I'm given the set $S=\{r\in \mathbb{Q}:r \ge \sqrt{2}\}$ and asked to find
a. $\operatorname{bd}S = [\sqrt{2},\infty)$
b. $S'= [\sqrt{2},\infty)$
c. $\operatorname{cl} S = [\sqrt{2},\infty)$
d. $\operatorname{bd}(\operatorname{cl} S) = \{\sqrt{2}\}$
My understandings of a boundary point of $S$ is every neighborhood of $x$ must have a point in $S$ and also a point not in $S$. By this logic it's easy to see every number n greater or equal to $\sqrt{2}$ is a boundary point for $S$.
For accumulation point it follows the same logic as boundary except every deleted neighborhood of $x$ should have a point in $S$ other than itself.
For the closure of the set I know $\operatorname{cl} S = S \cup S'= S \cup \operatorname{bd} S$ So it seems clear $\operatorname{cl}S=\operatorname{bd} S=S'$
Lastly for $\operatorname{bd}([\sqrt{2}, \infty))$ I can easily see that $\operatorname{bd}(\operatorname{cl} S)=\{\sqrt{2}\}$.
Am I correct?
Yes. You are correct.
So that this moves off the Unanswered-Q list.