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.)

This is a very interesting question (and I really want to see what other answers you receive). I do not know of any general metatheorems ensuring that what you ask (in particular, about consistency strength) is the case, at least under reasonable conditions. However, arguments establishing the proof theoretic ordinal of a theory $T$ usually entail this. You […]

This is false; take a look at https://en.wikipedia.org/wiki/Analytic_set for a quick introduction. For details, look at Kechris's book on Classical Descriptive Set Theory. There you will find also some information on the history of this result, how it was originally thought to be true, and how the discovery of counterexamples led to the creation of desc […]

This is open. In $L(\mathbb R)$ the answer is yes. Hugh has several proofs of this, and it remains one of the few unpublished results in the area. The latest version of the statement (that I know of) is the claim in your parenthetical remark at the end. This gives determinacy in $L(\mathbb R)$ using, for example, a reflection argument. (I mentioned this a wh […]

A classical reference is Hypothèse du Continu by Waclaw Sierpiński (1934), available through the Virtual Library of Science as part of the series Mathematical Monographs of the Institute of Mathematics of the Polish Academy of Sciences. Sierpiński discusses equivalences and consequences. The statements covered include examples from set theory, combinatorics, […]

There is a new journal of the European Mathematical Society that seems perfect for these articles: EMS Surveys in Mathematical Sciences. The description at the link reads: The EMS Surveys in Mathematical Sciences is dedicated to publishing authoritative surveys and high-level expositions in all areas of mathematical sciences. It is a peer-reviewed periodical […]

You may be interested in the following paper: Lorenz Halbeisen, and Norbert Hungerbühler. The cardinality of Hamel bases of Banach spaces, East-West Journal of Mathematics, 2, (2000) 153-159. There, Lorenz and Norbert prove a few results about the size of Hamel bases of arbitrary infinite dimensional Banach spaces. In particular, they show: Lemma 3.4. If $K\ […]

You just need to show that $\sum_{\alpha\in F}\alpha^k=0$ for $k=0,1,\dots,q-2$. This is clear for $k=0$ (understanding $0^0$ as $1$). But $\alpha^q-\alpha=0$ for all $\alpha$ so $\alpha^{q-1}-1=0$ for all $\alpha\ne0$, and the result follows from the Newton identities.

Nice question. Let me first point out that the Riemann Hypothesis and $\mathsf{P}$-vs-$\mathsf{NP}$ are much simpler than $\Pi^1_2$: The former is $\Pi^0_1$, see this MO question, and the assertion that $\mathsf{P}=\mathsf{NP}$ is a $\Pi^0_2$ statement ("for every code for a machine of such and such kind there is a code for a machine of such other kind […]

For brevity's sake, say that a theory $T$ is nice if $T$ is a consistent theory that can interpret Peano Arithmetic and admits a recursively enumerable set of axioms. For any such $T$, the statement "$T$ is consistent" can be coded as an arithmetic statement (saying that no number codes a proof of a contradiction from the axioms of $T$). What […]

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