Do we know if there exist true mathematical statements that can not be proven?

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
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.

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.

"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.