I know that this kind of question has been answered before in many places, but I'd like a proof with a little spin on it. In his book "A course of pure mathematics", Hardy defines a real number $\alpha$ as a section composed by a lower class (a) (in which a is rational and a$\lt$$\alpha$) and an upper class (A) (in which A is rational and A$\ge$$\alpha$).
I'd like to prove that there are infinitely many rational numbers between any given two real numbers definied in this way by using this idea of sets.
I'm not sure if it's possible, but I believe so, because Hardy says "All these results are immediate consequences of our definitions" after a series of examples I managed to prove except for this one.
It might be useful to tell you that, according to thoses definitions, the relations of magnitude between two real numbers are defined in this way:
$\alpha$$\lt$$\beta$ if, and only if, (a)$\subset$(b) and (A)$\supset$(B)
$\alpha$$\gt$$\beta$ if, and only if, (a)$\supset$(b) and (A)$\subset$(B)
(where (a), (A) and (b), (B) are the lower and upper classes of $\alpha$ and $\beta$).
Evidently, if $\alpha$$\lt$$\beta$ and x is a rational number between $\alpha$ and $\beta$, then x$\in$(A)$\cap$(b). Showing that there are infinitely many elements in (A)$\cap$(b) would show that there are infinitely many rational numbers between $\alpha$ and $\beta$, but I'm stuck at this point and cannot find a way to develop the proof from this approach.
Both hints and solutions are welcome.
Greetings!
Best Answer
Let $a,b\in\mathbb R$, WLOG $a< b$ then $b-a>0$. Let us assume that $x=b-a$ and $y=5$ then there exist (by Archimedean property) $n\in\mathbb N$ s.t $$nx>y$$ $$n(b-a)>5$$ $$nb-na>5$$. Since difference between $nb$ and $na$ is greater than $5$,then there exists atleast one $m\in\mathbb Z$ s.t $$na<m<nb$$ $$a<\frac mn<b$$ $$a<r<b$$,where $r=\frac mn\in\mathbb Q$ i.e there exist one rational number $r$ between $a$ and $b$.Again $a<r$,there exist $r_1\in\mathbb Q$ s.t $$a<r_1<r$$.Proceeding this way we can generate as many as rational number as we wish.
Hope this will help!!!