## The irrational numbers are dense

It suffices to solve the exercise in a bounded interval $$(a,b)$$, because every unbounded interval contains a bounded subinterval. Using the density of the rational numbers, we know that there is some rational number $$q$$ in the interval $$(a - \sqrt 2, b - \sqrt 2)$$. Then $$q + \sqrt 2$$ is an irrational number in the interval $$(a, b)$$.