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Does there exist an \( 1-1 \) and onto function \( f:\mathbb{Q} \rightarrow \mathbb{Q} \setminus \{0 \} \) that preserves provision in \( \mathbb{Q} \) ? (Meaning that whenever \( x<y \Rightarrow f(x)< f(y) \).

Hi Apostole. The right word is "order" not "provision".

I know the answer but it is too nice to spoil this problem by giving it. Let's hope that somebody who does not know the answer will work it out.