Let’s prove that if , then either is an integer, or else it is irrational. (Cf. Abbott, Understanding analysis, Exercise 1.2.1.) There are many proofs of this fact. I present three.

1.

The standard proof of this fact uses the prime factorization of : There is a unique way of writing as , where the are distinct primes numbers, and the are positive integers (the number corresponds to the empty product, but since is a square, we may as well assume in what follows that ).

We show that if is rational, then in fact each is even, so is actually an integer. Write where are integers that we may assume relatively prime. This gives us that .

Consider any of the primes in the factorization of . Let and be the largest powers of that divide and , respectively, say and where does not divide either of and . Similarly, write , where does not divide ( is what we called above). We have

The point is that since is prime, it does not divide or : If is a prime and divides a product (where are integers), then divides or it divides .

This means that either is even (as we wanted to show), so that , or else (upon dividing both sides of the displayed equation by the smaller of and ), divides one of the two sides of the resulting equation, but not the other, a contradiction.

2.

The above is the standard proof, but there are other arguments that do not rely on prime factorizations. One I particularly like uses Bézout theorem: If is the greatest common divisor of the positive integers and , then there are integers such that .

Suppose . We may assume that are relatively prime, and therefore there are integers such that . The key observation is that . This, coupled with elementary algebra, verifies that

but the latter is an integer, and we are done.

3.

Another nice way of arguing, again by contradiction, is as follows: Suppose that is not an integer, but it is rational. There is a unique integer with , so . Let be the least positive integer such that is an integer, call it . Note that , which gives us a contradiction if is again an integer. But this can be verified by a direct computation:

.

4.

As a closing remark, the three arguments above generalize to show that is either an integer or irrational, for all positive integers . Similarly, if is rational for some positive integers , then both are th powers. (It is a useful exercise to see precisely how these generalizations go.)

The argument you are looking for is given in Kanamori's book, see Theorem 28.15. For the more nuanced version of the lemma, see section 7D in Moschovakis's descriptive set theory book (particularly 7.D.5-8), or section 3.1 in the Koellner-Woodin chapter of the Handbook.

This problem is very much open. Cheng Yong calls Harrington's $\star$ the assumption that there is a real $x$ such that all $x$-admissible ordinals are $L$-cardinals. From the work of Yong we know that Second- and even Third-order arithmetic do not suffice to prove that Harrington's $\star$ implies the existence of $0^\sharp$. Whether this was poss […]

Georgii: Let me start with some brief remarks. In a series of three papers: a. Wacław Sierpiński, "Contribution à la théorie des séries divergentes", Comp. Rend. Soc. Sci. Varsovie 3 (1910) 89–93 (in Polish). b. Wacław Sierpiński, "Remarque sur la théorème de Riemann relatif aux séries semi-convergentes", Prac. Mat. Fiz. XXI (1910) 17–20 […]

What precisely do you mean by a standard model? An $\omega$-model? (That is, a model whose set of natural numbers is isomorphic to $\omega$.) Or a $\beta$-model? (That is, a model whose ordinals are well-ordered.) If the latter, the Mostowski collapse theorem tells us any such model is isomorphic in a unique way to a unique transitive model. If the former, t […]

This is Theorem 39 in the paper (see Theorem 4.(i) for a user-friendly preview). But the fact that $(2^\kappa)^+\to(\kappa^+)^2_\kappa$ is older (1946) and due to Erdős, see here: Paul Erdős. Some set-theoretical properties of graphs, Univ. Nac. Tucumán. Revista A. 3 (1942), 363-367 MR5,151d. (Anyway, it is probably easier to read a more modern presentation, […]

It is not possible to give a finitary proof of this fact. Consider, for instance, the statement that all Goodstein sequences terminate. This is an arithmetic claim, it is independent of $\mathsf{PA}$, and it is equivalent to the claim that some (very specific) computable non-increasing sequences of polynomial towers stabilize. Instead of Goodstein sequences, […]

Given any field automorphism of $\mathbb C$, the rational numbers are fixed. In fact, any number that is explicitly definable in $\mathbb C$ (in the first order language of fields) is fixed. (Actually, this means that we can only ensure that the rationals are fixed, I expand on this below.) Any construction of a wild automorphism uses the axiom of choice. Se […]

A proof can be found in Kanamori's book The higher infinite. The result follows from work of Solovay on the theory of uniform indiscernibles and work of Martin on projective scales. Look at sections 14, 28, and 30 of the book. A different approach (using the infinite partition properties of $\omega_1$ and $\omega_2$, themselves due to Solovay and Martin […]

Forcing with sufficiently homogeneous forcing that adds reals is enough to obtain the negation of $(*)$. The point is that if a formula $\phi$ defines a parameter-free well-ordering of $\mathbb R$, then for any ordinal $\alpha$, the statement "$x$ is the $\alpha$-th real in the well-ordering defined by $\phi$" uniquely characterizes $x$ in terms of […]

Assuming that $\gamma$ is finite, the argument is fairly simple: Suppose $\beta\to(\alpha)^\gamma_\delta$, and fix a bijection $f$ between $|\beta|$ and $\beta$. Consider a coloring $c:[|\beta|]^\gamma\to\delta$. Using $f$ , this gives us a coloring $c':[\beta]^\gamma\to\delta$ (here we used that $\gamma$ is finite). We want to argue that there is a $c$ […]

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