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Lets $S(\mathbb{Q}^+)=\{\log(a) \vert a \in \mathbb{Q}^+\}$. Is this set a group under addition?

  1. Closure if $a,b \in S$ , $a= \log(\frac{p}{q}), \, b= \log(\frac{m}{n}) , \quad a+b= \log(\frac{pm}{qn})$

  2. Associativity $(a+b)+c = \log(a^*b^*)+\log(c^*)= \log(a^*)+ \log(b^*c^*) = a+(b+c)$

  3. Identity element. $0= \log(1)$

  4. Inverse. $a= \log(\frac{p}{q})$ then $a^{-1} = \log(\frac{q}{p})$

Am I making a mistake here?

Is the set $S(\mathbb{R}^+)=\{\log(a) \vert a \in \mathbb{R}^+ \}$ also a group under addition?

If true, is this something interesting? Is there something I should read that looks at this sort of stuff?

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I could've specified the base with a symbol but I am not sure if the base matters. Everything will go through no matter what the base. – Amatya Feb 19 at 23:02
1  
Since no one has pointed this out yet, I think it's worth to note that you don't need to take the rationals in the form $\displaystyle \frac{p}{q}$.The same argument would have worked if you had replaced $\displaystyle \frac{p}{q}$ with $r$ and $\displaystyle \frac{q}{p}$ with $\displaystyle \frac{1}{r}$. Given this, the proof that $\Bigl(S(\Bbb R^+),+_\Bbb R\Bigr)$ is a group is the "same". – Git Gud Feb 19 at 23:08

3 Answers

up vote 5 down vote accepted

There's not much to it, really. The function $\log$ is a homomorphism from the group $\mathbb{R}^{+}$ under multiplication to the group $R$ under addition (an isomorphism, really).

It is a general fact that the image of a subgroup (like $\mathbb{Q}^{+}$ in your example) under a homomorphism is a subgroup of the codomain, that's all.

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Indeed. It's probably worth adding that, since it's an isomorphism, the group in the question is isomorphic to $\mathbb{Q^+}$ under multiplication. – Tara B Feb 19 at 23:07
@Andreas Thanks. Are there a class of functions which work as homomorphisms from $\{\mathbb{R}^+,*\}$ to $\{\mathbb{R},+\}$? Can I understand more about $\log$ because of this? Thanks. – Amatya Feb 19 at 23:10
@Amatya, all I can think of in these small hours are all the logs with respect to all possible bases. I don't see anything particularly work pursuing here, but I might be wrong. – Andreas Caranti Feb 19 at 23:16
@AndreasCaranti: The only continuous ones are the logarithms, but if we assume the Axiom of Choice, we can get many non-continuous wild ones by composing with arbitrary isomorphism of $(\mathbb R,{+})$. – Henning Makholm Feb 19 at 23:24
@HenningMakholm, oh, sure, thanks, it's definitely time for bed, I quoted Hamel bases and such myself only a few days ago here. – Andreas Caranti Feb 19 at 23:27
up vote 2 down vote

The map $x \mapsto e^x$ is an isomorphism from the group of reals under multiplication to the group of positive reals under addition. The map you are describing is just the inverse map of this isomorphism.

Isomorphisms take subgroups to subgroups, so the preimage of $\mathbb Q^+$ under exponentiation should be a multiplicative subgroup of $\mathbb R$.

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Please make it the other way around. – Andreas Caranti Feb 19 at 23:17
up vote 1 down vote

Your proof is correct. But I can't offer any literature on this group.

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