# Proof that there is a real number $x$ such that $x^2=2$

I am quite confused by this part of the proof:

"Let $S:=\{s\in\mathbb{R}: 0\leq s, s^2\leq 2\}$. Since $1\in S$, the set is not empty. Also $S$ is bounded above by 2, since if $t>2\Rightarrow t^2>4$ so $t\notin S$."

This last part here where they mention taking $t>2$ is what makes no sense to me. Our set is bounded by $s^2\leq 2$ and there are plenty of numbers smaller than $2$ whose squares are greater than $2$, i.e. take $t=\frac{3}{2}.$

• They are just establishing that $S$ is bounded. They could have used $100$. – John Douma Nov 6 '15 at 2:47
• I think the point is show that the set of upper bounds of $S$ is not empty. – cr001 Nov 6 '15 at 2:48

So they are just trying to show you for what ever number you picked, if it were to be an element of S is has to be less than or equal to $2$. That's why they took a $t>2$, and show that $t \notin S$; Which proves that all $t$ in $S$ has to be less than or equal to $2$.
And you are right that there are plenty of numbers smaller than $2$ whose squares are greater than $2$, but what they've shown is sufficient.