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.)

Yes. This is a consequence of the Davis-Matiyasevich-Putnam-Robinson work on Hilbert's 10th problem, and some standard number theory. A number of papers have details of the $\Pi^0_1$ sentence. To begin with, take a look at the relevant paper in Mathematical developments arising from Hilbert's problems (Proc. Sympos. Pure Math., Northern Illinois Un […]

I am looking for references discussing two inequalities that come up in the study of the dynamics of Newton's method on real-valued polynomials (in one variable). The inequalities are fairly different, but it seems to make sense to ask about both of them in the same post. Most of the details below are fairly elementary, they are mostly included for comp […]

Let $C$ be the standard Cantor middle-third set. As a consequence of the Baire category theorem, there are numbers $r$ such that $C+r$ consists solely of irrational numbers, see here. What would be an explicit example of a number $r$ with this property? Short of an explicit example, are there any references addressing this question? A natural approach would […]

Not necessarily. That $\mathfrak m$ is consistently singular is proved in MR0947850 (89m:03045) Kunen, Kenneth. Where $\mathsf{MA}$ first fails. J. Symbolic Logic 53(2), (1988), 429–433. There, Ken shows that $\mathfrak{m}$ can be singular of cofinality $\omega_1$. (Both links above are behind paywalls.)

Ignas: It is not possible to provide an explicit expression for a non-linear solution. The reason is that (it is a folklore result that) an additive $f:{\mathbb R}\to{\mathbb R}$ is linear iff it is measurable. (This result can be found in a variety of places, it is a standard exercise in measure theory books. As of this writing, there is a short proof here. […]

The question immediately reminded me of this. Here is an argument following the same basic idea at the beginning of that argument: First, consider $B=\{x\in A\mid A\cap(-\infty,x]$ is countable$\}$, and note that $B$ itself is countable: The point is that if $x\in B$ then $A\cap(-\infty,x]\subseteq B$. Now, if $B\ne\emptyset$, let $t=\sup B$, fix an increasi […]

What is the context? In the setting of analysis, $\sum_{i\in\mathbb N}x_i$ is defined as usual; other than that, the infinite sum $\sum_{i\in I}x_i$ is defined only when $\sum_i|x_i|$ is defined (so, we do not have a proper theory of conditionally convergent series). Assume then that the $x_i$ are non-negative, in which case $\sum_i x_i$ is defined as the un […]

There are several nice proofs of the result. The most intuitive I'm aware of is the following: We may as well assume that $A$ and $B$ are disjoint. Consider the directed graph whose set of vertices is $A\cup B$, in which you add an edge from $a$ to $b$ precisely if $a\in A$, $b\in B$, and $f(a)=b$, or $a\in B$, $b\in A$, and $g(a)=b$. Now consider the c […]

A forcing collapses cardinals iff (by definition) some cardinal of the ground model is no longer a cardinal in the forcing extension. Naturally, this means that there is some $\kappa$ in the ground model whose cardinality in the extension is strictly smaller than $\kappa$ (e.g., let $\kappa$ be the first cardinal that witnesses the definition above). Note th […]

$\mathrm{HOD}$ always contains $L$ because any inner model contains $L$, by absoluteness. How easy it is to exhibit a difference really depends on your background. For instance, $0^\sharp$, if it exists, is a real that always belongs to $\mathrm{HOD}$ but is not in $L$. If you are not too comfortable with large cardinals, but know forcing, you may enjoy prov […]

[…] Solution to 1.2.1. […]