Which irrationals become rational for some positive integer power?

Question

Related to Irrationals becoming rationals after being raised to some power. Let $r \in \mathbb{R} \setminus \mathbb{Q}$. True or false: there exists an $n \in \mathbb{N}$ (positive integers) such that $r^n = r \cdot \dots \cdot r \in \mathbb{Q}$. This is clearly true for some irrationals like $\sqrt{2}$ or $a^{1/n}$ (positive integer $a$ such that $a^{1/n} \notin \mathbb{Z}$; see How to prove: if $a,b \in \mathbb N$, then $a^{1/b}$ is an integer or an irrational number?). But is it true for ALL irrationals? If not, can we classify all the irrationals for which the statement is true?

Answer 1

Every number whose $n$'th power is rational is algebraic, with a minimal polynomial that divides $X^n-q$ for some positive integer $n$ and rational $q$. In particular, all its conjugates are $n$'th roots of the same $q$.

EDIT: The minimal polynomial (over the rationals) of an algebraic number $\alpha$ is a polynomial $P(X)$ of lowest possible degree such that the coefficients are rational numbers, the leading coefficient (i.e. the coefficient for the highest power of $X$) is $1$, and $P(\alpha) = 0$. For example, the minimal polynomial of $\sqrt{2}$ is $X^2 - 2$. The conjugates of $\alpha$ are all the roots of the minimal polynomial: in this example they are $\alpha$ and $-\alpha$. The minimal polynomial of $\alpha$ is the product of $X - r$ for each conjugate $r$ of $\alpha$.

Answer 2

The easiest way to see that not every irrational number $r \in \mathbb R\backslash\mathbb Q$ has a natural exponent $n$ with $r^n=q \in \mathbb Q$ is to remember that the set of ordered pairs $(n,q)$ with $n \in \mathbb Z, n > 0, q \in \mathbb Q$ is still a countable infinite set, that means has the same cardinality as the natural numbers, integers and rational numbers.

If you don't know that fact, take a look at several proofs that the rational numbers are indeed countable. The idea behind proof number 2 is easily generalizable for this case.

Now for each such pair $(n,q)$, there are at most 2 real numbers with $r^n=q$, because the function $y=x^n$ is monotonically increasing over the whole $\mathbb R$ for odd $n$, so given $n$ and $q$ there is at most one such $r$. For even $n$, it is decreasing for negative $x$ and increasing for positive $x$, so there might be 2 such $r$.

So each irrational number $r \in \mathbb R\backslash\mathbb Q$ that has a natural exponent $n$ with $r^n=q \in \mathbb Q$ can be found by going through the list of all pairs $(n,q)$, then looking at at most 2 numbers that become $q$ when taken to the $n$-the exponent.

In other words, the number of such $r$ is at most countably infinite.

Now the reals are not countable (see here), that means there are more (much more) real numbers than any countable set. So the number of "such" $r$ is countable, the rationals are countable, so together they are still countable, so there must be numbers missing. So, there are "plenty" of irrationals that when taken to any natural power > $0$ never yield a rational number as result.

Answer 3

You have to find the transcendental numbers like exponential number, π, sin(a) , sin h(a) etc. which make your statement wrong! One thing that you have to remind, every transcendental numbers are irrational numbers, but converse is not always true i.e, there can be irrationals which are not transcendental numbers, like the numbers what you have given and these are part of algebraic numbers, which are suitably described in 1st answer!
So, for your question, only algebraic irrational numbers can be transformed into rational numbers by repeated multiplication as many times you need. But transcendental irrational numbers can't be!
Also cardinality of the set of these transcendental irrational numbers is uncountable!
But on the other hand, cardinality of the set of these algebraic rational numbers is countable!