$2+2 = 5$? error in proof $$\begin{align} 2+2 &= 4 - \frac92 +\frac92\\
&= \sqrt{\left(4-\frac92\right)^2} +\frac92\\
&= \sqrt{16 -2\times4\times\frac92 +\left(\frac92\right)^2} + \frac92\\
&= \sqrt{16 -36 + \left(\frac92\right)^2} +\frac92\\
&= \sqrt {-20 +\left(\frac92\right)^2} + \frac92\\
&= \sqrt{25-45 +\left(\frac92\right)^2} +\frac92\\
&= \sqrt {5^2 -2\times5\times\frac92 + \left(\frac92\right) ^2} + \frac92\\
&= \sqrt {\left(5-\frac92\right)^2} +\frac92\\
&= 5 + \frac92 - \frac92 \\
&= 5\end{align}$$
Where did I go wrong

Clearly, $\sqrt{(4-\frac{9}{2})^2}$ is what? Well, $4-\frac{9}{2}$ is $-\frac{1}{2}$. But we all know $(-\frac{1}{2})^2$ is $\frac{1}{4}$ since $(-x)^2=x^2$. That means that $\sqrt{(4-\frac{9}{2})^2}$ is $\frac{1}{2}$. So the assumption that $\sqrt{(4-\frac{9}{2})^2} = 4 - \frac{9}{2}$ leads to $-\frac{1}{2}=\frac{1}{2}$ which is clearly wrong.
So the square root of a positive number squared is a positive number and the square root of a negative positive number squared is also a positive number. So if we assume $\sqrt{(-x)^2} = -x$ to $x = -x$ which is false.  Note: x is positive, not negative. This is the correct statement:
$\sqrt{(-x)^2} = \sqrt{x^2} = x \not= -x$ 
