We are now ready to multiply differentials. Starting from \begin{equation} du\,dv = \Jacobian{u}{v}{x}{y} \> dx\,dy \end{equation} first set $u=y$ and $v=x$ to obtain \begin{equation} dy\,dx = -dx\,dy \end{equation} Thus, multiplication of differentials is antisymmetric. Now set $u=v=x$, in which case the determinant is $0$, so that \begin{equation} dx\,dx = 0 \end{equation} Thus, differentials “square” to zero.

In fact, these two properties are equivalent. Clearly, antisymmetry implies that squares are zero, since only zero is unchanged when multiplied by $-1$. But the reverse is also true: If all squares are zero, then from \begin{equation} (dx+dy)(dx+dy)=0 \end{equation} and $dx\,dx=0=dy\,dy$ we obtain \begin{equation} dx\,dy + dy\,dx = 0 \end{equation} from which antisymmetry follows. (This technique is called polarization.)