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

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 $$(a^m)^n=a^{mn}$$?

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.

• What are $m$ and $n$ ? Mar 10, 2020 at 15:50
• Remember $1=(e^{2\pi i})^{\frac 12} \ne e^{\pi i}=-1$ but $e^{2\pi i}=1\in \mathbb R$ and $e^{\frac 12} \in \mathbb R$. Mar 10, 2020 at 15:55
• @MaximilianJanisch a, m and are all real numbers. Mar 10, 2020 at 15:57
• @fleablood I am just considering arithmetic on the real numbers. Mar 10, 2020 at 16:00
• "Sufficient" is a very strange way of putting it then. Among real numbers if $a^m, a^n$ is defined then $(a^m)^n= a^{mn}$ is a basic property. ... Okay, I guess there are issues if $a \le 0$ but.... okay.... if $a^m$ is defined (which it isn't if $a< 0$ and $m$ is fraction with an even denominator) then this holds. .... okay, it's just me but "sufficient" threw me for a loop. Mar 10, 2020 at 16:17

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.

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, $$0^{mn}=(0^m)^n$$.

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

• But $(-1)^{1/2}$ is not a real number, whereas the OP demands that $a^m$ and $a^n$ are real numbers.
– user239203
Mar 10, 2020 at 16:41