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Why can't we do this?

Example: $(-1) ^ {1/3}$

Math definitions are based on a definite logic. What is the logic here? Can you give me some examples where it violates the equation?

I'm just a high school student b.t.w.. It would be greatly appreciated if you could describe it as simply as possible. thanks.

Leucippus
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Parham Moieni
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    See https://math.stackexchange.com/questions/608023/what-is-1-frac23, https://math.stackexchange.com/questions/317528/how-do-you-compute-negative-numbers-to-fractional-powers, https://math.stackexchange.com/questions/1211/non-integer-powers-of-negative-numbers, https://math.stackexchange.com/questions/582737/negative-base-to-non-integer-power, etc. – Hans Lundmark Dec 13 '21 at 20:11
  • You can, but the answers are not real numbers, so the teachers say not to worry about it. For example what's $(-1)^{1/4}$? One of several answers is .707+.707i. By the way, if the denominator is odd there is no such restriction. (-1)^{1/3}=-1 since (-1)^{3}=-1. – TurlocTheRed Dec 13 '21 at 20:12

2 Answers2

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It's not that you can't do it (if you're willing to accept answers that are non-real complex numbers). It's that there are multiple possible answers and no obvious way to choose among those answers that preserves certain "nice" properties we want those answers to have.

That's not a problem with with positive reals. When we take a rational power $s=r^{\frac pq}$, we're really saying that $s^q=r^p$. There are multiple complex numbers $s$ (in fact, $q$ of them) that satisfy this equation, but exactly one of them will always be a positive real number and that's the "obvious" choice. Importantly, this choice lets us extend the definition of $r^t$, where $t \in \Bbb R \setminus \Bbb Q$, in a "nice" (continuous) way when $r$ is a positive real number.

If you try this when $r$ is a negative real number, there is often no obvious choice because there are no negative real solutions for $s$ unless $q$ is odd. And it turns out that there's no choice that always lets you extend the definition of exponentiation beyond rational exponents in a "nice" way, although any choice you make will allow you to extend the definition of exponentiation to almost all complex numbers. These wrinkles are usually taught in advanced undergraduate or graduate classes, so for high school classes by far the more prudent course is to simply say "Don't do it."

Robert Shore
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  • thanks for the detailed answer. I asked this question from my teacher and he told me we can write A) (-1) ^ (1/3) as B) ∛-1 but then I asked the same question from another teacher and he told me that we can convert B to A but not vice versa. I've got a little bit confused here. then I asked them if (-1) ^ (1/3) = ((-1) ^ 2) ^ (1/6) and again they gave me different answers. what do you think? – Parham Moieni Dec 13 '21 at 22:55
  • @ParhamMoieni Your approach is good, but I think slightly misdirected. I think that rather than worrying about notation you'll gain more understanding by realizing that these are questions of definition and mulling over how best to define these various terms. You probably don't yet have enough background to come up with a sensible answer in full generality, but you can almost certainly understand what the correct definition should be when dealing with rational exponents. – Robert Shore Dec 14 '21 at 00:01
  • I was a little bit worried to lose Score on exams. Currently if I see any of these in school exams I will go with the wrong answer ( ³√-1 ) and in the other exams I will assume it as undefined based on the textbook :) just to make sure, I can't write like this ( (-1) ^ (1/3) = ((-1) ^ 2) ^ (1/6) ) then, right? Thank you for the guidance. I really needed to tackle this problem. – Parham Moieni Dec 14 '21 at 10:21
  • Euler's formula allows arbitrary complex bases to be raised to arbitrary complex powers. – Michael Ejercito Feb 11 '24 at 20:19
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$-1$ can very well be raised to some rational exponents, namely those with an odd denominator.

E.g. $(-1)^{4/7}=1$ because $(-1)^4=1^7$ and $(-1)^{3/7}=-1$ because $(-1)^3=(-1)^7$.

Irrational exponents do not allow this "trick".