in a different formulation of mathematics, but that is a guess. I won't elaborate in why the facts of mathematics have to be as they are, but I think that there is general agreement that some of them have to be as they are.
Is either of those the question that you are asking, or are you asking something else? Robert McClenon (talk) 15:51, 6 December 2014 (UTC)[reply]
It has actually been an integer, namely 3, in at least one of the american states. At least according to state law. Source: Guinness book of records (old edition), don't remember details. YohanN7 (talk) 15:53, 6 December 2014 (UTC)[reply]
I also think the question isn't is clear enough. Is there a reason that pi isn't an integer (except it isn't)? YohanN7 (talk) 15:57, 6 December 2014 (UTC)[reply]
You're probably thinking about either the Indiana Pi Bill or the Alabama Pi hoax. If the latter, it is pure fiction. If the former, there are several errors in your account:
  • It was a bill put before Indiana legislators; it was never approved to achieve the status of law.
  • It didn't specify a value for Pi, but values for Pi were implied by its content.
  • More than one value was implied; Our article mentions 3.2, I'm not sure if 3 was one of the values implied.
-- Meni Rosenfeld (talk) 16:25, 6 December 2014 (UTC)[reply]
Thank you Meni! I'll check my sources better the next time. YohanN7 (talk) 17:13, 6 December 2014 (UTC)[reply]
There are non-trivial philosophical issues there, though. What qualifies as a mathematical reason? If we simply restate an existing proof that pi isn't an integer, does that suffice? This gets into issues like the relationship between mathematical implication and causality.--80.109.80.31 (talk) 16:38, 6 December 2014 (UTC)[reply]
Perhaps this is an answer: pi isn't an integer because we've chosen the euclidean metric. To forestall an annoying pedantic quibble: "pi" in my responses should be read as "the ratio of a circle's circumference to its diameter in the plane".--80.109.80.31 (talk) 16:47, 6 December 2014 (UTC)[reply]
Pi is defined by the Euclidean circle, even if it might appear in other geometries. Other ratios simply aren't called pi.--Jasper Deng (talk) 17:39, 6 December 2014 (UTC)[reply]


Guys, I mean that you have to calculate pi to know what it is (or you have to try to measure it.) When you start your calculation you don't know what you'll get. I know that for thousands of years we've gotten something like 3.14... My question is a priori, when you start calculating, how could you already know it won't be a whole number? What I'm asking is very similar to the question of the irrationality of pi. (Since all integers are rational, in fact the irrationality of pi is a fine answer to my question. These are proofs that pi is irrational and by extension not an integer. But my question is simpler than proving that it's not rational: give me a sentence or two that, without having to calculate pi, intuitively proves that it doesn't make sense for it to be an integer. (i.e. give me an intuitive, weaker version of the proof that it's irrational - produce an easy intuitive proof that it isn't integral.) 212.96.61.236 (talk) 17:45, 6 December 2014 (UTC)[reply]

Proof without words that 3 < π < 4.
Well, you could try a simple geometric argument, using a circle with an inscribed hexagon and circumscribed square, and the definition of π in terms of c = 2πr. If you accept the visibly obvious strict ordering of the lengths of the respective circumferences, you are left with 3 < π < 4, which means that it cannot be an integer. It is not clear whether this is what you're seeking, though. —Quondum 17:56, 6 December 2014 (UTC)[reply]
yes, sure, that will do it. If you had never tried it though couldn't you 'imagine' that in fact the lengths of the respective circumferences would coincide? If so it's not a good intuitive a priori argument, you really have to know how the world happens to work. I mean couldnt' they 'happen to' coincide, if pi 'happened to' be 3. (I know this can't be the case due to the proofs that it is irrational. but I'm looking for something simpler). Can you try to produce something that is obvious even if somebody doesn't try it? 212.96.61.236 (talk) 18:23, 6 December 2014 (UTC)[reply]
Here is a diagram of Quondum's suggestion. Is that not simple enough? Then again, if Igor Teper can imagine an integer between 3 and 4, perhaps you can as well. -- ToE 17:32, 8 December 2014 (UTC)[reply]
pi is defined as the least positive real number such that cos(x)=-1. For non-zero integer values of x, cos(x) is irrational. So pi cannot be an integer. Sławomir Biały (talk) 19:11, 6 December 2014 (UTC)[reply]
(ec) This is highly dependent on the questions "What would be intuitive?" or "What could you imagine?", which unfortunately have no clear answer. It also depends strongly on what definition of pi you use; in a modern context, in many mathematical fields its definition is unrelated to geometry other than historically. —Quondum 19:34, 6 December 2014 (UTC)[reply]

Compressed and encrypted data

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When uploading encrypted files you always get a bit of an existential crisis: Is it really encrypted? So I thought to create an automatic check, but to my dismay the standard randomness test, chi-square test, at least as implemented by ent, can't tell compressed data from encrypted with a good enough reliability. I ran ent against a bunch of files and their encrypted versions, and a good number of compressed files came out as more random than encrypted files (the offending cases are in bold):

As you can see xz compression is pretty indistinguishable from encrypted data. Limiting the test to the first 32 bytes of each file, where header data would reside, clearly separates the .xz files:

If combined with the values for full files, this is great for this limited set of file formats, considering for a chi-square distribution of 255 degrees of freedom (256 values in a byte) a value of more than 500 occurs with a probability around 10^-17. But if you don't include any special knowledge about where the non-random data is, say if you check every 32 byte block, at %50 false negative probability you could only upload roughly one exabyte (about a million raw hard disk images), which is less than stellar for whoever gets hit by that.

Now I'm sure if you play with the details you could might gain a few orders of magnitude (e.g. you get 6 if you raise the threshold to 550) but there could also be plenty of hiding weaknesses (e.g. I haven't even bothered to figure out a false positive rate (not that I'm looking for a silver bullet)). What I'd really like to ask is if you have any ideas that would decisively improve on this. Do you know a test other than chi-square that would perform better, or can you think of some modification to the chi-square scheme? One idea to square the false negative rate is to generate a somehow related-in-randomness-yet-independent-in-chi-square block if a block fails, and then also test the related block, but then I've got no idea if a thing like that exists. --Swedmann (talk) 20:25, 6 December 2014 (UTC)[reply]

Since the statistics of perfect compression and perfect encryption are both by definition indistinguishable from random data, they will clearly be indistinguishable. High-quality encryption is already indistinguishable from random data (aside, perhaps for an encapsulating protocol), and compression is approaching it, especially for source data of well-characterized statistics. I'd suggest that you should not be looking at a simple statistical measure, since this is going to be unreliable, aside from the fact that it would be wide open to being deliberately fooled. A better approach might be to simply parse and recognize all known file formats. —Quondum 20:58, 6 December 2014 (UTC)[reply]
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