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I need help proving that there is no isomorphism between the structures $(\mathbb{R},+)$ and $(\mathbb{R}^+,\cdot)$

I think I'm very close to the answer and I need some hint how to get there:

It has to do with the fact, that in $\mathbb{R}^+$, $y_1\cdot y_2= 1$ holds if and only if $y_1,y_2=1$.

For example, if $0$ gets mapped to $1$ (by the isomorphism),

then $\phi(x_1)\cdot\phi(x_2)=1\Rightarrow\phi(x_1+x_2)=1$

Therefore we have the conditions:

  1. $x_1 = x_2$
  2. $x_1 + x_2 = 0$

However, there are many elements in $\mathbb{R}$ which can be zero when added, for example $-3$ and $3$. Since these elements are not equal, we have a contradiction.

Am I on the right track with this? How should I proceed?

1 Answers 1

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The "fact" that you are using, isn't. What is ${1\over 2}\cdot 2$?

Indeed, the two groups are isomorphic. HINT: do you know a function which converts multiplication into addition?

  • 1
    Logarithm? For example $\log(ab)=\log(a)+\log(b)$. But is it applicable here? – 2017-01-18
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    @de_dust Yup! And yes, it is. Can you show that $\log$ (say, base $e$) is an isomorphism from $(\mathbb{R},\cdot)$ to $(\mathbb{R},+)$? – 2017-01-18