The set $S$ of rational numbers $x$ with $x^2 \lt 2$ has rational upper bounds but no least rational upper bound. Suppose that S has a least rational upper bound and call it $x_0$.
Show that $x_0^2 \lt 2$ is impossible by showing that if $x_0^2 \lt 2$, then there is a positive integer $n$ for which $(x_0 + \frac{1}{n})^2 \lt 2$. This places $x_0 + \frac{1}{n}$ in S and shows that $x_0$ cannot be an upper bound for $S$.
Apologies for asking such a simple question, but I don't follow the problem posed by the question there could be a "a positive integer $n$ for which $(x_0 + \frac{1}{n})^2 \lt 2$"
Why is this problematic? Where is $n$ coming from? Is this because $x_0$ is defined to be the least rational upper bound and adding any number to it would contradict $x_0$ being the least rational upper bound?
I still don't quite see why that strictly contradicts. Or why $n$ matters... Can someone clarify the question for me? I'm obviously misunderstanding something.
We are trying to prove that there is no rational least upper bound. This part proves that there cannot be a least upper bound $x_0$ with $x_0^2 \lt 2$. There will be another part to prove there cannot be a least upper bound $x_1$ with $x_1^2 \gt 2.$ Finally, you will prove there is no rational $x_2$ with $x_2^2=2$.
We want to show there is a rational in $S$ that is greater than $x_0$, which means $x_0$ is not an upper bound at all. If $x_0^2 \lt 2, \text{ then }x_0 \lt \sqrt 2$ by some amount. You need to choose an $n$ large enough that $x_0+\frac 1n$ is still less than $\sqrt 2$. That means $x_0+\frac 1n$ is in $S$, so $x_0$ is not an upper bound for $S$
If you were working in the reals, you could just let $y=\sqrt 2 -x_0$ and choose $n$ to be an integer greater than $\frac 1y$. Here we don't have $\sqrt 2$ so you need to find a rational between $x_0$ and $\sqrt 2$. Then $n$ can be any number greater than the inverse of the difference between your new rational and $x_0$.