117b – Undecidability and Incompleteness – Lecture 4

We showed that is Diophantine. This implies that the prime numbers are Diophantine. Therefore, there is a polynomial in several variables with integer coefficients whose range, when intersected with the natural numbers, coincides with the set of primes. Amusingly, no nonconstant polynomial with integer coefficients can only take prime values.

We proved the bounded quantifier lemma, the last technical component of the proof of the undecidability of Hilbert’s tenth problem. It implies that the class of relations definable by a formula in the structure coincides with the class of relations (i.e., the Diophantine ones).

To complete the proof of the undecidability of the tenth problem, we will show that any c.e. relation is Diophantine. For this, recall that a set is c.e. iff it is the domain of a Turing machine. We proceeded to code the behavior of Turing machines by means of a Diophantic representation. This involves coding configurations of Turing machines and it remains to show how to code the way a configuration changes into another one.

Advertisements

Like this:

LikeLoading...

Related

This entry was posted on Thursday, February 1st, 2007 at 2:51 pm and is filed under 117b: 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.

4 Responses to 117b – Undecidability and Incompleteness – Lecture 4

I was a bit confused about this in class, but it’s probably just because I was majorly sleep-deprived: “Therefore, there is a polynomial in several variables with integer coefficients whose range on the natural numbers coincides with the prime numbers. Amusingly, no nonconstant polynomial with integer coefficients can only take prime values.”

The distinction, then, is that there IS a polynomial whose range is all the primes AND a bunch of negatives, but there is NO polynomial whose range is primes only? Is that correct?

Yes, exactly. I think I may have actually written something incorrect on the board, so I will briefly review it on Tuesday. We have that the primes are Diophantine. From this, we saw how to build a polynomial P such that, if R is its range, then the intersection of R with the naturals is the set of primes. But any such R necessarily takes negative values (and 0?) as well. I rephrased the entry a little to make this clear.

If anyone’s interested, I remembered reading about this polynomial (though I didn’t know the historical context in which it was derived) a while back in the book The Music of the Primes. According to Amazon’s search-inside-this-book function, it’s on page 200, which I have downloaded from Amazon and is now stored on my web page. It’s good to see these things actually written out!

Thanks!
The same polynomial is shown in page 331 of the handout “Hilbert’s tenth problem. Diophantine equations: positive aspects of a negative solutions”, it is due to J. P. Jones. The polynomial we obtain if we simply apply the definitions shown in lecture is different than this one.
It is curious that this polynomial is written as the product of two polynomials, and yet its (positive) values are precisely the prime numbers.

I am not sure which statement you heard as the "Ultimate $L$ axiom," but I will assume it is the following version: There is a proper class of Woodin cardinals, and for all sentences $\varphi$ that hold in $V$, there is a universally Baire set $A\subseteq{\mathbb R}$ such that, letting $\theta=\Theta^{L(A,{\mathbb R})}$, we have that $HOD^{L(A,{\ma […]

A Wadge initial segment (of $\mathcal P(\mathbb R)$) is a subset $\Gamma$ of $\mathcal P(\mathbb R)$ such that whenever $A\in\Gamma$ and $B\le_W A$, where $\le_W$ denotes Wadge reducibility, then $B\in\Gamma$. Note that if $\Gamma\subseteq\mathcal P(\mathbb R)$ and $L(\Gamma,\mathbb R)\models \Gamma=\mathcal P(\mathbb R)$, then $\Gamma$ is a Wadge initial se […]

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 study of finite choice axioms is quite interesting. Besides the reference given in Asaf's answer, there are a few papers covering this topic in detail. If you can track it down, I suggest you read MR0360275 (50 #12725) Reviewed. Conway, J. H. Effective implications between the "finite'' choice axioms. In Cambridge Summer School in Mat […]

I feel this question may be a duplicate, because I am pretty certain I first saw the paper I mention below in an answer here. You may be interested in reading the following: MR2141502 (2006c:68092) Reviewed. Calude, Cristian S.(NZ-AUCK-C); Jürgensen, Helmut(3-WON-C). Is complexity a source of incompleteness? (English summary), Adv. in Appl. Math. 35 (2005), […]

The smallest such ordinal is $0$ because you defined your rank (height) inappropriately (only successor ordinals are possible). You want to define the rank of a node without successors as $0$, and of a node $a$ with successors as the supremum of the set $\{\alpha+1\mid\alpha$ is the rank of an immediate successor of $a\}$. With this modification, the smalles […]

The perfect reference for this is MR2562557 (2010j:03061) Reviewed. Steel, J. R.(1-CA). The derived model theorem. In Logic Colloquium 2006. Proceedings of Annual European Conference on Logic of the Association for Symbolic Logic held at the Radboud University, Nijmegen, July 27–August 2, 2006, S. B. Cooper, H. Geuvers, A. Pillay and J. Väänänen, eds., Lectu […]

I was a bit confused about this in class, but it’s probably just because I was majorly sleep-deprived: “Therefore, there is a polynomial in several variables with integer coefficients whose range on the natural numbers coincides with the prime numbers. Amusingly, no nonconstant polynomial with integer coefficients can only take prime values.”

The distinction, then, is that there IS a polynomial whose range is all the primes AND a bunch of negatives, but there is NO polynomial whose range is primes only? Is that correct?

Yes, exactly. I think I may have actually written something incorrect on the board, so I will briefly review it on Tuesday. We have that the primes are Diophantine. From this, we saw how to build a polynomial P such that, if R is its range, then the intersection of R with the naturals is the set of primes. But any such R necessarily takes negative values (and 0?) as well. I rephrased the entry a little to make this clear.

If anyone’s interested, I remembered reading about this polynomial (though I didn’t know the historical context in which it was derived) a while back in the book The Music of the Primes. According to Amazon’s search-inside-this-book function, it’s on page 200, which I have downloaded from Amazon and is now stored on my web page. It’s good to see these things actually written out!

Thanks!

The same polynomial is shown in page 331 of the handout “Hilbert’s tenth problem. Diophantine equations: positive aspects of a negative solutions”, it is due to J. P. Jones. The polynomial we obtain if we simply apply the definitions shown in lecture is different than this one.

It is curious that this polynomial is written as the product of two polynomials, and yet its (positive) values are precisely the prime numbers.