# Thread: Powers of transcendental numbers

1. ## Powers of transcendental numbers

Hi guys,

I'm having trouble proving that if $\pi$ is transcendent, then $\pi^{2k}$ and $\beta\pi$ (where $\beta$ is some non-zero rational constant) are transcendental.

I know that if $\beta\pi$ is algebraic, then it is the root of some polynomial, $X-\beta\pi \in \mathbb{Q}[X]$. Then, if this is a linear polynomial in $\mathbb{Q}[X]$, is it right to conclude that any roots will be in $\mathbb{Q}$? If that is the case, can we then set some other polynomial, $\beta X - \beta\pi \in \mathbb{Q}[X]$ and conclude a contradiction?

For the other one, if we have $\pi^{2k}$ is algebraic, then it is the root of $X-\pi^{2k} \in \mathbb{Q}[X]$. I don't know where to proceed from here, really.

HTale.

2. Originally Posted by HTale
Hi guys,

I'm having trouble proving that if $\pi$ is transcendent, then $\pi^{2k}$ and $\beta\pi$ (where $\beta$ is some non-zero rational constant) are transcendental.

I know that if $\beta\pi$ is algebraic, then it is the root of some polynomial, $X-\beta\pi \in \mathbb{Q}[X]$. Then, if this is a linear polynomial in $\mathbb{Q}[X]$, is it right to conclude that any roots will be in $\mathbb{Q}$? If that is the case, can we then set some other polynomial, $\beta X - \beta\pi \in \mathbb{Q}[X]$ and conclude a contradiction?

For the other one, if we have $\pi^{2k}$ is algebraic, then it is the root of $X-\pi^{2k} \in \mathbb{Q}[X]$. I don't know where to proceed from here, really.

HTale.
The algebraic numbers $\widehat{\mathbb{Q}}$ are an example of an algebraically closed field. If $\beta \pi$ is algebraic then $\tfrac{\beta \pi}{\beta} = \pi$ is algebraic, since $\beta \in \widehat{\mathbb{Q}}$. Also, algebraic numbers contain all $n$-th roots of an algebraic number. Therefore, if $\pi^n$ was algebraic then $\pi$ would be algebraic. Another way to see this is if $\pi^n$ was algebraic then consider the polynomial $X^n - \pi^n \in \widehat{\mathbb{Q}}[X]$. This polynomial splits over the algebraic numbers and so $\pi \zeta^j$ are contained in $\widehat{\mathbb{Q}}$ where $0\leq j\leq n-1, \zeta = e^{2\pi i/n}$, but this is a contradiction.

3. Here is an elementary proof. You need just to know the definition of the algebraic element

Let n be a natural number and t be a trancendent element. If t^n were an algebraic element;

there exists a non zero polynomial, say f(x), over rational numbers field Q such that f(t^n)=0

Say g(x):=f(x^n) (is a non zero polynomial over Q, because f(x) is non zero)

g(t)=f(t^n)=0. But this is a contradiction ( t is trancendent). Then t^n is not algebraic.....