Question

if a,b ∈ R with a < b, then there exists a rational number r ∈Qsuch that a < r <b.

Answer 1

By Dedekind's definition of a real number, a real number $x$ is characterised by all rational numbers that are smaller than this real number - $S_x=\{y\in\mathbb{Q}|y\leq x\}$ . As $a<b$, every rational number $p$ such that $p\leq a$ (which means $p\in S_a$) satisfies also $p\leq b$ by transitivity of an order (which means that $p\in S_b$) , so $S_a\subset S_b$. But as $b\neq a$, we should have $S_b\neq S_a$, so there is $r\in S_b, r\notin S_a$, which means $r$ is a rational number such that $a<r\leq b$. If $r\neq b, a<r<b$ and we have found the number we have been searching for. If $r=b, b\in \mathbb{Q}$. In this case we can apply the same reasoning as above to find a rational number $r'$ such that $-b<r'\leq-a$. This gives us either a rational number $r', -b<r'<-a$ and thus $a<-r'<b$, or $r'=-a$ and thus $a\in\mathbb{Q}$. But in the latter case $a,b$ are both rational and thus $\frac{a+b}{2}$ is a rational number such that $a<\frac{a+b}{2}<b$. Therefore we have found such a rational number in all the possible cases.