I wish to describe the exponential function as unique continuous group isomorphism $f:(\mathbb{R},+)\rightarrow(\mathbb{R}^+,\cdot)$ satisfying $f(1)=a$, $f\equiv a^x$.

Lemma 1: Assume $f:G\rightarrow H$ is a group isomorphism, then $f$ is unique if and only if $\operatorname{Aut}(G)$ is trivial group.

Lemma 2: Let $f$ be automorphism on $(\mathbb{R},+)$, then if $f$ is continuous, then $f(x)=rx$ for some $r\in\mathbb{R}$

Proof. $f$ is an automorphism, thus for any $a,b\in\mathbb{R}:f(a+b)=f(a)+f(b)$.

For $n\in\mathbb{N}$ $f(n)=n\cdot f(1)$, proceed by induction, base step is trivial. Assume (Induction step) $f(n)=n\cdot f(1)$, we wish to show that $f(n+1)=(n+1)\cdot f(1)$. We have $f(n+1)=f(n)+f(1)=n\cdot f(1)+f(1)=(n+1)\cdot f(1)$

For $z\in\mathbb{Z}$ is also $f(z)=z\cdot f(1)$, for $z=0$ we have $f(0)=f(0-0)=f(0)-f(0)=0$. Let $-z\in\mathbb{N}$, that is $z=-1,-2,-3,\ldots$. Then $-z\cdot f(1)=f(-z)=-z\cdot f(1)$. Now $f(-z)=f(0-z)=f(0)-f(z)=-f(z)=$, so claim holds for $z\in\mathbb{Z}$.

For $x\in\mathbb{Q}: x\cdot f(1)=f(x)$. Let $x=\frac{p}{q}$. We have that $p\cdot f(1)=f(p)=f(p\cdot \frac{q}{q})=q\cdot f(\frac{p}{q})$ thus $\frac{p}{q}\cdot f(1)=f(\frac{p}{q})$ so $x\cdot f(1)=f(x)$

Now, $\mathbb{Q}$ is dense in $\mathbb{R}$ so there is at most one continuous extension to $\mathbb{R}$, so let $f(x)=x\cdot f(1)$ be the function. Where $f(1)$ is some chosen number $r\in\mathbb{R}$

Now, we wish to make the final claim, that $a^x$ is unique.

Claim: $a^x$ as defined above, is unique.

Proof. By the 2nd lemma all continuous automorphisms $f\in\operatorname{Aut}(\mathbb{R},+)$ are of the form $f(x)=f(1)\cdot x$. We have fixed $f(1)=a$. So $\operatorname{Aut}(\mathbb{R},+)$ is trivial group, namely, it contains only automorphism $x\mapsto ax$. Thus by Lemma 1, $a^x$ is unique.

Is my proof valid and rigorous enough? Aren't there any crucial errors? Anyways, this can't be classified as proving something like $a^x$ exists, right? It's just pure description, I haven't proven anything like this actually exists.

  • $\begingroup$ Your lemma 1 does not really make sense. E.g., $(\Bbb R,+)$ has lots of automorphisms. Every non-trivial abelian group has at least one non-trivial automorpshism: $x\mapsto -x$ $\endgroup$ – Hagen von Eitzen Feb 8 at 2:30
  • $\begingroup$ So, using the density of $\mathbb{Q}$, you want to describe all the continuous automorphisms of $(\mathbb{R},+)$ in order to find, from one continuous isomorphism $(\mathbb{R},+) \to (\mathbb{R}_{> 0},\times)$, all the continuous automorphisms of $(\mathbb{R}_{> 0},\times)$ and all the continuous isomorphisms $(\mathbb{R},+) \to (\mathbb{R}_{> 0},\times)$. This is valid and rigorous enough. $\endgroup$ – reuns Feb 8 at 5:12

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