The Exponential Property

Question

Prove true or false for the statement: every $x \in \mathbb{R}$, holds $x^{\frac{6}{2}} = x^3$

The habit of what we did everyday when facing exponential forms like this creates confusion to prove whether it is true and holds for every real numbers or not. If we take $x = -1$, I am afraid that it will leads to fallacy, since $(-1)^{\frac{6}{2}} = ((-1)^6)^{\frac{1}{2}} = 1$, while the right side: $(-1)^3 = -1$, but it is obviously $1 \neq -1$.

So, the statement is wrong in my opinion. How about your ideas? Please share. Thanks

Answer 1

$\frac{6}{2} = 3$, so $x^{\frac{6}{2}} = x ^ 3$ unconditionally. The "law" that you are thinking about, i.e., $x^{ab} = (x^a)^b$ is the thing that needs some qualification: it only holds under suitable assumptions if $a$ and $b$ are not natural numbers.

Answer 2

As you well know $$(-1)^{\frac{6}{2}} = (-1)^3 = -1$$ and there is no confusion about it.

If you wanted $$(-1^6)^{1/2}$$ the answer was $1$

Answer 3

Note: I rewrote the answer extending the key points. My reasoning has not changed.

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$x^{\frac62}$ is not defined for $x<0$.

Before we can substitute $\frac62$ with 3, the expression $x^\frac62$ needs to have a meaning on its own. This means that we need to have a definition for elevating to a rational power that works for all $x\in\mathbb{R}$, including negative numbers. Without this, statement is not well formed, so it can't be either true or false.

There are a few ways to extend exponentiation to negative numbers but each approach has its limits. The statement can be either true or false, depending on which one you pick.

The definition you refer to, using root extraction, does not extend well for negative bases because it involves even root extraction.

One could argue that we do not need a definition for all rational exponents but just for $\frac62$. Your definition could only include those rational numbers that are integer, fall back to the integer exponent case and then your statement is true. Personally I doubt this is the definition the author of the question had in mind, because it would make the whole thing trivial and not worth asking in the first place.

A more promising approach uses complex numbers and the definition of $x^\alpha:=\exp(\alpha\log x)$ extending the logarithm outside $\mathbb{R}^+$ going around $0$. This is possible but is complicated by the fact that you get a different extension of $\log$ to $\mathbb{R}^-$ depending on which way around $0$ you go, making this definition hard to use to define a single-valued function.

About $\frac62=3$

Integer numbers have a canonical immersion $\mathbb{Z}\hookrightarrow\mathbb{Q}$ that sends an integer $x$ into $\frac{x}1$. They are not technically the same thing: the latter is an equivalence class of pairs of integers, which is not an integer.

As everyone knows, for most practical uses this differentiation is not needed. The whole immersion is ignored and we write that $3\in\mathbb{Q}$ with what is technically an abuse of notation. Schools do not enter in this kind of detail and their choice is most reasonable.

This question is obviously about subtleties like this or at least I see little point in writing something like that if not to explore how $\frac62$ might not be the same entity as $3$.