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Given the set of standard axioms (I'm not asking for proof of those), do we know for sure that a proof exists for all unproven theorems? For example, I believe the Goldbach Conjecture is not proven even though we "consider" it true.

Phrased another way, have we proven that if a mathematical statement is true, a proof of it exists? That, therefore, anything that is true can be proven, and anything that cannot be proven is not true? Or, is there a counterexample to that statement?

If it hasn't been proven either way, do we have a strong idea one way or the other? Is it generally thought that some theorems can have no proof, or not?

Bill Dubuque
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Jeremy
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    This may be an oversimplification, but Godel proved that there are true but unprovable mathematical statements. So the answer to the first question in your second paragraph is "no". You can expect a more complete answer, backed up by explanation and references, by someone who knows much more about logic than I do, very shortly. For the last question in your question: it is impossible to know what theorems are unprovable, because if we knew, presumably we would know somehow whether they were true or false without "proving" them. – Stefan Smith Jan 02 '14 at 20:10
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    Theorems have proofs. That is basically the *definition* of what is is to be a theorem! But not all true statements are theorems, that is the content of Gödel's incompleteness theorem. – Harald Hanche-Olsen Jan 02 '14 at 20:19
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    In fact, Goldbach's conjecture is not proven, but it is almost surely true due to statiscal evidence. But this is no guarantee, that a counterexample is impossible. – Peter Jan 02 '14 at 20:35
  • It is a very interesting question, if any true statement is provable in a theory strong enough without simply taking the statement as a new axiom, which would always be possible. – Peter Jan 02 '14 at 20:39
  • I made a typo. Please replace "statiscal" with "statistical" – Peter Jan 02 '14 at 20:42
  • Can I ask an extra question too? When there are no proof of something given a set of axiom, does that means: (a) the rules of inference are too weak to reach it, even though it have a definite truth value (this is akin to Abel-Ruffini: the arithmetic and surd operation are too weak to reach the root of the polynomial even though the root are there and exist); or (b) there are different equiconsistent models all satisfy the axioms, but the statement is true in one and false in another (this is akin to the parallel postulate in Euclidean and non-Euclidean geometry). So which one is it? – Gina Jan 03 '14 at 11:14
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    I highly recommend [this](http://www.amazon.co.uk/Godel-Escher-Bach-Eternal-Golden/dp/0465026567) book to anyone interested in this subject, as a clear (but not overly technical) explanation of Gödel's incompleteness theorem. – rlms Jan 03 '14 at 13:47
  • @Gina: Yes and yes. Gödel showed, that you can (simply put) formulate a sentence that says "this sentence is not provable". If arithmetic is consistent (which everyone believes in, I think), then the sentence is true, but obviously can't be proven or disproven. So you can build two different, equally consistent axiom systems by adding the Gödel sentence or its negation, respectively, as an axiom. (One of your questions was about *truth*. The other was about *consistency*.) – nikie Jan 03 '14 at 16:12
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    @Peter Considering there are an infinity of numbers, but we have only tested a finite number for the Goldbach Conjecture, I highly doubt we can say it is a statistically "almost surely" true. (almost surely does have a definition, btw...) – apnorton Jan 03 '14 at 16:25
  • @anorton An informal argument might run as follows: There are quite a few theorems which could have been disproved by a single counter-example, of which no counter-examples were found up to a size `n`. An example theorem would be Fermat's last. Out of those theorems, I think that most went on to be proved. Goldbach's conjecture fits into the first category of being proved accurate for many cases, so there is strong evidence to believe that it fits into the second, of being proven true. (On another subject, is the word theorem meant just to refer to true theories?) – rlms Jan 03 '14 at 16:37
  • On the other hand, because most theorems which seem to be true have in the long term been proven correct, it also seems likely that Goldbach's conjecture is likely to be provable (and not undecidable). – rlms Jan 03 '14 at 16:38
  • @nikie: thanks for the answer, but what I meant is, could (a) happen alone without (b). I meant, if (b) happened then (a) also happened. But let's say, hypothetically, that there is a statement that have a definite truth value ie. under all consistent model satisfying the assumption, that statement is true; yet somehow all rule of inference cannot reach it in a finite number of step. Is such hypothetical scenario possible? – Gina Jan 03 '14 at 20:11
  • @Gina: I'm not sure I can follow. If a statement isn't provable or refutable, then that means you can add the statement or it's negation as an axiom, without making the system inconsistent. (If it would make the system inconsistent, that would be a proof by contradiction.) And if it's an axiom, it's obviously true in every model of that new axiom system. So your question boils down to: do consistent but "false" axiom systems (like Peano Arithmetic + negation of the Gödel statement) have consistent models, right? (I'm afraid I don't know - but that might be an interesting answer on its own.) – nikie Jan 03 '14 at 20:21
  • @nikie: how should I say this...consider this analogy. Given any function from $\mathbb{N}$ to $\mathbb{R}$, we can always find a real number not in the range of the function. However, we cannot find a specific number once and for all so that it would work for all such function; instead each function have to have its own number. Luckily, Cantor's argument allow us to always find such number, tailored to each possible function. No w imagine if you do not have anything like Cantor's argument. Then if someone give you a function, you can find a number, but you can't show that no bijection exist. – Gina Jan 03 '14 at 21:04
  • @nikie: so back to the issue. Hypothetically, we could have a situation wherein every model have a contradiction somewhere, but it have to be tailored to each model; and there is no general way to produce a contradiction for all model. That could be one way wherein you cannot add a negation of the statement to the list of axiom, despite the fact that there is no proof of the statement. Now I think there is also another possible way, but I cannot think of any good analogy right now; so I might come back later. – Gina Jan 03 '14 at 21:08
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    Short note from the OP: Thanks everyone for your interest, discussions and thorough explanations. I'm a layman to mathematics, but I'll be reading through these over and over again until they make sense so that I can select the one that was most illuminating to me. – Jeremy Jan 03 '14 at 21:11
  • Related http://math.stackexchange.com/q/203076/8271 – leo Jan 08 '14 at 04:01
  • Goldbach's conjecture is most likely true, but heuristically arguments (which are all we have) show there is a non-zero probability that it is false. Interestingly, if some extra-terrestrial super mathematician gave us a an even 100 digit number claiming it is not the sum of any two primes, there is no way we could verify this. We would try to prove he is lying, and would expect to show this easily, but if he said the truth, we would have no way of proving it. – gnasher729 Mar 14 '14 at 14:00
  • sweeneyrod, the fact that Goldbach has been verified for many numbers doesn't mean it is true. For example, there is the claim "for any integers 2 < n < m, the set of integers from n to m doesn't contain more primes than the (equally large) set of integers from 2 to m - n + 2". There is very strong evidence that this is false, but there is no chance whatsoever to find a counterexample. What makes Goldbach likely true is that heuristically, the chance of a counterexample is ridiculously _tiny_. – gnasher729 Mar 14 '14 at 14:08
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    The comments here should be wiped. Saying things like "consistent but "false" axiom systems (like Peano Arithmetic + negation of the Gödel statement)" is unhelpful. Peano Arithmetic + negation of the Gödel statement is not in any way 'false'. – Miles Rout May 24 '14 at 12:16
  • From another point of view, Intuitionistic Mathematicians say statement $p$ is *TRUE*, iff they've a proof for $p$. And say it's *FALSE* iff every proof for $p$ reach to *contradiction*. They believe *Truth* is not an external concept beyond our minds, But this is an inner concept being assigned by our minds. In this manner, for every conjecture in mathematics we can NOT say it's *True* or *False*. But if you believe in *Platonic Realism*, then every statement is "True" or "False". To read more, study *Constructivism Schools* – Fardad Pouran Nov 13 '14 at 14:23

