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?

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    $\begingroup$ 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. $\endgroup$ – Stefan Smith Jan 2 '14 at 20:10
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    $\begingroup$ 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. $\endgroup$ – Harald Hanche-Olsen Jan 2 '14 at 20:19
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    $\begingroup$ 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. $\endgroup$ – Peter Jan 2 '14 at 20:35
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    $\begingroup$ I highly recommend this book to anyone interested in this subject, as a clear (but not overly technical) explanation of Gödel's incompleteness theorem. $\endgroup$ – rlms Jan 3 '14 at 13:47
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    $\begingroup$ @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...) $\endgroup$ – apnorton Jan 3 '14 at 16:25

11 Answers 11


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].


[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
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.

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    $\begingroup$ 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. $\endgroup$ – Stefan Smith Jan 3 '14 at 1:25
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    $\begingroup$ 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)." 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? $\endgroup$ – Dan Christensen Jan 3 '14 at 5:12
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    $\begingroup$ @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! $\endgroup$ – Jeppe Stig Nielsen Jan 3 '14 at 12:00
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    $\begingroup$ @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). $\endgroup$ – Bill Dubuque Jan 3 '14 at 15:01
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    $\begingroup$ This sort of things are what make worth to still visit math.se $\endgroup$ – leo Jan 8 '14 at 3:01

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.

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    $\begingroup$ If a logical argument allows you to end with "Hence it is true", how is that not a proof? $\endgroup$ – mbeckish Jan 3 '14 at 0:34
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    $\begingroup$ @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.) $\endgroup$ – Stephen Montgomery-Smith Jan 3 '14 at 0:37
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    $\begingroup$ I think we actually used $\Sigma_1$-soundness. $\:$ However, the (Gödel-)Rosser sentence works with just consistency. $\endgroup$ – user57159 Jan 3 '14 at 4:05
  • $\begingroup$ @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. $\endgroup$ – Stephen Montgomery-Smith Jan 3 '14 at 15:20

"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.

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    $\begingroup$ 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. $\endgroup$ – Stefan Smith Jan 3 '14 at 1:20
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    $\begingroup$ @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. $\endgroup$ – Carl Mummert Jan 3 '14 at 2:04
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    $\begingroup$ @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. $\endgroup$ – kinokijuf Jan 3 '14 at 10:12
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    $\begingroup$ 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, $\endgroup$ – Matt E Jan 3 '14 at 19:15
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    $\begingroup$ @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 ... $\endgroup$ – Matt E Jan 4 '14 at 2:25

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.


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.

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    $\begingroup$ What are decidable axioms ? $\endgroup$ – Shuzheng Jan 3 '14 at 12:54
  • $\begingroup$ 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" $\endgroup$ – kutschkem Jan 3 '14 at 16:07
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    $\begingroup$ @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). $\endgroup$ – Mario Carneiro Jan 7 '14 at 20:20

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.

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    $\begingroup$ 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. $\endgroup$ – Carsten S Jan 2 '14 at 20:41
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    $\begingroup$ Yes, things are quite difficult. Thererfore I began with : This is a try ... $\endgroup$ – Peter Jan 2 '14 at 21:20

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.

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    $\begingroup$ Is this an answer or a comment? $\endgroup$ – R R Jan 15 '14 at 23:53

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) dicover 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) ?


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.

  • $\begingroup$ So what is your stand regarding the continuum hypothesis? $\endgroup$ – Asaf Karagila Jan 3 '14 at 18:41
  • $\begingroup$ 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 $\endgroup$ – rocinante Jan 3 '14 at 18:51
  • $\begingroup$ 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 $\endgroup$ – rocinante Jan 3 '14 at 18:55
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    $\begingroup$ 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. $\endgroup$ – Asaf Karagila Jan 3 '14 at 19:05
  • $\begingroup$ Wolfram's explanation of it 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). $\endgroup$ – rocinante Jan 3 '14 at 19:14

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.

  • $\begingroup$ ? For the downvote, an explanatory comment would be helpful. $\endgroup$ – r.e.s. Jul 2 '14 at 13:30
  • $\begingroup$ 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). $\endgroup$ – gmatht Aug 28 '17 at 5:46
  • $\begingroup$ @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".) $\endgroup$ – r.e.s. Aug 28 '17 at 6:11
  • $\begingroup$ 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. $\endgroup$ – gmatht Aug 28 '17 at 10:34
  • $\begingroup$ @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 $"\Sigma(k)\le n"$. $\endgroup$ – r.e.s. Aug 29 '17 at 0:10

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.)

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    $\begingroup$ 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). $\endgroup$ – rlms Jan 4 '14 at 15:26

protected by Asaf Karagila Jan 4 '14 at 23:39

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