# Is $\sqrt[3]{-1}=-1$?

I observe that if we claim that $\sqrt[3]{-1}=-1$, we reach a contradiction.

Let's, indeed, suppose that $\sqrt[3]{-1}=-1$. Then, since the properties of powers are preserved, we have: $$\sqrt[3]{-1}=(-1)^{\frac{1}{3}}=(-1)^{\frac{2}{6}}=\sqrt[6]{(-1)^2}=\sqrt[6]{1}=1$$ which is a clear contradiction to what we assumed...

• You chose the wrong logarithm of $1$ to compute $\sqrt[6]{1}$. You should have taken $\pm 6\pi i$ instead of $0$. You only have an equality $(a^b)^c = a^{bc}$ if you choose fitting logarithms to compute the powers. – Daniel Fischer Aug 26 '15 at 11:23
• There are 6 sixth roots of $1$. One being -1. – Ali Caglayan Aug 26 '15 at 11:23
• $\left(x^{a}\right)^{b}= x^{ab} = \left(x^{b}\right)^{a}$ if valid if $x > 0$ is real or $a,b \in \mathbb{Z}.$ Otherwise the power functions are mult-valued. – gammatester Aug 26 '15 at 11:28
• The problem with your reasoning is that nth roots are multi valued functions. If you define them to be the 'principle root' then you end up breaking your exponent laws. You have used exponent laws where they are invalid. – Ali Caglayan Aug 26 '15 at 11:38
• The only thing this reasoning will show is that, if $x^{3} = -1$, then $x^{6} = 1$ (which is not a contradiction). – Morgan Rodgers Aug 26 '15 at 12:34

You say

since the properties of powers are preserved

but this is not true, and you have given a proof that exactly the opposite is true:

There is no rule that, for $a$ negative, one has $a^{bc} = (a^b)^c$. Indeed, we can see that this is not a rule even without using cube roots: $$-1 = -1^{1/1} = -1^{2/2} =" (-1^2)^{1/2} = \sqrt{1} = 1.$$ This rule is only true when $a$ is a positive real number.

• (To be clear, as the other answers are showing, there is a way to interpret the multiplicative "property of powers" as true if we interpret the power operation as a "multi-valued complex-valued function." But there is no good generalization of the rule if we want to think of power operations as honest-to-goodness functions (on the reals or complexes). – hunter Aug 26 '15 at 11:39
• Then how do you define the function $f(x)=\sqrt[3]{x}$ for x<0? – Jason Aug 26 '15 at 11:40
• $f(x)$ is the unique real number $y$ such that $y^3 = x$. – hunter Aug 26 '15 at 11:41
• To put it one more way: in your question, you formulated a contradiction by making two assumptions. The first was that $\sqrt[3]{-1} = -1$. The second was that $x^{ab} = (x^a)^b$, an identity which you know is true for $x$ positive, is still true when $x$ is negative. Since you deduced a contradiction from your assumptions, you correctly concluded that one of your assumptions must be wrong. However, you were mistaken in concluding that the first assumption is wrong; in fact, it's the second assumption that's wrong. – hunter Aug 26 '15 at 11:44
• @Jason interesting! That results from the definition that Wolfram picks of the ^(1/3) function (they extend it to the complex plane using the usual branch of the logarithm, if you know what this means). As a result, they define $(-1)^{1/3}$ to be a certain complex root of $-1$ and not $-1$ itself. I think what Wolfram is doing is highly non-standard, but it does make sense. In my personal experience, most mathematicians don't mean this unless they are explicitly considering complex-valued functions (and even then, they have to say which branch they're taking) so Wolfram is a bit of an outlier – hunter Aug 26 '15 at 13:14

The notation $\sqrt[3]{-1}$ is a little bit ambiguous, since there are exactly three third roots of $-1$ over the complex numbers (in general, there are exactly $n$ $n-$roots of any complex number $z$ so the notation $\sqrt[n]{z}$ is ambiguous too).

Since

$$(-1)^3 = -1$$

$-1$ is one of those roots, but there are other two, namely, the roots of the equation

$$x^2-x+1$$

Which arise from the factorization

$$x^3+1=(x+1)(x^2-x+1)$$

• The n-th root is not ambiguous, because if we require n-th root to be a function (as it is) then it must have only one value... – Jason Aug 26 '15 at 11:29
• @Jason You can define a function choosing one of the $n$ distinct $n$-th roots for nonzero numbers, fine. But you can't do the manipulations you did, $a^{b\cdot c} = (a^b)^c$ is generally not true. – Daniel Fischer Aug 26 '15 at 11:36
• Who said the n-th root IS a function? – YoTengoUnLCD Sep 11 '15 at 3:53