I am having a problem understanding this proof.
Suppose that $\sqrt{2}$ was rational.
Then there would exist positive integers $a,b$ such that $\sqrt{2}=\frac{a}{b}$. Consequently, the set $S=\{k\sqrt{2} : k \text{ and } k\sqrt{2}$ are positive integers $\}$ is nonempty set of positive integers. Therefore, by the well-ordering property, $S$ has a smallest element, say, $s = t \sqrt{2}$.
We have
$s \sqrt{2}-s = s\sqrt{2}-t\sqrt{2}=(s-t)\sqrt{2}$.
Because $s\sqrt{2} = 2t$ and $s$ are both integers,
$s\sqrt{2} – s = s\sqrt{2}-t\sqrt{2} = (s-t)\sqrt{2}$
must also be an integer.
Furthermore, it is positive, because $s\sqrt{2}-s = s(\sqrt{2}-1)$ and $\sqrt{2} > 1$.
It is less than $s$, because $\sqrt{2} < 2$ so that $\sqrt{2}-1 < 1$. This contradicts the choice of $s$ as the smallest positive integer in $S$. It follows that $\sqrt{2}$ is irrational.
My main problem here is how they got the set $S$. My second confusion is the algebra part $s \sqrt{2}-s$. How did they come up with this?
If someone can explain this proof to me it would be great! thank you
Best Answer
Observe that
$$\sqrt2=\frac ab\implies b\sqrt2=a\in\Bbb Z\implies b\in S$$
as defined.
The algebra: observe that the minimal element in $\;S\;$ is $\;\color{red}{s=t\sqrt2}\in\Bbb Z\;$ , thus
$$s\sqrt2-\color{red}s=s\sqrt2-\color{red}{t\sqrt2}=(s-t)\sqrt2\ldots\text{ and etc.}$$