My professor proved there exists a real number $\alpha\in\mathbb{R}$ such that $\alpha^2=2$. He proved the case where the set is bounded from above, namely $S=\{s\in\mathbb{R}:s^2<2\}$. He said a good exercise (not for hw/extra credit/assignment, i.e., for fun)
would be to try to prove it when it is bounded from below, so here is my attempt but I get stuck very fast.
Proof:
Consider $S=\{s\in\mathbb{R}:s^2>2\}$, for any $x\in\mathbb{R}$, $x^2\geq 0$, thus $S\not=\emptyset$ and is a nonnegative set, so it has a lower bound. By the Axiom of Completeness, there exists an $\alpha=\inf S$, we claim $\alpha^2=2$. We will proceed by contradiction by exhausting the cases $\alpha^2>2$ and $\alpha^2 <2$.
(If my logic is correct showing $\alpha^2>2$ contradicts $\alpha$ being a lower bound, and $\alpha^2<2$ contradicts $\alpha$ being the infimum.)
But this is where I get lost, I am not sure if for $\alpha^2>2$ do I look for an element in $S$ smaller than $\alpha$ namely ($\alpha -\frac{1}{n}$) that is larger than 2? As that would show $\alpha$ is not a lower bound, a contradiction to our original assumption.
What I mean if my above assumption is correct am I looking for
$$
\begin{align*}
\left(\alpha-\frac{1}{n}\right)^2=& \cdots\text{stuff}\\
>&\cdots\text{less stuff}
\end{align*}
$$
And then show that $(\dots\text{less stuff})>2$?
Your help, advice, and expertise would be appreciated.
Best Answer
Proof:
Consider $S=\{s\in\mathbb{R}:s^2>2\}$, for any $x\in\mathbb{R}$, $x^2\geq 0$, thus $S\not=\emptyset$ and is a nonnegative set, so it has a lower bound. By the Axiom of Completeness, there exists an $\alpha=\inf S$, we claim $\alpha^2=2$. We will proceed by contradiction by exhausting the cases $\alpha^2>2$ and $\alpha^2 <2$.
for the first case we assume $\alpha^2>2$ then $$ \begin{align*} \left(\alpha-\frac{1}{n}\right)^2=&\alpha^2 -\frac{2\alpha}{n}+\frac{1}{n^2}\\ >& \alpha^2-\frac{2\alpha}{n} \end{align*} $$ Since $\alpha^2>2$, let $\alpha^2-2=\epsilon >0$, then we have $\displaystyle\alpha^2-\frac{2\alpha}{n}-2=\epsilon -\frac{2\alpha}{n}$
$\displaystyle\epsilon -\frac{2\alpha}{n}>0\Leftrightarrow\epsilon>\frac{2\alpha}{n}\Leftrightarrow\frac{\epsilon}{2\alpha}>\frac{1}{n}$
By the Archimedean property, such $n\in\mathbb{N}$ exists, thus $\left(\alpha-\frac{1}{n}\right)\in S$, but $\left(\alpha-\frac{1}{n}\right) < \alpha$, contradicting our assumption that $\alpha$ is a lower bound.
For the second case we assume $\alpha^2<2$ then
$$ \begin{align*} \left(\alpha+\frac{1}{n}\right)^2=&\alpha^2 +\frac{2\alpha}{n}+\frac{1}{n^2}\\ <& \alpha^2+\frac{2\alpha}{n}+\frac{1}{n}\\ =&\alpha^2+\frac{2\alpha+1}{n} \end{align*} $$ Since we assumed $\alpha^2<2$ we can fix the $\displaystyle\frac{2\alpha+1}{n}$ term such that we are less than 2, that is by the Archimedean property we can find an $n_0\in\mathbb{N}$ such that: $$ \frac{1}{n_0}<\frac{2-\alpha^2}{2\alpha+1},\text{ which implies } \frac{2\alpha+1}{n_0}<2-\alpha^2 $$
Then $$ \left(\alpha+\frac{1}{n}\right)^2 < \alpha^2+(2-\alpha^2)=2$$
Thus $\displaystyle\left(\alpha+\frac{1}{n}\right)$ is a lower bound for $S$, but $\displaystyle \left(\alpha+\frac{1}{n}\right)>\alpha$, a contradiction to our assumption that $\alpha =\inf A$.