This post is essentially an adaptation to an answer I gave to a question on MathSE. In there, I gave another approach that may be worth checking: response in MathSE.

Initial observation: I'll assume that it is well-understood what is a number raised to a rational power.

Firstly, there are a lot of "motivational" approaches to the study of the exponential function. I'll adopt a purely mathematical motivation (specifically, algebraic) that is essentially in the nucleus of almost all others.

We know that there are two canonical algebraic structures in $ \mathbb{R}$: addition and multiplication. Formally, we consider the groups: $ (\mathbb{R},+)$ and $(\mathbb{R}_{> 0},\times)$. The question we ask is:

Now, let's try to extract some properties of what a function of this type, which we will call $\xi : (\mathbb{R},+) \rightarrow (\mathbb{R}_{> 0},\times)$ should have. Firstly, she should satisfy, by its definition, the following equation:

$\xi(x+y)=\xi(x).\xi(y)$

Since $\xi(0)=\xi(0+0)=\xi(0)\xi(0)$, $\xi(0)=1$.

Now, define $k:= \xi(1)$. What we will show now is that, with the value of $k$, we have the value of $\xi$ in all rationals.

We want to find a $\displaystyle \xi_1 (x)=\sum_{n=0}^{\infty}a_nx^n$ that is of the "type" of $ \xi$, meaning: $\xi_1 (x+y)=\xi_1(x)\xi_1(y)$. Do:

$ \displaystyle \xi_1(x+y)=\sum_{n=0}^{\infty}a_n(x+y)^n=\sum_{n=0}^{\infty}\sum_{k=0}^{n}a_n\frac{n! x^ky^{n-k}}{k!(n-k)!}=\sum_{n=0}^{\infty}a_nn!\sum_{k=0}^{n}\frac{ x^ky^{n-k}}{k!(n-k)!}$ , by the Binomial Theorem.

Notice that the last equality is

So, the function $\displaystyle \xi_1(x)=\sum_{n=0}^{\infty}\frac{x^n}{n!}$ satisfies all we want. (The series is well defined for all x real, which is easily verifiable by the ratio test. The series is obviously greater than zero is x is greater than 0, and the fact that $\xi_1(x).\xi_1(-x)=\xi_1(0)=1$ implies the series is greater than zero for all x real. Other observations also imply that it is bijective, since it is surjective (follows from the fact that it goes to infinity when x goes to infinity, to zero when x goes to -infinity, and it is continuous) and injective (because it is strictly increasing)

As we saw, it is enough knowing $\xi_1$ in $1$ to know its value in all rationals. But we know what it is in $1$:

$\displaystyle \xi_1(1)=\sum_{n=0}^{\infty}\frac{1}{n!}$

We will call this amazing number $e$.

So, $\xi_1(r)= e^r$ if $r$ is rational, in the usual sense we have for "a number raised to a rational power" (this follows from (1), (2) and (3)). Note that we already have an answer to our initial problem: we found a function $\xi_1$ which is our isomorphism between the groups. But it may be worth a while to try and say a few more things about it.

Since $\xi_1(r)=e^r$ if $r$ is rational, we

$e^x:=\xi_1(x)$

when $x$ is real, and this satisfies what we would wait of $e^x$ by previous observations. OK then, we found

Now, since $e^x$ is bijective, there is an inverse, which we will call $ln(x)$. This allows us to define, for an arbitraty $ a >0 $ :

$a^x:=e^{ln(a).x}$

Since, by definition $ a^1=a$ e $ a^{(x+y)}=a^x.a^y$, it follows from the same considerations as before that this function is what deserves to be called "raising $a$ to a real power".

Thus, we have:

$ (e^x) ^y=e^{(ln(e^x)).y}=e^{(x.y)}$

But going back to our previous question: what if we forget the constraint of a continuous extension: is there another function that coincides with $e^x$ on the rationals?

Yes, but we will need the following theorem to "show" a function of this kind:

Theorem: Every vector space has a basis.

