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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

Shane Dizzy Sukardy
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    Where are the parentheses in $x^{6/2}$? – AHusain Jul 11 '18 at 00:49
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    We're missing some context here. I guess that you have defined $x^3 = x \cdot x \cdot x$ and $x^{p/q} = \sqrt[q]{x^p}$ with $\sqrtq$ for $t \geq 0$ being the unique non-negative number $s$ such that $s^q = t.$ Is that correct? – md2perpe Jul 11 '18 at 06:31
  • The problem here is very formal so you should specify how you define the fractional powervand specifically how that definition works for a negative base. – Rad80 Jul 11 '18 at 09:32
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    Related: https://math.stackexchange.com/q/317528/96384 – Torsten Schoeneberg Jul 11 '18 at 09:41
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    I think this question is really about whether $\frac62$ and $3$ are the same thing, and thus can be applied equally to things like exponents. Obviously they represent the same number, but are they the same? This is more philosophical than mathematical in my opinion, and there is no single correct answer. – Arthur Jul 11 '18 at 11:46

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$\frac{6}{2} = 3$, 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.

Rob Arthan
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    You also have the reverse issue when you try to combine radicals, with $-1 = \sqrt{-1} \cdot \sqrt{-1} \ne \sqrt{-1 \cdot -1} = 1$. In general, roots are nasty. – Kevin Jul 11 '18 at 05:41
  • $(-1)^\frac62$ is not well defined, so I would not say it can be equal to $(-1)^3$ – Rad80 Jul 11 '18 at 07:45
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    $\frac{6}{2} = 3$, so $(-1)^{\frac{6}{2}} = (-1)^3 = -1$. There is no issue of undefinedness here. – Rob Arthan Jul 11 '18 at 08:01
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    Even if you evaluate it as $((-1)^\frac{1}{2})^6$ you still get -1 as result: $((-1)^\frac{1}{2})^6 = i^6 = -1$ – fabian Jul 11 '18 at 08:54
  • $\frac62$ is a rational number and is the same thing as$\frac31$. 3 is an integer. In most cases they can be seen as the same thing because the injection of Z into Q is not relevant but this is not the case here. – Rad80 Jul 11 '18 at 09:29
  • @Rad80, please write an answer so that we can vote on it. – Carsten S Jul 11 '18 at 09:45
  • @CarstenS I already did, it's down below – Rad80 Jul 11 '18 at 09:47
  • @Rad80: please explain why you think defining $q=\frac{6}{2}$ as opposed to $q=3$ makes any difference to the meaning of the expression $x^q$. – Rob Arthan Jul 12 '18 at 00:11
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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 $1$

Mohammad Riazi-Kermani
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    $(-1)^\frac62$ is not well defined, so I would not say it can be equal to $(-1)^3$ – Rad80 Jul 11 '18 at 07:44
  • @Rad80 What makes it not well defined? $6/2$ is an integer. And integer powers are well defined unless the base is zero and the exponent is non positive. – Kitegi Jul 12 '18 at 12:39
  • @Kitegi I expanded that part in my own answer, under "About $\frac62=3$". – Rad80 Jul 12 '18 at 12:57
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Note: I rewrote the answer extending the key points. My reasoning has not changed.

$x^{\frac62}$ 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 $3$.

Rad80
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    $\frac62$ is an integer, so you don't need to worry about fractional powers, as such – Izaak van Dongen Jul 11 '18 at 08:40
  • $\frac62$ might not be the same entity as $3$ — as soon as we compare $3$ to $\frac62$, it is implied, that the $3$ referred to here is that in $\mathbb{Q}$, i.e. the class $3={\frac31, \frac62, \frac93, …}$. The fact that we use the same notation for $3\in\mathbb{Q}$ and $3\in\mathbb{Z}$ doesn't indeed mean they are the same thing. The same applies to, say, $1\in\mathbb{R}$ and $1=1+0i\in\mathbb{C}$. However, by saying that $x\in\mathbb{R}$, the OP has already defined the domain of the variables, and indeed what is not defined there is $x^\frac12$ for $x\in\mathbb{R}^-$. – André Levy Jul 11 '18 at 10:48
  • The OP defined the domain for the base, not for the exponents. There is no common definition for exponentiation $\mathbb{R}\times\mathbb{R}\rightarrow\mathbb{R}$. The only way to include negative bases is to either restrict exponents to integers or use the multi-value complex definition. Neither applies here. – Rad80 Jul 11 '18 at 10:56
  • It's entirely ordinary in any context where we have proved that $y$ is an integer (including the sense of being in the image of the standard injection from $\mathbb{Z}$ to any other unitary ring) to write $x^y$ meaning the usual $R \times \mathbb{Z} \rightarrow R$ exponentiation, without explicit comment on that meaning. You are implying a less common meaning from the notation $a^{b/c}$, which might make sense in a context where that notation has been specifically defined. It's not clear which is a better context for the OP question. – aschepler Jul 11 '18 at 12:04
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    Oops. I should have said either $F \times \mathbb{Z} \rightarrow F$ for a field $F$ or $R \times \mathbb{N} \rightarrow R$ for a ring $R$, not $R \times \mathbb{Z} \rightarrow R$. – aschepler Jul 11 '18 at 12:21
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    The question is literally whether $x^\frac62$ is the same as $x^3$, so I find it hard to believe that whoever wrote it meant the definition of $x^\frac62$ to be $x^3$. What would be the point in asking? – Rad80 Jul 11 '18 at 12:25
  • What about $x^\binom32$? Is that defined? – Carsten S Jul 11 '18 at 13:31
  • @Rad80 In first instance and on that level of math I would worry more about the correctness of what OP wrote instead of set theoretic constructions of the rationals. I think this is over the head of OP. – M. Winter Jul 12 '18 at 08:17