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Most of the systems mathematicians are interested in are consistent, which means, by Gödel's incompleteness theorems, that there must be unprovable statements.

I've seen a simple natural language statement here and elsewhere that's supposed to illustrate this: "I am not a provable statement." which leads to a paradox if false and logical disconnect if true (i.e. logic doesn't work to prove it by definition). Like this answer explains: https://math.stackexchange.com/a/453764/197692.

The natural language statement is simple enough for people to get why there's a problem here. But Gödel's incompleteness theorems show that similar statements exist within mathematical systems.

My question then is, are there a simple unprovable statements, that would seem intuitively true to the layperson, or is intuitively unprovable, to illustrate the same concept in, say, integer arithmetic or algebra?

My understanding is that the continuum hypothesis is an example of an unprovable statement in Zermelo-Fraenkel set theory, but that's not really simple or intuitive.

Can someone give a good example you can point to and say "That's what Gödel's incompleteness theorems are talking about"? Or is this just something that is fundamentally hard to show mathematically?

Update: There are some fantastic answers here that are certainly accessible. It will be difficult to pick a "right" one.

Originally I was hoping for something a high school student could understand, without having to explain axiomatic set theory, or Peano Arithmetic, or countable versus uncountable, or non-euclidean geometry. But the impression I am getting is that in a sufficiently well developed mathematical system, mathematician's have plumbed the depths of it to the point where potentially unprovable statements either remain as conjecture and are therefore hard to grasp by nature (because very smart people are stumped by them), or, once shown to be unprovable, become axiomatic in some new system or branch of systems.

Michael Harris
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11 Answers11

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Here's a nice example that I think is easier to understand than the usual examples of Goodstein's theorem, Paris-Harrington, etc. Take a countably infinite paint box; this means that it has one color of paint for each positive integer; we can therefore call the colors $C_1, C_2, $ and so on. Take the set of real numbers, and imagine that each real number is painted with one of the colors of paint.

Now ask the question: Are there four real numbers $a,b,c,d$, all painted the same color, and not all zero, such that $$a+b=c+d?$$

It seems reasonable to imagine that the answer depends on how exactly the numbers have been colored. For example, if you were to color every real number with color $C_1$, then obviously there are $a,b,c,d$ satisfying the two desiderata. But one can at least entertain the possibility that if the real numbers were colored in a sufficiently complicated way, there would not be four numbers of the same color with $a+b=c+d$; perhaps a sufficiently clever painter could arrange that for any four numbers with $a+b=c+d$ there would always be at least one of a different color than the rest.

So now you can ask the question: Must such $a,b,c,d$ exist regardless of how cleverly the numbers are actually colored?

And the answer, proved by Erdős in 1943 is: yes, if and only if the continuum hypothesis is false, and is therefore independent of the usual foundational axioms for mathematics.

The result is mentioned in

Fox says that the result I described follows from a more general result of Erdős and Kakutani, that the continuum hypothesis is equivalent to there being a countable coloring of the reals such that each monochromatic subset is linearly independent over $\Bbb Q$, which is proved in:

A proof for the $a+b=c+d$ situation, originally proved by Erdős, is given in:

