# how to show that the set of nonzero real numbers under multiplication is not isomorphic to the set of real numbers under addition?

I think it wants me to show that there's no bijection f between the two sets.

I first tried to show that there's simply no bijection, then I realised that it doesn't work.

If I'm to show there's no bijective morphism that carries multiplication in nonzero real numbers to addition in real numbers, how am I supposed to do it? I'm thinking about doing something with 0, since it's the element of the first set but not the second set.

Thanks!

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There is no number $a\ne0$ such that $a+a=0$. But there is a number $a\ne1$ such that $a\cdot a=1$. (I won't post this as an answer since (in effect) someone's already done that.) –  Michael Hardy Sep 26 '11 at 2:20
I thought of that in the beginning, but I don't think that works because you can always use f to assign 0 to any element. –  Scharfschütze Sep 26 '11 at 23:19

In $(\mathbb{R}, +)$ there are no torsion elements (the group generated by any non-identity element is infinite cyclic). On the other hand, in $\mathbb{R}^{\times},$ the group $\langle -1 \rangle$ is finite.

That said, it is interesting to note $\mathbb{R}^{\times}/ \langle -1 \rangle \cong (\mathbb{R}_{+}, \cdot)$ which is isomorphic to $(\mathbb{R}, +)$ via the natural logarithm. In fact, $\mathbb{R}^{\times} \cong \mathbb{Z}/2 \oplus \mathbb{R}.$

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I haven't learnt groups yet, is it possible to do this problem using just the definition of morphisms? and by the way what does R × /⟨−1⟩ mean? thanks –  Scharfschütze Sep 26 '11 at 0:37
$\mathbb{R}^{\times}$ is the group of nonzero elements of $\mathbb{R}$ under multiplication and $\mathbb{R}^{\times}/ \langle -1 \rangle$ is a quotient group (don't sweat if you don't know what that means as it was an aside). –  jspecter Sep 26 '11 at 0:55
The hidden theorem used in my answer is that any isomorphism preserves order. Thus since $-1$ has finite order and all elements of $\mathbb{R}$ have infinite order there can be no isomorphism from $\mathbb{R}^{\times}$ to $\mathbb{R}.$ –  jspecter Sep 26 '11 at 0:56
@Scharfschütze: I'm not sure what you mean when you say you know about morphisms but not about groups. You don't just have "morphisms": you have "morphisms in a certain category" or, more concretely, "morphisms preserving a certain structure". So when you say isomorphic you need to specify a certain structure. The two objects given are isomorphic as sets, i.e., there is a bijection between them. They are not isomorphic as groups, but if you don't know what a group is, how is it that you have been asked to show this?? –  Pete L. Clark Sep 26 '11 at 1:27
@Pete L. Clark: I'm going to learn about groups this week but this is a homework problem I got last week(on semigroups and monoids). We haven't learnt isomorphism of groups yet nor have we looked at cyclic groups. –  Scharfschütze Sep 26 '11 at 6:16
Hint: there is something special about $-1$.