Q:

$\sqrt{2}$ cannot represent a rational number

I am not asking for a proof that shows me that $\sqrt{2}$ cannot represent a rational number, because I have already seen one by contradiction, which was quite simple, but I have problems in understanding the following proof:

By Rational Zeros Theorem, the only rational numbers that could possibly be solutions of $x^2 - 2 = 0$ are $\pm1, \pm2$. [Here n = 2, $a_2 = 1$, $a_1 = 1$, $a_0 = -2$. So rational solutions must have the form of $\frac{p}{q}$, where $p$ divides $a_0 = -2$, and $q$ divides $a_2 = 1$.] One can substitute each of the four numbers $\pm1, \pm2$ into the equation $x^2 - 2 = 0$ to quickly eliminate them as possible solutions of the equation. Since $\sqrt{2}$ represents a solutions for $x^2 - 2 = 0$, it cannot represent a rational number.

I have a few questions:

Where does this equation $x^2 - 2 = 0$ come from?
I have read the Rational Zeros Theorem, but I don't understand completely why $\pm1, \pm2$ are the only solutions. 
We want to show that $\sqrt{2}$ cannot represent a rational number, that the proof finishes by saying "Since $\sqrt{2}$ represents a solutions for $x^2 - 2 = 0$, it cannot represent a rational number.", which sincerely I am not seeing well the point. Ok, $\sqrt{2}$ is a solution, but that it cannot represent a rational number from this proof, this seems like not connected at all.

I am know I am missing something that you guys understand on the fly, but of course I am not you :D

A:

HINT: $\alpha=\pm\sqrt{2}$ if and only if $\alpha$ is a solution of $x^2-2=0$. The Rational Zeros Theorem states the following:

If $p/q$ (where $p$ and $q$ are relatively prime integers) is a root of the polynomial $a_0+a_1x+\cdots +a_nx^n$ (where each $a_i$ is an integer), then $p$ must be a divisor of $a_0$ and $q$ must be a divisor of $a_n$

If you apply this result to the polynomial $x^2-2=0$, you get that any rational root of it must be of the form $p/q$ where $p$ is a divisor of $-2$ and $q$ is a divisor of $1$, what are the possibilities for the rationals roots then? can you find a rational root of this polynomial?

A:

Below I rewrite the quoted proof to clarify the points that you ask about..
$\color{brown}{\rm Suppose}$ $\,\sqrt 2\,$ is rational, i.e. $\, \sqrt 2 = p/q\,$ for some integers $\,p,q\neq 0.\,$ By cancelling any common factors we may assume wlog that $\,p,q\,$ are $\color{#90f}{\rm coprime}$. Since $\,x=\sqrt 2\,$ $\Rightarrow$ $\,x^2 = 2,\,$ we deduce that $\,\sqrt 2 = \color{#0a0}p/\color{#c00}q\,$ is a $\rm\color{#90f}{reduced}$ rational root of $\,\color{#c00}1\cdot x^2 - \color{#0a0}2.\,$ Therefore, upon applying the Rational Root Test $ $  (or Theorem), $ $ we deduce that $\ \color{#c00}{q\mid 1}\,$ and $\color{#0a0}{p\mid 2}.\,$ Considering all possible factors of $\,p\,$ and $\,q\,$ implies that the root $\, x = p/q\,$ is either $\,\pm1\, $ or $\,\pm 2,\,$  a contradiction, since neither squares to $2.\,$ This contradiction proves false our initial $\color{brown}{\rm hypothesis}$ that $\,\sqrt 2\,$ is rational.
Remark $\ $ Generally, if $\,f(x)\,$ is a nonzero polynomial with integer coefficients whose leading coefficient $\color{#c00}{= 1},\,$ then the Rational Root Test implies that the only possible rational roots are $\rm\color{#c00}{integers}\,$ (and such integers must divide its constant term $\,\color{#c0a0}{f(0)}).\,$ This property (integrally closed) plays a key role in generalizing divisibility theory to other domains.

