Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Am I incorrect in believing that the following exercise is not possible?

Prove that the additive group $( \mathbb{R}, + )$ of real numbers is isomorphic to the multiplicative group $( P , \cdot )$ of positive reals.

My reasoning is that if we had $\phi \colon \mathbb{R} \to P$ as our isomorphism, then we have

$$\phi(\tfrac{1}{3}) = \tfrac{1}{3}$$ and $$\phi(-3) = \tfrac{1}{3}$$

Am I missing something?

share|cite|improve this question
Why would we have that? – Qiaochu Yuan Dec 15 '12 at 9:51
@QiaochuYuan Do we not have to map inverses to inverses? That along with the fact that $\frac{1}{3} \in \mathbb{R}^+$ and $P$. – providence Dec 15 '12 at 10:00
Both of those things are true, but they don't imply what you're concluding. If you write out your argument in more detail it should be easier to see the mistake. – Qiaochu Yuan Dec 15 '12 at 10:05
@QiaochuYuan Oh, right, reals. I've just come off a binge of problems dealing with $\mathbb{Z}$ and $\mathbb{Q}$ exclusively. I see my error. – providence Dec 15 '12 at 10:14
possible duplicate of Isomorphism between groups of real numbers – Johanna Apr 20 '15 at 20:28
up vote 4 down vote accepted

Your reasoning is a little faulty since you are assuming that $\phi ( \frac{1}{3} ) = \frac 13$ (since a priori there is not reason to think that $\frac 13$ must be a fixed point of such an isomorphism). But even if $\phi ( \frac 13 ) = \frac 13$, then this would tell us that $$\begin{align} \phi ( 3 ) &= \phi (\tfrac 13+\tfrac 13+\tfrac 13+\tfrac 13+\tfrac 13+\tfrac 13+\tfrac 13+\tfrac 13+\tfrac 13) \\ &= \phi (\tfrac 13)\cdot\phi(\tfrac 13)\cdot\phi(\tfrac 13)\cdot\phi(\tfrac 13)\cdot\phi(\tfrac 13)\cdot\phi (\tfrac 13)\cdot\phi(\tfrac 13)\cdot\phi(\tfrac 13)\cdot\phi(\tfrac 13) \\ &= \tfrac 13 \cdot \tfrac 13 \cdot \tfrac 13 \cdot \tfrac 13 \cdot \tfrac 13 \cdot \tfrac 13 \cdot \tfrac 13 \cdot \tfrac 13 \cdot \tfrac 13 \\ &= 3^{-9} \end{align}$$ and therefore $\phi ( -3 ) = ( \phi ( 3 ) )^{-1} = ( 3^{-9} )^{-1} = 3^9$.

If you recall the following rule of exponentiation: $$a^{x+y} = a^x \cdot a^y$$ you should begin to think that a mapping of the form $x \mapsto a^x$ looks "homomorphism-ish," and it is not too difficult to show that if $a > 0$, then such a mapping is an isomorphism between $( \mathbb{R} , + )$ and $( P , \cdot )$.

share|cite|improve this answer

Your Answer


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.