How does the exponent of a function effect the result?

The $x^{2/2}$ can be represented by these ways: \begin{align} x^{2\over2}=\sqrt{x^2} = |x|\\ \end{align} And
\begin{align} x^{2\over2}=x^{1} = x\\ \end{align} Which one is correct? And what is the domain of $x^{2 \over 2}$?

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The second form is deceptive: the function $\ x^{2/2} \$ is not formally the same as the function $\ x \ .$ There is a domain issue that must be considered for the implied square-root in $\ x^{2/2} \ ,$ just as there would be for $\ x^{3/2} \ .$ –  RecklessReckoner Mar 19 '14 at 6:27
$2/2=1$. So $x^{\frac 2 2}$ and $x^1$ are the same thing. –  Emanuele Paolini Mar 19 '14 at 7:40
–  mrf Mar 19 '14 at 7:57
After some sleep, I withdraw some part of my comments and have now written an answer. On further (and hopefully better) thought, I realized there is a point I have been inconsistent about. This has also been addressed by many other posters here already. –  RecklessReckoner Mar 19 '14 at 15:55

$2/2 = 1$, so $x^{2/2} = x^1 = x$. Always. What is confusing you is that this may not be the same as $(x^2)^{1/2}$ or $(x^{1/2})^2$ when $x$ is not positive. In mathematics, an expression such as $x^p$ depends on the values of $x$ and $p$, not on the way they are represented.

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$(x^2)^{1/2}$ is defined for every real $x$, but $(x^{1/2})^2$ is defined for $x \ge 0$

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if it is presented as $x^\frac{2}{2}$, then simplify the exponent first. –  Joshua Biderman Mar 19 '14 at 6:21
@xs21 What is "wrong" with that is that fractional exponents have the specific meaning $\ x^{p/q} = \ \sqrt[q]{x^p} \ .$ The domains of such functions must be considered when proposing to evaluate them for chosen values of $\ x \ .$ Also, the square-root operation is defined as giving the positive square-root. That is why $\ \sqrt{x^2} = |x| \ .$ –  RecklessReckoner Mar 19 '14 at 6:30
But the OP has asked about $(x^2)^{\frac 1 2}$ with respect to $x^1$ which are both defined for all $x$. –  Emanuele Paolini Mar 19 '14 at 7:30
@RecklessReckoner: No. Fractional exponents do not have to satisfy $x^{p/q} = \sqrt[q]{x^p}$. This will hold if $x$ is a positive real number; if not, one or both sides could be undefined, independently, and even if both are defined their values could be different. $(-1)^{6/2}=(-1)^3=-1\neq 1=\sqrt{(-1)^6}$. –  Marc van Leeuwen Mar 19 '14 at 13:51
@MarcvanLeeuwen I realized later that my definition is incomplete. The interpretation I should have written is $\ x^{p/q} \ = \ \sqrt[q]{x^p} \ = \ (\sqrt[q]{x})^p \ .$ Had I done so at the time, I'd have noticed I was being inconsistent; I have hopefully said this better in the answer I've now written. –  RecklessReckoner Mar 19 '14 at 15:59

The rule $$x^{pq} = (x^p)^q$$ is not always valid if $p$ or $q$ are not integers and $x<0$. As you have noticed $$-1 = (-1)^1 = (-1)^{\frac 2 2} \neq ((-1)^2)^{\frac 1 2} = 1^{\frac 1 2} = 1.$$

So, unfortunately, $x^{\frac 2 2} \neq \sqrt{x^2}$ for $x<0$. For the last question: $x^{\frac 2 2} = x^1 = x$ is defined for all $x$.

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There is freedom to play $$-1 = (-1)^1 = (-1)^{ \frac{ 2}{2}} = ((-1)^{ \frac{ 1}{2}})^2 = (\pm i)^2 = -1$$ –  neofoxmulder Mar 19 '14 at 7:32

I'm amazed to ctrl-f 'order of operations' and come up with nothing here!

Other answers do well to explaining what's going on here, but recognize that the confusion is about perceived ambiguity with respect to order of operations. In this case, there is an invisible (but understood) set of brackets in the exponent $x^{(\frac{2}{2})}$, the same way that $\frac{5+1}{2}$ is understood to be the same as $\frac{(5+1)}{2}$.

All of this is just convention, which of course is malleable (even in mathematics). Exercise care to express what you mean in a way that is unambiguous to yourself and to whoever you are trying to communicate with.

