I know there is something wrong with this but I don't know where. It's some kind of a math fallacy and it is driving me crazy. Here it is: $$-1= (-1)^3 = (-1)^{6/2} = \sqrt{(-1)^6}= 1?$$
2026-03-29 18:30:22.1774809022
On
On
On
On
On
On
On
On
On
On
On
math fallacy problem: $-1= (-1)^3 = (-1)^{6/2} = \sqrt{(-1)^6}= 1$?
7.6k Views Asked by Bumbble Comm https://math.techqa.club/user/bumbble-comm/detail At
13
There are 13 best solutions below
7
On
The problem here is that the square root function, $\sqrt{-},(-)^\frac{1}{2}$, is not a single-valued function.
As PVAL says, it is a two-valued function, meaning you have two consistently choose which square root you're talking about. That's why you often will have problems when you have chains of equalities as above.
4
On
There is a simpler version of this fallacy: $-1 = (-1)^{2/2} = \sqrt{(-1)^2} = \sqrt{1} = 1$. The mistake comes from the fact that the function $f(x)=x^2$ is not invertible so you cannot conclude that for any real number $x$ it is the case that $x = \sqrt{x^2}$.
There is a version of the same mistake that uses the fact that $log$ is not invertible on $\mathbb{C}$ to prove that all numbers equal 1:
$x = e^{\ln(x)} = e^{\ln(x) * (2\pi i) / (2\pi i)} = (e^{2\pi i})^{\ln(x)/2\pi i} = (\cos(2\pi)+i \sin(2\pi))^{\ln(x) / 2\pi i} = 1^{\ln(x) / 2\pi i} = 1$
1
On
As other say the square root may be two valued. However you use it as a function, so it's single valued. It really depends on your definition of used functions. I think that $(-1)^{6/2} = -1$ but $\sqrt{(-1)^6} = ((-1)^6)^{1/2} = 1$. So @mrf is right that $(a^b)^c ≠ a^{bc}$ in general and the third equality of your equation is the one that doesn't hold.
6
On
What is wrong here is assuming that $\sqrt{x^2} = x$ when the fact is $\sqrt{x^2} = |x|$. Let $x=−1$ and use $\sqrt{x^2} = |x|$ in the problem above, you should arrive at a valid equation.
6
On
Hi this is not a fallacy but this problem overlooks some facts i.e. there will be two square roots to every number like. $$ \sqrt{4} = +2 \text{ as well as} -2 $$ similarly $$ \sqrt{(-1)^6} = +1 \text{ or } -1 $$
So I think writing $$\sqrt{(-1)^6} = 1$$in this proof is wrong
And also people who are saying that $$(a^b)^c != a^{b*c}$$ for a <0 or something like that please calculate this in your calculator so that it proves you worng $$10^{2.5*log10(-1)} $$
0
On
$\sqrt{x^2} = +x$ or $-x$
The fault with the "proof" is the false assumption that you can choose the positive root and still have everything hold. In fact, you need to select the correct root based on context or accept two possible answers.
0
On
The correct use of ${\sqrt{}}$ in this context would be
$$ −1 = (−1)^3 = (−1)^{6 / 2} = -\sqrt{(-1)^6} = -\sqrt{1} = -1$$
and this is simply a consequence of the inverse of $x^2$ being $-\sqrt{x}$ not $+\sqrt{x}$ when $x < 0$.
$$ -1 = \sqrt{(-1)^2} = \sqrt{1} = 1$$
is an equally invalid chain of equalities becuase the square root function is not injective.
If we start at one value, and apply a function whose inverse is not injective, we can easily (by choice) end up at a different value; ie if $f^-1(x)$ is not injective, then
$$f^-1 \circ f (x) \in S$$
where S contains values other than $x$. For mathematical rigor we have to specify in which domain we are working, so that we don't simply "choose" the inverse value. If from the start we have said $x < 0$, then the inverse of $x^2$ is $-\sqrt{x^2}$ not $+\sqrt{x ^2 }$ and the fallacy would be avoided.
2
On
The problem is in step: $$(-1)^3 = (-1)^{6/2} $$ You are squaring a number and taking root. As many people pointed out: $$ (x^2)^{1/2} $$ can be +x or -x.
0
On
Think about operator associativity in $$\sqrt{(-1)^6}$$. This is constructed in following way
-1 --> -1^6 --> (-1^6)^0.5
let us solve this
(-1^6)^0.5 = 1^0.5 = 1
for more http://en.wikipedia.org/wiki/Operator_associativity
0
On
The rule $a^{b/c}=\sqrt[c]{a^b}$ is always true when $a > 0$. If $a$ is negative, it may not be true.
0
On
Your error $$(-1)^{6/2} = \sqrt{(-1)^6}\tag✗$$ can be explained in either of two ways:
- in real analysis, for negative $a,$ the definition $$a^{\frac xy}:=\sqrt[y]{a^x}$$ conventionally requires $x$ and $y$ to be coprime such that $y$ is positive and odd;
- for nonzero complex $a,$ the law/theorem $$a^{xy}=(a^x)^y$$ requires $x$ and $y$ to be integers.
Taking square roots is in a sense a two valued function, because every non-zero complex number $z$ has two distinct complex numbers $w_1, w_2$ for which $w_1^2=w_2^2=z$.