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?

Let $A=\Bbb{Q}\cap (a,b)$. Of course, $b\geq x$ for all $x\in A$. Suppose that the supremum of $A$ is $m<b$. By density, there exists $r\in\Bbb{Q}\cap(m,b)$. Hence, $r\in A$ and $r>m$, which contradicts the maximality of $m$.
