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Mathematics 17 Online
OpenStudy (loser66):

In both\(\mathbb Z\) and \(\mathbb R\) 1) if \(a,b\in S, a< b,~~then~~a\neq b\) 2)\(if ~~a

OpenStudy (loser66):

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?

OpenStudy (loser66):

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.

ganeshie8 (ganeshie8):

it must be one-to-one too right and integers and reals have different cardinalities idk what that means for present problem..

OpenStudy (loser66):

Modern geometry is my course. This is the very first section on the book and I got stuck everywhere. hihihi.....

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