Proving $\sqrt{x}$ is not a rational function I've got a monster of a proof by contradiction going where to finish I need to show that there is no $\alpha\in \mathbb{Q}(x)$ (the field of fractions of $\mathbb{Q}[x]$) such that $\alpha^2=x$. Obviously this is equivalent to showing that $\sqrt{x}\not\in\mathbb{Q}(x)$, which seems trivial enough, but I'm stuck.
Here's what I've done so far:
If $\sqrt{x}=\frac{p(x)}{q(x)}$ where $q(x)=a_0+...+a_nx^n$, then $p(x)=a_0x^{1/2}+...+a_nx^{n/2}$. It seems like this is a contradiction, but we are of course not prevented from choosing $a_0,...,a_n$ to be nonzero only for even terms greater than two, so it isn't.
Is there a quick algebraic proof of this fact?
 A: If $p(x)$ and $q(x)$ are polynomials, let $m=\deg p(x)$ and let $n=\deg q(x)$. Then$$\lim_{x\to\infty}\frac{\sqrt x}{p(x)/q(x)}=\begin{cases}\infty&\text{ if }m-n\leqslant0\\0&\text{ if }m-n>0.\end{cases}$$Therefore, $\sqrt x$ is not a rational function; if it was equal to $\frac{p(x)}{q(x)}$, that limit would be $1$.
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Suppose there exists $\alpha\in\mathbb{Q}(x)$ such that $|\alpha(x)|=\sqrt{x}$ for all $x\in\mathbb{Q}$ with $x\ge 0$.

Then in particular, we have $|\alpha(2)|=\sqrt{2}$, contradiction, since $\alpha(2)$ can't be irrational.
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Here's another way . . .

Suppose there exists $\alpha\in\mathbb{Q}(x)$ such that $\alpha^2=x$.

Then in particular, we have $\bigl(\alpha(-1)\bigr)^2=-1$, contradiction, since $\bigl(\alpha(-1)\bigr)^2$ can't be negative.
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Here's still another way, which has the advantage of working for $K(x)$, where $K$ is any field . . .

Let $K$ be a field, and suppose there exists $\alpha\in K(x)$ such that $\alpha^2=x$.

Then writing $\alpha={\large{\frac{p}{q}}}$ for some nonzero $p,q\in K[x]$, we get
$$
p^2=xq^2
$$
contradiction, since $\deg(p^2)$ is even, whereas $\deg(xq^2)$ is odd.
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