11 Answers11

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Relatively recent discoveries yield a number of so-called 'natural independence' results that provide much more natural examples of independence than does Gödel's example based upon the liar paradox (or other syntactic diagonalizations). As an example of such results, I'll sketch a simple example due to Goodstein of a concrete number theoretic theorem whose proof is independent of formal number theory PA (Peano Arithmetic) (following [Sim]).

Let $\,b\ge 2\,$ be a positive integer. Any nonnegative integer $n$ can be written uniquely in base $b$ $$\smash{n\, =\, c_1 b^{\large n_1} +\, \cdots + c_k b^{\large n_k}} $$

where $\,k \ge 0,\,$ and $\, 0 < c_i < b,\,$ and $\, n_1 > \ldots > n_k \ge 0,\,$ for $\,i = 1, \ldots, k.$

For example the base $\,2\,$ representation of $\,266\,$ is $$266 = 2^8 + 2^3 + 2$$

We may extend this by writing each of the exponents $\,n_1,\ldots,n_k\,$ in base $\,b\,$ notation, then doing the same for each of the exponents in the resulting representations, $\ldots,\,$ until the process stops. This yields the so-called 'hereditary base $\,b\,$ representation of $\,n$'. For example the hereditary base $2$ representation of $\,266\,$ is $$\smash{266 = 2^{\large 2^{2+1}}\! + 2^{2+1} + 2} $$

Let $\,B_{\,b}(n)$ be the nonnegative integer which results if we take the hereditary base $\,b\,$ representation of $\,n\,$ and then syntactically replace each $\,b\,$ by $\,b+1,\,$ i.e. $\,B_{\,b}\,$ is a base change operator that 'Bumps the Base' from $\,b\,$ up to $\,b+1.\,$ For example bumping the base from $\,2\,$ to $\,3\,$ in the prior equation yields $$\smash{B_{2}(266) = 3^{\large 3^{3+1}}\! + 3^{3+1} + 3\quad\ \ \ }$$

Consider a sequence of integers obtained by repeatedly applying the operation: bump the base then subtract one from the result. For example, iteratively applying this operation to $\,266\,$ yields $$\begin{eqnarray} 266_0 &=&\ 2^{\large 2^{2+1}}\! + 2^{2+1} + 2\\ 266_1 &=&\ 3^{\large 3^{3+1}}\! + 3^{3+1} + 3 - 1\ =\ B_2(266_0) - 1 \\ ~ \ &=&\ 3^{\large 3^{3+1}}\! + 3^{3+1} + 2 \\ 266_2 &=&\ 4^{\large 4^{4+1}}\! + 4^{4+1} + 1\qquad\! =\ B_3(266_1) - 1 \\ 266_3 &=&\ 5^{\large5^{5+1}}\! + 5^{5+1}\phantom{ + 2}\qquad\ =\ B_4(266_2) - 1 \\ 266_4 &=&\ 6^{\large 6^{6+1}}\! + \color{#0a0}{6^{6+1}\! - 1} \\ ~ \ &&\ \textrm{using}\quad \color{#0a0}{6^7\ -\,\ 1}\ =\ \color{#c00}{5555555}\, \textrm{ in base } 6 \\ ~ \ &=&\ 6^{\large 6^{6+1}}\! + \color{#c00}5\cdot 6^6 + \color{#c00}5\cdot 6^5 + \,\cdots + \color{#c00}5\cdot 6 + \color{#c00}5 \\ 266_5 &=&\ 7^{\large 7^{7+1}}\! + 5\cdot 7^7 + 5\cdot 7^5 +\, \cdots + 5\cdot 7 + 4 \\ &\vdots & \\ 266_{k+1} &=& \ \qquad\quad\ \cdots\qquad\quad\ = \ B_{k+2}(266_k) - 1 \\ \end{eqnarray}$$

