# Possible description/definition of exponential function (proof verification/review)

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.

• 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: $x\mapsto -x$ – Hagen von Eitzen Feb 8 at 2:30
• 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. – reuns Feb 8 at 5:12