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I read that "the set of integers and the set of rational numbers (with the standard ordering) do not have the same order type, because even though the sets are of the same size (they are both countably infinite), there is no order-preserving bijective mapping between them."

I am still new to set theory, I see that integers can have a bijection with the set of rational numbers, but why is there no order-preserving bijection mapping?

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Assume, for contradiciton, that $f:\Bbb Z\to \Bbb Q$ is an order-preserving bijection. Now consider $f(1)$ and $f(2)$. Since $f$ is order-preserving, we must have $f(1)<f(2)$. But more than that, we have $$ f(1)<\frac{f(1) + f(2)}2<f(2) $$ That number in the middle is a rational number, and $f$ is a bijection, so there must be an integer $n$ such that $f(n) = \frac{f(1) + f(2)}{2}$. Which is to say that $$ f(1)<f(n)<f(2) $$ But $f$ is order-preserving, meaning we have $$ 1<n<2 $$ which is impossible. Thus, by contradiciton, there cannot be any order-preserving bijections between the integers and the rationals.

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If $f:\mathbb Z \to \mathbb Q$ is an order preserving bijection then consider $r=f(1)$ and $s=f(2)$. Between any two rationals there is another one. If $t$ is between $r$ and $s$ then the order preserving nature of $f$ shows that there must be an integer (namely $f^{-1}(t)$) between $1$ and $2$ which is a contradiction.

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