What is wrong with $x=x^{2/2}=\sqrt {x^2}=\lvert x\rvert$ The title says it all, $x=x^{2/2}=\sqrt {x^2}=\lvert x\rvert$ cannot possibly be true, so what am I missing?
I think there is nothing wrong with $x=x^{2/2}$. The order of operatins say you should evaluate $2/2$ first. So this is the same as $x=x^1$.
The problem is with $x^{2/2}=\sqrt{x^2}$. You have changed the order of execution. The right side is the same as $\left(x^2\right)^{1/2}$. So this comes down to whether $x^{(2/2)}$ is the same as $\left(x^2\right)^{1/2}$. You are familiar with the identity $\left(x^a\right)^b=x^{ab}$, which you might think applies here, but that identity only holds when $x$ is non-negative and/or when both $a$ and $b$ are integers.
For example, $(-1)^{2/2}=(-1)^1=-1$, but $\left((-1)^2\right)^{1/2}=\left(1\right)^{1/2}=1$.
$x^{\frac{2}{2}}=\sqrt{x^2}$ only if $x\ge0$,
in fact $x^{\frac{n}{m}}$ (where $n,m\in \mathbb{N}-\left\{0\right\}$) is defined as $\sqrt[m]{x^n}$ only for any $x\ge0$,
so $x=|x|$ is not wrong because $x$ cannot be negative if we use the equality $x^{\frac{2}{2}}=\sqrt{x^2}$.