Prove that there is an infinite number of rationals between any two reals I just stumbled upon this question: Infinite number of rationals between any two reals..
As I' not sure about my idea of a proof, I do not want to post this as an answer there, but rather formulate as a question.
My idea is as follows:


*

*$\mathbb{Q} \subset \mathbb{R}$

*$\forall a,b \in \mathbb{R}$ with $a>b, \exists q_0 \in \mathbb{Q}$ s.t. $a > q_0 > b$ (which is proven e.g. on Proofwiki)

*As $\mathbb{Q} \subset \mathbb{R}, q_0 \in \mathbb{R}$

*For $a, q_0$, repeat step 2 to find $q_1 \in \mathbb{Q}$ s.t. $a > q_1 > q_0 > b$

*Repeat ad infinitum


Thus, there have to be infinitely many rationals between any two reals.
Can you argue like this, or is there anything wrong in my line of reasoning?  
 A: Since $q_0$ has been found such that $a > q_0 > b$, you can use induction proof:
For all integer $n$, let $P(n)$ be :
there exist $q_0, \cdots, q_n \in \Bbb Q$ such that $a>q_0> \cdots >q_n >b$.
Then:
(i)  $P(0)$ is true.
(2)  Let us suppose $P(n)$ true for any $n\in \Bbb N$. Let $q_{n+1} \in \Bbb Q$  such that $ q_n > q_{n+1} > b$ , then $a > q_0 > \cdots > q_n >q_{n+1} > b$  and $P(n+1)$ is true.
A: I was also thinking along similar lines : Take a rational upper bound A for a, and a rational lower bound B for b so that $a<A<B<b$, then use the fact that the mean M of two rational numbers is itself rational, since $\mathbb{Q}$ forms a ring with addition (+) and multiplication ($\cdot$) , and then apply it with regards A & M and B & M to find the two new means $M_A$ and $M_B$, and so on and so forth, ad infinitum.
A: you could use the axiom of archimedes
A: Let $\alpha, \beta$ be two real numbers such that $\alpha < \beta$. If we use Dedekind cuts for these numbers say $\alpha = \langle A, B \rangle, \beta = \langle C, D \rangle $ then we can see that $\alpha < \beta $ means $A $ is a proper subset of $C$ and hence there are infinitely many rationals lying in $C - A$ and these are all lying between $\alpha $ and $\beta$.
It is bit more difficult to show that there are infinitely many irrationals between $\alpha$ and $\beta$.