MJD
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    @Andres I hope you like this answer better, although I feel that it falls foul of the same criticism you leveled at the other one: it has nothing at all to do with Gödel's theorems. – MJD Dec 05 '14 at 02:35
  • Do you have a reference to this problem or Erdős's paper? 5 minutes of Googling didn't turn it up, but that may be because your (nice) presentation of it isn't the canonical one. – dfan Dec 05 '14 at 14:29
  • This is just really really nice. Thank you, Fox, Jacob and anyone else who contributed to it. –  Dec 05 '14 at 14:41
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    Wonderful. I really like this example. Although it seems like to really get this you have to understand some of the subtleties of Reals vs Integers and countable vs uncountable. That's probably straining the attention span of my audience (my family/friends) a bit. – Michael Harris Dec 05 '14 at 19:14
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    Thanks all for consistently correct spelling Erdős and Gödel. – Hagen von Eitzen Dec 06 '14 at 12:28
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    ^ It takes one person to start a good thing. The rest copy and paste. – Soham Chowdhury Dec 06 '14 at 18:55
  • Would a proof that no such 4-tuple $(a,b,c,d)$ exists amount to a proof of the continuum hypothesis? – Soham Chowdhury Dec 06 '14 at 18:56
  • @Sodham: Yes, except (1) it's the other way around, and (2) there is no such proof, because CH is known to be independent of ZFC. The best you can say is that if a particular theory is strong enough to prove CH, then that theory is strong enough to prove the existence of a coloring for which there is no monochromatic $a,b,c,d$, and vice versa. – MJD Dec 06 '14 at 21:33
  • Do you happen to know the answer to http://math.stackexchange.com/questions/1057206/a-question-about-infinities-and-pots-of-paint ? –  Dec 09 '14 at 11:26
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    I assume you mean "four distinct real numbers". – DanielV Nov 23 '16 at 13:47
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    Oh, to rule out $a+a=a+a$? I suppose there must be a condition that rules that out. If I find out I will make a correction. – MJD Nov 23 '16 at 14:01
  • A small question for my understanding: when we prove that some statement is unprovable within a particular axiomatic system AS, do we always have to introduce additional axioms not in AS? Otherwise, if you prove that "a statement S is true but not provable", haven't you just proven S? – mercury0114 Mar 24 '24 at 07:41
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Any statement which is not logically valid (read: always true) is unprovable. The statement $\exists x\exists y(x>y)$ is not provable from the theory of linear orders, since it is false in the singleton order. On the other hand, it is not disprovable since any other order type would satisfy it.

The statement $\exists x(x^2-2=0)$ is not provable from the axioms of the field, since $\Bbb Q$ thinks this is false, and $\Bbb C$ thinks it is true.

The statement "$G$ is an Abelian group" is not provable since given a group $G$ it can be Abelian and it could be non-Abelian.

The statement "$f\colon\Bbb{R\to R}$ is continuous/differentiable/continuously differentiable/smooth/analytic/a polynomial" and so on and so forth, are all unprovable, because just like that given an arbitrary function we don't know anything about it. Even if we know it is continuous we can't know if it is continuously differentiable, or smooth, or anything else. So these are all additional assumptions we have to make.

Of course, given a particular function, like $f(x)=e^x$ we can sit down and prove things about it, but the statement "$f$ is a continuous function" cannot be proved or disproved until further assumptions are added.

And that's the point that I am trying to make here. Every statement which cannot be always proved will be unprovable from some assumptions. But you ask for an intuitive statement, and that causes a problem.

The problem with "intuitive statement" is that the more you work in mathematics, the more your intuition is decomposed and reconstructed according to the topic you work with. The continuum hypothesis is perfectly intuitive and simple for me, it is true that understanding how it can be unprovable is difficult, but the statement itself is not very difficult once you cleared up the basic notions like cardinality and power sets.

Finally, let me just add that there are plenty of theories which are complete and consistent and we work with them. Some of them are even recursively enumerable. The incompleteness theorem gives us three conditions from which incompleteness follows, any two won't suffice. (1) Consistent, (2) Recursively enumerable, (3) Interprets arithmetics.

There are complete theories which satisfy the first two, and there are complete theories which are consistent and interpret arithmetics, and of course any inconsistent theory is complete.

