bijection between $\mathbb{Q}$ and $\mathbb{N}$ that preserve the order.

Question

I know there is a bijection between $\mathbb{Q}$ and $\mathbb{N}$. But is there a bijection $\mathbb{Q}\xrightarrow{f}\mathbb{N}$ that preserves the order? Intuitively I think this is not possible. What I would think of is enumerating the rationals in a form $\{q_n\}_{n \in \mathbb{N}}$ where $q_n < q_m$ if $n<m$ (I think this is not possible but why?) and then map to elements of $\mathbb{N}$.

Answer 1

The reson is that if you have natural number n then you can find the NEXT number (=n+1), but if you have Rational number q then you cannot find NEXT number - because between any two different rational numbers is infinite number of other rational numbers.

If such bijection $f$ exists then when you find $q=f(n)$ then you shoud be able to find next rational number $p=f(n+1)$ but because $q\neq p$ then between $q$ and $p$ exist infinite number of other rational numbers e.g. $r=(p+q)/2$

Answer 2

Let $n \in \mathbb{N}$ and $q_n,q_{n+1} \in \mathbb{Q}$ where $q_n < q_{n+1}$.

Assume there does not exist another rational number $q_k$ where $q_n<q_k<q_{n+1}$ (i.e., we can order the rationals). Since $\mathbb{Q}\subset \mathbb{R}$, this implies that the rationals are not dense in the reals, which is false. Therefore, our assumption is false.