Prove that $\sqrt2$ is the least upper bound

Question

Consider the set $A=\{\,x\in\mathbb{Q} : x^2<2\,\}.$

I want to prove that $\sqrt2$ is the least upper bound for $A$.

To do this, I think I need to prove that if I take any $\alpha\in\mathbb{R}$ with $\alpha<\sqrt2$, then there exists $y\in A$ with $y>\alpha$.

I'm not sure how to choose $y\in A$ to satisfy this property.

Answer 1

I'll expand my comment.

WLOG one may assume that $\alpha>0$. If $\alpha\le 0$ it is clear that $y:=\frac{1}{2}\in A$ and $\alpha<\frac12<\sqrt2$.

Assume $0<\alpha<\sqrt2$ is an upper bound of $A$. Archimedean property (or the density of rationals) guarantees the existence of $y\in\Bbb Q$ such that $\alpha<y<\sqrt2$, so (since $0<y<\sqrt2$) $y^2<2$ and $y\in A$.

Answer 2

Let $0\lt d\lt\sqrt2$, where $d\in\mathbb{Q}$.

Then we can prove the existence of a rational between $d$ and $\sqrt2$ by noting that such a rational, $\frac{a}{b}$, satisfies:

$$d\lt \frac{a}{b}\lt \sqrt2$$ $$bd\lt a\lt b\sqrt2$$

and we can make $b$ as large as required in order to find such an $a$.