What is wrong with this false proof? $-1=(-1)^1=(-1)^\frac{2}{2}=((-1)^2)^\frac{1}{2}=\sqrt{(-1)^2}=\sqrt{1}=1$ $-1=(-1)^1=(-1)^\frac{2}{2}=((-1)^2)^\frac{1}{2}=\sqrt{(-1)^2}=\sqrt{1}=1$
This proof bugs me for the following reasons:


*

*Mathematicians have defined the symbol $\sqrt{}$ (Named the principal square root) to mean 'take only the positive square root of the number under the radical'

*This makes it so that the $y = \sqrt{x}$ is a function and therefore for every x-value (input) there is only 1 y-value (output).

*Therefore $\sqrt{x^2}=|x|$ due to the above definition.

*But then this means $\sqrt{(-1)^2} = |-1| = 1!!!!$
I really don't understand what's wrong with the above proof, the only way this could make sense is to define $\sqrt{a} = \pm k$ where the symbol know gives rise to two solutions to the equation, but then y = $\sqrt{x}$ is not a function and is simply a relation. 
If we have $\sqrt{a}=\pm k$ then $\sqrt{(-1)^2}$ can be equal to -1 as well and then $-1=(-1)^1=(-1)^\frac{2}{2}=((-1)^2)^\frac{1}{2}=\sqrt{(-1)^2} = -1 = -1$ and everything works out fine.
Side Note: Why do we even want y = $\sqrt{x}$ to be a function anyway? What's the harm in defining $\sqrt{x} = \pm k$ where y = $\sqrt{x}$ is not a function? 

Proof for $\sqrt{x^2}=|x|$
Let k = $\sqrt{x^2}$ where k is a constant.
$k^2=(\sqrt{x^2})^2=(((x^2))^\frac{1}{2})^2$
$k^2=x^2$
$k^2-x^2=0$
$(k+x)(k-x)=0$
$k = x$ or $k=-x$
But $\sqrt{a}>0$ for $a>0$ by definition of the principal square root.
$\therefore k \neq-x$
In order to make this identity work properly, an absolute value sign is required.
$\therefore \sqrt{x^2}=|x|$

Btw, my question is not the same as why $\sqrt{-1\times-1}\neq(\sqrt{-1})^2$ as I am not using the property $\sqrt{a}\sqrt{b}=\sqrt{ab}$ here and I am not bringing up the concept of imaginary numbers.
 A: The problem lies in this equality:
$$(-1)^{\frac{2}{2}}=((-1)^{2})^{\frac{1}{2}}$$
which uses the identity $a^{\frac{m}{n}}=(a^m)^{\frac{1}{n}}$. The identity is true if $a$ is positive, but is false in general, and in some cases where $a^m$ is negative, the right hand side is not even defined.
I guess the lesson is that one should pay attention with the notations, their meanings and their conditions-to-apply. 
There are ways to make sense of the set of $1/n$th-roots, but I think it is not the central story in this example. However, I suggest you learn more about holomorphic functions, their domains of definition, the analytic continuation of these functions, monodromy obstruction and Riemann surfaces. In these contexts, first we get out of the line $\mathbb{R}$ and come into the whole plane $\mathbb{C}$. Then we discover that the function $\sqrt{z}$ has a singularity at $0$, and each time we analytically prolong it around the point $0$, it changes its size, and the correct way to define it is to change the domain: it's no longer a function $\mathbb{C}\rightarrow\mathbb{C}$, but from a $2$-fold branched covering with a ramification point at $0$, to $\mathbb{C}$. 
Another path may be to come back to the definition of $a^\frac{1}{n}$ in classical analysis book (I would recommend Rudin's Principles of mathematical analysis. There are situations where $a^{\frac{m}{n}}=(a^m)^{\frac{1}{n}}$ is true even for $a$ negative.In fact you can find all the triplet of real numbers $(a,m,n)$ such that $a^{\frac{m}{n}}=(a^m)^{\frac{1}{n}}$, and certainly $(-1,2,2)$ is not one of them.

A: The fallacy is due to the two-valued 'function', namely square root.
Denote by $y$, the square of a real number $x$.
That is $x^2= y$.
We also know $ (-x)^2= y$.
From this concluding $x= -x$ is the wrong logic. Only conclusion that can be drawn is $(-x)^2= x^2$, which is not really a new statement.
A: If I may be extreme, let me say once more:
In Mathematics, Nothing is True, unless there is a proof that it’s true.
The formula $(a^b)^c=a^{bc}$ when $a$ is a positive real number, and $b,c$ are real numbers, is true because there’s a proof. It can easily fail when $a\not>0$, and there is no proof when $a$ is not positive real. 
