Alternative proofs that square root is irrational function

Question

My work is to show alternative proof that $\sqrt x$ is irrational function.

The simplest proof might be to find a "counter-example". I mean, it is enough to show that $\sqrt 2=\dfrac{f(2)}{g(2)}$ is irrational.

I found the argument based on this argument I said in M.SE.

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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The argument that one of the functions $f^2(x)$ and $xg^2(x)$ has a even degree and the other has a odd degree is also valid.

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Now I want to know whether these two proofs that I have made are valid too.

Let, $f(x)$ and $g(x)$ be polynomials, such that

$$\sqrt x=\frac{f(x)}{g(x)} \Longrightarrow x=\frac{f^2(x)}{g^2(x)}$$

Then,

$$\deg \left (f^2(x)-g^2(x)\right)=\deg x=1$$

$$2 \deg \left (f(x)-g(x)\right)=1$$

$$\deg \left (f(x)-g(x)\right)=\frac 12 \not\in\mathbb Z^{+}$$

Thus, we can deduce such $f(x)$ and $g(x)$ doesn't exist.

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By accepting

$$\sqrt x=\frac{f(x)}{g(x)}$$

$$\deg f(x)=m ~~\text{and}~~ \deg g(x)=n, $$

$$m,n\in\mathbb Z^{+}∪\left\{0\right\}$$

Then,

$$\lim_{x\to\infty}\frac{f(x)}{g(x)}=\infty \implies m-n≥1$$

Then let $x≠0$, we have

$$\frac{\sqrt x}{x}=\frac{f(x)}{xg(x)}$$

$$\frac{1}{\sqrt x}=\frac{f(x)}{xg(x)}$$

$$\lim_{x\to\infty}\frac{f(x)}{xg(x)}=0 \implies n+1-m≥1$$

Finally we have,

$$\begin{cases} m-n≥1\\ n+1-m≥1 \end{cases} \implies \begin{cases} m-n≥1 \\m-n≤0\end{cases} $$

$$\implies \left\{m,n\right\}\in \emptyset.$$

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Are there any points / steps I missed?

Which proofs / arguments are valid and which are not?

_{Thank you.}

Answer 1

solution-verification

I think that the first proof that you have made is not valid.

I think that there is an error in the following step :

$\deg \left (f^2(x)-g^2(x)\right)=\deg x=1$

$2 \deg \left (f(x)-g(x)\right)=1$

It is not always true that $$\deg \left (f^2(x)-g^2(x)\right)=2 \deg \left (f(x)-g(x)\right)$$

For example, if $f(x)=x+1$ and $g(x)=x$, then we have $$\deg \left (f^2(x)-g^2(x)\right)=1\not= 0=2\deg \left (f(x)-g(x)\right)$$

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I think that the second proof that you have made is valid.

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Added :

Regarding the argument you found: I found no error, so I think that it is valid.

Answer 2

What abouth this proof : I start with $$x\cdot g^2(x)=f^2(x)$$ for some polynomials $f(x)$ and $g(x)$.

We can assume that $f(x)$ and $g(x)$ are not both divisible by $x$, otherwise we can divide both $f(x)$ and $g(x)$ by $x$ until we have the desired representation.

On the other hand, we have $f(0)=0$ , hence $f(x)$ must be disivible by $x$.

Since $f^2(x)$ is divisible by $x^2$, we have that $g^2(x)$ is divisible by $x$ which implies that $g(x)$ is divisible by $x$.

This is the desired contradiction.