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When n and m are both positive integers, to show that $a^n a^m = a^{n+m}$ will be pretty straightforward:

$$a^n a^m = (\underbrace{a\cdot a\cdots a}_{n\text{ copies}})(\underbrace{a\cdot a\cdots a}_{m\text{ copies}}) = (\underbrace{a\cdot a\cdots a}_{n+m\text{ copies}})= a^{n+m}$$

Now suppose that, for example, $n \in \mathbb R$ and $0 < n < 1$ and $m \in \mathbb R$

How do you show that $a^na^m = a^{n+m}$ in that scenario too?

Stokolos Ilya
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    It's not so easy for real numbers. Actually, in quaternions it's just wrong. – Michael Rozenberg Jul 18 '19 at 18:13
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    You're also going to want $a>0$ otherwise it doesn't hold again. – Ian Jul 18 '19 at 18:14
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    It really depends on how you define $a^x$ when $x\in\mathbb R.$ It is usually defined as $e^{n\ln a}$ when $a>0.$ Then you need to show one theorem that $e^{u}e^{v}=e^{u+v}$ for any real $u,v.$ You can define $e^{u}$ in terms of a power series or a limit $(1+u/n)^n$ as $n\to\infty.$ Alternatively, you can define $a^p$ for $a>0$ and any $p$ rational, then show that $a^x$ can be defined and made continuous by using the density of $\mathbb Q$ in $\mathbb R.$ So, really, it depends heavily on the definition of $a^x.$ – Thomas Andrews Jul 18 '19 at 18:26

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