Hey guys!

So I have two questions which are similar, but not the same. The first asks me to prove that between any two distinct rational numbers there exists an irrational number - I haven't managed to do this. The question after however, which asks me to show that between any two real numbers there exists an irrational number, I've had an attempt at but I'm not sure whether the proof is adequate enough - it is written below (please bear in mind that I have already proven that between any two real numbers lies a rational number, so the first statement of my proof follows from that theorem, and in addition that $\displaystyle \sqrt2 \notin \mathbb{Q}$):

Consider $\displaystyle \dfrac{a}{\sqrt2} < \dfrac{p}{q} < \dfrac{b}{\sqrt2}$ where $\displaystyle p,q \in \mathbb{Z}$ with $\displaystyle q\not= 0$ and $\displaystyle a,b \in \mathbb{R}$. That is, by definition, $\displaystyle \dfrac{p}{q} \in \mathbb{Q}$. Multiplying this inequality by $\displaystyle \sqrt2 > 0$ means that the inequality still holds, hence $\displaystyle a < \dfrac{p\sqrt2}{q} < b$ and therefore it follows that between any two real numbers there lies an irrational number as $\displaystyle \dfrac{p\sqrt2}{q} \notin \mathbb{Q}$.