Define $S\equiv\{ x\in \mathbb{Q}\mid x^2<2\}$. Show that $\sup S=\sqrt{2} $.

Solution 1:

Hint: Here is an elementary argument.

If you take any $x \in S$ and a rational number $r < 1$ such that $$0 < r < (2 - x^2)/(2x + 1)$$

show that $x + r \in S$. This shows that $S$ has no maximum element.

Consider $T = \{y \in \mathbb Q; y^2 > 2\}$ and similarly show that there is no minimum element in $T$.

Conclusion: If $a = \sup S$ then $a^2 = 2$, $a > 0$. Use a simple contradiction.

Solution 2:

Since $x^2<x2\implies -\sqrt 2<x<\sqrt 2$, $\sqrt 2$ is an upper bound of $S$.

Let $\epsilon>0$, and since $\Bbb Q$ is dense in $\Bbb R$, there is an $x\in Q$ such that $\sqrt 2-\epsilon<x<\sqrt 2$. So $x^2<2$ implies $x\in S$, and so $\sup S=\sqrt 2$.