Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

As the title says. I think a proof by contradiction is the most natural thing. Suppose $\pi \in Q(\pi^3)$. Then \begin{equation} \pi = \frac{a_n(\pi^3)^n+\cdots+a_1\pi^3+a_0}{b_m(\pi^3)^m+\cdots+b_1\pi^3+b_0} \text{.} \end{equation} Not sure how to proceed from here though.

I also have a related question, that is to show $\sqrt{2} \notin Q(\pi)$. I think if I can solve either one of them, I can solve the other. Any help would be much appreciated.

share|cite|improve this question
Multiply the denominator over, and show that there is no rational coefficient $c$ such that $\pi^{3m+1} = c \pi^{3n}$ for any $m,n \in \Bbb Z$. – Emily Dec 17 '12 at 21:54
Are you familiar with the fact that $\pi$ is transcendental? To solve this exercise you must know that, me thinks. – Jyrki Lahtonen Dec 17 '12 at 21:54
@JyrkiLahtonen Yes, I do know that $\pi$ is transcendental. – Aden Dong Dec 17 '12 at 21:58
@EdGorcenski Why is this sufficient for showing the problem? – Aden Dong Dec 17 '12 at 21:58
Ok. Then you can follow Hagen's solution (+1) below. – Jyrki Lahtonen Dec 17 '12 at 22:00
up vote 15 down vote accepted

From $$\pi = \frac{a_n\pi^{3n}+\ldots+a_1\pi^3+a_0}{b_m\pi^{3m}+\ldots +b_1\pi^3+b_0}$$ with $a_n\ne0$, $b_m\ne0$ we obtain a polynomial equation for $\pi$: $$\tag1(b_m\pi^{3m+1}+\ldots +b_1\pi^4+b_0\pi)-(a_n\pi^{3n}+\ldots+a_1\pi^3+a_0)=0.$$ If $n>m$, this is of degree $3n$ with leading coefficient $-a_n\ne 0$, if $n\le m$ this is of degree $3m+1$ with leading coefficient $b_m\ne0$. Hence $(1)$ shows that $\pi$ is algabraic, which it isn't.

From $$\sqrt 2=\frac{a_n\pi^{n}+\ldots+a_1\pi+a_0}{b_m\pi^{m}+\ldots +b_1\pi+b_0},$$ we obtain $$2=\frac{a_n^2\pi^{2n}+\ldots+2a_0a_1\pi+a_0^2}{b_m^2\pi^{2m}+\ldots +2b_0b_1\pi+b_0^2},$$ hence $$\tag2(a_n^2\pi^{2n}+\ldots+a_0^2)-2(b_m^2\pi^{2m}+\ldots +b_0^2)=0.$$ Since $\pi$ is transcendental, this must be the zero polynomial, i.e. everything cancels. Especially, we must have $n=m$ and $a_n^2-2b_m^2=0$. But then $\sqrt 2=\left\vert\frac{a_n}{b_m}\right\vert\in\mathbb Q$.

share|cite|improve this answer
Awesome! This is such a simple proof that I can't believe I missed it. Thanks a lot. – Aden Dong Dec 17 '12 at 22:01

The following general result might interest you.
If $k$ is a field and if $\phi(x)=\frac {f(x)}{g(x)}\in k(x)$ is a rational function with $f(x),g(x)\in k[x]$ relatively prime polynomials (not both constant), then the extension of fields $k(\phi (x))\subset k(x)$ has degree $$[k(x):k(\phi (x))]=\text {max}\:(\text {deg} \; f(x),\text {deg} \; g(x))$$ This of course implies (if you know that $\pi$ is transcendental and thus may play the role of the indeterminate $x$) that $[\mathbb Q(\pi):\mathbb Q(\pi^3]=3$ and thus a fortiori that $\pi \notin \mathbb Q(\pi^3)$ .

The displayed formula can be found in Theorem 8.36, page 614 of Jacobson's Basic Algebra II.

share|cite|improve this answer

I will prove the problem as a simple case of what Georges describes.

Let $k$ be a field. Let $L = k(x)$ be the rational function field with one variable $x$(we may take $x = \pi$ when $k = \mathbb{Q}$). Let $K = k(x^3)$. Then $L = K(x)$. Let $X^3 - a \in K[X]$, where $a = x^3$. Clearly $X^3 - a$ cannot have a root in $K$ considering the degree of a root if any. Hence it is irreducible over $K$. Hence $[L\colon K] = 3$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.