# Why is $(-1)^3=(-1)^{6/2}=((-1)^6)^{1/2}=1^{1/2}=1$ wrong? [duplicate]

Why is this wrong? $$(-1)^3=(-1)^{6/2}=((-1)^6)^{1/2}=1^{1/2}=1$$ It seems logical but I know it's wrong.

• When are you allowed to use which exponentiation laws? Be completely exact: for all laws with $a^b$ et cetera, $a$ is always positive. Right? The reason for this you see in your question. – Moritz Mar 4 '15 at 20:10

Hint: $1$ has more than one possible square root, and you need to check them all when you do a calculation like you are doing.

• Could you please explain this a little more? – Eng Myt Mar 4 '15 at 20:21

The problem is that $$(-1)^3\neq((-1)^6)^{1/2}$$

The well-known identity $$a^{mn}=(a^m)^n$$ stands when $a>0$. It also stands when $m,n\in\Bbb Z$ (provided that $a\neq 0$), but it is false if $a<0$ and the exponents are not integer.

Hint:

$$\sqrt{9}=3 \land -3$$

People tend to forget these simple things...

• People also forget that $\sqrt{9}=3 \land -3$ means $\sqrt{9}$ equals $3$ and $-3$ simultaneously. Technically it should be $\sqrt{9}=3 \vee -3$, or more correctly, $\sqrt{9} = \pm 3$. – graydad Mar 4 '15 at 20:41