Wikipedia:Reference desk/Archives/Mathematics/2012 July 30

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July 30

suppose a man produces a proof P that statement Q cannot be proven under Some Axioms but definitely has a certain Truth Value (unknown) under those axioms.

A man produces a proof, P, that under a certain set of axioms, TCA (the cool axioms), there is a statement, Q, that cannot be proven under TCA, but definitely has a concrete truth value (unknown) under those axioms and so assuming it is independent would certainly be an error as it would lead anyone to DEFINITELY produce an inconsistent system wherenever the wrong value was chosen for the "independent" axiom - it's not independent at all. 84.3.160.86 (talk) 01:38, 30 July 2012 (UTC)

the question here is, some people are saying this isn't possible. why not? why couldn't I prove that it is definitely unprovable under standard axioms whether there is an odd perfect number, but that this also DEFINITELY has a truth-value that no one can prove under these axioms? Why would provability be conflated with truth? I find it quite easy to imagine that an axiomatic system has a truth that cannot be proven; so assuming its opposite would be false and inconsistent. 84.3.160.86 (talk) 01:39, 30 July 2012 (UTC)

You're right that provability is not the same as truth. What's confusing, then, is talking about truth "under a set of axioms". Axioms don't determine truth, only provability (at least in the ordinary first-order logic paradigm).

It is quite possible for a statement to have a definite truth value in spite of being independent of some given set of axioms. Which statements you think those are is likely to depend on your philosophy of mathematics.

For example, ZFC neither proves nor refutes Con(ZFC), the statement that ZFC is consistent (unless ZFC is actually inconsistent, in which case ZFC both proves and refutes Con(ZFC)). The large majority of mathematicians would accept Con(ZFC) as a statement with a definite truth value. (Whether it's an "unknown" truth value is another matter and depends on what you mean by "known".)

I personally would go much further. I would assert, for example, that the continuum hypothesis is a statement with a definite, but still unknown, truth value. --Trovatore (talk) 01:52, 30 July 2012 (UTC)

You say that axioms don't determine truth, but how can they be? Any axioms that determine Pi do determine a billionth, trillionth, and so on digit. If it can't be proven whether Pi switches at a certain point to being the digits of e and then continues thereon like that, doesn't mean that this isn't actually true or actually false for actual pi: what does provability have to do with truth? Of course there are truths beyond provabilities - why would you say axioms don't have truth when they clearly do, under that system of assumptions some things are true and others false. 84.3.160.86 (talk) 02:13, 30 July 2012 (UTC)

I didn't say axioms don't have truth. I said they don't determine truth. From the mathematical realist point of view, mathematical objects are real things, existing without reference to our reasoning about them. We choose axioms that we believe to be true about those objects, and then use them to prove other statements that must also be true, assuming we didn't make an error in the choice of axioms in the first place. So mathematical truth comes before the axioms, not after. --Trovatore (talk) 02:29, 30 July 2012 (UTC)

It's certainly possible -- that's just what Gödel's theorem does. Gödel proved that in any axiom system that meets certain basic conditions, it is possible to construct a statement that is (a) true, and (b) impossible to prove within that axiom system. Looie496 (talk) 02:05, 30 July 2012 (UTC)

Yes, but this "true" is in an absolute sense; it is not relative to the axioms. So that would be my quibble with the way the OP put it. --Trovatore (talk) 02:07, 30 July 2012 (UTC)

You miss that I require that it be provable that it's unprovable, from within the axiomatic system. You don't add that - we are talking about the proof of not some general "must-be-some-true-statement-can't-be-proven" we are talking about "I just proved that THIS statement is unprovable and that (THIS statement xor OPPOSITE THIS STATEMENT) is true." Okay, I just felt like a complete idiot writing that out. Obviously (THIS statement xor OPPOSITE THIS STATEMENT) is always true. What I mean is that one of them is definitely false and it's not independent, you can't just assume one of them. 84.3.160.86 (talk) 02:13, 30 July 2012 (UTC)

In other words: I just made a proof that THIS statement is unprovable but that (either assuming THIS statement produces an inconsistent new set xor assuming OPPOSITE THIS statement produces an inconsistent new set): in other words, IT'S NOT INDEPENDENT AT ALL, but dependent in some unprovable way on the other axioms. 84.3.160.86 (talk) 02:16, 30 July 2012 (UTC)

You keep equating "inconsistent" with "doesn't describe intuitive reality", but they're very different concepts. If TCA + X is inconsistent, then TCA proves "not X". This is the basic proof by contradiction of classical logic. So in the above you are supposing a contradiction; it's no wonder strange consequences are following.--119.225.45.242 (talk) 02:21, 30 July 2012 (UTC)

