I'm currently in the beginning of a journey of self-studying real analysis, and I tried to prove the following statement:
Prove that the following set does not have a rational supremum: $$ E=\{x \in \mathbb{Q} \text{ : } x^2 < 2\} $$
Here is my proof and I would appreciate comments and constructive critics of what I was able to achieve at the moment:
Let $\sup{E} = \frac{a}{b}$ be the rational supremum of $E$. First consider the possibility that $\sup{E} \in E$, which implies that $\frac{a^2}{b^2} < 2$. Now let's look to the distance from $\sup{E}$ to $2$: $$ \frac{a^2}{b^2} < 2 \rightarrow \mid \frac{a^2}{b^2} - 2 \mid > 0 $$ Let's call that distance $d=\mid \frac{a^2}{b^2} - 2 \mid $. With that, we can build a new supremum: $$ k =\sup{E} + \frac{d}{2} $$ And clearly $k > \sup{E}$, and that contradicts the fact for $\sup{E}$ to be a supremum.
Now consider the case where $\sup{E} \notin E$, which implies that $\frac{a^2}{b^2} \geq 2$. It's a proven fact that $\sqrt{2}$ is not rational, therefore, by the same reasoning, we define a distance $d=\mid 2 - \frac{a^2}{b^2} \mid $. Still in the same reasoning as before, we can conclude that: $$ k' = \sup{E} - \frac{d}{2} $$ And clearly $k' < \sup{E}$, which is a contradiction.
Hence, the conclusion is that the set $E$ does not have a rational supremum.
Am I correct? If not, can you please point out what would you change in order to improve my proof?
Thank you, and any help is highly appreciated!