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Exercise: $f: \mathbb Q^2\to\mathbb R$. Where $\mathbb Q$ is the set of rational numbers.

$f$ is strictly increasing in both arguments.

Can $f$ be one-to-one?

This question is related to many questions like: Nonexistence of an injective $C^1$ map between $\mathbb R^2$ and $\mathbb R$. However, I don't assume continuity and I only need to embed $\mathbb Q^2$ not $\mathbb R^2$.

My try:

I am thinking about construction. Since $\mathbb N$ and $\mathbb Q$ are usually considered similar, I will start with $f:\mathbb N^2\to \mathbb R$.

Define:

$f(x,y)=x-1/y$

This is clearly an injective map. However, I don't know how to turn this into a map from $\mathbb Q^2$.

Dietrich Burde
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    Why do you use $Q$, which is usually used for the rationals, instead of $N$? What is the difference between them? Your $f$ is a fine answer, strictly increasing in both arguments, unless $0$ is in the naturals. It is easy to fix if it is. – Ross Millikan Jun 08 '24 at 19:39
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    @RossMillikan Sorry for the typo. Q does mean rationals in this problem. I start working with $N$, though. – High GPA Jun 08 '24 at 19:42
  • Yes, there is such a map. You can equip $\Bbb Q^2$ with the lexicographic order and then use this argument (in fact, that can give you an order-isomorphism between $\Bbb Q^2$ and $\Bbb Q$!) – Izaak van Dongen Jun 08 '24 at 19:45
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    Or $f(x, y) = x + y \sqrt{2}$... – Adayah Jun 08 '24 at 19:50

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