{{ tocSubheader }}
| {{ 'ml-lesson-number-slides' | message : article.intro.bblockCount}} |
| {{ 'ml-lesson-number-exercises' | message : article.intro.exerciseCount}} |
| {{ 'ml-lesson-time-estimation' | message }} |
Jonas (Diskussion | bidrag) | TemplateBot (Diskussion | bidrag) | ||
Rad 21: | Rad 21: | ||
\node at (0,0.43) {$\dfrac{5}{6}$}; | \node at (0,0.43) {$\dfrac{5}{6}$}; | ||
\node at (0,-0.33) {$\dfrac{3}{2}$}; | \node at (0,-0.33) {$\dfrac{3}{2}$}; | ||
− | \node at (1.1,0.43) {$\dfrac{5}{6} \ | + | \node at (1.1,0.43) {$\dfrac{5}{6} \t \textcolor{blue}{\dfrac{2}{3}}$}; |
− | \node at (1.1,-0.33) {$\dfrac{3}{2} \ | + | \node at (1.1,-0.33) {$\dfrac{3}{2} \t \textcolor{blue}{\dfrac{2}{3}}$}; |
− | \node at (2.45,0.43) {$\dfrac{5}{6} \ | + | \node at (2.45,0.43) {$\dfrac{5}{6} \t \dfrac{2}{3}$}; |
\node at (2.45,-0.17) {$1$}; | \node at (2.45,-0.17) {$1$}; | ||
\draw [line width=0.26mm] (-0.21,0.052)--++(0:0.42); | \draw [line width=0.26mm] (-0.21,0.052)--++(0:0.42); | ||
Rad 31: | Rad 31: | ||
\node at (1.77,0.05) {$=$}; | \node at (1.77,0.05) {$=$}; | ||
\node at (3.16,0.05) {$=$}; | \node at (3.16,0.05) {$=$}; | ||
− | \node at (3.74,0.08) {$\dfrac{5}{6} \ | + | \node at (3.74,0.08) {$\dfrac{5}{6} \t \dfrac{2}{3} $}; |
\end{tikzpicture} | \end{tikzpicture} | ||
</PGFTikz> | </PGFTikz> |
ba/dc=ba⋅cd
Man kan visa varför regeln fungerar genom att förlänga med nämnarens inverterade bråk. När man gör det blir produkten av bråken i nämnaren lika med 1 vilket innebär att bråkstrecket i mitten kan tas bort eftersom 1a=a. Nedan visas exemplet 65/23.