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 $\mathbb{N}$.
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And what of $q_0$ ( or $q_1$)? That is, there is a least natural number, but not a least rational. – Sarvesh Ravichandran Iyer Jan 24 '19 at 16:36
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1Your question is equivalent to this one, since defining an order-preserving bijection $\mathbb{Q} \to \mathbb{N}$ is equivalent to listing the rational numbers in increasing order. – Clive Newstead Jan 24 '19 at 16:37
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Compare also with this question here. – Dietrich Burde Jan 24 '19 at 16:38
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@CliveNewstead Thanks, that was actually easyer than I thought. – roi_saumon Jan 24 '19 at 16:46
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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.
Annika
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If we assume there does not exist another rational number $q_k$ with with $q_n<q_k<q_{n+1}$, does this preclude some irrational, say $q_{k+1}$ with $q_n<q_{k+1}<q_{n+1}$? – isaac Feb 07 '24 at 04:58
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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. $r=(p+q)/2$
