what are the problems with the followings "equations"?

Question

what are the problems with the followings "equations"?

A) In the complex number field consider the following:

$-1=i^2=(i^4)^{\frac{1}{2}}=1^{\frac{1}{2}}=1$.

B) In $\Bbb R$,

$0=(1-1)+(1-1)+\cdots=1-[(1-1)+(1-1)+\cdots]=1-[0+0+\cdots]=1-0=1$.

Answer 1

A) The answer has already been given by Simon. I'd like to add something I noticed about such fallacies.

Whenever you want to avoid such fallacies, you should always consider the fractional powers first.

$(i^4)^{\frac12}=(i^\frac12)^4$

Now, $i=\cos \pi/2+i\sin \pi/2=e^{2ik\pi+i\frac{\pi}{2}}$

So, $i^\frac12=e^{ik\pi+i\frac{\pi}{4}}$ where $k=\{0,1\}$

You can clearly see that $(i^\frac12)^4=-1$

B) What you are doing is rearranging the series. You pair the elements and say

$S=1-S$ and you use $S=0$ which is of course not true. (where $S$ denotes the sum of the series)

I think I read somewhere(please correct me if wrong) that such a rearrangement can be done only when the series is absolutely convergent? Which it clearly isn't.

The series doesn't even converge.

$S_n=0$ when $n$ is even.

$S_n=1$ when $n$ is odd.

$S=\lim_{n\to\infty}S_n$ does not exist.

Answer 2

These chestnuts again.

A) Square root is not uniquely defined for complex numbers, unlike positive real numbers. In particular $1^{1/2}$ is $1$ or $-1$. So it is not case that the second or last equalities hold.

B) In the second equality, try to write out the intermediate steps for the second equality. There's an implicit assumption that $1 - 1 + 1 - 1 + \cdots \ $ makes sense. It doesn't.