mit dem kehrwert mal nehmen :
$${\frac{\left({\frac{{\mathtt{3}}}{{\mathtt{8}}}}\right)}{\left({\frac{{\mathtt{1}}}{{\mathtt{4}}}}\right)}} = \left({\frac{{\mathtt{3}}}{{\mathtt{8}}}}\right){\mathtt{\,\times\,}}\left({\frac{{\mathtt{4}}}{{\mathtt{1}}}}\right)$$