2

I have been studying maths for a couple years now, but even though this is a very basic question, I am surprised that it never crossed my mind before.

I have been taught in school, that $(a^m)^n$ = $(a^n)^m$

Now, here’s where i am confused. Suppose that $a=-1$ and $m=2$ and $n=\frac{1}{2}$ Then,

$$\left(a^m\right)^n=\left((-1)^2\right)^{1/2} = \left(1\right)^{1/2} = 1$$

But then also,

$$\left(a^n\right)^m=\left((-1)^{1/2}\right)^2=i^2=-1$$

But shouldn’t the answer be the same in both cases? I don’t understand where I went wrong. I just used the rules that I was taught in school to get this result. Please help me out.

Thanks in advance.

Blue
  • 75,673
Srinidhi kabra
  • 169
  • 6
  • 5
    Commutative only for positive reals. – Ishraaq Parvez Aug 10 '21 at 03:20
  • Cf. this question: Fractional powers of negative numbers are not uniquely defined, and the "general rule" $(a^m)^n=a^{m×n}$ does not always work when $m$ and $n$ are not integers – J. W. Tanner Aug 10 '21 at 03:21
  • Arbitrary exponentiation is only defined for positive bases, so if you want to take exponents that are not integers, you need to restrict to $a\gt 0$. If you want to allow $a\lt 0$, then you need to restrict $n$ and $m$ to integers (or to only certain fractions, and that gets too annoying). – Arturo Magidin Aug 10 '21 at 03:23
  • 1
    Negative basis make exponents not well and singularly defined. For $b> 0$ OR $m,n \in \mathbb Z$ then then $(b^n)^m = (b^m)^n = b^{mn}$ . But if $b < 0$ and $m,n \not \in \mathbb Z$ the complicated things happen not because $(-1)^{\frac 12})^2 \ne (-1)^2)^{\frac 12}$ but because if we allow negative bases then $1^{\frac 12}$ could be either equal to $-1$ or to $+1$. There's no way to distinguish. $(1^2)^{\frac 12} = 1^{\frac 12}$ and one of the values of $1^{\frac 12} = -1$. And $(-1)^{\frac 12})^2 = (\pm i)^2 = -1$. No conflict. But not well defined. – fleablood Aug 10 '21 at 03:35
  • @fleablood $(-1)^{\frac 12}$ is not a real number. So, this may be an exact reason of restriction with integers. Thus, $a^{mn}$ will not work, I think. – lone student Aug 10 '21 at 04:16
  • 1
    @lonestudent true. – fleablood Aug 10 '21 at 05:29
  • Search tip: https://math.stackexchange.com/questions/tagged/fake-proofs?tab=Frequent – Hans Lundmark Aug 10 '21 at 09:03
  • I don't know who marked this as a duplicate, but it's not a duplicate. The other question is about how $\sqrt{ab}=\sqrt a\sqrt b$ can fail when $a$ and $b$ are negative. This question is about how $(a^m)^n=a^{mn}$ can fail when $a$ is negative and $n$ is fractional. This question should be easy to find, if for no other reason than that multiple published textbooks actually make the false claim that the latter identity holds. – Toby Bartels Aug 13 '21 at 05:20
  • I suppose you could mark it as a duplicate of https://math.stackexchange.com/questions/3219025/why-22-5-isnt-equal-to-2251-10 instead. – Toby Bartels Aug 13 '21 at 05:24
  • I think you should read this post that I wrote. – user21820 Jan 20 '22 at 17:13

1 Answers1

2

This is probably not your fault, since I've seen way too many math textbooks with outright errors about this.

As @fleablood noted in their comment to your question, exponentiation is not always well-defined, but you can make it well-defined by making some more or less arbitrary choices. Textbooks that stick to the real numbers usually make choices so that $ a ^ n $ has a single well-defined value whenever $ a $ is positive, $ a $ is zero and $ n $ is positive (or maybe zero), or $ a $ is negative and $ n $ is rational with an odd denominator in lowest terms. Then if they introduce complex numbers, they'll allow the denominator of $ n $ to be $ 2 $ in the last case (and sometimes consider more general values). All of this is fine as far as it goes.

