For suppose that \[ A(\mu\nu) B^{\nu} \] is always a covariant vector for any choice of the contravariant vector~$B^{\nu}$. Then by~\Eq{(23.12)} \[ \{A'(\mu\nu) B'^{\nu}\} = \frac{\dd x_{\alpha}}{\dd x_{\mu}'}\, \{A(\alpha\beta) B^{\beta}\}. \Tag{(24.2)} \] But by \Eq{(23.11)} applied to the reverse transformation from accented to unaccented coordinates \[ B^{\beta} = \frac{\dd x_{\beta}}{\dd x_{\nu}'}\, B'^{\nu}. \] Hence, substituting for~$B^{\beta}$ in~\Eq{(24.2)}, \[ B'^{\nu} \left(A'(\mu\nu) - \frac{\dd x_{\alpha}}{\dd x_{\mu}'}\, \frac{\dd x_{\beta}}{\dd x_{\nu}'}\, A(\alpha\beta)\right) = 0. \] Since $B'^{\nu}$~is arbitrary the quantity in the bracket must vanish. This shows that $A(\mu\nu)$~is a covariant tensor obeying the transformation law~\Eq{(23.22)}. We shall cite this theorem as the ``rigorous quotient theorem.'' \PageSep{55} \Section{25.}{The fundamental tensors} \index{Fundamental velocity!tensors}% It is convenient to write \Eq{(22.1)} as \[ ds^{2} = g_{\mu\nu} (dx)^{\mu} (dx)^{\nu} \] in order to show explicitly the contravariant character of $dx_{\mu} = (dx)^{\mu}$. Since $ds^{2}$~is independent of the coordinate-system it is an invariant or tensor of zero rank. The equation shows that $g_{\mu\nu} (dx)^{\mu}$ multiplied by an arbitrarily chosen contravariant vector $(dx)^{\nu}$ always gives a tensor of zero rank; hence $g_{\mu\nu} (dx)^{\mu}$ is a vector. Again, we see that $g_{\mu\nu}$~multiplied by an arbitrary contravariant vector~$(dx)^{\mu}$ always gives a vector; hence $g_{\mu\nu}$~is a tensor. This double application of the rigorous quotient theorem shows that $g_{\mu\nu}$~is a tensor; and it is evidently covariant as the notation has anticipated. Let $g$~stand for the determinant \[ \left\lvert \begin{array}{@{}cccc@{}} g_{11} & g_{12} & g_{13} & g_{14} \\ g_{21} & g_{22} & g_{23} & g_{24} \\ g_{31} & g_{32} & g_{33} & g_{34} \\ g_{21} & g_{42} & g_{43} & g_{44} \\ \end{array}\right\rvert. \] Let $g^{\mu\nu}$~be defined as the minor of~$g_{\mu\nu}$ in this determinant, divided by~$g$\footnotemark.\footnotetext {The notation anticipates the result proved later that $g^{\mu\nu}$~is a contravariant tensor.}