Bounded Set for Rational Number $$A = \{x \in \Bbb Q: x^2 &lt; 11\}$$

I am looking for the upper and lower bounds of set $A$. I know they should be the greatest and the smallest rational number within $(- \sqrt{11}, \sqrt{11})$ 

So the difficulty is how to get those rational numbers in exact value. 
Thanks
As you correctly noted, $A = (-\sqrt{11}, \sqrt{11}) \cap \mathbb Q$. However $\pm\sqrt{11}\notin \mathbb Q$ and $\mathbb Q$ is dense in $\mathbb R$. Therefor $A$ is dense in $(-\sqrt{11}, \sqrt{11})$ and we can find a sequence in $A$ converging to any of $\pm\sqrt{11}$. This proves that no upper or lower bound exists in the set (i.e. $\min A$ and $\max A$ are undefined). However, $\sup A$ and $\inf A$ are defined and are least upper and greatest lower bounds for $A$. Since we already have a convergent sequence to $\pm\sqrt{11}$ and both are bounds on $A$, we can see that
$$\sup A = \sqrt{11}\\
\inf A = -\sqrt{11}$$
As you correctly noted, $A = (-\sqrt{11}, \sqrt{11}) \cap \mathbb Q$. However $\pm\sqrt{11}\notin \mathbb Q$ and $\mathbb Q$ is dense in $\mathbb R$. Therefor $A$ is dense in $(-\sqrt{11}, \sqrt{11})$ and we can find a sequence in $A$ converging to any of $\pm\sqrt{11}$. This proves that no upper or lower bound exists in the set (i.e. $\min A$ and $\max A$ are undefined). However, $\sup A$ and $\inf A$ are defined and are least upper and greatest lower bounds for $A$. Since we already have a convergent sequence to $\pm\sqrt{11}$ and both are bounds on $A$, we can see that
$$\sup A = \sqrt{11}\\
\inf A = -\sqrt{11}$$