19
$\begingroup$

I'm trying to use the fact that $\pi$ is transcendental to show that $\sqrt \pi$ is also transcendental over $\Bbb{Q}$ . I don't know any theorems about algebraic and non-algebraic numbers so I don't have much in my tool box. Here's what I have done so far :

Let $P(x) \in \Bbb{Q}[x]$ and write $P(x)$ in the following way $$P(x)= \underbrace{ \left ( \sum_{\text{$j$ even}} a_j x^j +a_0 \right ) }_{Q(x)}+\underbrace{\sum_{\text{$i$ odd}} a_i x^i }_{G(x)}$$ Then $P(\sqrt \pi )= Q (\sqrt \pi) + G(\sqrt \pi)$ , $ Q (\sqrt \pi) =\displaystyle \sum_{\text{$j$ even}} a_j \sqrt\pi ^j+a_0 =\sum_{\text{$j$ even}} a_j \pi^{\frac j 2} +a_0$ hence $Q (\sqrt \pi) \not= 0$ since $\pi$ is transcendental, Similarly I showed $G(\sqrt \pi)$ is also non zero by taking $\sqrt \pi$ as a common factor. I'm now stuck on showing that $G(\sqrt \pi) \not= -Q(\sqrt \pi) $ . If can prove that then $P(\sqrt \pi) \not= 0$. How can it be done?

$\endgroup$
7
  • 22
    $\begingroup$ Am I missing something? The square of an algebraic number is algebraic, so if $\sqrt{\pi}$ were algebraic, $\pi$ would be too. $\endgroup$
    – vadim123
    May 6, 2013 at 14:10
  • $\begingroup$ That is why I wrote that I know nothing about them except the definitions. $\endgroup$
    – user10444
    May 6, 2013 at 14:13
  • 1
    $\begingroup$ You should be able to derive @vadim123's comment from the definitions. This will probably be a more fruitful approach than what you have tried so far. $\endgroup$
    – jwg
    May 6, 2013 at 14:15
  • 1
    $\begingroup$ Hint: If $\alpha$ is algebraic over $K$, then $K(\alpha)$ is a finite dimensional extension of $K$. $\endgroup$
    – awllower
    May 6, 2013 at 14:15
  • 2
    $\begingroup$ It's much easier to prove directly from the definitions that a square root of an algebraic number is algebraic than it is to prove that the set of algebraic numbers forms a field. To say that $\alpha$ is algebraic is to say that it is a root of a particular kind of polynomial, $f(x)$. Assuming that this is the case, and that $\sqrt{\alpha}$ is a square root of $\alpha$, can you think of a polynomial $g(x)$ of which $\sqrt{\alpha}$ is a root? $\endgroup$ May 6, 2013 at 14:35

4 Answers 4

32
$\begingroup$

Suppose $\sqrt \pi$ is algebraic : there is a polynomial $P(X)$ with rational coefficients such that $P(\sqrt \pi)=0$.

Develop the product $P(X)P(-X)$. You will observe that the odd powers of $X$ cancel each other, so you can write it as a polynomial in $X^2$ : There is a polynomial $Q$ (again with rational coefficients) such that $P(X)P(-X) = Q(X^2)$.

Now, $Q(\pi) = Q((\sqrt \pi)^2) = P(\sqrt \pi)P(- \sqrt \pi) = 0$, hence $\pi$ is algebraic.

$\endgroup$
1
  • 3
    $\begingroup$ The definition of $P$ should be a polynomial with $\sqrt \pi$ as root, not $\pi$. And to see that $P(X)P(-X)$ has only even terms, note that it is even as a function, and then equate coefficients of both sides. $\endgroup$
    – ronno
    May 6, 2013 at 14:40
3
$\begingroup$

Suppose $\sqrt{\pi}$ were algebraic over $\Bbb{Q}$, i.e. $\Bbb{Q}\left(\sqrt{\pi}\right)/\Bbb{Q}$ is algebraic. This gives us the intermediate extension $\Bbb{Q}(\pi)$: $$ \Bbb{Q}\left(\sqrt{\pi}\right)/\Bbb{Q}\left(\pi\right)/\Bbb{Q}. $$ In particular, this implies that both $\Bbb{Q}\left(\sqrt{\pi}\right) / \Bbb{Q}(\pi)$ and $\Bbb{Q}(\pi)/\Bbb{Q}$ are algebraic, which is a contradiction to the fact that $\pi$ is transcendental (every element of $\Bbb{Q}\left(\sqrt{\pi}\right)$ is algebraic, and $\Bbb{Q}(\pi)\subseteq\Bbb{Q}\left(\sqrt{\pi}\right)$).

Edit: The argument about the degrees of the extensions actually also works, after noting awllower's hint in the comments. $$ \infty > n = \left[\Bbb{Q}\left(\sqrt{\pi}\right) : \Bbb{Q}\right] = \left[\Bbb{Q}\left(\sqrt{\pi}\right) : \Bbb{Q}(\pi)\right]\cdot\left[\Bbb{Q}\left(\pi\right) : \Bbb{Q}\right], $$ implying $\Bbb{Q}(\pi)$ is finite, and therefore algebraic.

$\endgroup$
3
$\begingroup$

I found another way to solve it, suppose that $\sqrt \pi$ is a root of $P(x)$ which is defined in the question then we have$$ \left ( \sum_{\text{$j$ even}} a_j \pi^{\frac j 2} +a_0 \right ) +\sqrt \pi \sum_{\text{$i$ odd}} a_i \pi^{i-1} = 0 $$ then $Q(\pi)=-\sqrt \pi G(\pi)$ upon squaring both sides we get $Q^2(\pi)= \pi G^2(\pi)$ then $Q^2(\pi)- \pi G^2(\pi)=0$ is a polynomial in $\pi$ equal to zero which is a contradiction since $\pi$ is transcendental.

$\endgroup$
0
$\begingroup$

Let $\mathbb Q[\sqrt \pi]$ denote the smallest ring containing $\mathbb Q$ and $\sqrt\pi$. Consider $\mathbb Q[\sqrt \pi]$ as a vector space over $\mathbb Q$ and prove that it is finite dimensional by assuming that $\sqrt \pi$ is algebraic (Hint: use division algorithm to exhihit a finite generating set). Now, let the dimension of $\mathbb Q[\sqrt \pi]$ be n. Then $\pi^n.\pi^{n-1},\ldots, 1$ are linearly dependent vectors in $\mathbb Q[\sqrt \pi]$. Therefore, $\pi$ is algebraic, a contradiction.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .