Is $(-3)^\frac 23$ a valid number in $\mathbb{R}$?

Question

According to the definition, if $a ∈ \mathbb{R}^+$, $m ∈ \mathbb{Z}$, and $n ∈ \mathbb{Z}^+$, then: $a^ \frac mn = \sqrt[n] {a^m}$

I got why there must be a bound for $a$ $($and $m,n)$, but I wonder if this expression is valid for $a<0$ too.

Take $(-3)^\frac 23$. If I use the expression above, I have:

$(-3)^\frac 23 = \sqrt[3] {(-3)^2} = \sqrt[3] {9} ≃ 2.08$

When I input $(-3)^\frac 23$ into my CASIO hand calculator, I got the same result.

But now let's consider $(-3)^ \frac 22$. If I use the expression above, I have:

$(-3)^\frac 22 = \sqrt[2] {(-3)^2} = \sqrt[2] {9} = 3$

But we know that $(-3)^\frac 22 = (-3)^1 = - 3$

This can only be understood as either:

(1) $(-3)^\frac 22$ is a valid number but the expression $a^ \frac mn = \sqrt[n] {a^m}$ doesn't apply.

(2) $(-3)^\frac 22$ is not a valid number at all if I were to assume that $a^ \frac mn = \sqrt[n] {a^m}$ is true for all valid numbers $a^ \frac mn$.

When I input $(-3)^\frac 22$ into the same calculator, it yielded $-3$. Apparently, it simplified $\frac 22$ into 1 and computed $(-3)^1$, which is $-3$.

This makes me think that in the $(-3)^ \frac 23$ example above, it seems the calculator didn't compute the final value using the expression $a^ \frac mn = \sqrt[n] {a^m}$. It seems to have simplified $\frac 23$ into $0.666...7$ and computed $(-3)^{0.666...7}$ instead, which makes me wonder if $(-3)^ \frac 23$ is a valid number at all in $\mathbb{R}$. Please advise.

Thank you!

EDITED

Answer 1

The "square root" (and rational exponents in general) is a bit of a misused term at times, but for a relatively good reason. The term actually has two meanings, and as long as you know which one you are talking about, and remain consistent, then everything is fine. Once you start mixing the two meanings, you risk getting yourself into trouble.

The two meanings stem from two worlds, one being the real numbers and the other complex numbers.

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In the world of real numbers, we have the following rules:

1. For any positive real number $a$ and any integer $n$, $\sqrt[n]{a}$ is defined as the unique positive solution of the equation $x^n=a$.
2. For any positive real number $a$ and any positive rational number $\frac pq$, the value $a^\frac pq$ is defined as $\sqrt[q]{a^p}$, or, equivalently, as $\sqrt[q]{a}^p$. Note that the two definitions match on all positive real numbers.
3. For any positive real number $a$ and any positive real number $y$, the value $a^y$ is defined as the limit $$\lim_{n\to\infty} a^{x_n}$$ where $x_n$ is any sequence of positive rational numbers, converging to $y$. It is possible to show, though not entirely trivial, that the limit is independent of the exact values of the sequence $x_n$ (so long as the limit of the sequence is $y$).
4. For any positive real number $a$ and any real number $y$, $a^y$ is defined as $\frac{1}{a^{-y}}$.
5. For any positive real number $a$, $a^0$ is by definition equal to $1$.

Note that the rules above ensure that $a^y$ is defined whenever $a$ is a positive number. It also has some neat properties, such as $a^x \cdot a^y = a^{x+y}$, or $(a^x)^y = a^{xy}$.

For negative numbers, $a^y$ is not defined in the real numbers, and none of the neat properties applies.

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In the world of complex numbers, the definitions are different. There, typically, $z^w$ is defined to be $$e^{w\cdot \log(z)},$$ where $\log(z)$ is the principal branch of the complex logarithm function. Note that this is a whole can of worms in itself, with initial further reading available here.

Answer 2

When negative bases are raised to non-integer exponents, they are considered multivalued (over $\mathbb{C}$).

For example, consider the following cases that counter your argument of $(-3)^\cfrac{2}{3} \in \mathbb{R}$

$$(-3)^\cfrac{2}{3} = 3^\cfrac{2}{3} \cdot i^\cfrac{4}{3} = 3^\cfrac{2}{3}\left(\dfrac{-1}{2} + \dfrac{\sqrt{3}}{2} i \right) \space \neq \space ^3\sqrt{9}$$

$$(-3)^\cfrac{2}{3} = (-3)^\cfrac{4}{6} = \space \left(^6\sqrt{-3}\right)^4 = \dots \text{ wait, it does not exist}$$