Remark

When the ordered field $K$ contains the « square roots » of a positive element $a$ of $K$ (roots of the polynomial $X^2 - a$) then the notation $\sqrt{a}$ is generally reserved for the positive square root. If $K$ does not contain the square roots $b$ and $-b$ of $a$ in the field $E$, then the latter can be made an ordered extension of $K$ in two ways, each induced from the other via the $K$-automorphism which sends $b$ to $-b$; the choice of one of these orderings determines $\sqrt{a}$: it is whichever of the elements $b$ and $-b$ is positive.

If $a$ and $a'$ are two positive elements of $K$, whose square roots are in $K$, then $\sqrt{aa'} = \sqrt{a} \sqrt{a'}$, which follows from the definition of $\sqrt{a}$ and the sign rule.