Is $a^m \in \mathbb{R}$ and $a^n \in \mathbb{R}$ sufficient for $(a^m)^n=a^{mn}$?

Question

Suppose $a, m, n \in \mathbb{R}$. In real-number arithmetic, is $a^m \in \mathbb{R}$ and $a^n \in \mathbb{R}$ sufficient for $(a^m)^n=a^{mn}$?

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Edit: Symbolically, is it true that $$\forall a, m, n \in \mathbb{R} : [a^m, a^n \in \mathbb{R} \implies (a^m)^n=a^{mn}]$$ for the usual exponentiation on $\mathbb{R}$ (a partial binary function that is undefined for some combinations of base and exponent values, e.g. for $0^0$ and $(-1)^{1/2}$)?

If true, this would allow for negative bases in applications of this rule.

Answer 1

If $a^b, (a^b)^c, (a^{bc})$ are defined and we are only using real numbers then $(a^b)^c = a^{bc}$.

It is completely fundamental that $a^{b+k} = a^ba^k$ and so if $bc = \underbrace{b+b+....+b}_{c\text{ times}}$ then $(a^b)^c = a^{bc}$. ("But what if $c$ isn't an integ..." whack.... don't pay any attention to that voice under the carpet.)

If any of them are undefined for any reason, say $a<0$ and $b$ is a fraction with an even denominator in its lowest term, or if $a=b=0$ then it's not so much that $(a^b)^c = a^{bc}$ is false (it isn't) but that the statement is meaningless as it has undefined terms.

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With complex number we can define all $a^b$ (except for $a=b=0$). But that thing is it is multivalued. What is $a^{\frac 12}$? Well, it is the number $k$ so that $k^2 = a$. But... if $k^2 = a$ then $(-k)^2 = a$ also... which one is $a^{\frac 12}$. Well, the answer is... the both are. Or more $a^{\frac 12}$ is a set of two values so that $k^2 = a$.

So can we say $(a^b)^c = a^{bc}$. Well, the issue isn't that it isn't true. It's that there are multiple values of $a^b$ and of $(a^b)^c$ and $a^{bc}$ and for at least one set they will synch up but others can be notational trickery and not make sense.

There's an old paradox. $i = \sqrt{-1}$. (which isn't really true; $i^2 = -1$ but $\sqrt{-1} = $ the set of both $i$ and $-i$.)

Then $1 = \sqrt{1*1} = \sqrt{-1*-1} = \sqrt{-1}*\sqrt{-1} = i*i=i^2 = -1$.

If you ask must people where the error is they'll say the rule $(ab)^{k} = a^kb^k$ only holds for positive numbers. Which is partially true.

But more to the point there are multiple values that $\sqrt{}$ can be.

$1$ can be one thing: 1. And $\sqrt{1}=\sqrt{1*1} =\sqrt{-1*-1}$ can be two things: $1, -1$. And $\sqrt{-1}$ can be two things: $i$ and $-i$. And $\sqrt{-1}*\sqrt{-1}$ can be four things: $i*i = -1; i*(-i) = 1; (-i)*i=1; (-i)*(-i) = -1$. and as inclusions... it's true: $1 \in \{1,-1\}$ but we can't reverse the direction.

Answer 2

Yes. If $a>0$, then the identity holds universally. If $a<0$, then you are implying that $m$ and $n$ are rational numbers with odd denominator, in which case $a^{mn}=(a^m)^n$. A quick way to see this is by using the fact that, for $q$ odd, $b\ne 0$ and $h$ integer, $(-\lvert b\rvert)^{h/q}=(-1)^h\lvert b\rvert^{h/q}$. Therefore $$((-\lvert a\rvert)^{h_1/q_1})^{h_2/q_2}=((-1)^{h_1}\lvert a\rvert^{h_1/q_1})^{h_2/q_2}=(-1)^{h_1h_2}\lvert a\rvert^{h_1h_2/(q_1q_2)}\\ (-\lvert a\rvert)^{h_1h_2/(q_1q_2)}=(-1)^{h_1h_2}\lvert a\rvert^{h_1h_2/(q_1q_2)}$$ If $a=0$, depending on the convention, you are implying that $m$ and $n$ are strictly positive real numbers (or, if $0^0=1$, non-negative real numbers). Either way, $0^{mn}=(0^m)^n$.

Answer 3

The "rule" $(a^b)^c = a^{bc}$ doesn't necessarily hold when $a < 0$.