How to show that $\sup(\mathbb{Q} \cap (a,b)) = b$? 
Let $a<b$ be two real numbers. Show that $\sup(\mathbb{Q} \cap (a,b)) = b$ and $\inf(\mathbb{Q} \cap (a,b)) = a$.

This intuitively makes sense. Since a sequence of rationals will infinitely approach $a$ and $b$, it makes sense that the question is true. How do I prove it rigorously?

Just use the definition of supremum. We need to show that
(1) $b$ is an upper bound of $\mathbb Q\cap(a,b)$, and
(2) $\forall b'<b$ is not an upper bound of $\mathbb Q\cap(a,b)$.
We only need to prove (2). Without loss of generality, we can suppose $b'>a$.
Since $b'<b$ and $\mathbb Q$ is dense in $\mathbb R$, i.e. we can find $q\in\mathbb Q$ such that $q\in (b',b)$. Thus, $q\in\mathbb Q\cap(a,b)$ and larger than $b'$.
Hence, $b'$ is not an upper bound of $\mathbb Q\cap(a,b)$.
