Jakayla Benton

2022-04-24

Show that $\sqrt[3]{\pi}$ is transcendental

ritmesysv

Beginner2022-04-25Added 12 answers

Step 1

Suppose $a=\sqrt[n]{\pi}$ is the root of a polynomial $P\left(x\right)$ with rational coefficients of positive degree.

Writing $P\left(a\right)=0$ and using that ${a}^{n}=\pi$, powers $a}^{k$ with $k>n$ can be replaced with ${a}^{k}={\pi}^{\frac{k}{n}}\xb7{\sqrt[n]{\pi}}^{k\mathrm{mod}n}$

It follows that there exist rational polynomials in $\pi \text{}{a}_{j}\mid {,}_{j=0,1,\cdots ,n-1}$ such that:

1) ${a}_{n-1}{\sqrt[n]{\pi}}^{\u200an-1}+{a}_{n-2}{\sqrt[n]{\pi}}^{\u200an-2}+\cdots +{a}_{1}\sqrt[n]{\pi}+{a}_{0}=0$

Multiply the above by $\sqrt[n]{\pi}$ successively $n-1$ times:

2) ${a}_{n-2}{\sqrt[n]{\pi}}^{\u200an-1}+{a}_{n-3}{\sqrt[n]{\pi}}^{\u200an-2}+\cdots +{a}_{0}\sqrt[n]{\pi}+{a}_{n-1}\pi =0$

${a}_{n-3}{\sqrt[n]{\pi}}^{\u200an-1}+{a}_{n-2}{\sqrt[n]{\pi}}^{\u200an-2}+\cdots +{a}_{n-1}\pi \u200a\sqrt[n]{\pi}+{a}_{n-2}\pi =0\cdots $

${a}_{0}{\sqrt[n]{\pi}}^{\u200an-1}+{a}_{n-1}\pi \u200a{\sqrt[n]{\pi}}^{\u200an-2}+\cdots +{a}_{2}\pi \u200a\sqrt[n]{\pi}+{a}_{1}\pi =0$

Considering the n equations (1) and (2) as a linear system in $1,\text{}\sqrt[n]{\pi},\text{}{\sqrt[n]{\pi}}^{2},\text{}\cdots ,{\sqrt[n]{\pi}}^{n-1}$ it is a homogeneous system with a non-trivial solution, so its determinant must be zero. But all coefficients are rational polynomials in $\pi$ which implies $\pi$ is algebraic. The contradiction means that the original assumption cannot hold true, so $\sqrt{n}\left\{\pi \right\}$ is transcendental.

The same line of proof works for the $n}^{th$ root of any transcendental number, not just $\pi$.

(The above is essentially proving that $P\left(\sqrt[n]{\pi}\right)=0\Rightarrow R\left(\pi \right)=0$ where R is the resultant $R\left(z\right)=\text{res}(P\left(x\right),,{x}^{n}-z,,x)$, only without assuming prior knowledge of resultants or other higher algebra results.)

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