I noticed that there are two solutions for $(-1)^{14/2}$:

  1. $((-1)^{14})^{1/2} = 1$
  2. $(-1)^{14/2}=(-1)^7=-1$

What am I doing wrong?

  • 1
    $\begingroup$ This is mainly an issue of missing or misplaced parentheses. To make the expression unambiguous, parentheses might well be placed around the exponent, indicating the second evaluation is intended. $\endgroup$ – hardmath Jan 14 '16 at 17:22
  • 1
    $\begingroup$ @hardmath I agree that it may matter wether $[(-1)^{14}]^{1/2}$ or $[(-1)^{1/2}]^{14}$. But taken it with usual convention there is one way only $(-1)^{(14/2)}=(-1)^7=(-1)\cdots(-1)=-1$ apart from $[(-1)^{14}]/2=[(-1)\cdots(-1)]/2=(+1)/2=1/2$. $\endgroup$ – C-Star-W-Star Jan 14 '16 at 18:04
  • 1
    $\begingroup$ @Freeze_S: I agree, and by implication since the OP uses the same expression in (2) as at the outset, the OP must largely think so as well. I was trying to address "[w]hat am I doing wrong?". $\endgroup$ – hardmath Jan 14 '16 at 18:10
  • $\begingroup$ @hardmath: Yep, true that. $\endgroup$ – C-Star-W-Star Jan 14 '16 at 18:11
  • 11
    $\begingroup$ In your first equation, why do you think that $(+1)^\frac12=+1$ and not $-1$? $\endgroup$ – Gyro Gearloose Jan 14 '16 at 18:47

In real numbers, the standard definition of rational exponents only permits fully reduced fractions in the exponent. Example definition from Sullivan's College Algebra:

Definition. If $a$ is a real number and $m$ and $n$ are integers containing no common factors, with $n \geq 2$, then $$a^{m/n}=\sqrt[n]{a^m}=\left(\sqrt[n]a\right)^m$$ provided that $\sqrt[n]{a}$ exists.

We have two comments about this equation:

  1. The exponent $\frac mn$ must be in lowest terms and $n$ must be positive.
  2. In simplifying the rational expression $a^{m/n}$, either $\sqrt[n]{a^m}$ or $\left(\sqrt[n]a\right)^m$ may be used, the choice depending on which is easier to simplify. Generally, taking the root first, as in $\left(\sqrt[n]a\right)^m$, is easier.

The first suggested transformation, $((-1)^{14})^{1/2} = 1$, is therefore invalid under this definition, because it disallows separating the numerator and denominator when they have a common factor.

Therefore only the second transformation, $(-1)^{14/2}=(-1)^7=-1$, which reduces the rational exponent first, is valid under a real-number definition like this.

  • $\begingroup$ so if I replace "2" with "3" then by this rule, I can seperate and the 2 solutions are valid. Am I right? $\endgroup$ – Stav Alfi Jan 14 '16 at 17:54
  • $\begingroup$ @StavAlfi: Well, unambiguously $(-1)^{14/3} = \sqrt[3]{(-1)^{14}} = \sqrt[3]{1} = 1$ under this real-valued definition. But there are no "2 solutions" because there is no reduction possible as per the 2nd transformation in your question above. $\endgroup$ – Daniel R. Collins Jan 14 '16 at 20:53
  • $\begingroup$ Lesson: When one speaks of something being "well-defined", that means it's constructed to have a single, unique, unambiguous, resolution to it. $\endgroup$ – Daniel R. Collins Jan 14 '16 at 20:57
  • $\begingroup$ The image is barely legible and is also inaccessible to search engines and visually impaired people using either a screen reader or magnifier. Please transcribe the relevant sections into LaTeX. $\endgroup$ – David Richerby Jan 15 '16 at 1:15

You're not doing anything wrong, per se; the issue is that in order to define square root as a single-valued function on $\mathbb{R}_{\geq 0}$, we (somewhat arbitrarily) choose the positive value. So both $-1$ and $1$ square to $1$, but the square root of $1$ is only $1$, not $-1$. This becomes a bit more transparent if we replace $14$ by $2$:

$$ [(-1)^2]^{1/2} = 1^{1/2} = 1 \not= -1 = (-1)^1 = (-1)^{2/2} $$

Accordingly, the identity $(a^b)^c = a^{bc}$ does not hold generally for negative $a$, if $b$ and $c$ are not both integers.

  • $\begingroup$ For that matter, it also tends to break down for complex $a$. But then, complex is more general than negative. $\endgroup$ – Kevin Jan 15 '16 at 6:56

The root of the problem is that $\sqrt{ 1\;}=+1$, unambiguously by definition, but $1^\frac12=(e^{2\pi i k})^{\frac12}=e^{\pi i k}\;\;\forall k\in\mathbb Z$ is not, it results to $\pm1$ depending on $k$ being odd or even.