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I think this is the clearest answer. It's about those implied parentheses. –  Ryan Mar 19 '14 at 15:20
I don't agree... There should be no confusion on which is the order of operations in $(-1)^{\frac 2 2}$. The problem is that the rule $x^{\frac a b} = (x^a)^{\frac 1 b}$ is so strong in our mind, that we don't notice that actually we should first compute the fracion $\frac a b$ and then the power. –  Emanuele Paolini Mar 20 '14 at 7:22
" $\frac {5+1}{2}$ is understood to be the same as $\frac {(5+1)}{3}$ " Typo in the denominator –  user80551 Mar 20 '14 at 10:34

When you put an exponent in fractional form, you run into some problems. $x^{2/2}=|x|$, but $x^1=x$. You cannot simply cancel out fractions in exponents, because you may forget the restrictions.

A fake proof that $\sqrt{-1}=1$ uses this. The proof goes like so: $$\sqrt{-1}=(-1)^{1/2}$$ $$=(-1)^{2/4}$$ $$=\sqrt[4]{(-1)^2}$$ $$=\sqrt[4]{1}$$ $$=1$$ $$\boxed{i=1}$$ This proof is wrong because you cannot say that $x^{m/n}=\sqrt[n]{x^m}$ if $x < 0$ and $n$ is even. In this case, $x$ is negative and $n$ is even, therefore the proof is false.

A related problem is this:

Find the root(s) of $3x^2+9x=0$

Many people absentmindedly make an $x$ go away by saying $x(3x+9)=0$, $3x+9=0$. Then they say that the only root is $-3$, which is wrong. The roots are $0$, $3$. You have to be mindful about the steps you take to solve a problem.

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There are 4 4th roots of 1, One of the roots is $i$ , the right answer. What would really be devastating is if we used 'seemingly legitimate' algebra to get 4 roots and $i$ was not among them! :) Also , there is no problem if n is odd. –  neofoxmulder Mar 19 '14 at 7:05
What? $x^{\frac 2 2} = |x|$ is horrible... –  Emanuele Paolini Mar 19 '14 at 7:28
IMHO the interpretation, the presentation AND the intermediate steps are important. $$\frac{ 2}{2 } = 2 - 1$$ $$x^{ \frac{ 2}{2}} = x^{2 - 1} = \frac{x^2} {x}$$ Zero has been removed from the domain so this 'path' gives a ddifferent answer than the two cases asked by the OP. –  neofoxmulder Mar 19 '14 at 8:02

The second one, by definition the symbol $$x^\dfrac{m}{n} \qquad\qquad m,n\in\mathbb{Z}, n\neq 0$$ is not defined if $x<0$, it should be considered meaningless.

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You mean that $(-8)^{1\over3}=-2$ is meaningless? –  Taher Rahgooy Mar 19 '14 at 7:06
@xs21, strictly speaking, yes. It is the reason, someone prefer to say $\sqrt[3]{-8}={-2}$ and differ $\sqrt{}$ and rational power. –  sas Mar 19 '14 at 7:13
you mean that $(-1)^2$ is not the same as $(-1)^{\frac 4 2}$? –  Emanuele Paolini Mar 19 '14 at 7:27
Every symbol you use should have a proper definition. When you say $a^{\frac{m}{n}}$ and $a\geq 0, n\neq 0$, I understand, what you mean. When $a<0$ I do not. And I know, it is possible to have different definitions. So to not make conflicts I'd rather prefer to avoid notation like $(-1)^{\frac{4}{2}}$ when it is not about complex numbers. –  sas Mar 19 '14 at 7:50
The power is defined for negative $x$ if $\frac mn\in\Bbb Z$. –  Marc van Leeuwen Mar 19 '14 at 13:56

If $n$ is even, then the function $$x^{\frac{m}{n}},$$ is defined ONLY for $x>0$.

Otherwise, it might not be equal to a real numbers.

In general the identity $$x^\frac{km}{kn}=x^{\frac{m}{n}},$$ holds ONLY for $x>0$. If $x<0$, then it still holds if $k,n$ are odd numbers.

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So you are saying that $(-1)^{\frac 4 2}$ is not the same as $(-1)^2$? –  Emanuele Paolini Mar 19 '14 at 7:31
@EmanuelePaolini: I DID NOT SAY THIS! I said it simply does not always hold! –  Yiorgos S. Smyrlis Mar 19 '14 at 7:35
$-1<0$ and $2$ is even. So, you are saying that $(-1)^{\frac 4 2}$ is not defined. –  Emanuele Paolini Mar 19 '14 at 7:37
In the high-school textbooks of my country, it is stressed that operations with fractions in exponents with negative bases are not allowed, as they may lead to contradictions. –  Yiorgos S. Smyrlis Mar 19 '14 at 7:41
However you would agree that $(-1)^2$ IS defined... So the conclusion of my first comment. –  Emanuele Paolini Mar 19 '14 at 7:41

The discussion on this question has been an education to me, as it is clear that a fair amount of "sloppiness" has crept into the typical presentation of fractional exponents in introductory courses or general practice. I, for one, was shown a long time ago that $\ \sqrt{x^2} \ = \ |x| \ .$ [In fact, in the early days of computer languages before "library subroutines" such as ABS(X) were available, a standard way of obtaining absolute value was to write SQRT(X*X).]