In general, if we start this procedure at the integer $\,n\,$ then we obtain what is known as the Goodstein sequence starting at $\,n.$

More precisely, for each nonnegative integer $\,n\,$ we recursively define a sequence of nonnegative integers $\,n_0,\, n_1,\, \ldots ,\, n_k,\ldots\,$ by $$\begin{eqnarray} n_0\ &:=&\ n \\ n_{k+1}\ &:=&\ \begin{cases} B_{k+2}(n_k) - 1 &\mbox{if }\ n_k > 0 \\ \,0 &\mbox{if }\ n_k = 0 \end{cases} \\ \end{eqnarray}$$

If we examine the above Goodstein sequence for $\,266\,$ numerically we find that the sequence initially increases extremely rapidly:

$$\begin{eqnarray} 2^{\large 2^{2+1}}\!+2^{2+1}+2\ &\sim&\ 2^{\large 2^3} &\sim&\, 3\cdot 10^2 \\ 3^{\large 3^{3+1}}\!+3^{3+1}+2\ &\sim&\ 3^{\large 3^4} &\sim&\, 4\cdot 10^{38} \\ 4^{\large 4^{4+1}}\!+4^{4+1}+1\ &\sim&\ 4^{\large 4^5} &\sim&\, 3\cdot 10^{616} \\ 5^{\large 5^{5+1}}\!+5^{5+1}\ \ \phantom{+ 2} \ &\sim&\ 5^{\large 5^6} &\sim&\, 3\cdot 10^{10921} \\ 6^{\large 6^{6+1}}\!+5\cdot 6^{6}\quad\!+5\cdot 6^5\ \:+\cdots +5\cdot 6\ \ +5\ &\sim&\ 6^{\large 6^7} &\sim&\, 4\cdot 10^{217832} \\ 7^{\large 7^{7+1}}\!+5\cdot 7^{7}\quad\!+5\cdot 7^5\ \:+\cdots +5\cdot 7\ \ +4\ &\sim&\ 7^{\large 7^8} &\sim&\, 1\cdot 10^{4871822} \\ 8^{\large 8^{8+1}}\!+5\cdot 8^{8}\quad\!+5\cdot 8^5\ \: +\cdots +5\cdot 8\ \ +3\ &\sim&\ 8^{\large 8^9} &\sim&\, 2\cdot 10^{121210686} \\ 9^{\large 9^{9+1}}\!+5\cdot 9^{9}\quad\!+5\cdot 9^5\ \: +\cdots +5\cdot 9\ \ +2\ &\sim&\ 9^{\large 9^{10}} &\sim&\, 5\cdot 10^{3327237896} \\ 10^{\large 10^{10+1}}\!\!\!+5\cdot 10^{10}\!+5\cdot 10^5\!+\cdots +5\cdot 10+1\ &\sim&\ 10^{\large 10^{11}}\!\!\!\! &\sim&\, 1\cdot 10^{100000000000} \\ \end{eqnarray}$$

Nevertheless, despite numerical first impressions, one can prove that this sequence converges to $\,0.\,$ In other words, $\,266_k = 0\,$ for all sufficiently large $\,k.\,$ This surprising result is due to Goodstein $(1944)$ who actually proved the same result for all Goodstein sequences:

Goodstein's Theorem $\ $ For all $\,n\,$ there exists $\,k\,$ such that $\,n_k = 0.\,$ In other words, every Goodstein sequence converges to $\,0.$

The secret underlying Goodstein's theorem is that hereditary expression of $\,n\,$ in base $\,b\,$ mimics an ordinal notation for all ordinals less than epsilon nought $\,\varepsilon_0 = \omega^{\large \omega^{\omega^{\Large\cdot^{\cdot^\cdot}}}}\!\!\! =\, \sup \{ \omega,\, \omega^{\omega}\!,\, \omega^{\large \omega^{\omega}}\!,\, \omega^{\large \omega^{\omega^\omega}}\!,\, \dots\, \}$. For such ordinals, the base bumping operation leaves the ordinal fixed, but subtraction of one decreases the ordinal. But these ordinals are well-ordered, which allows us to conclude that a Goodstein sequence eventually converges to zero. Goodstein actually proved his theorem for a general increasing base-bumping function $\,f:\Bbb N\to \Bbb N\,$ (vs. $\,f(b)=b+1\,$ above). He proved that convergence of all such $f$-Goodstein sequences is equivalent to transfinite induction below $\,\epsilon_0.$

One of the primary measures of strength for a system of logic is the size of the largest ordinal for which transfinite induction holds. It is a classical result of Gentzen that the consistency of PA (Peano Arithmetic, or formal number theory) can be proved by transfinite induction on ordinals below $\,\epsilon_0.\,$ But we know from Godel's second incompleteness theorem that the consistency of PA cannot be proved in PA. It follows that neither can Goodstein's theorem be proved in PA. Thus we have an example of a very simple concrete number theoretical statement in PA whose proof is nonetheless independent of PA.

Another way to see that Goodstein's theorem cannot be proved in PA is to note that the sequence takes too long to terminate, e.g.

$$ 4_k\,\text{ first reaches}\,\ 0\ \,\text{for }\, k\, =\, 3\cdot(2^{402653211}\!-1)\,\sim\, 10^{121210695}$$

In general, if 'for all $\,n\,$ there exists $\,k\,$ such that $\,P(n,k)$' is provable, then it must be witnessed by a provably computable choice function $\,F\!:\, $ 'for all $\,n\!:\ P(n,F(n)).\,$' But the problem is that $\,F(n)\,$ grows too rapidly to be provably computable in PA, see [Smo] $1980$ for details.

