Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I think it wants me to show that there's no bijection 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!

share|improve this question
    
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
add comment

2 Answers 2

up vote 6 down vote accepted

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, $\mathbb{R}^{\times} \cong \mathbb{Z}/2 \oplus \mathbb{R}.$

share|improve this answer
    
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
1  
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
1  
@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
show 4 more comments

Hint: there is something special about $-1$.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.