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The Wikipedia page on exponentiation suggests that the following identity holds provided the base $b$ is non-zero:

$$(b^m)^n = b^{mn}$$

Consider the following function:

$$y = (x^2)^{0.5}$$

According to the identity above the following should hold:

$$y = (x^2)^{0.5} = x^1$$

However consider the plots of the two functions:

$$y = x$$

$$y = (x^2)^{0.5}$$

The functions are equal for $x \geq 0$ however for $x < 0$ there is a discrepancy. Could you comment on the use of exponentiation rules? Are there other circumstances that similar discrepancies can be found when applying exponentiation rules?

Cameron Williams
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egg
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  • I added the link, please verify it's the right one. –  Aug 01 '16 at 19:25
  • These rules hold for nonnegative bases no matter what but they do not necessarily hold for negative bases, as you have observed. When you move to the complex numbers you find out that in general exponentiation is not uniquely defined; one must always choose a branch cut and a branch from within that branch cut. – Ian Aug 01 '16 at 19:27
  • Unfortunately, using fractional exponents opens the possibility of the function no longer technically being a function but instead being multivalued. Often-times, this fact is glossed over before you reach complex-analysis, but for example $1^{0.25}={1,-1,i,-i}$. In order to force it to be a function instead, we will choose a branch cut so that we get only one output instead of several. That is what happened here, it chose to give the positive output instead of the negative. – JMoravitz Aug 01 '16 at 19:28
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    Note that the wiki page claims the above identity only for integer exponents. Further, if you scroll down to the section on rational (and then real) exponents, it addresses the issue you describe. Aside - as someone who's never written one, I find that the people who write the maths pages on wikipedia are surprisingly conscientious, and their work of a fairly high quality. I strongly encourage you to give the page a close read and thought before you make claims like the one in the question ("Wikipedia page ... suggests..."; emphasis added.) – stochasticboy321 Aug 01 '16 at 19:28
  • Notice that $\sqrt{x}$ is undefined for $x < 0$ (real). $x^2$ however, is positive for all $x$. Exponentiation when the base is negative is not necessarily defined. –  Aug 01 '16 at 19:28

3 Answers3

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The page you refer to actually says the following. (The emphasis below is mine.)

The following identities hold for all integer exponents, provided that the base is non-zero

$0.5$ is not an integer, so the property does not apply.

But as you've seen, $(x^2)^{0.5} = x$ is true for $x \ge 0$. The reason it fails for $x < 0$ is because raising to the exponent $0.5$ is the same as taking the square root (as long as we're working only with real numbers, which I'm assuming is the case here). That is, $y^{0.5} = \sqrt{y}$. And when we simply say $\sqrt y$, it is understood that we are talking about the positive square root. For example, $\sqrt9 = 3$, not $-3$.

In general when we take the square root of a square, we have $\sqrt{x^2} = |x|$. For example, if $x = -4$ then we have $$x^{0.5} = ((-4)^2)^{0.5} = (16)^{0.5} = 4 \ne x.$$

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    $y^{0.5}$ is not necessarily the same thing as $\sqrt{y}$. – JMoravitz Aug 01 '16 at 19:30
  • @JMoravitz it normally is, see here for example: http://math.stackexchange.com/questions/537383/why-is-x-frac12-the-same-as-sqrt-x – Zain Patel Aug 01 '16 at 19:33
  • @JMoravitz, when is $y^{0.5} \ne \sqrt{y}$? The only thing that comes to my mind currently is the concept of branch cuts but based on the question I don't believe that OP is learning math at a level where explaining that would be constructive. I generally try to keep my answers at a level on par with their respective questions, and this question really doesn't say "branch cuts" or "complex analysis" to me. –  Aug 01 '16 at 19:45
  • Also, would the downvoter care to explain? Or was that @JMoravitz? –  Aug 01 '16 at 19:48
  • By assuming fractional powers give single results, strange things can happen, such as $1=e^{0\pi i} = e^{2\pi i}$ but $1^{0.5} = e^{0\pi i\cdot 0.5} = e^{0\pi i} = 1$ as well as $1^{0.5}=e^{2\pi i\cdot 0.5} = e^{\pi i} = -1$. In general, they are multivalued. It happened to be that the software chose the positive branch cut, but that need not be the case. Explaining everything about branch cuts isn't necessary, but at least alluding to the discrepancy would be helpful as opposed to saying something that is untrue. (and yes, that was me) – JMoravitz Aug 01 '16 at 19:49
  • @JMoravitz, I agree only to a limited extent. When I read this question I had a flashback to my Intermediate Algebra/Precalc teaching days, & I know from experience that mentioning things beyond the current topic, especially as exceptions to rules, is detrimental. Nothing in the question even hinted at $\Bbb{C}$ so I operated under the assumption that OP cares only about $\Bbb{R}$ in this context. I suppose clarifying that couldn't hurt (unless the OP doesn't even know there are non-real numbers, which likely isn't the case given the question's phrasing) which is why I updated my response. –  Aug 01 '16 at 20:04
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The identity you stated only holds for $x \geq 0$ (as noted in your wikipeda link) so, for general $x \in \mathbb{R}$ we have $(x^2)^{1/2} = |x| \neq x$ for $x <0$, hence the discrepancy

A clarification; you have two options:

  1. Either the identity holds for non-zero bases provided the exponents are integers

  2. The identity holds for real exponents provided the base is non-negative.

Zain Patel
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Traditionally, exponentiation of real numbers is ill defined when the base is negative and the exponent is not an integer. That is because of the problem you yourself just found.

A rational number can be represented in multiple ways, and the denominator can be odd or even depending on your choice of representation. If, say, you defined that $x^{\frac{a}{b}}=\sqrt[b]{{\left(x^a\right)}}$, and $x$ is negative, the parity of $a$ changes the sign of what's inside the radical.

For irrational numbers, things get even more confusing. When the basis is positive, one can extend exponentiation via continuity (since these sign problems don't exist); after all, any irrational can be approximated by rationals. The thing is there's no 'canonical approximation' and even if there were, why should it in principle be such that the numerators of the approximations are all the same parity?

When you dive into complex analysis, these problems are dealt with via logarithm branches. Even then, it still essentially means that there is no 'unique' or best way to do so.

Fimpellizzeri
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