Okay, you've gotten at the crux here. You're saying "If [you can prove] TCA + X is inconsistent, then TCA proves "not X"". But my whole point is you CAN'T prove TCA + X is inconsistent, and you CAN'T prove TCA + Not X is inconsistent...you can PROVE that you CAN'T prove these things. But, at the same time, you can also prove that ONE of them must be false just under your previous axioms. (They can't both be false.) 84.3.160.86 (talk) 08:46, 30 July 2012 (UTC)

A trivial example: Some of the numbers 2,3,4,5,6,7,8,9,... are composite (such as 2×2=4 and 2×3=6). The others are prime. Now let's make this true statement an axiom: "Even numbers beyond 2 are composite". This axiom is insufficient. It cannot prove that 9 is composite even if it is true that 9=3×3 is composite. Nor can it prove the negation, that 9 is prime. It is the rule rather than the exception that axioms are insufficient. Bo Jacoby (talk) 07:30, 30 July 2012 (UTC).

I'm trying to follow your argument. You give an example of a true statement being taken as an axiom; but my question is about a false statement that you can't prove. Can you take it or its opposite as an axiom, even though it is false under the rest of the axioms in an unprovable way? Indeed is that possible? FOr example, in a system that is too weak to prove that 7 is prime, but strong enough to prove under itself that it SURELY CAN'T prove that, under such a system it would seem that "7 is prime" is independent. But in fact we know that it is NOT prime. So it has a definite truth-value. Here, let me give you a case that is apparently as simple as that; http://wiki.riteme.site/wiki/Transcendental_number#Numbers_which_may_or_may_not_be_transcendental

Catalan's_constant is a number we can 'count to'. Suppose that we prove that we can't prove whether it's irrational. That doesn't mean it's ACTUALLY not rational!! So, people here are saying hte moment you prove you can't prove whether Catalan's constant is irrational, you can take "Catalan's constant is irrational" as an axiom and go about your way as it's completely independent. But in fact that doesn't change that, there it is, perhaps it is rational and has a very nice integer numerator and very nice integer denominator that are simply outside of our system's ability to prove. So, we have an unbeknonst-truethness, that it's rational, coupled with a proof that we can't prove it way way or another. It's not really independent; it's just unprovable. If you pick the wrong side as an axiom you've created an unbeknownst-false-system. (Since some of your axioms imply something that you take the opposite of as a new axiom.). Do you see what I'm getting at? Proving that Catalan's constant can't be proved irrational or rational doesn't mean that it definitely ISN'T one of these. Isn't there a problem if you assume something that is false under the rest of your arithmetic, merely because you've proven you can't prove it false or true? 84.3.160.86 (talk) 09:08, 30 July 2012 (UTC)

specific example

For a long time a proof was sought for Euclidean parallel postulate, until it was proved independent. In other words, if Euclidean parallel postulate was "x", then it was proven that neither TCA + X nor TCA + Not X was inconsistent.

Now imagine that if instead of this proof, for a similar postulate this was proved: "1. You cannot prove TCA + X is consistent or inconsistent and you cannot prove TCA + Not X is consistent or inconsistent, under TCA. 2. Secondly, through some other means you prove that one of them is in fact (Unprovably) inconsistent. [but obviously non-constructively; you can't prove which, just that one of the two are inconsistent. obviously if you could prove which it would contradict (1).]"

Obviously if #2 specified WHICH one was inconsistent, it would be tantamount to a proof. BUt without specifying WHICH one is inconsistent, just that ONE of them is (unprovably), in an unconstructive way that doesn't tell you which one is false. In other words: it's dependent but unprovable how. 84.3.160.86 (talk) 08:53, 30 July 2012 (UTC)

So that is the mistake in the above poster: ". If TCA + X is inconsistent, then TCA proves "not X". This is the basic proof by contradiction of classical logic." -- I didn't say you prove that TCA + X is incosistent; only that you prove that either TCA + X is inconsistent or TCA + Not X is inconsistent, and, therefore, it's not indendependent. However the proof is not a constructive one: you are only able to prove that there is a truth-value inside the system. You can't prove that truth-value; and in fact you proceed to prove that you can't prove it. So, please explain to me why this state of affairs is not possible. 84.3.160.86 (talk) 08:52, 30 July 2012 (UTC)

Standard proof for identity:

Is there a standard proof for this identity?: ${\sqrt {a^{2}+b^{2}}}\leq |a|+|b|\leq {\sqrt {2}}{\sqrt {a^{2}+b^{2}}}$. Intuitively it makes sense and you can prove it using polar coordinates if you assume some trigonometric identities, but there is probably a better way. --130.56.93.90 (talk) 04:29, 30 July 2012 (UTC)

1. Show all three expressions are non-negative.
2. As they are nonnegative, you can square them and inequalities will hold, so you'll get $a^{2}+b^{2}\leq \left(|a|+|b|\right)^{2}\leq 2(a^{2}+b^{2})$.
3. HTH --CiaPan (talk) 05:26, 30 July 2012 (UTC)

To write out a formal proof, you need to start from an identity and derive the given expression (although informal exploration often proceeds in the opposite direction). There are two parts to your given expression, so your proof comes in two parts:

1. Start from $(|a|+|b|)^{2}=a^{2}+2|a||b|+b^{2}\,$ and derive ${\sqrt {a^{2}+b^{2}}}\leq |a|+|b|\,$.
2. Start from $(|a|-|b|)^{2}\geq 0\,$ and derive $|a|+|b|\leq {\sqrt {2}}{\sqrt {a^{2}+b^{2}}}\,$. Gandalf61 (talk) 06:49, 30 July 2012 (UTC)

If you have a series of statements connected by "iff"'s, then it doesn't matter which direction you write them in. They are logically equivalent. --Tango (talk) 20:33, 31 July 2012 (UTC)

we just proved we can't prove it one way or another or know which one it is, but this statement is definitely false or true - not independent.

what's wrong with this statement? It looks fine to me. — Preceding unsigned comment added by 84.3.160.86 (talk) 09:35, 30 July 2012 (UTC)

What's wrong with it is that "independent" does not mean "not false or true". Rather, "independent" means "not provable nor refutable". You seem to have picked up a distorted impression of what "independence" means — not hard to do; the popularizers make this sort of mistake all the time. --Trovatore (talk) 10:29, 30 July 2012 (UTC)

Could you be more explicit. So, it can be independent while being false not in a physical sense or in our world, but in the world of a few axioms which are enough to "make" it true? So, for example, in some very simplistic system that is barely able to count 1, 2, 3, 4, 5, it can be "independent" whether there is another integer between 1 and 2 - even though, since it successfully enumerated the integers, it has produced a world in which that statement is very much physically false. So, we could add an axiom to that system that assumes an integer between 1 and 2, even though it is physically false EVEN within that simplistic system? 84.3.160.86 (talk) 11:03, 30 July 2012 (UTC)

I also find it very interesting that we can do a heuristic/statistical analysis on, e.g. the GOldbach conjecture, "Every even integer greater than two 2 can be expressed as the sum of two primes". Suppose someone proves it independent. You're saying it could still be true, and that the system where it's opposite is taken as an independent axiom is simply false mathematics - it assumes something that is false according to actual reality as established by the rest of the axioms? 84.3.160.86 (talk) 11:03, 30 July 2012 (UTC)

There's some interesting discussion of what it would mean for the Goldbach conjecture to be undecidable here. AndrewWTaylor (talk) 16:59, 30 July 2012 (UTC)

Trovatore: okay, what if I emend the statement to: "{we just proved we can't prove it one way or another or know which one it is, but this statement is definitely false or true - but our proof of this fact is not constructive and we don't know which." Is there anything wrong with it then? What if I were to add... "so INSOFAR as it is independent (not saying it is), there is a right/true and a wrong/false value for it as a new axiom. We just don't know which is which, but we have the non-constructive proof that it definitely has a specific truth-value we can't prove." — Preceding unsigned comment added by 84.3.160.86 (talk) 11:07, 30 July 2012 (UTC)

example of an axiom that is statistically probably false

have there been examples of an axiom for which the previously accepted axioms were eventually sufficient to give a high statistical/heuristic likelihood that it had a particular truth value without defining it as such by taking it axiomatically? — Preceding unsigned comment added by 84.3.160.86 (talk) 11:17, 30 July 2012 (UTC)

popular description of result of independence

in the popularization of mathematics, it sounds like when a mathemtician proves - usually surprisingly - that a statement is 'independent', it is taken to mean that it could be true or false without leading you to prove anything false by assuming either one. But that isn't right then is it - you could very much prove wrong results, using independent axioms, if in fact you picked an axiom that happened to actually be false, right? In other words: INDEPENDENCE guarantees your proofs won't be inconsistent: not that you won't prove something that is false. Do I have it right? 84.3.160.86 (talk) 14:09, 30 July 2012 (UTC)