Asaf Karagila
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    I'm not a fan of "Everything should be explainable to the layperson". If everything was explainable to the layperson, why do we have to spend so many years studying before we can understand something correctly, even at the intuitive level? Why is it that when someone explains to me the gist of something in my own field I know that there's a good chance that I have no idea what it really does? Thinking that a few examples make it clearer is wishful thinking. From both parties involved. – Asaf Karagila Dec 05 '14 at 08:19
  • I agree with that, but I do still think there is considerable value in trying to explain complex concepts in a manner accessible to laypeople - not just to the laypeople it is being explained to, but also to the experts doing the explaining. – David Z Dec 05 '14 at 08:49
  • David, I'm not disagreeing with the considerable value of this. If only because when done right, the layperson will walk away with a sense of wonder, awe and perhaps enough curiosity to start studying more seriously themselves. – Asaf Karagila Dec 05 '14 at 08:51
  • Yep; I was also targeting my comment at others who might come along and misinterpret yours. It seems to be a common misinterpretation. No offense intended! (in case there was any perceived) – David Z Dec 05 '14 at 08:53
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    @AsafKaragila, there is a considerable difference in explaining something to a layperson in such a way that they are satisfied and understand why we are doing it, and explaining it so that they understand it in the same rigor like us, or in your words, "correctly". The latter requires a maths degree. But the former is something I do on a regular basis to people without mathematical background, and I do quantum gravity and category theory. And it's not an 'inferior' or 'worse' way of explaining, it's just explaining less. – Turion Dec 05 '14 at 15:18
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    @Turion: I find that explaining in very broad strokes and skipping the details you end up either giving the wrong impression or lying through your nose. I do that too, of course, and I tell people that I'm currently misleading them and there is a lot more to it. But I often feel that the oversimplification is running against the ultimate cause, which is to get people interested more in whatever you're telling them. – Asaf Karagila Dec 05 '14 at 15:28
  • Right, if you tell them that this is a simplification, it's ok. I don't think oversimplifying puts off people as long as one doesn't create the impression "This guy must think I'm too stupid to understand the actual matter.". On the other hand, I noticed just today that including 'all the details' can be very off-putting. The ideal simplification, I find, is one that gives the audience the feeling of having understood some essential bits, while creating an impression of how much details are missing. – Turion Dec 05 '14 at 15:37
  • @Turion: I definitely agree that too many details are off putting. It takes an artist to balance between the simplification and the details in order to keep this at the appropriate level. – Asaf Karagila Dec 05 '14 at 15:59
  • @AsafKaragila and others. True, I agree that watering something down for the sake of a snappy, easy explanation can, and often does confuse things. At the same time, I like math, and I'm fascinated by it. I would like to share some of that fascination. My goal is not to elucidate Godel's incompleteness theorem to the layperson, (I probably don't understand it all that well myself), instead, I want to illustrate an interesting concept that could open the door to further interest. Like when you say something like "This statement is false." It gets people thinking in a new way about logic. – Michael Harris Dec 05 '14 at 19:29
  • @Michael: That's a reasonable cause. But I should tell you that the more I try to think about easily explainable examples of incompleteness, the less happy I get. I used to think that things like the example about $\sqrt2$ was a good one, but still in order to understand it you need to know about the completeness theorem, and about models, and proofs and whatnot. These are not things you can just explain over coffee that easily. These take more time, and more effort to understand. [...] – Asaf Karagila Dec 05 '14 at 19:37
  • @Michael: [...] And if you think about examples which don't require you to understand these things, for example Goodstein sequences, or whatever "nice" unprovable statements from $\sf PA$ which are true in $\Bbb N$, to understand those you still need to be able and understand the idea of a non-standard model, and all that shizzle. On top of all that, you run the risk of whomever is listening just telling you "But if it's true in $\Bbb N$, why do you care if it's not provable?" and shutting the door for the rest of the discussion. – Asaf Karagila Dec 05 '14 at 19:40
  • Having received the only downvote on the page, I wonder if there is something wrong with my answer, or with the person who posted it. – Asaf Karagila Dec 06 '14 at 07:58
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    I am starting to suspect there is something personal about these downvotes. – Asaf Karagila Dec 08 '14 at 07:02
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The one I find most intuitive, as an unprovable statement from ZF without Axiom of Choice, is that for any two sets X and Y, either there's an injective function from X to Y, or there's one from Y to X.

Roughly, and informally, I read this as: either X is at least as big as Y, or Y is at least as big as X.

