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In both\(\mathbb Z\) and \(\mathbb R\) 1) if \(a,b\in S, a< b,~~then~~a\neq b\) 2)\(if ~~a
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I feel like they are not one to one but don't know how. I establish \(f:\mathbb Z\rightarrow \mathbb R\) , \(f: (a,b)\mapsto a/b\) I can see that f is not one to one. But how to put this logic into the problem?
On the other hand, if the relation is "less than", then, it works well for both set, hence they are isomorphic. Ha!!!!! I confused myself.
it must be one-to-one too right and integers and reals have different cardinalities idk what that means for present problem..
Modern geometry is my course. This is the very first section on the book and I got stuck everywhere. hihihi.....
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