Prove that there is an infinite number of rationals between any two reals

Question

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?

When you write that down, you ought to use a lot more English words and a lot less symbols. Otherwise it can work. Depending on your context, you may or may not need to flesh out "repeat ad infinitum" more precisely. — hmakholm left over Monica, Nov 14 '13 at 18:53
@bonifaz: It makes your argument more difficult to read than writing things out in prose would be. — hmakholm left over Monica, Nov 14 '13 at 19:05
@vadim123, I added a link for a proof of #2. Thanks for pointing out. — bonifaz, Nov 14 '13 at 19:07
In my opinion, all the difficulty is hidden in step 2. All the rest is a formality. So I would definitely write it up as opposed to just relying on a reference. — Adam, Nov 14 '13 at 20:32

score 1 · Answer 1 · edited Jun 12 '20 at 10:38

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.

score 0 · Answer 2 · answered Nov 14 '13 at 20:50

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.

score 0 · Answer 3 · answered Nov 14 '13 at 20:52

0

you could use the axiom of archimedes

answered Nov 14 '13 at 20:52

Raul

938

Could you elaborate a bit on this? – amWhy Nov 14 '13 at 21:40
@amWhy ok so do you know what the axiom states? with that, if we want to find a rational between say a and b, we can find a natural n s.t. a-b>1/n from the axiom, do you see how to proceed from here.....we can say there is a q s.t. q/n is in-between a and b being careful in regards to what is q then we have found a rational. – Raul Nov 15 '13 at 00:36

score 0 · Answer 4 · answered Nov 17 '13 at 16:13

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 $\beta$.

Prove that there is an infinite number of rationals between any two reals

4 Answers4