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You are right to view the Goldbach conjecture as having a particularly simple logical form. Such statements of the form "for every $n$, property $P(n)$ holds", where $P$ is a particularly simple state …

Nice question, Erin. Here is one quick easy thing to say. If $\pi$ and $e$ disagree in infinitely many digits, then there are continuum many choices of the particular subset of those digits to swap, …

The other answers and comments are fascinating, particularly about the irrationality measure, but allow me to give a little more information along the lines of Mark Sapir's answer by mentioning that t …

Goedel did indeed use the Chinese remainder theorem in his proof of the Incompleteness theorem. This was used in what you describe as the `boring' part of the proof, the arithmetization of syntax. Con …

The integers can indeed be defined in the rational field, but not in the real field. $\newcommand\Q{\mathbb{Q}}\newcommand\Z{\mathbb{Z}}\newcommand\R{\mathbb{R}}$ The question can be made precise by i …

Perhaps an example of the kind of circularity you mention arises with the self-reference phenomenon that arises in connection with the incompleteness theorems and related applications. Specifically, G …

To my way of thinking, the two notions of undecidability are closely related, and the associated undecidability phenomenon and independence phenomenon, which are both pervasive in mathematics, are dee …

Yes, this line of thought is perfectly fine. A set is decidable if and only if it has complexity $\Delta_1$ in the arithmetic hiearchy, which provides a way to measure the complexity of a definable s …

I have heard that Jack Silver's discovery of zero sharp ($0^\#$) was part of his attempt to show measurable cardinals inconsistent. Instead of finding the long-sought-after contradiction, however, he …

One method, used in proving the Paris-Harrington result, a statement of Ramsey theory that is independent of PA, works roughly as follows. The statement to be proved has the form $\forall n\ \exists k …

Presburger arithmetic admits elimination of quantifiers, if one expands the language to include truncated minus and the unary relations for divisibility-by-2, divisibility-by-3 and so on, which are de …

I believe that there are many instances of this phenomenon in set theory, where an elaborate theory is developed over a period of years by many people, even though the theory is not viewed ultimately …

The MRDP solution of Hilbert's 10th problem establishes that the integer solution sets $\{\, n\, |\, \exists\vec n\, p(n,\vec n)=0\}$ of diophantine equations $p(n,\vec n)=0$ are exactly the computabl …

The answer is no; this generalization is inconsistent, even with just two colors, and with $F=\{\omega,\omega^2\}$ of size two. Theorem. There is a coloring of ordinals with two colors, such that for …

Consider the arithmetic progression game, a two-player game of perfect information, in which the players take turns playing natural numbers, or finite sets of natural numbers, all distinct, and the fi …