# Prove that Q under addition is not isomorphic to R+

1. Oct 20, 2004

### afirican

How do I prove that Q under addition is not isomorphic to R+ under multiplication?

2. Oct 20, 2004

### matt grime

They cannot be isomorphic as groups because they are not even in bijective correspondence as <insert one word to get the answer>

3. Oct 20, 2004

### afirican

Isn't it f(x) = exp(x) a bijection between Q and R+?

4. Oct 20, 2004

### jcsd

No an isomorphism must be onto.

5. Oct 20, 2004

### afirican

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?

6. Oct 20, 2004

### jcsd

If r is irrational is r in Q? is er in R+?

7. Oct 20, 2004

### afirican

You're totally right. Then is there any way to show that Q and R+ are not isomorphic?

8. Oct 20, 2004

### afirican

I think I know the answer. If I say that any map between Q and R+ is not onto, is that enough?

9. Oct 20, 2004

### jcsd

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).