Well, I've been searching through this fórum and I know this question has been answered many times. But the answers I see, are kinda circular (I think). Let's start by the natural case.
- Natural case
As the name says, this is natural to think. We can
define$a^x \cdot a^y$ as $a^{x+y}$ because it is something we can observe and that is definitively true. When we think about the $0$ case for na expoent, and define it by being $1$ for every number except $0$ itself (not gonna enter in the $0^0$ discussion in this post) we can think in many ways. The best way to define this strongly as being equals to one, should be by taking the limit from both sides, but we don't have a definition yet, for a continuous function $f(x) = a^x$ so we're closed in the natural case. If we observe this behavior:
$$ 2^3 = 8 $$ $$ 2^2 = 4 $$ $$ 2^1 = 2 $$ Then maybe $2^0$ should be $1$ because in the left side we're subtracting $1$ for the exponent, and in the right side we are dividing by the base of the exponent. So let's proceed to other definitions and see if we can still stay with the definition of the zero power.
- Integer case
If we think the same way, we can create a definition that works for both the natural case, and our new case: the integers. If we keep subtracting that table in the left side, we're gonna get to the negative case. So if we accept: $$2^0 = 1$$ And subtract 1 from the exponent on the left side, and divide by the base in the right side, we reach: $$2^{-1} = \frac{1}{2}$$ And if we continue, we'll see that we can define, for every integer exponent:
$$a^y = a\cdot a \cdots a \qquad\text{a times itself $y$ times, for every $y>0$, $y \in \mathbb Z $}$$ $$a^y = 1 \qquad\text{for $y = 0$}$$ $$a^y = \frac{1}{a\cdot a \cdots a} \qquad\text{$1$ over $a$ times itself $y$ times, for every $y<0$, $y \in \mathbb Z$}$$
- Rational case
Well, here's what I've been reading:
Define$a^{\frac{1}{n}} = c$ as the n-th root of $c$. So, for example, $a^{\frac{1}{2}} = \sqrt{a}$ This definition only makes sense to me if weACCEPTthat $$a^{b} \cdot a^{c} = a^{b+c}\tag {for $b$ and $c$ $\in R$}$$ then $$a^{\frac{1}{n}} = b$$ so $$(a^{\frac{1}{n}})^n = c^n$$ then $$a = c^n$$ As we can see, $a$ was the n-th root of $c$, but this only makes sense if we accept that the law for natural numbers hold for the rational case. We can, of course, 'take the limit' (we cannot take the limit yet, because it's not in the reals) of both sides and verify that by our acception, when the exponent $\frac{1}{n}$ gets closer to $1$ (in other words, $n$ gets closer to $1$) we get the number that multiplied by itself $1$ times is equal to itself, wich is just the number itself. In other words: $a^{\frac{1}{n}} = a$ when $n$ gets closer and closer to $1$. So it seems that we havegood reasonsto believe that the sum property holds for rational exponentes. But since math is highly rigorous we cannot ever accept this only because we have good reasons. Just because something don't makes sense under some sets, we cannot define it. So is there something that really proves that the sum rule rolds forrationals, and evenrealexponents? And what aboutimaginaryandcomplexexponents? What can we say about it?
