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.)

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|

(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 […]

Here is a nice example I recently discussed in lecture. Recall that a function $ f:\mathbb R\to \mathbb R $ is Baire one iff it is the pointwise limit of a sequence of continuous functions. These functions do not need to be continuous, but Baire proved that they always have uncountably many points of continuity; in fact, the set of points of continuity of $ […]

Prove first that given any $\alpha$ large enough, and any $T\subset V_\alpha$ of size less than $\kappa$, there is a $Y\prec V_\alpha$ with $T\subset Y$, $|Y|

Here are three examples: Let $C$ be the Cantor set. For each interval $(a,b)$ contiguous to $C$, define $f$ on $[a,b]$ by $$ f(x)=\frac{2(x-a)}{b-a}-1, $$ so $f$ maps the interval to $[-1,1]$. Otherwise, let $f(x)=0$. Write each $x\in(0,1)$ in binary: $x=0.a_1a_2a_3\dots$, not terminating in a string of $1$s, and define $$f(x)=\limsup_{n\to\infty} \frac{a_1+ […]

Let $C_0=X$ and $C_{n+1}=f(C_n)$ for all $n$. By induction, note that $C_n\supseteq C_{n+1}$ for all $n$. Let $C=\bigcap_n C_n$, and note that $C$ is nonempty, being the intersection of a decreasing sequence of nonempty compact sets. To check that $f(C)=C$, consider a point $x\in C$. Since $x\in C_{n+1}$ for all $n$, then there are points $x_n\in C_n$ with $ […]

Show: Every open set is union of balls with rational radius and rational center. Every open ball is a countable union of closed balls. This gives (ii). For (i), given two points in your open set, say that they are equivalent iff there is a continuous path between them, completely contained in the open set. Argue that this is indeed an equivalence relation, a […]

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