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.

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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 thought about this question a while ago, while teaching a topics course. Since one can easily check that $${}|{\mathbb R}|=|{\mathcal P}({\mathbb N})|$$ by a direct construction that does not involve diagonalization, the question can be restated as: Is there a proof of Cantor's theorem that ${}|X|

First of all, note (as Monroe does in his question) that if $\mathbb P,\mathbb Q$ are ccc, then $\mathbb P\times\mathbb Q$ is $\mathfrak c^+$-cc, as an immediate consequence of the Erdős-Rado theorem $(2^{\aleph_0})^+\to(\aleph_1)^2_2$. (This is to say, if $\mathbb P$ and $\mathbb Q$ do not admit uncountable antichains, then any antichain in their product ha […]

The technique of almost disjoint forcing was introduced in MR0289291 (44 #6482). Jensen, R. B.; Solovay, R. M. Some applications of almost disjoint sets. In Mathematical Logic and Foundations of Set Theory (Proc. Internat. Colloq., Jerusalem, 1968), pp. 84–104, North-Holland, Amsterdam, 1970. Fix an almost disjoint family $X=(x_\alpha:\alpha

At the moment most of those decisions come from me, at least for computer science papers (those with a 68 class as primary). The practice of having proceedings and final versions of papers is not exclusive to computer science, but this is where it is most common. I've found more often than not that the journal version is significantly different from the […]

The answer is no in general. For instance, by what is essentially an argument of Sierpiński, if $(X,\Sigma,\nu)$ is a $\sigma$-finite continuous measure space, then no non-null subset of $X$ admits a $\nu\times\nu$-measurable well-ordering. The proof is almost verbatim the one here. It is consistent (assuming large cardinals) that there is an extension of Le […]

R. Solovay proved that the provably $\mathbf\Delta^1_2$ sets are Lebesgue measurable (and have the property of Baire). A set $A$ is provably $\mathbf\Delta^1_2$ iff there is a real $a$, a $\Sigma^1_2$ formula $\phi(x,y)$ and a $\Pi^1_2$ formula $\psi(x,y)$ such that $$A=\{t\mid \phi(t,a)\}=\{t\mid\psi(t,a)\},$$ and $\mathsf{ZFC}$ proves that $\phi$ and $\psi […]

A notion now considered standard of primitive recursive set function is introduced in MR0281602 (43 #7317). Jensen, Ronald B.; Karp, Carol. Primitive recursive set functions. In 1971 Axiomatic Set Thoory (Proc. Sympos. Pure Math., Vol. XIII, Part I, Univ. California, Los Angeles, Calif., 1967) pp. 143–176 Amer. Math. Soc., Providence, R.I. The concept is use […]

The power of a set is its cardinality. (As opposed to its power set, which is something else.) As you noticed in the comments, Kurepa trees are supposed to have countable levels, although just saying that a tree has size and height $\omega_1$ is not enough to conclude this, so the definition you quoted is incomplete as stated. Usually the convention is that […]

The key problem in the absence of the axiom of replacement is that there may be well-ordered sets $S$ that are too large in the sense that they are longer than any ordinal. In that case, the collection of ordinals isomorphic to an initial segment of $S$ would be the class of all ordinals, which is not a set. For example, with $\omega$ denoting as usual the f […]

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.