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 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. […]

MR2449474 (2009j:03067) Woodin, W. Hugh. A tt version of the Posner-Robinson theorem. Computational prospects of infinity. Part II. Presented talks, 355–392, Lect. Notes Ser. Inst. Math. Sci. Natl. Univ. Singap., 15, World Sci. Publ., Hackensack, NJ, 2008. The proof is nice, invoking both recursion-theoretic and set-theoretic tools. Hugh uses a Prikry-like f […]

There is a general argument without choice: Suppose ${\mathfrak m}+{\mathfrak m}={\mathfrak m}$, and ${\mathfrak m}+{\mathfrak n}=2^{\mathfrak m}$. Then ${\mathfrak n}=2^{\mathfrak m}.\,$ This gives the result. The argument is part of a nice result of Specker showing that if CH holds for both a cardinal ${\mathfrak m}$ and its power set $2^{\mathfrak m}$, th […]

By Fermat's little theorem, $t^p\equiv t\pmod p$ for any $t$, so if $p\mid x^p+y^p$ then in fact $p\mid x+y$ (or $y\equiv -x\pmod p$). Now, since $p$ is odd, $$x^p+y^p=(x+y)(x^{p-1}-x^{p-2}y+x^{p-3}y^2-\cdots+y^{p-1}).$$ The first term is a multiple of $p$, as explained above. The second is $$\sum_{k=0}^{p-1}(-1)^k x^{p-1-k}y^k=\sum_{k=0}^{p-1}x^{p-1-k} […]

Your idea is sound, but it requires more work. As pointed out, you have only described so far a very small subcollection of the Borel sets. Instead, show that you can associate to each Borel set a code that keeps track of the "history" of its construction starting from basic open sets, and then count the number of such codes. There is a lot of leew […]

Recently, I was reading Hardy's Orders of Infinity (available here or here): Godfrey Harold Hardy. Orders of infinity. The Infinitärcalcül of Paul du Bois-Reymond. Reprint of the 1910 edition. Cambridge Tracts in Mathematics and Mathematical Physics, No. 12. Hafner Publishing Co., New York, 1971. MR0349922 (50 #2415). The book discusses this result, so […]

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