Taking roots, you have to choose a branch, like the commonly accepted branch that $\sqrt x \ge 0$ for $x\ge0$. We could just as well have chosen the other branch, but you get in big trouble if you mix those.

  • $\begingroup$ i c wat u did there $\endgroup$ – user1717828 Jan 15 '16 at 4:49

(One of) the rigourous definitions, and the simplest, of $\;a^{\tfrac{14}2}$ is it's $\;\mathrm e^{\tfrac{14}2\ln a}$ so that it supposes $a>0$ – and avoids this kind of misbehaviour. $a^x$ for a negative $a$ is defined only for integer exponents, because it has an intuitive meaning.

  • 1
    $\begingroup$ Actually if we allow complex numbers and stick to the principal branch of the natural logarithm, then we can define $a^x$ for negative $a$ and all $x$ $\endgroup$ – ASKASK Jan 14 '16 at 17:45
  • 1
    $\begingroup$ And we even can use Riemann surfaces, but definitely not at the college level! $\endgroup$ – Bernard Jan 14 '16 at 17:50
  • 1
    $\begingroup$ @ASKASK: Since the principal branch of the natural logarithm excludes negative real numbers (branch cut), we cannot do as you suggest. Of course one can place the branch cut differently, but this requires a decision to be articulated. $\endgroup$ – hardmath Jan 14 '16 at 19:30

The domain of the function $f(x) = (-1)^x$ is the set of all integers. If $x$ is not an integer, the function returns a non-real value.

Specifically, we cannot use the standard rules of exponentation for this problem because the base is negative. As @hardmath has clarified, we should "force" an integer exponent on a value, as fractional exponents are not allowed. Thus, the second solution given in the original post is correct, and the first is invalid.

  • $\begingroup$ I cant see the connection for what i asked. Can you be more specific? $\endgroup$ – Stav Alfi Jan 14 '16 at 17:24
  • 3
    $\begingroup$ While it doesn't per se raise -1 to a fractional power, the manipulation of a fraction as exponent on -1 in the first evaluation (see Question) leads to the inconsistent result. If we coerce the exponent to an integer at the beginning, the result is unambiguous. $\endgroup$ – hardmath Jan 14 '16 at 17:27
  • $\begingroup$ Yes. That is a good clarification. Thank you @hardmath $\endgroup$ – K. Jiang Jan 14 '16 at 17:30
  • $\begingroup$ @hardmath: That is what needed to be expanded on, yes! $\endgroup$ – Brian Tung Jan 14 '16 at 17:35
  • $\begingroup$ @K.Jiang: Feel free to use (verbatim if you like) my Comment to clarify your Answer (which I already upvoted). $\endgroup$ – hardmath Jan 14 '16 at 17:51

Formally speaking, $a^{1/2}$ is defined to be such a special number $b$ that $b^2 =a$. That means $1^{1/2}$ can formally be either $+1$ or $-1$, we just choose the $+1$ for convenience...


The standard rules for exponentiation in $\mathbb R$ only always work for positive real number bases, and $((-1)^{14})^{1/2} = 1 \ne (-1)^{\frac{14}{2}}$ is an example of that.

The rules of order of operation reduce $((-1)^{14})^{1/2}$ to 1. To claim that this is not true would mean that the rules of order of operation are faulty. Because of that, I disagree with any one who wants the answer to be $-1$.

The Standard Rules

If $x,y,m,n \in \mathbb R$ with $x,y \gt 0$, then

  • $x^m x^n = x^{m+n}$
  • $(x^m)^n = x^{mn}$
  • $(xy)^n = x^n y^n$
  • $x^0 = 1$
  • $x^1 = x$
  • $\sqrt[n]{x^m} = x^\frac{m}{n}$

When $x\lt 0$, then the standard rules do not always work. It's possible to come up with special cases that do work; but, in my mind, it's not worth the effort.

  • $\begingroup$ you didnt mention the rules.. $\endgroup$ – Stav Alfi Jan 15 '16 at 10:43
  • $\begingroup$ Note that the question is not inherently about $((-1)^{14})^{1/2}$. The question is actually about the value of $(-1)^{14/2}$ (and whether or not that is the same as $((-1)^{14})^{1/2}$; which we can agree is not the case). $\endgroup$ – Daniel R. Collins Jan 17 '16 at 18:28
  • 1
    $\begingroup$ @DanielR.Collins - expanding your universe to complex numbers does a better job of explaining exactly what the problem is. But that feels like using a moving van to pop over to the store and bring home a half gallon of milk. $\endgroup$ – steven gregory Jan 23 '16 at 3:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.