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.)

As suggested by Gerald, the notion was first introduced for groups. Given a directed system of groups, their direct limit was defined as a quotient of their direct product (which was referred to as their "weak product"). The general notion is a clear generalization, although the original reference only deals with groups. As mentioned by Cameron Zwa […]

A database of number fields, by Jürgen Klüners and Gunter Malle. (Note this is not the same as the one mentioned in this answer.) The site also provides links to similar databases.

As the other answer indicates, the yes answer to your question is known as the De Bruijn-Erdős theorem. This holds regardless of the size of the graph. The De Bruijn–Erdős theorem is a particular instance of what in combinatorics we call a compactness argument or Rado's selection principle, and its truth can be seen as a consequence of the topological c […]

Every $P_c$ has the size of the reals. For instance, suppose $\sum_n a_n=c$ and start by writing $\mathbb N=A\cup B$ where $\sum_{n\in A}a_n$ converges absolutely (to $a$, say). This is possible because $a_n\to 0$: Let $m_0

Consider a subset $\Omega$ of $\mathbb R$ of size $\aleph_1$ and ordered in type $\omega_1$. (This uses the axiom of choice.) Let $\mathcal F$ be the $\sigma$-algebra generated by the initial segments of $\Omega$ under the well-ordering (so all sets in $\mathcal F$ are countable or co-countable), with the measure that assigns $0$ to the countable sets and $1 […]

Sure. A large class of examples comes from the partition calculus. A simple result of the kind I have in mind is the following: Any infinite graph contains either a copy of the complete graph on countably many vertices or of the independent graph on countably many vertices. However, if we want to find an uncountable complete or independent graph, it is not e […]

I think that, from a modern point of view, there is a misunderstanding in the position that you suggest in your question. Really, "set theory" should be understood as an umbrella term that covers a whole hierarchy of ZFC-related theories. Perhaps one of the most significant advances in foundations is the identification of the consistency strength h […]

I'll only discuss the first question. As pointed out by Asaf, the argument is not correct, but something interesting can be said anyway. There are a couple of issues. A key problem is with the idea of an "explicitly constructed" set. Indeed, for instance, there are explicitly constructed sets of reals that are uncountable and of size continuum […]

The question seems to be: Assume that there is a Vitali set $V$. Is there an explicit bijection between $V$ and $\mathbb R$? The answer is yes, by an application of the Cantor-Schröder-Bernstein theorem: there is an explicit injection from $\mathbb R$ into $\mathbb R/\mathbb Q$ (provably in ZF, this requires some thought, or see the answers to this question) […]

If a set $X$ is well-founded (essentially, if it contains no infinite $\in$-descending chains), then indeed $\emptyset$ belongs to its transitive closure, that is, either $X=\emptyset$ or $\emptyset\in\bigcup X$ or $\emptyset\in\bigcup\bigcup X$ or... However, this does not mean that there is some $n$ such that the result of iterating the union operation $n$ […]

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