From Georg Kreisel‘s review of The decision problem for exponential diophantine equations, by Martin Davis, Hilary Putnam, and Julia Robinson, Ann. of Math. (2), 74 (3), (1961), 425–436. MR0133227 (24 #A3061).

This paper establishes that every recursively enumerable (r.e.) set can be existentially defined in terms of exponentiation. […] These results are superficially related to Hilbert’s tenth problem on (ordinary, i.e., non-exponential) Diophantine equations. The proof of the authors’ results, though very elegant, does not use recondite facts in the theory of numbers nor in the theory of r.e. sets, and so it is likely that the present result is not closely connected with Hilbert’s tenth problem. Also it is not altogether plausible that all (ordinary) Diophantine problems are uniformly reducible to those in a fixed number of variables of fixed degree, which would be the case if all r.e. sets were Diophantine.

Of course, my favorite quote in relation to the tenth problem is from the Foreword by Martin Davis to Yuri Matiyasevich’s Hilbert’s tenth problem.

During the 1960s I often had occasion to lecture on Hilbert’s Tenth Problem. At that time it was known that the unsolvability would follow from the existence of a single Diophantine equation that satisfied a condition that had been formulated by Julia Robinson. However, it seemed extraordinarily difficult to produce such an equation, and indeed, the prevailing opinion was that one was unlikely to exist. In my lectures, I would emphasize the important consequences that would follow from either a proof or a disproof of the existence of such an equation. Inevitably during the question period I would be asked for my own opinion as to how matters would turn out, and I had my reply ready: “I think that Julia Robinson’s hypothesis is true, and it will be proved by a clever young Russian.”

Advertisements

Like this:

LikeLoading...

Related

This entry was posted on Sunday, August 11th, 2013 at 6:06 pm and is filed under Computability theory. You can follow any responses to this entry through the RSS 2.0 feed.
You can leave a response, or trackback from your own site.

Craig: For a while, there was some research on improving bounds on the number of variables or degree of unsolvable Diophantine equations. Unfortunately, I never got around to cataloging the known results in any systematic way, so all I can offer is some pointers to relevant references, but I am not sure of what the current records are. Perhaps the first pape […]

Yes. Consider, for instance, Conway's base 13 function $c$, or any function that is everywhere discontinuous and has range $\mathbb R$ in every interval. Pick continuous bijections $f_n:\mathbb R\to(-1/n,1/n)$ for $n\in\mathbb N^+$. Pick a strictly decreasing sequence $(x_n)_{n\ge1}$ converging to $0$. Define $f$ by setting $f(x)=0$ if $x=0$ or $\pm x_n […]

(1) Patrick Dehornoy gave a nice talk at the Séminaire Bourbaki explaining Hugh Woodin's approach. It omits many technical details, so you may want to look at it before looking again at the Notices papers. I think looking at those slides and then at the Notices articles gives a reasonable picture of what the approach is and what kind of problems remain […]

The description below comes from József Beck. Combinatorial games. Tic-tac-toe theory, Encyclopedia of Mathematics and its Applications, 114. Cambridge University Press, Cambridge, 2008, MR2402857 (2009g:91038). Given a finite set $S$ of points in the plane $\mathbb R^2$, consider the following game between two players Maker and Breaker. The players alternat […]

Yes. This is a consequence of the Davis-Matiyasevich-Putnam-Robinson work on Hilbert's 10th problem, and some standard number theory. A number of papers have details of the $\Pi^0_1$ sentence. To begin with, take a look at the relevant paper in Mathematical developments arising from Hilbert's problems (Proc. Sympos. Pure Math., Northern Illinois Un […]

It is easy to see without choice that if there is a surjection from $A$ onto $B$, then there is an injection from ${\mathcal P}(B)$ into ${\mathcal P}(A)$, and the result follows from Cantor's theorem that $B

Only noticed this question today. Although the selected answer is quite nice and arguably simpler than the argument below, none of the posted answers address what appeared to be the original intent of establishing the inequality using the Arithmetic Mean-Geometric Mean Inequality. For this, simply notice that $$ 1+3+\ldots+(2n-1)=n^2, $$ which can be easily […]

First of all, $f(z)+e^z\ne 0$ by the first inequality. It follows that $e^z/(f(z)+e^z)$ is entire, and bounded above. You should be able to conclude from that.

Yes. The standard way of defining these sequences goes by assigning in an explicit fashion to each limit ordinal $\alpha$, for as long as possible, an increasing sequence $\alpha_n$ that converges to $\alpha$. Once this is done, we can define $f_\alpha$ by diagonalizing, so $f_\alpha(n)=f_{\alpha_n}(n)$ for all $n$. Of course there are many possible choices […]

I disagree with the advice of sending a paper to a journal before searching the relevant literature. It is almost guaranteed that a paper on the fundamental theorem of algebra (a very classical and well-studied topic) will be rejected if you do not include mention on previous proofs, and comparisons, explaining how your proof differs from them, etc. It is no […]