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What is the proof that $x^{a+b} = x^a x^b$ ?

I believe this is called the exponentiation identity

I have known about it for a long time, but don't recall if I have ever seen a proof of this.

There is a proof which works for integer values of $a$, $b$ which is the following:

  • $x^a$ = $\underbrace{x\cdot x\cdot x \cdot\dots\cdot x}_{\text{a times}}$
  • $x^b$ = $\underbrace{x\cdot x\cdot x \cdot\dots\cdot x}_\text{b times}$

therefore

  • $\begin{align*} x^a x^b &= \underbrace{\underbrace{ x \cdot\dots\cdot x}_\text{a times}\underbrace{\cdot x\cdot \dots \cdot x}_\text{b times}}_\text{a+b times}\\ &=x^{a+b}\end{align*}$

Solution

I personally feel that all the (linked) answers to this question are very unclear, so I will provide a short explanation here to make it exactly clear:

  • $x^ax^b=x^{a+b}$ is true for integers because we defined $x^a$ to mean $x$ multiplied by itself $a$ times
  • It makes sense to write down on paper the symbol $x$ multiplied by itself an integer number of times
  • It does not make sense to write down on paper the symbol $x$ multiplied by itself a non-integer number of times
  • But this is not a problem, because we use this as an abstract concept
  • To circle back: Really what we are doing is this:

We define $x\times x$ to be $x^2$

We then move forward and define $x \times x \times x\hat{=} x^3$ where the hat symbol means is defined to be

We continue for arbitrary integer $n$ such that $\underbrace{x\cdot x\cdot x \cdot\dots\cdot x}_\text{n times}\hat{=}x^n$

Finally, we abstract upwards and say that if we could write down an expression on paper for $x$ multiplied by itself $y$ times, where $y$ is a non-integer number, then we would write this as $x^y$, which happens to be something we can write on paper.

Conclusion

There is no real proof here as such, it is more like by definition this is what $x^y$ means, for $y\in\mathbb{R}$.

Additional

From here it is relatively easy to prove other identities. For example:

  • $x^a=x^{a+0}=x^a x^0$ and from this $x^0\hat{=}1$

Now we have an identity for $x^0$ we can use this to get an identity for division.

  • $x^0=x^{0+b-b}=x^0 x^b x^{-b}=1\times x^b x^{-b}$ and since $x^0=$ we have $1=x^b\times x^{-b}$ and from this we define $x^{-b}=\frac{1}{x^b}$

  • Further: If $b=1$ we have $x^{-1}=\frac{1}{x^1}=\frac{1}{x}$ hence from the above bullet point $x^{-b}=\left(x^b\right)^{-1}$

  • Or, if you don't like that argument, we can do $x^{-b}=x^{-1\times b}$. At this point we need to introduce a new identity, which I do not show a proof of here but the reader can surely derive for himself, $x^{a\times b}=\left(x^a\right)^b=\left(x^b\right)^a$, therefore $x^{-b}=\left(x^b\right)^{-1}$

user3728501
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    Your integer proof can easily be adapted to prove it for rational numbers. And then you can invoke continuity of exponentiation to prove it for real numbers. – TonyK Sep 22 '24 at 11:10
  • @TonyK at least for positive rational numbers. – GEdgar Sep 22 '24 at 11:49
  • @GEdgar: it is easy to generalize from naturals to integers. –  Sep 22 '24 at 15:25
  • What is your question? –  Sep 22 '24 at 15:26
  • Your conclusion is quite insufficient. Because stating $x^{a+b}=x^ax^b$ for real $a,b$ does in no way say what $x^a$ is in fact. –  Sep 27 '24 at 08:33
  • @YvesDaoust It does, if taken in combination with a "proof" of $x^ax^b$ for integer a, b – user3728501 Sep 27 '24 at 08:34
  • The standard extention of exponentiation to rationals and reals is not "true", nor "proven", it is just an accepted convention. But in all cases the definition is natural, because it remains compatible with the rule $x^{a+b}=x^ax^b$. (In the complex numbers, the latter rule does not hold anymore.) –  Sep 27 '24 at 08:35
  • No. How much is $2^\pi$, using the fact that $x^{a+b}=x^ax^b$ ? –  Sep 27 '24 at 08:36

  • it is just an accepted convention - yes, this is exactly what I have stated in my question/answer

     – user3728501 Sep 27 '24 at 08:39

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