Goodstein's theorem was one of the first examples of so-called 'natural independence phenomena', which are considered by most logicians to be more natural than the metamathematical incompleteness results first discovered by Gödel. Other finite combinatorial examples were discovered around the same time, e.g. a finite form of Ramsey's theorem, and a finite form of Kruskal's tree theorem, see [KiP], [Smo] and [Gal]. [Kip] presents the Hercules vs. Hydra game, which provides an elementary example of a finite combinatorial tree theorem (a more graphical tree-theoretic form of Goodstein's sequence).

Kruskal's tree theorem plays a fundamental role in computer science because it is one of the main tools for showing that certain orderings on trees are well-founded. These orderings play a crucial role in proving the termination of rewrite rules and the correctness of the Knuth-Bendix equational completion procedures. See [Gal] for a survey of results in this area.

See the references below for further details, especially Smorynski's papers. Start with Rucker's book if you know no logic, then move on to Smorynski's papers, and then the others, which are original research papers. For more recent work, see the references cited in Gallier, especially to Friedman's school of 'Reverse Mathematics', and see [JSL].

References

[Gal] Gallier, Jean. What's so special about Kruskal's theorem and the ordinal $\Gamma_0$?
A survey of some results in proof theory,
Ann. Pure and Applied Logic, 53 (1991) 199-260.

[HFR] Harrington, L.A. et.al. (editors)
Harvey Friedman's Research on the Foundations of Mathematics, Elsevier 1985.

[KiP] Kirby, Laurie, and Paris, Jeff. Accessible independence results for Peano arithmetic,
Bull. London Math. Soc., 14 (1982), 285-293.

[JSL] The Journal of Symbolic Logic,* v. 53, no. 2, 1988, jstor, cambridge.org
This issue contains papers from the Symposium "Hilbert's Program Sixty Years Later".

[Kol] Kolata, Gina. Does Goedel's Theorem Matter to Mathematics?
Science 218 11/19/1982, 779-780; reprinted in [HFR]

[Ruc] Rucker, Rudy. Infinity and The Mind, 1995, Princeton Univ. Press.

[Sim] Simpson, Stephen G. Unprovable theorems and fast-growing functions,
Contemporary Math. 65 1987, 359-394.

[Smo] Smorynski, Craig. (all three articles are reprinted in [HFR])
Some rapidly growing functions, Math. Intell., 2 1980, 149-154.
The Varieties of Arboreal Experience, Math. Intell., 4 1982, 182-188.
"Big" News from Archimedes to Friedman, Notices AMS, 30 1983, 251-256.

[Spe] Spencer, Joel. Large numbers and unprovable theorems,
Amer. Math. Monthly, Dec 1983, 669-675.

Martin Sleziak
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Bill Dubuque
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    the answer is probably in your references, and is suggested in your answer itself, but which set-theoretic axioms (beyond PA, which seems obviously "true") do you need to prove Goodstein's Theorem? (just the basic version). I really like your answer, by the way. – Stefan Smith Jan 03 '14 at 01:25
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    From WikiPedia: "In mathematical logic, Goodstein's theorem is a statement about the natural numbers, proved by Reuben Goodstein in 1944, which states that every Goodstein sequence eventually terminates at 0. Kirby & Paris 1982 showed that it is unprovable in Peano arithmetic (but it can be proven in stronger systems, such as second order arithmetic)." http://en.wikipedia.org/wiki/Goodstein%27s_theorem If the parenthetic remark is true, what's the big deal? Doesn't nearly everyone (every number theorist anyway) implicitly use some form of second order arithmetic? – Dan Christensen Jan 03 '14 at 05:12
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    @DanChristensen I think the "big deal" is simply that here is a theorem/postulate that *(1)* lives inside PA, *(2)* is true inside PA (which we know only because we know "stronger" systems like second order arithmetic), and *(3)* has no proof inside PA. So it is a natural example of a theorem in a system which is true in that system but has no proof in that system. If we didn't have "stronger" logics, we couldn't *know* if the theorem was true or not! – Jeppe Stig Nielsen Jan 03 '14 at 12:00
  • This answer appears to "prove" the axiom that length=quality (and yes I'm just joking here about the number of upvotes -- but ask any PublishOrPerish victim!) – Carl Witthoft Jan 03 '14 at 13:33
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    So, there are serious limitations as to what can be accomplished with first-order PA. Is this what Gödel was talking about? Or are there limitations on any 2nd-order axioms for the natural numbers as well? If so, what examples from number theory do we have other than variations of the Liar Paradox such as "this sentence is unprovable?" – Dan Christensen Jan 03 '14 at 13:56
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    @Dan The "big deal" is that PA is strong enough to develop all of what is normally deemed to be number theory, but it is not strong enough to prove Goodstein's theorem (which, unlike Gödel's "artificial" statement, is a statement that might occur to one "naturally" while investigating arithmetical properties of natural numbers). – Bill Dubuque Jan 03 '14 at 15:01
  • @BillDubuque Moving from 1st-order to 2nd-order PA seems like a minor tweaking of the axioms to me if you are using any kind of set theory as well. So, if switching to 2nd-order axioms solves the GT problem, it doesn't seem like a "big deal" to me. BTW, I would still like to know if there are any known 2nd-order axioms that can get around the limitations that seem to have been imposed by Godel. – Dan Christensen Jan 03 '14 at 15:32
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    @Dan Second-order arithmetic (a.k.a. "analysis") is much stronger than PA since it allows quantification over *sets* of naturals, so one can develop a theory of real numbers and much of classical analysis. – Bill Dubuque Jan 03 '14 at 15:54
  • @BillDubuque So, are there known "true statements" about the natural numbers (other than variations of the Liar Paradox) that are unprovable in 2nd-order arithmetic? – Dan Christensen Jan 03 '14 at 16:07
  • @TimParenti Thanks much for finishing up the $TeX$'ing. I had planned to do that, but ran out of spare time. The plain-text is from an old [sci.math post, 11 Dec 95](http://www.math.niu.edu/~rusin/known-math/95/independence) – Bill Dubuque Jan 03 '14 at 18:21
  • @JeppeStigNielsen, do you know maybe how proof that the theorem can't be proved within PA relates to Turing's theorem that such a proof can not exist? – bbozo Jan 05 '14 at 12:35
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    This sort of things are what make worth to still visit math.se – leo Jan 08 '14 at 03:01
  • @JeppeStigNielsen Re: your (2): but it isn't necessarily true in PA, it's true in second order arithmetic. Who says that that means it's true in PA? – Miles Rout May 24 '14 at 05:11
  • Hey, there is a minor typo for the number of steps needed for $4_k$ to first get to $0$, Wikipedia has $3\cdot 2^{402653211} − 2$ :) – no lemon no melon Aug 09 '21 at 21:49
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Gödel was able to construct a statement that says "this statement is not provable."

