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.)

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|

(I am replacing prior nonsense with a completely different suggestion. I am also turning this into CW so details can be added by somebody with time (which, sadly, most likely won't be me). Comments prior to Feb. 9, 2011, refer to said prior nonsense.) Start with $V=L$ and force to add a Mathias real $s$. Let $W$ be the resulting extension. Let $A$ be th […]

(1) Patrick Dehornoy gave a nice talk at the Séminaire Bourbaki explaining Hugh Woodin's approach. It omits many technical details, so you may want to look at it before looking again at the Notices papers. I think looking at those slides and then at the Notices articles gives a reasonable picture of what the approach is and what kind of problems remain […]

It is open whether the continuum hypothesis for an infinite set $E$ implies the well-orderability of $E$. Of course, if $CH(E)$ holds, then the assumption in your (first) statement holds. ($CH(E)$ is the statement that any subset $A$ of $\mathcal P(E)$, either $A$ injects into $E$, or else $A$ is in bijection with $\mathcal P(E)$.) This is a question that da […]

I saw a while ago in a book by Clifford Pickover, that whether $\displaystyle \sum_{n=1}^\infty\frac1{n^3\sin^2 n}$ converges is open. I would think that the question of its convergence is really about the density in $\mathbb N$ of the sequence of numerators of the standard convergent approximations to $\pi$ (which, in itself, seems like an interesting quest […]

As indicated in the other answer, there are countably many undecidable statements (the largest possible number), and this is an immediate consequence of the incompleteness theorem. The version of incompleteness that Andrews is using is that if $T$ is a consistent, recursive set of axioms that interprets a modicum of arithmetic, then the set of $T$-decidable […]

Yes, this is inconsistent, and your argument is a way of proving this. In fact, $\mathsf{AD}^*$ is an overkill, and we can prove that there is an undetermined game with $A=\omega_1$. To see this, note that either there is an undetermined game on integers (and we are done), or else $\mathsf{AD}$ holds, so $\omega_1$ does not inject into $\mathbb R$. (The reas […]

For each $i=1,\dots,l$, you have a function $n_i:\{1,\dots,k_i\}\to\{1,\dots,k\}$ where $k_i$ is the arity of $g_i$. Given $x_1,\dots,x_k$, you then define $$ x^i_m=x_{n_i(m)}. $$ So, for instance, you could have $E$-rud functions $h,g,j$ and define $f$ by $$ f(x_1,x_2,x_3,x_4,x_5)=h(g(x_1,x_3,x_5),j(x_2,x_2,x_2,x_1)), $$ with the relevant schema asserting t […]

The answer to your question is negative. The reason is that there are monotone continuous functions $f$ that are singular, that is, for a.e. $x$ we have that $f'(x)$ exists and equals $0$. An example is the Cantor function $f:[0,1]\to[0,1]$: This function is continuous and increasing, with $f(0)=0$, $f(1)=1$, and $f'(x)=0$ for all $x$ in the comple […]

And Alexander Soifer ( uccs.edu/asoifer/index.…) seems to be working on a book that will have a comprehensive list. But this may be years away. 9 hours ago

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