16
$\begingroup$

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?

$\endgroup$
  • 17
    $\begingroup$ 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 '13 at 14:10
  • $\begingroup$ That is why I wrote that I know nothing about them except the definitions. $\endgroup$ – user10444 May 6 '13 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 '13 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 '13 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? $\endgroup$ – Manny Reyes May 6 '13 at 14:35
27
$\begingroup$

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.

$\endgroup$
  • 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 '13 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.

$\endgroup$
2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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