The proof is something like this. First create an enumeration scheme of written documents. Then create a statement in number theory "$P(x,y,z)$", which means "if $x$ is interpreted as a computer program, and we input the value $y$, then the value $z$ is the output." (This part was quite hard, but intuitively you can see it could be done.)

Then write a computer program that checks proofs. Creating proofs is undecidable, and it is hard to create a program to do that. But a program to check a proof can be created. Let's suppose this program becomes the literal number $n$ in our enumeration scheme. Then we can create a statement in number theory "$Q(x)$"${}={}$"$\exists y:P(n,\text{cat}(x,y),1)$". Here $\text{cat}(x,y)$ concatenates a written statement in number theory $x$ with its proof $y$. So $Q(x)$ says "$x$ is provable."

Now construct in number theory a formula $S(x,y)$, which means take the statement enumerated by $x$, and whenever you see the symbol $x$ in it, substitute it with the literal number represented by $y$.

Now consider the statement "$T(x)$"${}={}$"$\text{not} \ Q(S(x,x))$". Let's suppose this enumerates as the number $m$.

Then "$T(m)$" is a statement in number theory that says "this statement is not provable."

Now suppose "$T(m)$" is provable. Then it is true. But if it is true, then it is not provable (because that is what the statement says).

So "$T(m)$" is clearly not provable. Hence it is true.

I know I am missing some important technical issues. I'll answer them as best I can when they are asked. But that is the rough outline of the proof of Gödel's incompleteness theorem.

MJD
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Stephen Montgomery-Smith
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    If a logical argument allows you to end with "Hence it is true", how is that not a proof? – mbeckish Jan 03 '14 at 00:34
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    @mbeckish Because in our proof that it is not provable, we used an addition axiom in addition to the axioms we were given. Namely, that the system is consistent. (If it is inconsistent, then we every statement is provable.) That is where Goedel's second incompleteness statement comes from: you cannot prove a sufficiently complex set of axioms is consistent within itself. (Sufficiently complex means: you can represent computer programs in it.) – Stephen Montgomery-Smith Jan 03 '14 at 00:37
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    I think we actually used [$\Sigma_1$-soundness](http://en.wikipedia.org/wiki/Ω-consistent_theory#Definition). $\:$ However, the [(Gödel-)Rosser sentence](http://en.wikipedia.org/wiki/Rosser%27s_trick) works with just consistency. –  Jan 03 '14 at 04:05
  • @RickyDemer That was one of the technical issues I was referring to. Another thing - when Goedel developed his proof, the theory of computer programming was all theoretical. So he didn't have available the natural intuition that many of us have today. That makes his achievement all the more impressive. – Stephen Montgomery-Smith Jan 03 '14 at 15:20
  • Doesnt this just prove that the true statements are not countable? Why do they have to be countable. With the same argument I could say:"there are real numbers which do not exists". Proof: lets write down all real numbers that do exists and then show one is not on the list. But of course this doesnt show the statement. – lalala May 28 '21 at 09:22
  • @lalala The total number of statements is countable, since we have enumerated them. So the number of true statements must also be countable. – Stephen Montgomery-Smith May 28 '21 at 14:35
  • @StephenMontgomery-Smith ok. This relies on the set of all finite sequences to be countable? – lalala May 28 '21 at 15:04
  • @lalala Yes it does. – Stephen Montgomery-Smith May 28 '21 at 20:24
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"Can't be proven" is an inappropriately vague notion for the question you want to ask. Proven from what axioms? In a logical system that includes Goldbach's conjecture as an axiom, the proof of Goldbach's conjecture is only one line long. So to have the question make sense, you can't just say "proven"; you have to say "proven from such-and-so axioms".

There is a standard set of axioms for arithmetic, called the Peano axioms. We like these axioms because they are intuitive and simple, and also because they seem to be powerful enough to prove almost all of the things we'd like to prove about arithmetic.

However, it is known that there are particular true statements of arithmetic that are not provable from the Peano axioms; Goodstein's theorem is an example.

Gödel's famous incompleteness theorem states that any system of axioms that is expressive enough to prove all true statements of arithmetic must also prove some false statements of arithmetic. Conversely, any system of axioms that proves only true statements of arithmetic must fail to prove some true statements of arithmetic. The proof is constructive; starting from the given axioms, it constructs a (highly artificial) statement of arithmetic $G$ which is true if and only if there is no proof of $G$ from the axioms. Either $G$ is false and has a proof, or it is true and it has no proof.

Martin Sleziak
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MJD
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    I upvoted your answer because I liked it, especially your links (I'd like to learn more about the Peano axioms and I never heard of Goodstein's theorem until just now). But unlike, say, the Axiom of Choice or the Continuum Hypothesis, Goldbach's conjecture is either true or false. If you haven't proven or disproven it, there is the danger that it is false, and it doesn't make sense to me why anyone would consider including it in a logical system, because if it is false you can "prove" anything. I'm not a logician, so feel free to correct anything I wrote, but please be gentle. – Stefan Smith Jan 03 '14 at 01:20
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    @Stefan Smith: if a statement is not provable nor disprovable from a consistent set of axioms, then when you add it to the set of axioms as a new axiom, the resulting set of axioms is still consistent, meaning that it won't prove everything. It is perfectly possible to have a consistent set of axioms that includes false statements. But more crucially, the is no "absolutely unprovable" *true* statement, since that statement itself could be used as a (true) axiom. A statement can only be provable or unprovable *relative to* a given, fixed set of axioms; it can't be unprovable *in and of itself*. – Carl Mummert Jan 03 '14 at 02:04
  • This should not be surprising; it's not really different than the fact that a function is not continuous in and of itself, it is only continuous relative to a given topology. For any function between two sets, we can of course define a topology which makes that function continuous. – Carl Mummert Jan 03 '14 at 02:16
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    @Stefan Smith: Goldbach's conjecture is no different (conceptually) to the Axiom of Choice or the Continuum Hypothesis. Like these theorems, you cannot determine if it is true or false within first order logic. You could adopt the converse of Goodstein's theorem as an axiom, and not derive a contradiction. – Peter Webb Jan 03 '14 at 09:00
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    @Stefan Smith: Goodstein's theorem, like the Continuum Hypothesis and axiom of choice, is independent of the normal rules of arithmetic. If you add appropriate axioms, you can form consistent arithmetic systems where Goodstein's theorem is true; you can also add axioms to make it false. Same as the axiom of choice and continuum hypothesis. An axiom which makes Goodstein true could include transfinite induction, which ultimately is the (additional) axiom used to prove it. An axiom which makes Goodstein false would be "Goodstein is false", which cannot possibly lead to a contradiction. – Peter Webb Jan 03 '14 at 09:15
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    @PeterWebb If you add the negation of Goodstein’s theorem as an axiom, the resulting system will no longer have **the** natural numbers as a model. – kinokijuf Jan 03 '14 at 10:12
  • In a criminal investigation, the police could (at a certain point before the case is solved) have a set of working hypothesis (axioms) that is consistent (within itself, they could possibly have a model) but nevertheless is inconsistent with the preferred model, which of course is the resolved case. – kjetil b halvorsen Jan 03 '14 at 12:04
  • @CarlMummert : thanks for your comment. I am confused. You write that a consistent set of axioms "can't prove everything", but also that a consistent set of axioms can include false statements. But a false antecedent implies any consequent, so if your set of axioms includes a false statement, doesn't that mean you can use it to prove anything? I am an analyst, so I normally think of mathematical statements as being "true" or "false". Can you recommend a good book for studying logic on my own? I think I took two semesters of set theory + logic in grad... – Stefan Smith Jan 03 '14 at 17:45
  • ...school a long time ago, and I got lost somewhere in the middle of the second semester. – Stefan Smith Jan 03 '14 at 17:46
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    @Stefan Smith: a false statement only implies arbitrary consequents if its (true) negation is (assumed or) provable. In other words a *provably false* statement, if assumed, implies an arbitrary consequent. By definition, a theory is consistent if there is no formula $\phi$ such that both $\phi$ and its negation are provable - therefore, trivially, each consistent theory has statements it cannot prove, namely at least one of each pair $\phi$, $\lnot \phi$. If neither $\phi$ nor $\lnot \phi$ is provable in a theory, then the result of assuming either one as an extra axiom is still consistent. – Carl Mummert Jan 03 '14 at 18:36
  • The issue here is that there are many examples of a theory $T$ and a pair of formulas $\phi$, $\lnot \phi$ neither of which is provable in $T$, even if, say, $\lnot \phi$ is true. So $T$ plus $\phi$ is still consistent, even though it contains a false axiom $\phi$. – Carl Mummert Jan 03 '14 at 18:37
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    Dear MJD, You may want to add that in Godel's theorem, the axioms have to be assumed to be recurvisely enumerable, or something like this. (Otherwise we could just take the axioms to be the set of all true results for a given model, e.g. the standard naturals, and such a theory is complete.) Best wishes, – Matt E Jan 03 '14 at 19:15
  • @Matt Thanks, but I almost always leave that out. There are a lot of technicalities around the incompleteness theorem that a complete explanation must include, but a short description has to omit the less important ones. If I were making this one longer I would point out that the underlying language must be sufficiently expressive before I would mention that the axioms must be RE. – MJD Jan 03 '14 at 20:47
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    @MJD: Dear MJD, I guess I think of this as being perhaps the *key* statement. (An analogy I learnt from Manin is that Godel's theorem can be thought of as saying that the theory of $\mathbb N$ can't be recursively generated, which is analogous to exhibiting a group that can't be finitely generated.) But this is because I believe in the standard model of $\mathbb N$ as an object existing prior to PA, ZFC, or any other attempts to describe it, and so the theory of $\mathbb N$ is, for me, a real thing. I can appreciate that for others, with a different perspective, other aspects are more ... – Matt E Jan 04 '14 at 02:25
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    ... important to emphasize. Regards, – Matt E Jan 04 '14 at 02:26
  • @CarlMummert : Thanks. I know enough that I can tell you know what you're talking about. – Stefan Smith Jan 04 '14 at 03:04
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Gödel's incompleteness theorem is one of those widely misunderstood results.

Roughly it means in the context of arithmetic you can only have two of the following:

  • Decidable axioms
  • Consistency
  • Completeness

The "truths that cannot be proven" is an abbreviation for the context of choosing decidable axioms, consistency, but a lack of completeness. This means there are sentences P for which there is no proof of P or not P.

