What's wrong with this demonstration? (1 = -1)

Question

What's wrong with this demonstration?:

$$A \iff 1 = 1^1$$ $$A \implies 1 = 1^\frac{2}{2}$$ $$A \implies 1 = (1^2)^\frac{1}{2}$$ $$A \implies 1 = ((-1)^2)^\frac{1}{2}$$ $$A \implies 1 = (-1)^\frac{2}{2}$$ $$A \implies 1 = (-1)^1 = -1$$

So finaly we get $$1 = -1$$

Answer 1

The property $\forall p\in \mathbb Z\forall q\in \mathbb N\left(a^{p/q}=(a^p)^{1/q}\right)$ doesn't work for negative $a$.

Howevever $\forall p\in \mathbb Z\forall q\in \mathbb N\left(q\text{ is odd}\implies a^{p/q}=(a^p)^{1/q}=(a^{1/q})^p\right)$ is true for all $a\in \mathbb Z\setminus \{0\}$.

Answer 2

One massive problem is that you assume that $\left(x^2\right)^{1/2} \equiv x$.

Here's a counter example: If $x=-2$ then $x^2=(-2)^2=+4$. Finally $(+4)^{1/2} = \pm 2$.

Clearly $-2$ is not equal to both $+2$ and $-2$ at the same time. There must be a problem!

Note that there is a big difference between $x^{1/2}$ and $\sqrt{x}$. The symbol $x^{1/2}$ represents the solutions to the equation $y=x^2$ while the symbol $\sqrt{x}$ represents the unique, real, positive solution to $y=x^2$. In a similar way $\sqrt[4]{1}=1$ while $1^{1/4} = \{\,\pm 1,\pm\operatorname{i}\,\}.$

Answer 3

The core problem: the squaring function defined on the line is not 1-1.