# 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?

• Hint: try to adapt the proof that $\sqrt 2\not \in \mathbb Q$. Start with $xq^2(x)=p^2(x)$.
– lulu
Apr 5, 2021 at 15:31
• rational function where the polynomials are real? Apr 5, 2021 at 15:31
• More simply, one could just note that, if you had suitable $p,q$, then $\sqrt 2 =\frac {p(2)}{q(2)}\in \mathbb Q$.
– lulu
Apr 5, 2021 at 15:33
• You can prove by using the algebraic properties that a rational function whose square is $x$ must satisfy, or by looking at the growth rates of rational functions. Apr 5, 2021 at 15:38
• Assuming it is, you can also get a contradiction about the degrees of the polynomials.
– leo
Apr 5, 2021 at 15:45

## 2 Answers

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 $$1$$.

• Can you answer my little question? Is my claim correct? If $\text{deg}~f(x)>\text{deg}~g(x)$, then $\text{deg}~f'(x)>\text{deg}~g'(x)$. Thank you! Apr 5, 2021 at 17:41
• @lonestudent Yes, for non-constant polynomials. Apr 5, 2021 at 17:42

$$\text{------------------------------------------------------------------------------------------------------}$$ 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. $$\text{------------------------------------------------------------------------------------------------------}$$ 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. $$\text{------------------------------------------------------------------------------------------------------}$$ 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. $$\text{------------------------------------------------------------------------------------------------------}$$