Now, consider $ \mathbb{R}$ as a vector space over $ \mathbb{Q}$, and take a basis $ \{x_i\}_{i \in \Lambda}$ that has $1$. Define:

$f(1):=e$

$ f(x_i):=e^{x_i/2}$ if $ x_i \neq 1$

Since we defined the function on the basis and we want that $f(x+y)=f(x).f(y)$, we define:

$ f(x):=f(x_1)^{\lambda_1}f(x_2)^{\lambda_2}...f(x_n)^{\lambda_n}$,

where $ x=\lambda_1x_1+\lambda_2x_2+...+\lambda_nx_n$ is the representation of $x$ as a linear combination of elements of the basis (keep in mind that $\lambda_i$ are rationals!), and everything is well defined since$\{x_i\}$ is a basis. The fact that this function satisfies $f(x+y)=f(x).f(y)$ follows from definition, and the non-continuity of the function follows from the fact that it coincides with the exponentiation on the rationals but is not the exponentiation on the reals (she takes different values on $ x_i$, for example).

We see that requiring continuity is not only very reasonable, but also necessary. Otherwise, we would have a lot of "non-constructive exponentials" to consider.

Initial observation: I'll assume that it is well-understood what is a number raised to a rational power.

Firstly, there are a lot of "motivational" approaches to the study of the exponential function. I'll adopt a purely mathematical motivation (specifically, algebraic) that is essentially in the nucleus of almost all others.

We know that there are two canonical algebraic structures in $ \mathbb{R}$: addition and multiplication. Formally, we consider the groups: $ (\mathbb{R},+)$ and $(\mathbb{R}_{> 0},\times)$. The question we ask is:

- Is there an isomorphism between these two structures?

Now, let's try to extract some properties of what a function of this type, which we will call $\xi : (\mathbb{R},+) \rightarrow (\mathbb{R}_{> 0},\times)$ should have. Firstly, she should satisfy, by its definition, the following equation:

$\xi(x+y)=\xi(x).\xi(y)$

Since $\xi(0)=\xi(0+0)=\xi(0)\xi(0)$, $\xi(0)=1$.

Now, define $k:= \xi(1)$. What we will show now is that, with the value of $k$, we have the value of $\xi$ in all rationals.

- Firstly, if $n \in \mathbb{N}$, we have$\xi(n)=\xi(1)...\xi(1)$ $ n$ times=$\xi(1)^n$
- If $-n$ is a negative integer, we have $ 1=\xi(0)=\xi(n-n)=\xi(n)\xi(-n)$, which implies: $\displaystyle \xi(-n)=\frac{1}{\xi(n)}=\frac{1}{\xi(1)^n}=\xi(1)^{-n}$
- If $a=\frac{p}{q} \in \mathbb{Q}$, $ \xi(p)=\xi(q.\frac{p}{q})=\xi(\frac{p}{q}+\frac{p}{q}+...+\frac{p}{q})$ $ q$ times $ =\xi(p/q)^q$. So, $ \xi(\frac{p}{q})=\xi(p)^{1/q}=(\xi(1)^p)^{1/q}$

We want to find a $\displaystyle \xi_1 (x)=\sum_{n=0}^{\infty}a_nx^n$ that is of the "type" of $ \xi$, meaning: $\xi_1 (x+y)=\xi_1(x)\xi_1(y)$. Do:

$ \displaystyle \xi_1(x+y)=\sum_{n=0}^{\infty}a_n(x+y)^n=\sum_{n=0}^{\infty}\sum_{k=0}^{n}a_n\frac{n! x^ky^{n-k}}{k!(n-k)!}=\sum_{n=0}^{\infty}a_nn!\sum_{k=0}^{n}\frac{ x^ky^{n-k}}{k!(n-k)!}$ , by the Binomial Theorem.

