View Full Version : Isomorphism
afirican
Oct20-04, 12:17 PM
How do I prove that Q under addition is not isomorphic to R+ under multiplication?
matt grime
Oct20-04, 12:26 PM
They cannot be isomorphic as groups because they are not even in bijective correspondence as <insert one word to get the answer>
afirican
Oct20-04, 02:22 PM
Isn't it f(x) = exp(x) a bijection between Q and R+?
Isn't it f(x) = exp(x) a bijection between Q and R+?
No an isomorphism must be onto.
afirican
Oct20-04, 03:04 PM
Why f:Q -> R+, f(x) = exp(x) is not onto?
For all r of R+, there exists r' = lnr in Q such that r = exp(lnr) = exp(r') = f(r'). Where do I go wrong?
If r is irrational is r in Q? is er in R+?
afirican
Oct20-04, 03:14 PM
You're totally right. Then is there any way to show that Q and R+ are not isomorphic?
afirican
Oct20-04, 03:16 PM
I think I know the answer. If I say that any map between Q and R+ is not onto, is that enough?
Yes, you just need to look at the two sets Q and R+ to see that the two groups cannot be isomorphic (as Matt grime indicated).
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