You can throw in more axioms of arithmetic so that every sentence P has a proof of P or not P. That will give completeness, consistency, but the axioms will necessarily be undecidable because of Gödel's incompleteness theorem.

A point that is often missed in the statement "truths that cannot be proven" is that it is meaningful to speak of undecidable, complete, consistent axioms of arithmetic where every true sentence can be proven. But it comes at the cost of undecidable axioms which is why its not particularly useful.

user782220
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    What are decidable axioms ? – Shuzheng Jan 03 '14 at 12:54
  • As far as i understand the role of axioms, axioms describe the set of possible models. Undecidable axioms would be things like "Every model that suffices this axiom is a halting computer program" – kutschkem Jan 03 '14 at 16:07
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    @user111854 Decidable axioms means that you could write a program that would tell you in a finite time whether a given statement is an axiom or not. If you have a finite number of axioms this is easy - just check it against the list - and even some "parameterized" axioms like the induction schema in PA are decidable by writing a program that can check substitutions. But a simple example of an undecidable axiom system is "the set of all statements true in $\Bbb N$". To test if a given statement is an axiom, now you need to know if it is true, which is undecidable by Godel (see answers). – Mario Carneiro Jan 07 '14 at 20:20
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Amongst the many excellent answers you have received, nobody appears to have directly answered your question.

Goldbach's conjecture can be true and provable, true but not provable using the "normal rules of arithmetic", or false.

There are strong statistical arguments which suggest it is almost certainly true.

Whether it is provable using the "normal laws of arithmetic" - like those used to prove Fermat's Last Theorem or the Prime Number Theorem and everything you learned in high school maths - is not known. Assuming it can't be proven is a complete dead-end. To be interested at all you have to either assume it is true and be looking for a proof, or assume it is false and be looking for a counter example.

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This is a try to summarize the most important results of Gödel's theorems.

  • A statement is provable within a theory if and only if it is true for any interpretation allowed in this theory.

  • If a statement is true for some interpretation (model) and false for some other, then it is independent of the theory and undecidable within the theory.

  • But the fact, that a statement is undecidable within a theory, cannot be proven within the theory itself. A stronger theory might prove this undecidability, or might not.

  • Any theory, that is strong enough, that the representation theorem holds for it, is incomplete, that means, that there are true statements, not provable within it.

  • Finally, a theory cannot prove its own consistency.

MJD
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Peter
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    I understand that this is not meant to be exact, but still you might want to add „consistent“ to the last three bullet points. And then the third point is true, but perhaps slightly misleading, because of course a consistent theory to which the second incompleteness theorem applies cannot even prove that it cannot prove statements that it refutes, it may however be able to prove that it cannot prove a certain statement if it itself is consistent. – Carsten S Jan 02 '14 at 20:41
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    Yes, things are quite difficult. Thererfore I began with : This is a try ... – Peter Jan 02 '14 at 21:20
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It is disturbing that in the extensive discussion on this topic, Gödel's work (his Incompleteness Theorem) has been referenced several times but there has been no mention of Alan Turing (the Halting Problem) or Emil Post (his production systems). All three independently proved the same thing, that in any proof system there are some true statements that cannot be proven (incompleteness) or else the proof system will also prove some statements that are not true (inconsistency). Curiously, these three developments were accomplished independently at very roughly the same time. As Robert Heinlein posited, when the time comes that railroading is possible it will arise independently in multiple places.

user118987
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Given the set of standard axioms [of some mathematical theory], do we know for sure that a proof exists for all unproven theorems? For example, I believe the Goldbach Conjecture is not proven even though we "consider" it true.

In addition to all the interesting discussion about Godel's and Goodstein's Theorems, I want to suggest also another "thread" of discussion, regarding epistemology of mathematical knowledge.

During the '60s and '70s, the philosophy of science debate was concerned with the distinction between :

  • the Context of Discovery and the Context of Justification.

Roughly speaking, the context distinction regards : how science (e.g. physics) discover a new fact or law; the second is: how science explain it (ref.Paul Hoyningen-Huene, On the Varieties of the Distinction between the Context of Discovery and the Context of Justification, 2002).

Applied to mathematics, this points to the difference between :

the discovery of a new math idea or concept vs the proof of a theorem.

As far as I know, very few philosophers of mathematics are concerned with this kind of issue ; the only book I've read about something similar was Lakatos' Proofs and Refutations, (1976).

The connection I see is this :

when we don't have a proof of a mathematical "fact" , what are the gorund for asserting or believing it ?

Here some comments about comments in the above debate :

a) "not all true statements are theorems, that is the content of Gödel's incompleteness theorem"

They are not theorems of the formal arithmetic in question (i.e. first-order PA) but THEY ARE proved via Godel's "construction" provided by G's Theorem itself (i.e.proved in the meta-theory): isn't it ?

b) "Goldbach's conjecture is not proven, but it is almost surely true due to statistical evidence"

Are there research about "inductive" grounds for unproven mathematical facts ?

A single contradiction con destroy a theory (Russell's Paradox in front of Frege's system) but how many years (?) of absence of contradiction can support our sound belief in a theory (e.g.ZFC) ?

Martin Sleziak
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Mauro ALLEGRANZA
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Let $\Sigma$ be Rado's Busy Beaver function.

There exist specific positive integers $k$ and $n$ such that $"\Sigma(k) = n"$ is true-but-unprovable in ZFC.