Notice that the last equality is

**BEGGING**for the Cauchy Product Formula. If $ a_nn!$ was $1$... the last side would be $\displaystyle \left(\sum_{k=0}^{\infty}\frac{x^k}{k!}\right)\left(\sum_{m=0}^{\infty}\frac{y^m}{m!}\right)$. But we have the power to choose who the $a_n$ are! We are searching a candidate to the series, after all. So, take $a_nn!=1$, which implies $ \displaystyle a_n=\frac{1}{n!}$. Now, stare at our luck! Since $\displaystyle a_n=\frac{1}{n!}$, what we have in the last equality is precisely the product $ \xi_1(x).\xi_1(y)$.So, the function $\displaystyle \xi_1(x)=\sum_{n=0}^{\infty}\frac{x^n}{n!}$ satisfies all we want. (The series is well defined for all x real, which is easily verifiable by the ratio test. The series is obviously greater than zero is x is greater than 0, and the fact that $\xi_1(x).\xi_1(-x)=\xi_1(0)=1$ implies the series is greater than zero for all x real. Other observations also imply that it is bijective, since it is surjective (follows from the fact that it goes to infinity when x goes to infinity, to zero when x goes to -infinity, and it is continuous) and injective (because it is strictly increasing)

As we saw, it is enough knowing $\xi_1$ in $1$ to know its value in all rationals. But we know what it is in $1$:

$\displaystyle \xi_1(1)=\sum_{n=0}^{\infty}\frac{1}{n!}$

We will call this amazing number $e$.

So, $\xi_1(r)= e^r$ if $r$ is rational, in the usual sense we have for "a number raised to a rational power" (this follows from (1), (2) and (3)). Note that we already have an answer to our initial problem: we found a function $\xi_1$ which is our isomorphism between the groups. But it may be worth a while to try and say a few more things about it.

Since $\xi_1(r)=e^r$ if $r$ is rational, we

**define****:**$e^x:=\xi_1(x)$

when $x$ is real, and this satisfies what we would wait of $e^x$ by previous observations. OK then, we found

**one**function that extends what we call "exponentiation" of rational powers to real powers. But this function is analytic. Is there another less smooth function, for example $ C^{\infty}$ or even only continuous that could extend exponentiation? The answer is not, since this function would coincide in rational points, and the rationals are dense in $\mathbb{R} $ (Two continuous functions that coincide in a dense subset coincide in all $\mathbb{R}$). Then, this extension is unique if we require it to be continuous. This is enough to call $\xi_1(x)$ "**the**exponential", and have no regrets for calling it $e^x$.Now, since $e^x$ is bijective, there is an inverse, which we will call $ln(x)$. This allows us to define, for an arbitraty $ a >0 $ :

$a^x:=e^{ln(a).x}$

Since, by definition $ a^1=a$ e $ a^{(x+y)}=a^x.a^y$, it follows from the same considerations as before that this function is what deserves to be called "raising $a$ to a real power".

Thus, we have:

$ (e^x) ^y=e^{(ln(e^x)).y}=e^{(x.y)}$

But going back to our previous question: what if we forget the constraint of a continuous extension: is there another function that coincides with $e^x$ on the rationals?

Yes, but we will need the following theorem to "show" a function of this kind:

Theorem: Every vector space has a basis.

Now, consider $ \mathbb{R}$ as a vector space over $ \mathbb{Q}$, and take a basis $ \{x_i\}_{i \in \Lambda}$ that has $1$. Define:

$f(1):=e$

$ f(x_i):=e^{x_i/2}$ if $ x_i \neq 1$

Since we defined the function on the basis and we want that $f(x+y)=f(x).f(y)$, we define:

$ f(x):=f(x_1)^{\lambda_1}f(x_2)^{\lambda_2}...f(x_n)^{\lambda_n}$,

where $ x=\lambda_1x_1+\lambda_2x_2+...+\lambda_nx_n$ is the representation of $x$ as a linear combination of elements of the basis (keep in mind that $\lambda_i$ are rationals!), and everything is well defined since$\{x_i\}$ is a basis. The fact that this function satisfies $f(x+y)=f(x).f(y)$ follows from definition, and the non-continuity of the function follows from the fact that it coincides with the exponentiation on the rationals but is not the exponentiation on the reals (she takes different values on $ x_i$, for example).

We see that requiring continuity is not only very reasonable, but also necessary. Otherwise, we would have a lot of "non-constructive exponentials" to consider.