I mean, what's the alternative? They're both bigger than each other?!

chiastic-security
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    There's a second alternative: neither $X$ nor $Y$ is bigger than the other! – murray Dec 07 '14 at 17:46
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    @murray At least as big.. if $X$ and $Y$ are "of the same size", then it is true that $X$ is at least as big as $Y$ and the viceversa is also true – Ant Dec 09 '14 at 16:44
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    Why is this statement unprovable? – Marco Flores Dec 27 '14 at 05:13
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    @MarcoFlores it's provably equivalent to the axiom of choice, which is provably independent of the axioms of ZF. – chiastic-security Dec 27 '14 at 08:09
  • But if it's equivalent to the Axiom of Choice, how can you say it's a true statement? Don't we have to assume the Axiom of Choice? – Ryder Rude Mar 17 '20 at 14:39
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    @RyderRude I can't prove it's true, of course: that's the point. The question was: Are there a simple unprovable statements, that would seem intuitively true to the layperson? And this one seems intuitively true to me, for the reasons I gave. – chiastic-security Mar 22 '20 at 22:30
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Most laypersons will understand the following. If you have a file containing random data that is larger than 1 GB in size, then it's unlikely that it could be compressed to a self-extracting file of size less than 1 MB. The probability is astronomically small that such a self-extracting program could generate your file.

But while the vast majority of such files cannot be compressed to under 1 MB in size, you can never have a mathematical proof of that fact for any specific file. This is because you can generate all such proofs recursively using a small program. Have this program stop at the first proven theorem that says that such and such a file larger than 1 GB cannot be compressed, and output that large file. If the program outputs, that means the program is the small self-extracting program (compressed) version of its output, the very output that the theorem says cannot be compressed, which is a contradiction.

Dan Getz
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Count Iblis
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    Interesting. Though, of course, practically, this would take $\mathcal{O}(\text{way too long})$ time to execute. – wchargin Dec 06 '14 at 17:19
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    How does one know that the theorem itself will be small? – Soham Chowdhury Dec 06 '14 at 18:59
  • @SohamChowdhury The size of the theorem doesn't matter, as long as the program to search for it is under 1 MB. – Gordon Davisson Dec 07 '14 at 08:54
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    Um, how do we know it will? – Soham Chowdhury Dec 07 '14 at 12:59
  • I think there are proof checker algorithms that are just a few hundred lines of codes long, so it should be doable using a lot less than 1 MB. – Count Iblis Dec 08 '14 at 03:32
  • It depends on the Kolmogorov (algorithmic) complexity of your axiomatic system, i.e. the length of the shortest program that decides whether a given string is a valid proof in your system. For example, if you make "K_complexity() = 10^11 bits" an axiom, you can prove that at least one string is highly complex; but then your axiomatic system itself is going to have complexity over 10^11 bits itself (assuming that axiom is true, that is). This is the basis of Chaitin's incompleteness theorem. – Gordon Davisson Dec 08 '14 at 04:35
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    Is your proof enumeration program supposed to output every >1GB file which is not compressible, or just a single >1GB file that is not compressible? Because if it is supposed to do the second, how do you plan to make the program 1MB while letting it know which 1GB program to stop at? Or are you considering the large file an input to the program? If that's the case then it's not a self extracting program. – DanielV Nov 23 '16 at 14:06
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    @DanielV If the program would halt, it would output the first such file larger than 1 GB. But note that if it were to ever halt and output some file, that would lead to a paradox. Therefore, the program can never halt, which means that there doesn't exist a proof for incompressibility to lower than 1 MB for any file larger than 1 GB at all. – Count Iblis Nov 23 '16 at 19:25
  • Amazing example, but I would suggest rewording it because I kept reading the 2nd paragraph while imagining a specific file. Instead, the 2nd paragraph is generating a 1GB contradiction in a Halting-Problem fashion, and not generating a given 1GB file. – Nicholas Pipitone Jan 15 '19 at 14:38
  • Will your.program halt?? Seems undecidable! – Travis Wells Jul 26 '20 at 16:11
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In $\mathsf{ZF}$ (i.e. Zermelo–Fraenkel's set theory axioms, without the Axiom of Choice) the following statements (among many many others) are unprovable:

  1. Countable union of countable sets is countable.
  2. Every surjective function has a right-inverse.
  3. Every vector space has a basis.
  4. Every ring has a maximal ideal.