Indeed, is it possible that out of the 5 axioms we hold most dear, some of them are actually false in every logical world where the others are true? But it simply can't be proven as it's independent, and in fact even though false (more specifically, you can prove false things), the story remains "consistent". 84.3.160.86 (talk) 14:09, 30 July 2012 (UTC)

The Axiom of Choice is false. Count Iblis (talk) 17:21, 30 July 2012 (UTC)

Can you also specify the relationship between this fact and its independence, and how you surmise that it is false? Also is it false in a physical sense, i.e. "our" geometry or the "best" world-approximating axioms we might pick, or false on some deeper level and probably notwithstanding a little fidging with what particular axioms we're working with? How do you know, or think you know, about it's falseness? This gets to the center of my questions above and perhaps I should have just begun with this example. 84.3.160.86 (talk) 17:42, 30 July 2012 (UTC)

See this article. Count Iblis (talk) 18:05, 30 July 2012 (UTC)

I think the question I have to ask in all of these is "independent of what?" - independence is not a term that makes sense of a statement on its own. Straightontillmorning (talk) 18:44, 30 July 2012 (UTC)

zfc or other normal axiomatic systems that are strong enough to prove many questions about the same things that are of interest to us. 84.3.160.86 (talk) 20:27, 30 July 2012 (UTC)

The thing is, a lot of things that are independent of ZFC are not independent of natural strengthenings of ZFC. The large-cardinal axioms provide a natural hierarchy of such strengthenings. You could potentially identify a category of statements that are independent of ZFC plus all known large-cardinal axioms (for example, the continuum hypothesis), but it's still a little artificial; there is no guarantee that all future natural strengthenings will behave the same way. --Trovatore (talk) 23:51, 31 July 2012 (UTC)

sigh. "a lot of things that are independent of ZFC are not independent of natural strengthenings of ZFC". Is that because they're already true under ZFC as well, so it's natural to strengthen it so they can be proven instead of just sitting there all true? 84.3.160.86 (talk) 12:47, 1 August 2012 (UTC)

"True under ZFC" is a category error. ZFC determines provability and refutability, not truth and falsity. The relevant concept is "true in the intended interpretation of ZFC", that intended interpretation being the von Neumann universe.

Note that any statement expressible in the language of set theory is either true or false in the intended interpretation. --Trovatore (talk) 16:34, 1 August 2012 (UTC)

The axiom of choice, of course, is true. It is self-evident; anyone who understands the intended interpretation (and agrees that the objects of that interpretation actually exist) is almost forced to agree that it is true. I take it that Iblis's point is, he doesn't agree that those things actually exist (I know that's what Zeilberg thinks). They're wrong, of course, but this is discovered by a more a posteriori process. --Trovatore (talk) 23:39, 31 July 2012 (UTC)

This is exactly what attracts me in mathematics, and I think this is the true heart of math. People arguing by bold statement over what we should definitely believe a priori. The rest just follows. 84.3.160.86 (talk) 12:50, 1 August 2012 (UTC)

Indeed. :-) One bold statement is worth another. There are only two philosophies of mathematics (or anything else): The pro-mathematics philosophy & the anti-mathematics philosophy. The second usually consists of some complicated, obscure way of saying: You aren't allowed to do, think or say that, because me & my "-ism" (Teddy-bear) say so.

Zeilberger's point of view, exulting that the true, ultimate, Teddy Bear, "the big computer in the sky", "the digital Messiah" "has arrived" is a bit odd for physicists. "Continuous analysis and geometry [which poor, ignorant physicists & geometers have been using for quite some time] are just degenerate approximations to the discrete world, made necessary by the very limited resources of the human intellect." Teddy says the human notion of continua is BAD. Digital messiah Teddy is in the sky, his indisputable truths untainted by & beyond the limited human intellect.

Is a point of view, a notion of meaningfullness that a priori characterizes Andrew Wiles's `proof' of FLT, as (a) "a crowning human achievement," but also (b) "alleged" and "not rigorous, since it uses continuous analysis, which is meaningless." - a crowning human achievement?

