How does the exponent of a function effect the result?

Question

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 $x^{2 \over 2}$?

Answer 1

$2/2 = 1$, 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.

Answer 2

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

Answer 3

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.

Answer 4

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.

Answer 5

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 $x$.

Answer 6

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.

Answer 7

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.

Answer 8

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!