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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?

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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
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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
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@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 at 20:28

1 Answer 1

up vote 3 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 $( P , \cdot )$.

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