Showing $H=\{\log a : a\in \Bbb Q, a>0\}$ is a subgroup of $\left \langle \Bbb R,+ \right \rangle$ The Problem: Let $G=\left \langle \Bbb R,+ \right \rangle$ and $H=\{\log a : a\in \Bbb Q, a>0\}$
Show that $H$ is a subgroup of $G$.

My Proof:
Let $\log a$ and $\log b$ be elements in $H$ where $a,b \in \Bbb R$ and $a,b>0$. Then $\log a + \log b=\log ab \in H$ since $ab\in \mathbb Q$ and $ab>0$. Thus $H$ is closed.
Let $-\log a =\log a^{-1} =\log({1\over a}) \in H$, then $\log a + (-\log a)=\log a - \log a=0$. Thus $-\log a \in H$ and $H$ has additive inverses and $H$ is a subgroup of $G$.

Note: I am specifically trying to prove $H$ is a subgroup by showing $H$ is closed and has inverses. What would happen to this result if $a$ was only allowed to be a natural number?
 A: More or less, this proof is correct. My main nitpick is with your wording of showing inverses. You should say

Let $\log a \in H$. Then [...] and thus $-\log a$ (the inverse of $\log a$ under $+$) is in $H$.

As worded, it seems to presuppose that this inverse is in $H$. Some other nitpicks could arise, such as (unless you know $\left \langle \Bbb R,+ \right \rangle$ is an abelian group specifically) needing to show the inverse works on each side.
Also, a minor thing: you should have that $\log a, \log b \in H$ because $a,b$ are specifically in $\Bbb Q^+$ (i.e. are positive rationals, not just real numbers). Perhaps it is also worth noting that you are assuming $\Bbb R, \Bbb Q$ are fields in this proof, with the usual properties - for instance, assumping that $a,b \in \Bbb Q \implies a \cdot b \in \Bbb Q$.

Also, note that you can use the one-step subgroup test instead for a slightly slicker proof, in the sense that
$$\log a + (- \log b) = \log \left( \frac a b \right)$$
which is in $H$ since $a,b > 0 \implies b \ne 0$ and $a/b > 0$.


What would happen to this result if $a$ was only allowed to be a natural number?

Then it would not work since you do not have inverses for every element. For instance, $2 \in \Bbb N \implies \log 2 \in H$ but it has inverse
$$-\log 2 = \log \left( \frac 1 2 \right)$$
which is not in $H$, since $1/2 \not \in \Bbb N$.