I realized a while after writing my comments (far too late at night) that I was not even clear about how such exponents are read. The way I have presented this notation, which is somewhat cumbersome, is:

As a function for real numbers, if we are to have the interpretation $\ x^{p/q} \ = \ \sqrt[q]{x^p} \ = \ (\sqrt[q]{x})^p \$ , then, with integers $\ p \ \ne \ 0 \ , \ q \ > \ 0 \ ,$ and the ratio $\ \frac{p}{q} \$ not to be reduced,

with $\ p \ > \ 0 \ ,$

for $\ q \$ odd: domain and range are $\ \mathbb{R} \ ,$

for $\ q \$ even: domain is $\ x \ \ge \ 0 \ ,$ range is $\ y \ \ge \ 0 \ ;$

with $\ p \ < \ 0 \ ,$

for $\ q \$ odd: domain and range are $\ \mathbb{R} - \{0\} \ ,$

for $\ q \$ even: domain is $\ x \ > \ 0 \ ,$ range is $\ y \ > \ 0 \ .$

So saying, it is now clear that it is not consistent for me to have said that $\ x^{2/2} \ = \ \sqrt{x^2} \ = \ |x| \ .$ $^* \$ (So I will stop going around saying that...) It should rather be said that the absolute value issue simply does not arise because writing $\ x^{2/2} \$ requires that $\ x \$ be non-negative. So the complete statement would be more like $\ x^{2/2} \ = \ x \ ,$ for $\ x \ \ge \ 0 \$ only (and not defined for $\ x \ < \ 0 \$). One should not "reduce" the fraction, as that changes the intent of the notation.

$^* \$ What I should be saying to students is: the "square-root" operation does give $\sqrt{x^2} \ = \ |x| \$ , but this is not the same as $\ x^{2/2} \ .$

I note in reading this thread that mathematicians apparently avoid the ambiguity entirely by restricting the domain to $\ x \ > \ 0 \ .$ It is clear that writing fractional exponents should be done advisedly.

EDIT -- In light of the excellent points raised, I see that there is a certain "friction" between the overly-broad use of fractional exponents and the need for compatibility with the exponential function $\ a^x \ , \ a > 0 \ .$ The definition I proposed above ends up having too many conditions to be particularly useful, so I see that I should reject it and limit the domain to $\ x > 0 \ .$ Thank you to everyone for a helpful discusssion!

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I think that those who are willing to use a definition like the one you gave, would do so only after insisting that $p/q$ be a fraction in reduced form. Then the defintion makes sense, and leaves $x^{2/2}=x^{1/1}=x$ a valid function with domain $\Bbb R$. (Also, everybody wants to exclude $x=0$ from the domain if $p/q<0$, and some people also want to exclude $x=0$ when $p/q=0$; this is however an unrelated issue.) –  Marc van Leeuwen Mar 19 '14 at 16:17
I should amend my definition to indicate that $\ \frac{p}{q} \$ is not to be reduced, since I would interpret $\ x^{4/6} \$ as not the same as $\ x^{2/3} \$ . And, yes, I should address the matter of negative integer $\ p \$ and $\ q \$ . This certainly contains pitfalls for the unwary... Thank you for your comments! –  RecklessReckoner Mar 19 '14 at 16:28
However I find very dangerous (and formally wrong) to introduce a notation where $x^{4/6} \neq x^{2/3}$. What happens if you have $x^a$ with $a=4/6$? You should conclude that $x^a \neq x^{4/6}$. –  Emanuele Paolini Mar 19 '14 at 19:04
That certainly adds to the thorniness of this notation, because I would not interpret $\ \sqrt[6]{x^4} \$ as meaning the same thing as $\ \sqrt[3]{x^2} \ .$ This is far more discussion of the issues than I've seen before, and I understand better now why one would want to limit the domain to positive values of $\ x \ .$ –  RecklessReckoner Mar 19 '14 at 22:05
(I see I was in a bad way yesterday... Too bad comments can't be edited after 5 min.) The example I wanted was to compare $\ \sqrt[3]{x} \$ with $\ \sqrt[6]{x^2} \$ . –  RecklessReckoner Mar 20 '14 at 15:37