Also is it false in a physical sense, i.e. "our" geometry or the "best" world-approximating axioms we might pick, or false on some deeper level and probably notwithstanding a little fidging with what particular axioms we're working with? Felix Hausdorff's Hausdorff paradox, developed into the better known Banach-Tarski Paradox, depends on (weak versions of) the Axiom of Choice. The fact that we don't see such phenomena in physics, that we don't seem to be able to cut balls into pieces & rearrange them into a bigger ball, argues against fundamental usefulness of set theory with the axiom of choice & the naive application to the physical world, to "our" geometry. So is ZFC or its intended model(s) really the ultimate Teddy either? :-) John Z (talk) 09:49, 4 August 2012 (UTC)

Well, I don't think that's a very strong argument, really. The axiom of choice certainly implies that there are sets of points in R^3 that don't correspond to the space occupied by physical objects, but you don't actually need AC for that. Basically any useful notion of set is going to have that property. Physical objects are too limited to realize all the sets that we can see must exist. Mathematics can describe physical reality but is not limited to it, and the imposition of such a limitation is artificial and reduces mathematics' power. --Trovatore (talk) 16:40, 4 August 2012 (UTC)

I was mainly answering the "false in a physical sense" question I italicized above, which I think is a very good question. Hausdorff-Banach-Tarski phenomena are utterly unlike anything that has ever had physical use AFAIK. Asserting that "we can see must exist" is an even weaker argument - especially if not everyone can see this, agrees on this. Does Zeilberger? Why "must" rather than "can"? If one can concoct theories that don't require such "must exists", or even contradict these "must exists" & these theories are more applicable to physics, crystallize "intuition" better, why must ZFC be the One True Teddy? Of course we should not artificially limit mathematics to that which can or currently does describe physical reality, but the point is that we should not limit mathematics' power to exclude mathematics that can describe physical reality either, by making weak assertions that "we can see must exist" to support AxC either, or insisting that ' "our" geometry / physics / physical mathematics' must somehow be founded on the merely-asserted-to-be more fundamental ZFC. Not saying that ZFC is not a good Teddy, but why must thoughts on existence be based on Teddies, on dogmatic -isms? I was criticizing the whole modern project of the last century odd of thinking of philosophy as the search for the One True Teddy -ism. To paraphrase Laozi, the Teddy that can be clutched that way is not The Teddy. That's my Teddy, anyways. (To paraphrase Friedrich Engels).John Z (talk) 00:13, 5 August 2012 (UTC)

AC does not in any way limit mathematics' power by excluding mathematics that describes physical reality. The world view that makes AC a clear intuitive consequence is more inclusive than any that does not, and includes the ones that do not as a special case. --Trovatore (talk) 07:52, 7 August 2012 (UTC)

Integral proof

$\int _{0}^{\infty }{\frac {\ln(x)}{1+e^{x}}}dx={\frac {-1}{2}}\ln ^{2}(2)$

This is according to Wolfram Alpha, but no closed form of the indefinite integral is given. How can I derive this? 149.169.218.127 (talk) 18:01, 30 July 2012 (UTC)

Looks pretty nasty but it is obviously a contour integration problem. I believe you'll need a special 'trick' to generate logs by substituting x^t for the log, solving that general problem and then differentiate by t to generate the log x. On a quick try there seems a couple of other problems but I believe that should get the answer somehow. Dmcq (talk) 18:55, 30 July 2012 (UTC)

This is number 4.354 in Gradshteyn-Ryzhik, a table of integrals that always gives references to places where proofs can be found. They give the reference GW (324) (86a), but I can't tell you what that means from the Google Books snippet. See here for the book. —Kusma (t·c) 19:14, 30 July 2012 (UTC)

Some more searching gives this book as a source. It probably contains a proof. —Kusma (t·c) 19:26, 30 July 2012 (UTC)

Kusma, looking for the answer in books is cheating :(

Dmcq's suggestion seems to work, but you need to do additional work here as you need to consider the limit of the derivative of (1-1/2^s)Zeta(s+1) Gamma(s+1) for s to 0. Count Iblis (talk) 19:34, 30 July 2012 (UTC)

$\int _{0}^{\infty }{\frac {\log(x)}{1+e^{x}}}dx=-{\frac {(\log 2)^{2}}{2}}$. (clarifying $\ln ^{2}(2)$). Bo Jacoby (talk) 07:39, 31 July 2012 (UTC).

Made all the "ln"s in this thread operators for consistency. OCD and all... --Kinu ^t/_c 09:24, 31 July 2012 (UTC)

Um yeah that's right Count Iblis but I'm afraid I funked out after that and did a google search of 'contour integral logarithm exponential' and the second entry after the wikipedia one of contour integrals was [1] which is this exact problem with a complete solution! The main delay was typing the query. I'm getting seriously scared that google is both able to read my mind and do maths far better than me ;-) The thing is contour integrals weren't used directly but I wouldn't have found that discussion if I hadn't stuck it in the query. Dmcq (talk) 09:57, 31 July 2012 (UTC)