Das Komma kann man erst einmal vernachlässigen. Berechne das Produkt mit Hilfe der schriftlichen Multiplikation.

$\begin{array}{c}\underset{‾}{4\cdot \mathrm{13,75}}\\ 20\\ 280\\ 1200\\ \underset{‾}{\phantom{9\cdot ,}4000}\\ 5500\end{array}\backslash def\backslash arraystretch\{1.25\}\; \backslash begin\{array\}\{r\}\backslash underline\{4\backslash cdot13\{,\}75\}\backslash \backslash 20\backslash \backslash 280\backslash \backslash 1200\backslash \backslash \backslash underline\{\backslash hphantom\{9\backslash cdot\{,\}\}4000\backslash \backslash \}\backslash \backslash 5500\backslash end\{array\}$

Zähle nun die Anzahl der Nachkommastellen der beiden Faktoren. In diesem Fall sind es 2 Nachkommastellen, Setze im Ergebnis das Komma so, dass es genauso viele Nachkommastellen hat.

$4\cdot \mathrm{13,75}=\mathrm{55,00}4\backslash cdot13\{,\}75=55\{,\}00$

$\begin{array}{c}\phantom{-}503,74\\ \underset{‾}{-{}_1{}_1{}_{1}7,83}\\ \phantom{-}495,91\end{array}\backslash def\backslash arraystretch\{1.25\}\; \backslash begin\{array\}\{l\}\backslash hphantom\{-\}\backslash ;5\backslash ;0\backslash ;3,\backslash ;7\backslash ;4\backslash \backslash \backslash underline\{-~\_1~\_1~\_17,\backslash ;8\backslash ;3\}\backslash \backslash \backslash hphantom\{-\}\backslash ;4\backslash ;9\backslash ;5,\backslash ;9\backslash ;1\backslash end\{array\}$