The root sign and it's relation with 1/2. $\sqrt{4}=2$
But is it same as writing...
$4^{1/2}=2$?
Basically I do not understand why $\sqrt{}$-sign equals $1/2$?
 A: Basically it is just a definition of the $\sqrt{}$-sign. Note, that the $\sqrt{}$-sign is one of the few examples (maybe the only one?) in mathematics, where you write "nothing" (since $\sqrt{x}=\sqrt[2]{x}$) and actually mean $2$.
So yes. It is $\sqrt{4}=4^{1/2}=2$
To make it more clear, we have $\sqrt[n]{x}=x^{1/n}$, where $\sqrt[n]{x}$ notes the $n-th$ root of $x$. For example $\sqrt[3]{8}=2$, because $2\cdot 2\cdot 2=2^3=8$.
A: Unfortunately, I think, roots were studied before powers with arbitrary exponents were dared explored, so that the (ugly, in my opinion) symbol $\sqrt{(\cdot)}$ came much earlier than the other notation $(\cdot)^{1/2},$ which is a more natural notation as it falls out of the extension of exponentiation to rational numbers.
Specifically, a motivation for why $(\cdot)^{1/2}$ is considered to be the same as the operator $\sqrt{(\cdot)}$ is this. The (positive) square root $x$ of a positive number $r$ is the number (why this $x$ is unique is proved in calculus) satisfying the relation $x^2=r.$ Once we grasp this characteristic property of square roots, it becomes easy to see that if we want to extend exponentiation to rational exponents so that the usual laws (especially in this case that $(a^m)^n=(a^n)^m=a^{mn}$ and $a^1=a$) are preserved, then we must have $$(x^2)^{1/2}=r^{1/2},$$ whose left hand side becomes $x$ under our assumptions, so that we have the square root $x$ of $r$ given by $r^{1/2}.$
