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.)

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, […]

This is a nice problem. Here is what I know. (Below, I refer to the Handbook. This is the Handbook of Set Theory, Foreman, Kanamori, eds., Springer, 2010.) First of all, the consistency of the failure of diamond at a weakly compact cardinal seems open. Woodin has asked this explicitly, I do not know if the question itself is due to him. Of course, $\diamonds […]

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|

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 $c$ […]

An example I like is this: Say two sets of natural numbers are almost disjoint iff they are infinite but their intersection is finite. Suppose $\mathcal F$ is an uncountable family of almost disjoint sets. Then the collection of characteristic functions of the sets in $\mathcal F$ is an example. (If $A$ is a subset of $\mathbb N$, then its characteristic fun […]

The set $\mathrm{Def}(X)$ consists of all subsets of $X$ first-order definable in the structure $(X,\in)$ from parameters. What matters here is that there are only countably many formulas, and only $|X^{

There is no slowest divergent series. Let me take this to mean that given any sequence $a_n$ of positive numbers converging to zero whose series diverges, there is a sequence $b_n$ that converges to zero faster and the series also diverges, where "faster" means that $\lim b_n/a_n=0$. In fact, given any sequences of positive numbers $(a_{1,n}), (a_{ […]

Yes. In fact, by a counting argument, most dense co-dense sets are neither $G_\delta$ nor $F_\sigma$. The point is that there are exactly as many $G_\delta$ or $F_\sigma$ as there are real numbers, but there are as many dense co-dense sets as there are sets of real numbers. In somewhat more detail: There are only countably many rationals, so there are counta […]

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