Proving arithmetical properties for non-natural numbers

Question

Sorry if my question is dumb but here it is:

I know how to prove all of the arithmetical properties such as

$(a^{m})^{n}=a^{mn}$ and $a^{m}a^{n}=a^{m+n}$ and $a(b+c)=ab+ac$ etc.

For numbers that are natural. I kind of accept negative integers too, but they bug me a little bit.

The thing is, I know all these properties work and apply to all the real numbers. But how does one go about proving them for all real numbers?

What I mean by not being able to prove for all real numbers is this, consider the following:

$a^{m}a^{n}=a^{m+n}$ Because if I were to count all $a's$ I can see they are exactly $m+n$ when $m$ and $n$ are natural numbers. But what about fractional numbers? I have an intuition of what's going on but it's not absolutely obvious...

Should these rules be taken like axioms? Such that if they weren't true "all hell would break loose" or whatever. Are these like, the laws of numbers? Such that math is more like a science in that we've "experimentally" seen these rules work well?

Thanks, and sorry if my question is dumb, but it's bugging me a lot right now.

Answer 1

Great question.

One of the goals, when we define the meaning of new symbols, is to make them easy to manipulate consistently. One property that $a^n$ has is that $(a^n)^k = a^{nk}$ when $n$ and $k$ are integers. So if we're trying to define $a^n$ when $n$ is a rational number like $2/3$, it makes sense to make the same "property" hold. So we want, for instance, $$ (a^{2/3})^3 = a^{\frac{2}{3} \cdot 3}= a^2 $$ So it makes sense to define $a^{2/3}$ as "the number whose cube is $a^2$".

There are a couple of difficulties: one is that there might be multiple numbers whose square, or cube, or whatever, is the thing we're looking for. For instance, if you said something was "the number whose square is 4", you wouldn't really have defined it, because it could be $2$ or $-2$. This is, in fact, the only problem that can arise, so we say $a^{2/3}$ is "the positive number whose cube is $a^2$".

Having done that, we still have to check that all the other exponent rules still hold -- that we haven't "broken" anything by making this new rule. For instance, do we get the same thing when we compute $a^{2/3}$ and $a^{4/6}$? We do, but you have to check that this always works before committing to this new definition.

This stuff also cannot be made to work nicely if $a$ is negative, but it works so well for positive values of $a$ that we go ahead an use it in that case.

What about raising $a^r$, where $r$ is not a rational number, but something like $\pi$? In that case, we need to use calculus to come up with a viable definition, and in doing so it turns out that things get briefly harder, and then a great deal simpler. So my main answer here is "wait and see" for this irrational-number case.

Answer 2

For $(a^m)^n=a^{(mn)}$ at least two arguments (but undoubtedly more) are possible: $$ (a^m)^n = (\bullet)^n = \underbrace{\bullet\cdots\cdots\bullet}_{\text{$n$ factors}} = \underbrace{(a^m)\cdots(a^m)}_{\text{$n$ factors}} = \underbrace{(\ \underbrace{a\cdots\cdots a}_{\text{$m$ factors}}\ )\cdots(\ \underbrace{a\cdots\cdots a}_{\text{$m$ factors}}\ )}_{\text{$n$ factors}} = \underbrace{a\cdots\cdots a}_{\text{$mn$ factors}} $$ (Remember that "factor" means something that gets multiplied by something.)

One could also write a combinatorial argument: $a^m$ is the number of functions from a set of size $m$ into a set of size $a$. And $(a^m)^n$ is the number of functions from a set of size $n$ into that set of functions. So you would seek a one-to-one correspondence between the last-mentioned set of functions and the set of all functions from a set of size $mn$ into a set of size $a$.

$$ a^{m+n} = \underbrace{a\cdots\cdots\cdots\cdots a}_{\text{$m+n$ factors}} = \underbrace{a\cdots\cdots a}_{\text{$m$ factors}}\ \underbrace{a\cdots\cdots a}_{\text{$n$ factors}} = a^m a^n. $$