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.)

You assume $\omega_\alpha\subseteq M$ and $X\in M$ so that $X$ belongs to the transitive collapse of $M$ (because if $\pi$ is the collapsing map, $\pi(X)=\pi[X]=X$. You assume $|M|=\aleph_\alpha$ so that the transitive collapse of $M$ has size $\aleph_\alpha$. Since you also have that this transitive collapse is of the form $L_\beta$ for some $\beta$, it fol […]

Perhaps the following may clarify the comments: for any ordinal $\delta$, there is a Boolean-valued extension of the universe of sets where $2^{\aleph_0}>\aleph_\delta$ holds. If you rather talk of models than Boolean-valued extensions, what this says is that we can force while preserving all ordinals, and in fact all initial ordinals, and make the contin […]

I do not know of any active set theorists who think large cardinals are inconsistent. At least, within the realm of cardinals we have seriously studied. [Reinhardt suggested an ultimate axiom of the form "there is a non-trivial elementary embedding $j:V\to V$". Though some serious set theorists found it of possible interest immediately following it […]

There is a fantastic (and not too well-known) result of Shelah stating that $L({\mathcal P}(\lambda))$ is a model of choice whenever $\lambda$ is a singular strong limit of uncountable cofinality. This is a consequence of a more general theorem that can be found in 4.6/6.7 of "Set Theory without choice: not everything on cofinality is possible", Ar […]

In set theory, definitely the notion of a Woodin cardinal. First, it is not an entirely straightforward notion to guess. Significant large cardinals were up to that point defined as critical points of certain elementary embeddings. This is not the case here: Woodin cardinals need not be measurable. If $\kappa$ is Woodin, then $V_\kappa$ is a model of set the […]

The precise consistency strength of the global failure of the generalized continuum hypothesis is somewhat technical to state. As far as I know, it has not been published, but I think we have a decent understanding of what the correct statement should be. The most relevant paper towards this result is MR2224051 (2007d:03082). Gitik, Moti Merimovich, Carmi. P […]

There are integrable functions that are not derivatives: Any function that is continuous except at a single point, where it has a jump discontinuity, is an example. (Derivatives have the intermediate value property.) More interestingly, we can ask whether the existence of an antiderivative ensures integrability. The answer depends on what integral you are co […]

$0^¶$ is the sharp for an inner model with a strong cardinal in the same sense that $0^\dagger$ is the sharp for an inner model with a measurable cardinal. In terms of mice, this is the first mouse containing two overlapping extenders. The effect of this is that, by iterating its top measure throughout the ordinals, you extend the bottom extender in a variet […]

Clearly $\omega^\omega\le(\omega+n)^\omega$. Also, $(\omega+n)^\omega\le (\omega^2)^\omega=\omega^{(2\cdot \omega)}=\omega^\omega$, and the equality follows. If you do not feel comfortable with the move from $(\omega^2)^\omega$ to $\omega^{(2\cdot \omega)}$, simply note the left-hand side is $\omega\cdot\omega\cdot\omega\cdot\dots$, where there are $2\cdot\o […]

A function $f:\mathbb N\to\mathbb R$ is $2^{O(n)}$ if and only if there is a constant $C$ such that for all $n$ large enough we have $f(n)\le 2^{Cn}$. We can think of the $O$ notation as decribing a family of functions. So, $2^{O(n)}$ would be the family of functions satisfying the requirements just indicated. In contrast, a function $f$ is $O(2^n)$ if and o […]

[…] Solution to 1.2.1. […]