...Use these and the Well-Ordering Principle to prove that no such $a$ and $b$exist. From this it follows that $\sqrt{3} \notin \mathbb{Q}$
I have no idea where well-ordering principle comes in for this question?
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$\begingroup$fairly common to demand the $a^2 = 3 b^2 $ be the solution in positive integers that has the smallest possible value of $a+b$
But then they point out $(3b-a)^2 = 3 (a-b)^2 .$ You are required to show that $a-b, 3b-a$ are both positive integers, and then show that$$ 3b-a + a-b < a+b, $$which is a contradiction of the assumption that $a+b$ was minimal
$\endgroup$ $\begingroup$Typical trick due to Hardy:
\begin{align} \sqrt{3} &= \frac{a}{b} \tag{assuming in lowest terms} \\ \frac{3}{\sqrt{3}}&= \frac{3b}{a} \\ \sqrt{3} &= \frac{3b-a}{a-b} \tag{Componendo et Dividendo} \end{align}
Since $1<\sqrt{3}<2$,$$b<a<2b \implies 0<\color{blue}{a-b}<\color{red}{b}$$
Now $\dfrac{3b-a}{\color{blue}{a-b}}$ is lower than $\dfrac{a}{\color{red}{b}}$, contradiction.
See similar trick on another post of mine here.
Also papers here, here and also here.
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