These statements are not exactly "intuitively true to the layperson", but seem natural to many mathematicians. In particular, (2) is probably taught in every math university during the first week of the first year.

If you are interested in models of $\mathsf{ZF}$ in which (1),(2),(3) or (4) don't hold, you can start taking a look at Axiom of Choice, by Horst Herrlich. It has a very nice and well organised Appendix where you can look for models depending on which (main) statements they satisfy.

aerdna91
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  • Great list. #2 and #3 I especially like. They are all great examples, but like you pointed out not necessarily accessible to the layperson. Most people who have taken a university math class could get #2, but to see it's relationship to incompleteness you would have to understand the Axiom of Choice pretty well, and therefore set theory and how integers and reals are built from sets, right? – Michael Harris Dec 05 '14 at 19:20
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    @MichaelHarris It depends on what you meen by "relationship". Look at Proof 1 here: https://proofwiki.org/wiki/Surjection_iff_Right_Inverse. It is a very simple proof, and it's easy to understand why it relies on the Axiom of Choice. So it's clear that $\mathsf{ZFC}$ can prove (2). Of course now you might ask a different question: are we sure that it is impossible to prove (2) in $\mathsf{ZF}$? The answer is YES, but the proof of this fact is all but easy, since it uses quite an advanced tool from set theory, namely forcing. – aerdna91 Dec 05 '14 at 22:07
  • And to understand how forcing works is much more difficult than understanding the axioms of $\mathsf{ZF}$. Or how $\mathbb Z$ and $\mathbb R$ are built (which I actually think is not even strictly needed to understand forcing). – aerdna91 Dec 05 '14 at 22:09
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There is no natural number, $n$, such that $n$'s interpretation as an ASCII string is a proof of this statement.

Aaron Golden
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  • Ah, simplicity! Elegance, too! +1 – TobiMcNamobi Dec 08 '14 at 14:11
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    Explaination? It doesn't make much sense how this is unproveble – user118161 Jul 24 '20 at 07:32
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    Which statement? You actually didn't state or propose anything. – Rodrigo Amaral May 27 '21 at 00:07
  • The phrase "interpretation as an ASCII string" simply means that we write down naturals numbers in the conventional way. $1, 2$ and $3$ are all examples of natural numbers. $5021$ is also an example of a natural number. There are unconventional ways of writing down numbers. I think @aron was just trying to say that we are not allowed convert numbers into English. For example, the number 9 billion five-hundred and thirty eight isn't somehow written as "2 is the only even prime number." – Toothpick Anemone May 31 '21 at 05:35
  • The conjecture is equivalent to "For any whole number $n$, $n$ is not a proof of this statement." The conjecture is very self-referential. Actually, I think you could prove the conjecture. One lemma helpful is that "For any whole number $n$ and any true statement $S$, $n$ is NOT a proof of statement $S$". – Toothpick Anemone May 31 '21 at 05:38
  • Isn’t this the same proposition as in Godels original proof? (except the “formal” system here is the English language, but that’s fair if the question is about something intuitive). – roim Aug 08 '21 at 09:36
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The problem with your goal is this: There are conjectures that are currently unknown (i.e., with our current knowledge, they could be provably right, or provably wrong, or undecidable from the generally accepted axiom systems) and simple enough to formulate so that a layperson can understands their underlying concepts. Some examples are