Also see this question and answer.

r.e.s.
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  • ? For the downvote, an explanatory comment would be helpful. – r.e.s. Jul 02 '14 at 13:30
  • I didn't downvote, but you made a very strong statement. It sounds doable to prove e.g. that k=2 and n=4 work in ZFC. The sources you cite seem to make the much weaker claim that we cannot compute an upper bound for n (given sufficiently large k). – gmatht Aug 28 '17 at 05:46
  • @gmatht - The claim is about *proving* a value or an upper bound, not about *computing* one; also, "cannot prove an upper bound" is a *stronger* claim, not a weaker one, than is "cannot prove the value".) – r.e.s. Aug 28 '17 at 06:11
  • OK, without the word "compute": Proving that Σ(2)=4, and thus that `There exist specific positive integers k and n such that "Σ(k)=n"' sounds much easier to me than proving something about all k. – gmatht Aug 28 '17 at 10:34
  • @gmatht - As discussed in the posted links, for some specific $k$ and *for all* $n$, $"\Sigma(k)\le n"$ is unprovable in ZFC; consequently, for some specific $k$ and *for some specific* $n$, $"\Sigma(k)\le n"$ is unprovable in ZFC. But if $"\Sigma(k)\le n"$ is unprovable, it follows that $"\Sigma(k)= n"$ is unprovable, because $"\Sigma(k)= n"$ implies $"\Sigma(k)\le n"$. – r.e.s. Aug 29 '17 at 00:10
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do we know for sure that a proof exists for all unproven theorems?...Phrased another way, have we proven that if a mathematical statement is true, a proof of it exists?

Perhaps this is a semantic issue, but I don't think these two questions are identical, even though you may have intended them to be so.

If a mathematical statement has been proven as true, and the proof is correct, then yes we have proven that a proof exists. It doesn't mean that it's the only proof possible. In the book "Proofs Without Words" there's around 13(?) proofs of the Pythagorean Theorem with no words at all, just diagrams.

For statements that haven't been proven yet, there are two possibilities. Somebody can prove that the statement can't be proved by pointing out a flaw in the reasoning used to arrive at that statement. Alternatively, if you can prove that a contradictory statement to be true, you have proven the original statement false - therefore no proof can exist.

rocinante
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  • So what is your stand regarding the continuum hypothesis? – Asaf Karagila Jan 03 '14 at 18:41
  • I don't know if I am qualified enough to have a respectable opinion on it. I think that the Cantor's Continuum Hypothesis holds true because it rests on the *convention* that the natural numbers are the counting numbers when we're talking about counting the number of elements in a set. Paul Cohen's claim that no contradictions would arise if the negation of the hypothesis were added to set theory is another beast altogether 1/2 – rocinante Jan 03 '14 at 18:51
  • No contradictions would arise if the hypothesis was false. However the *convention* that the natural numbers are the counting numbers would be violated. Just like there is a convention of sorts to frame "everything" in terms of set theory since Cantor, there are proponents who argue that alternatives to the set theory "convention". Currently I think the community views them as cranks, but they exist just the same and just suffer from not having their standard widespread. 2/2 – rocinante Jan 03 '14 at 18:55
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    I'm not sure what CH has to do with the convention that the natural numbers are the counting numbers. That much doesn't change at all, regardless to CH or even the axiom of choice. The continuum hypothesis is a statement on sets larger than the natural numbers. – Asaf Karagila Jan 03 '14 at 19:05
  • Wolfram's explanation of it http://mathworld.wolfram.com/ContinuumHypothesis.html claims that the validity of hypothesis depends on what version of set theory is used. If you can pick your version of set theory, there is convention at play. (And for what it's worth, I do think that which version of set theory you pick does depend on how you define counting numbers. However, as I said I don't know if I am qualified to have respectable opinion on the subject). – rocinante Jan 03 '14 at 19:14
  • Yes, the truth of the continuum hypothesis depends on the universe of set theory, which itself usually is taken to satisfy a particular extension of $\sf ZFC$. Some of these extensions satisfy CH, others don't -- that is what is meant by "version of set theory". Both have merits in modern set theory, and both are used commonly. – Asaf Karagila Jan 03 '14 at 19:17
  • If both extensions are acceptable, then it still does not really invalidate CH. CH is true if you adopt a certain set of assumptions and false otherwise. That does not invalidate the truth of CH or violate the principle that a statement can either be true or false. Every statement has a context. What I was trying to say is that you can prove that *within that context* a statement can be shown true or false (but not both). The minute you depart that context, of course things get murky. – rocinante Jan 03 '14 at 19:28
  • Yes, the point about the context wasn't clear enough in your answer. – Asaf Karagila Jan 03 '14 at 19:33
  • I tend to think of proof questions in terms of court verdicts. Just because somebody is declared "not guilty" doesn't mean that they're innocent of wrong-doing. It just means that there is not enough evidence to convict them of the *specific* charges laid against them. They may very well be guilty if the prosecutor would only have filed different charges. – rocinante Jan 03 '14 at 19:37
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Euclid's fifth postulate:

Two lines are drawn which intersect a third in such a way that the sum of the inner angles on one side is less than two right angles, then the two lines inevitably must intersect each other on that side if extended far enough. This postulate is equivalent to what is known as the parallel postulate.

Euclid's fifth postulate cannot be proven as a theorem, although this was attempted by many people. Euclid himself used only the first four postulates ("absolute geometry") for the first 28 propositions of the Elements, but was forced to invoke the parallel postulate on the 29th. In 1823, Janos Bolyai and Nicolai Lobachevsky independently realized that entirely self-consistent "non-Euclidean geometries" could be created in which the parallel postulate did not hold. (Gauss had also discovered but suppressed the existence of non-Euclidean geometries.)

Prankster
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    I think this is an answer to a slightly different question. Euclid's fifth postulate isn't really "true" as such, it just happens to be the fifth postulate that is necessary for working in a geometry that seems to correspond to the universe we live in (although it actually doesn't). – rlms Jan 04 '14 at 15:26