# A problem in algebra: how does $-1=1$? [duplicate]

I have algebra problem from a friend, that is 1=-1!!! because

$$-1=-1^{3}=-1^{^{\frac{6}{2}}}=\sqrt{(-1)^6}=\sqrt{1}=1$$

I can not see what is wrong with this? I will appreciate it any help.

• Try a simpler problem $1^2=(-1)^2 \not\Rightarrow 1=-1$. – user Jan 10 '19 at 10:26
• Notice that 1 has two roots: $1$ and $-1$. – denklo Jan 10 '19 at 10:41

$$\;-1^3=(-1)^3=(-1)^{6/2}\color{red}{\stackrel {!!}=}\left[(-1)^{1/2}\right]^6$$
as $$\;(-1)^{1/2}=\sqrt{-1}\;$$ cannot be done within the real numbers (and this is also another reason why the above mentioned properties don't apply to complex numbers...)
• All $-1^3$, $(-1)^3$ and $(-1)^{6/2}$ are equal! The problem is that $a^{b/c} = \sqrt[c]{a^b}$ is not always true. – Christoph Jan 10 '19 at 10:33
• I'd still disagree that this is the problem. Over the complex numbers $(-1)^{1/2}$ is defined as the principal square root of $-1$, which is $i$ and $i^6=-1$. The problem is that this is different from $\left[ (-1)^6 \right]^{1/2}$, which is $1$. – Christoph Jan 10 '19 at 11:30
• @Christoph Over the complex numbers it is false in general: $$1=\sqrt1=\sqrt{(-1)(-1)}=\sqrt{-1}\sqrt{-1}=i\cdot i=-1$$ You can't do the above, certainly. We're talking here of general propieties . And in the complex number I don't know of any definition a priori of "principal or not" branches, though many times one usually choses the "usual" branches. – DonAntonio Jan 10 '19 at 12:50