For most such conjectures there is numerical evidence for them to hold up to the very limits of computational search - but does that make them "intuitively true" for the layperson? If any of these should ever turn out to be undecidable, we'd be in the position to add one of two conflicting axioms to our inventory, e.g., one of "There exists a natural number $N$ that cannot be written as sum of two primes" or "Goldbach is true". When picking the first choice, we'd know that $N$ cannot be reached in finite time by counting $1,2,3,\ldots$ as for any number reachable this way in finite time can also be checked in finite time if it can or cannot be written as sum of two primes. Therefore, we'd know intuitively that the "real-life" natural numbers do not contain such $N$, i.e. the only "correct" extension of our axiom system is that Goldbach is true. This argument works for some others: If "all perfect numbers are even" is independent, then it is true. In general, any independent $Π_1$-sentence is true in the natural numbers. This fact is equivalent to the $Σ_1$-completeness of first-order Peano Arithmetic. Currently, the twin prime conjecture and Collatz conjecture and irrationality of $e+π$ are only known to be equivalent to $Π_2$-sentences, so we cannot apply this argument to them.

Besides the above possibilities, other possibly independent statements include:

  • Those specifically constructed to be independent (of the "I am unprovable" kind) that are totally dull and unrelated to any real-life mathematics.
  • Those that worthy of being added (positively or negatively) to $\mathsf{ZFC}$ without there being a very natural preference in one direction or the other (such as the Continuum Hypothesis; or already the Axiom of Choice if you start with $\mathsf{ZF}$). I doubt that any of these could be considered intuitively clear for the layperson.
user21820
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Hagen von Eitzen
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  • I just wonder, if the statement is about natural numbers and of the kind that a counterexample in principle is possible, wouldn't then "no possible proof" be equivalent with true? Since no possible proof means no counterexample? As a natural and necessary assignment? – Lehs Dec 06 '14 at 17:57
  • So I take this to mean that an open question can go three ways: unprovable, true, or false. If true or false then no problem. If it turns out to be unprovable, then we will likely adopt it as axiomatic in some new system or branch of systems. But whatever that question is, the very fact that it remains an open question means it is likely too complicated for easy understanding. And once something has been shown to be unprovable in the current system, it ceases to be unprovable because mathematicians immediately shift focus to taking it as a new assumtion (axiom) in one direction or another. – Michael Harris Dec 06 '14 at 20:29
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    @MichaelHarris I doubt anyone would want axioms like "Goldbach is true" (or false) if G. should turn out unprovable. An axiom should never be so tailored to a specific problem. Also, once CH was shown independent, this didn't caus either $CH$ or $\neg CH$ to be adopted as axiom. Instead, people tend to formulate results relying on either of these statements as $If CH then (my result)" or "If not CH then (my result)", thus somwhat keeping the "open" status of CH. But I indeed guess that known to be unprovable in ZF(C) (witout being tailored to be unprovable) implies not suiable for laypersons. – Hagen von Eitzen Dec 07 '14 at 18:47
  • I think the point about Goldbach should be reinforced: this is a case, and essentially all other number-theory conjectures will behave likewise, where 'unprovable' implies 'functionally true' because any counterexamples would have to be non-standard numbers. This is a problem with almost all natural unprovable statements. – Steven Stadnicki Dec 08 '14 at 01:11
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    Why is it that rationality of $e+π$ is true if unprovable? Also, both the twin prime conjecture and $(3n+1)$ conjecture are $Π_2$-sentences, so even if unprovable it does not imply that it is true. So please fix your answer. =) – user21820 Feb 24 '17 at 12:46