But they also want to state some identities about exponentiation, such as $ ( a ^ m ) ^ n = a ^ { m n } $ (which implies that $ ( a ^ m ) ^ n = ( a ^ n ) ^ m $, since $ m n = n m $). This identity is valid in the sense that you can always make appropriate choices in the definition of exponentiation to make the equation true[*]. Unfortunately, the choices in the previous paragraph will not always work! With those choices, the identity holds if $ a $ is positive, it holds if $ m $ and $ n $ are both integers, and it holds if $ m $ is $ 1 / 2 $ and $ n $ is an integer; but it fails in other cases, such as if $ a $ is negative, $ m $ is an even integer, and $ m n $ is odd (which can happen if $ n $ is fractional).

In particular, if $ a = - 1 $, $ m = 2 $, and $ n = 1 / 2 $, then the identity fails; $$ ( a ^ m ) ^ n = \big ( ( - 1 ) ^ 2 ) ^ { 1 / 2 } = 1 ^ { 1 / 2 } = 1 \text , $$ while $$ a ^ { m n } = ( - 1 ) ^ { 2 \cdot 1 / 2 } = ( - 1 ) ^ 1 = - 1 \text . $$ (On the other hand, if $ a = - 1 $, $ m = 1 / 2 $, and $ n = 2 $, then the identity holds; $ ( a ^ m ) ^ n $ and $ a ^ { m n } $ are both $ - 1 $ in that case.)

[*] If $ a $ is zero and $ b $ is negative, then you have to go beyond the complex numbers to the extended complex numbers to define $ a ^ b $, but you can still make this identity hold.

Toby Bartels
  • 4,679
  • Thank you very much. This clears my question for me. Can you please recommend me some resources where i can find the rules of exponents for real and complex numbers in more detail ? – Srinidhi kabra Aug 10 '21 at 04:14
  • If the base $a$ is positive and you're working only with real numbers, then everything is fine; there is one clear best definition of $a^n$ (which always makes $a^n$ positive too), and all identities hold. If your base might be negative but your exponents are all rational, and you still want to stick to real numbers, then I recommend not using fractional exponents at all, but only roots; if $p/q$ is in lowest terms, change $a^{p/q}$ to $\sqrt[q]{a^p}$. At http://tobybartels.name/MATH-1100/2020FA/rootrules/ you can find some notes that I give in my classes about identities for these. – Toby Bartels Aug 10 '21 at 08:54
  • If you allow square roots of negative numbers, then you have to keep in mind that $\sqrt{ab}=\sqrt a\sqrt b$ is now no longer valid either. But to go much beyond that in complex numbers, you may as well go back to using fractional exponents and define $a^n$ as $\exp(n\ln a)$, where $\exp$ and $\ln$ are natural exponential and logarithmic functions. This gets very complicated because there are multiple ways to define $\ln$ on the complex numbers. It's best to work as much as possible with $\exp$ instead (with $\exp z = e^z$), which is well behaved and satisfies all of its usual identites. – Toby Bartels Aug 10 '21 at 09:00
  • This stuff also requires trigonometry to really get anywhere, in the form of Euler's Formula $\exp(x+\mathrm{i}y) = \exp x(\cos y+\mathrm{i}\sin y)$. You can read a lot about this and the corresponding logarithms at https://en.wikipedia.org/wiki/Complex_logarithm and the references therein. [Last comment] – Toby Bartels Aug 10 '21 at 09:03
  • Thank you for your help. I tried reading these articles and others that were mentioned, however I don’t have a background in complex analysis, i am not even a college student yet. So, i am not bale to understand these things in depth. So for now i think, i will just remeber that the laws of exponents are only valid for a positive base. – Srinidhi kabra Aug 11 '21 at 04:15
  • It sounds like you should save the stuff about complex exponentiation for later then. Hopefully my first comment is still useful for you when you want to deal with negative numbers. Otherwise, stick to positive numbers, and then all the identities work, no problem. – Toby Bartels Aug 12 '21 at 03:18
  • Yep, thank you. – Srinidhi kabra Aug 12 '21 at 04:41