Defining exponential function using functional equations Find continuous functions f: $\mathbb R$->$\mathbb R$ which satisfy the following properties: $f(x+y)=f(x)f(y) \forall x,y \in \mathbb R$, and $f(1)=a$.
I've been struggling with the proof of the theorem.
My attempt:
Let $f(a)=1$.
For $y=0$, we have $f(x+0)=f(x)f(0) \implies  f(0)=1 \text{ or } f(0)=0$ . However, if $f(0)=0$, then $f(1)=f(1)(0)=0$ which is a contradiction since we have $f(1)=a$. Therefore, we conclude $f(0)=1$. Now using induction we can easily deduce that for $n\in \mathbb N,$ we get $f(n)=f(1)^n$, also since $1=f(0)=f(1+(-1))=f(1)f(-1)$, we can find the value of $f(-1)$ and uniquely define it on $\mathbb Z$, and using similar methods on $\mathbb Q$. But the problem is defining it on $\mathbb R$. I get that this is the hard part where we need the continuity to 'fill' the gaps. So how do you proceed from this?

Well you have already showed essentially that
$$f|_{\mathbb{Q}}\implies\,~\,~f(q)=a^q\,.$$
The denseness of $\mathbb{Q}$ in $\mathbb{R}$ and the continuity of $f$ then guarantees a unique such $f$; indeed, for any $r$, we can find a sequence of rational numbers $\{q_n\}_{n\ge 1}$ converging to $r$ such that
$$\lim_{n\to\infty}f({q_n})= f(\lim_{n\to\infty}{q_n})\,,$$
where the continuity requirement allows the limit and the function to commute (that is, to take the limit “inside” the function). However, the left hand side just becomes
$$\lim_{n\to\infty}a^{q_n}=a^r\,$$ implying that the only such function is $f(r)=a^r$.