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    @StevenStadnicki: My above comment contradicts your implicit claim that undecidable universal sentences are true (in the standard model). We can only make that claim for $Π_1$-sentences. For instance, if "every even number greater than $2$ is the sum of two primes" (a $Π_1$-sentence) is undecidable, then it is true and hence "for every natural $n \ge 2$ there is an even number greater than $n$ that is not the sum of two primes" (a $Π_2$-sentence) is also undecidable but false. (Note that I accidentally used "unprovable" in my earlier comment but should have used "undecidable".) – user21820 Feb 24 '17 at 13:40
  • @user21820: $e+\pi \notin \mathbb{Q}$ is true-if-unprovable, if $\frac{a}{b} = e + \pi$ is provably true or false $\forall\frac{a}{b} \in \mathbb{Q}$. But that doesn't look obviously provable to me... I suppose you could construct some kind of Cauchy sequence for the RHS and check whether the LHS eventually falls within it? But that still doesn't prove true, so I give up. – Kevin Aug 16 '17 at 05:08
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    @Kevin: I don't think you understand the "true if unprovable" phenomena. Please read my earlier comments. Also, Hagen, I do not understand why you insist on not fixing your answer, when it is mathematically wrong. – user21820 Nov 05 '17 at 13:16
  • @user21820: I did read your comments when I wrote that comment. I am not going to do that a second time months later. Good day. – Kevin Nov 05 '17 at 14:43
  • @user21820 I never "insisted on not fixing" my answer. Instead, I thank you for correcting any issues it had with the claim about $e+\pi$. I am however unsure why you also removed the reference to Goodstein's Hydra game theorem - I think it did illustrate some aspects of what the OP was after (layman-understandable statement, unprovable in a certain system) – Hagen von Eitzen Dec 13 '17 at 11:16
  • @HagenvonEitzen: Sorry if I misinterpreted you, because I noticed you did not reply twice. And thanks for accepting my correction. I removed the paragraph for 2 reasons: (1) The leading statement was wrong for so many of the statements in your list that it was no longer 'meaningful' to leave that 'intuition' behind. Actually, given a random undecidable sentence over any system with length at most $n$, I am pretty sure that the likelihood that it is true would be $1/2$ as $n \to \infty$. (2) The transfinite induction used to show that the Hydra game ends is very very unlikely to be related. – user21820 Dec 13 '17 at 12:14
  • To expand on (2), ZFC's ordinals are to some extent an artifact of ZFC, so there is no reason (according to my intuition) to think that they may be of use in proving statements about natural numbers and functions from naturals to naturals (which suffice to express those statements in your list). Note that the Hydra game can be proven in 2 different ways, one by using a well-ordering of length $ε_0$, and the other by simply using higher-order arithmetic and ordinary induction. It can be argued that the proof using $ε_0$ is in fact less convincing. – user21820 Dec 13 '17 at 12:20
  • To expand on my last statement, there is no reasonable way of justifying that there is a well-ordering of length $ε_0$ other than by, guess what, ordinary induction in higher-order arithmetic. So the point is that those statements in your list, if ever provable, are very unlikely to require transfinite induction to prove them. Also see Friedman's grand conjecture and especially David Robert's answer. – user21820 Dec 13 '17 at 12:31
  • Wait, maybe some comments here have been deleted, but I don't see what's wrong with the argument "if $e+\pi$'s irrationality is unprovable, then there cannot exist $p/q$ that $e+\pi$ equals (otherwise it would be a disproof), thus it must be irrational". – Abhimanyu Pallavi Sudhir Jul 23 '20 at 16:22
  • @AbhimanyuPallaviSudhir: From your other post, it is clear that you don't actually understand the incompleteness theorems, which may be because you read some popular but wrong account of it (most of them are incorrect). If you want a proper understanding, read this post, at least the first few sections. Also, note that halrankard's answer is correct and completely in line with my statements above; "$e+π$ is rational" is not known to be Σ1 because the obvious way to express it as an arithmetical sentence gives a Σ2-sentence. – user21820 Jul 24 '20 at 16:07
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You asked

