Finding the value of the supremum of a set Consider the set $S = \{x \in \mathbb R: x &lt; \frac2x\}$. Determine the value of $sup$ $S$ (if it exists).  

Here is my attempt:
Firstly, $S = \{x \in \mathbb R: 0 &lt; x &lt; \sqrt 2$ $\lor$ $x &lt; -\sqrt2 \}$.
Since $1 \in S$, $S \ne \emptyset$.
Also, $\forall x \in S$, $x &lt; \sqrt2$ and so $S$ is bounded above.
By the least upper bound property of $\mathbb R$, $sup$ $S$ exists.
Suppose $\exists y \in \mathbb R: y &lt; \sqrt2 \land y$ is an upper bound of $S$.
Since $1 \in S$, $1 \le y &lt; \sqrt2 \Rightarrow y \in S$.
...  

How do I show that $sup$ $S$ $= \sqrt 2$?
To find $S$ the equation $x&lt;\frac2{x}$ or equivalently $\frac{(x-\sqrt2)(x+\sqrt2)}{x}&lt;0$ must be solved. 

This leads to $S=(-\infty,-\sqrt2)\cup(0,\sqrt2)$. 

Here $\sqrt2$ is an upper bound of $S$ and proving that $\sup S=\sqrt2$ comes to showing that every $y&lt;\sqrt2$ is not an upper bound of $S$. That means: showing that $y&lt;s$ is true for some $s\in S$.
You shall prove that the supremum is not greater nor less than $\sqrt 2$. ie by supposing so in both cases and getting a contradiction.