1
vote
2answers
54 views

Simplifying $x^i$ to real numbers

I have been studying power functions, and started to think about imaginary powers. Take the function $x^i$. Because I don't know how to multiply a number $i$ times, I tried to simplify the equation ...
10
votes
1answer
510 views

What is wrong with this funny proof that 2 = 4 using infinite exponentiation?

Out of boredom, I decided to recall the following equation: $$x^{x^{x\cdots}} = 2.$$ Which, I simply rewrote like this: $x^2 = 2$, and therefore $x = \sqrt{2}$. Then I took a look at the more ...
6
votes
6answers
453 views

How is this proof flawed?

$\sqrt{x}=-1$ $\sqrt{x}^2=(-1)^2$ $x=1$ Now substitute it into the original equation $\sqrt{1}=-1$ $1=-1$
0
votes
4answers
111 views

Square and square root and negative numbers [duplicate]

Are they equal? -5 = $\sqrt{(-5)^2}$
10
votes
4answers
531 views

Which step in this process allows me to erroneously conclude that $i = 1$

I was playing around with imaginary numbers and exponents and came up with this: $$ i = \sqrt{-1} $$ $$ \sqrt{-1} = (-1)^{1/2} $$ $$ (-1)^{1/2} = (-1)^{2/4} $$ $$ (-1)^{2/4} = ((-1)^{2})^{1/4} ...
1
vote
3answers
320 views

A contradiction involving exponents

Where is the error in the following statement: $i^2=(i^2)^{\frac{4}{4}}=(i^4)^{\frac{2}{4}}=(1)^{\frac{1}{2}}=1$? I feel the error is in the first equality, because $(i^2)^{\frac{4}{4}}$ is in fact ...