Can someone give a good example you can point to and say "That's what Godel's incompleteness theorems are talking about"?

I think you can cut right to the heart of the matter by explaining the Peano axioms, which anyone can understand and which seem as self-evidently true as anything in mathematics. And you can explain how one can use the Peano axioms to prove theorems that are true of the natural numbers, such as $\forall a\,b. a+b = b+a$ and the like.

Now, we might like some assurance that these axioms are sound; that is, how do we know that the Peano axioms don't somehow allow us to “prove” something that is actually false?

Perhaps we could prove that using the Peano axioms themselves, some sort of self-bootstrapping demonstration of the soundness of the axioms? Mathematicians at the beginning of the 20th century hoped for just such a demonstration, and their hope was dashed in 1931.

Gödel's second incompleteness theorem says that the axioms themselves might indeed be able to prove that the axioms were sound—but that if they do, it is only because they are not sound and can prove anything at all, true or false. Sound axioms cannot prove their own soundness.

MJD
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  • This explains the idea of incompleteness very well (as far as I understand it), but it doesn't give a concrete example of what a particular statement that is unprovable is or what it looks like. I could certainly see using this as an introduction to explain the concept, followed up by a concrete example. I suppose what I'm asking might be impossible. By it's very nature, incompleteness seems to only apply to more complex examples, otherwise it wouldn't have taken so long for mathematicians to figure it out. – Michael Harris Dec 05 '14 at 23:45
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    If you want a concrete example of incompleteness, very simple examples are available. Take the axioms of Euclid's geometry. Remove the parallel postulate, which says that if $L$ is a line and $p$ is a point not on $L$, there is exactly one line $L'$ through $p$ that is parallel to $L$. The parallel postulate is not provable from the remaining axioms. Or: Take the Peano axioms for the natural numbers. Delete the axiom that says that 0 is not the successor of any natural number. This axiom is not provable from the remaining axioms. – MJD Dec 06 '14 at 00:15
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Luckily Gödels proof is constructive and so provides an example itself. This is nicely explained in "Gödel, Escher, Bach".

In short (from what I humbly understand and remember): You can assign a number to every logical statement in your formal context, let's call this the Gödel number of that statement. Then you can construct statements that make propositions about other statements (by referencing the Gödel number, e.g. "the statement with the number xyz is true"). Those statements of course have associated Gödel numbers, too. Now construct a statement "The statement with Gödel number xyz is unprovable" so that the Gödel number of the statement is xyz.

The Wikipedia entry for this (here) is more detailed but maybe still considered popular science.

Fornost
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TobiMcNamobi
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  • Isn't this exactly the same as saying "This statement is false."? It's not a problem if it is, but that's what I understand. – Michael Harris Dec 05 '14 at 23:36
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    @MichaelHarris Well, Gödel built a statement (in a formal language!) that could be interpreted as "This statement is false". Still there are two important points here I think: First, Gödels proof does not only show the pure existence of such a statement, and second the simple natural language sentence can indeed be formally expressed to serve as the wanted example. – TobiMcNamobi Dec 06 '14 at 12:48
  • Late to the party but for the record (also @MichaelHarris): your comment is incorrect in a subtle-but-important way. The Godel sentence absolutely can't be interpreted as "this statement is false." Its interpretation is "this statement is unprovable in T" (or, Rosser's improvement: "for every T-proof of this statement there is a shorter T-proof of this statement") where T is our theory in question. Per Tarski, we can't actually express "is false" in arithmetic (hopefully). – Noah Schweber Mar 10 '19 at 17:57
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The parallel postulate can't proved from the other axioms of Euclid's geometry.

You don't even have to drag in Gödel's theorem when you explain this, because it's not relevant. That's a big plus.

Similarly, there are plenty of incomplete axiom systems. Take the Peano axioms for arithmetic, and leave one out, say the axiom that says that there's no $n$ of which zero is the successor. This axiom can't be proved from the other four axioms. (How can you show this? By exhibiting a model that satisfies the other axioms but not that one. For example, just take $SSS0=0$.)

Or take the axioms for the real numbers and leave out the well-ordering principle.

Or take the axioms for set theory and delete one, say the axiom of regularity. You can't prove the deleted axiom from the remaining ones. (There are a few exceptions to this; you can prove the axiom of the empty set from the others.)

MJD
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An example of a problem that in general cannot be solved is the existence of solutions to Diophantine equations http://en.wikipedia.org/wiki/Diophantine_equation. Hilbert's 10th problem was to find an "effective method" (that is, a computer program) to solve Diophantine equations. It was shown in work by Yuri Matiyasevich and Julia Robinson that there is no such effective method. (As I recall, Julia did most of the work, leaving one lemma that needed to be proved. Yuri then proved the lemma. Yuri's paper is rather short, and you need to read Julia's paper to get the full picture.)

Stephen Montgomery-Smith
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