3

Consider the "obvious" equality $$(a^m)^n = a^{mn}.$$ In particular, if I only want to consider real numbers, for what values of $a,m,n$ is the above displayed equality true? And where can I find the precise statement written down? My first instinct was that as long as $a$, $a^m$, and $(a^m)^n$ were real numbers, then the equality would hold. This would have ruled out examples like $((-1)^{1/2})^2$. But we also have $$((-1)^2)^{1/2} \neq (-1)^1.$$

I'm pretty sure the displayed equation is always true if we require $a$ to be positive, but it's also true for values like $a = -1$, $m = 1/3$, and $n = 3$.

My guess is that the most general statement (again, only considering real numbers) requires $a$, $a^m$, $(a^m)^n$, and also $a^n$ to be real numbers. Is this "easy" stuff written down in something like Rudin's Principles of Mathematical Analysis?

Thanks a lot!

CJD
  • 420
  • $a>0$ and $m,n \in \mathbb R$. – Em. Dec 31 '15 at 00:33
  • I agree that the equation holds for the values probablyme wrote down, but it also holds for other values like in the example I gave with m=1/3. What's the most general statement that is true, and where can I find it written down? Thank you! – CJD Dec 31 '15 at 00:43
  • 2
    @CJD: This is a really good question -- I don't know the answer, but it seems to me like your question really centers around negative numbers raised to fractional exponents. Perhaps this will help? http://math.stackexchange.com/questions/317528/how-do-you-compute-negative-numbers-to-fractional-powers – Eli Rose Dec 31 '15 at 01:19
  • 2
    Wikipedia also highlights the breakdown (or inconsistent interpretation) of this identity for negative bases. – Frentos Dec 31 '15 at 03:07
  • I actually taught the rules to my calculus class; let me know if the above WikiPedia link does not answer your question. – Alex Dec 31 '15 at 03:23
  • Thanks everyone! These answers all look good and make sense. I'm still interested in a reference like a textbook or article if anyone knows of one, but it's more for curiosity than for an actual urgent need for one. (I'm totally happy ignoring base 0.) – CJD Dec 31 '15 at 15:46

1 Answers1

1

With $a,m,n \in \mathbb{R}$, the taxonomy from Hurkyl's answer in the question linked by Eli Rose and avoiding complex intermediates and exponentiation, the values that make the identity true are:

Case $a>0$: no restrictions on $m,n$.

Case $a=0$: requires $m,n \ge 0$ if you allow/define $0^0 \in \mathbb{R}$ (it depends on convenience and context), $m,n > 0$ otherwise.

Case $a<0$: The continuous real exponentiation operator forbids $a < 0$. The discrete real exponentiation operator (which allows $x^{1/3}$ etc. to be defined over all $\mathbb{R}$) requires $m=\dfrac{p_m}{q_m} \in \mathbb{Q}$ with $q_m$ odd (includes any $m \in \mathbb{Z}$ since then $m=\dfrac{m}{1}$). Similarly, the RHS of the identity requires $mn = \dfrac{p_{mn}}{q_{mn}} \in \mathbb{Q}$ with $q_{mn}$ odd. Then $n = \dfrac{mn}{m} = \dfrac{p_{mn}q_m}{q_{mn}p_m}$. If $p_m$ is odd, $a^m < 0$ and we require the denominator of $n$ to be odd, and this is the case since $q_{mn}, p_m$ are odd by assumption. If $p_m$ is even, $a^m > 0$ and there are no further constraints on $n$.

The most compact expression I can devise, if you like $x^{1/3}$ etc. defined over all $\mathbb{R}$, is: $(a^m)^n = a^{mn}$ is true/meaningful for $a,m,n \in \mathbb{R}$ if $a > 0$, or if $a = 0$ and $m,n > 0$ (or $\ge 0$ if $0^0$ defined), or if $a < 0$ and $m,n \in\mathbb{Q}$ with the denominators of $m$ and $mn$ odd in lowest terms.

I've had no luck finding book-based references to the $a<0$ case. The analysis and calculus texts I have at hand deal with the case $a>0$. Good elementary algebra texts have the rule that only odd (integral) roots of negative numbers are meaningful and give rules like $\sqrt[q]{a^x} = a^{x/q}$, but I haven't found a clear treatment anywhere.

Community
  • 1
Frentos
  • 3,041