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I'm trying to prove that $\mathbb{Q}^{+}$ has the same cardinality with $\mathbb{N}$ by using the theorem Schroder-Bernstein.

So I just have to prove that $\mathbb{Q}^{+}\preceq \mathbb{N}$ and $\mathbb{N}\preceq \mathbb{Q}^{+}$.

For the first one we have to find a function $f:\mathbb{Q^{+}}\rightarrow \mathbb{N}$ one-to-one.Is this the same like showing that there exists $f:\mathbb{N}\times \mathbb{N}\rightarrow \mathbb{N}$ one-to-one ???Because $\mathbb{Q^{+}}$ it's of the form $\left \{ \frac{p}{q},p,q\in\mathbb{N} \right \}$.

Also the same idea for $\mathbb{N}\preceq \mathbb{Q^{+}}$,is it true that is the same with finding an one-to-one function $f:\mathbb{N}\rightarrow \mathbb{N}\times\mathbb{N}$???

Any advise will be helpful.

Andrés E. Caicedo
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Jonathan1234
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    Yes, if you could find a 1-1 map from $\mathbb{N} \times \mathbb{N}$ to $\mathbb{Q}^+$, that would be a good start. Be mindful that you need to consider that $p$ and $q$ should be relatively prime. Perhaps if you can find surjective maps from $\mathbb{Q}^+$ to $\mathbb{N}$ and $\mathbb{N} \times \mathbb{N}$ to $\mathbb{Q}^+$, this would be enough. – Matt Oct 14 '17 at 12:12
  • @Matt: No, surjective maps do not work for invoking Schröder-Bernstein. – hmakholm left over Monica Oct 14 '17 at 14:02

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You need an injective map from $\Bbb Q^+ \to \Bbb N$. Yes, it is enough to find an injection $\Bbb {N \times N \to N}$ because you have an injection $\Bbb {Q^+ \to N \times N}$ by taking the lowest term representation of each rational. Then you need an injection $\Bbb {N \to Q^+}$ but the identity supplies that.

Ross Millikan
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