Bruch-zu-Dezimal-Rechner

Your input: convert $\frac{900}{90}$ into a decimal.

Write the problem in the special format:

$\require{enclose}\begin{array}{rlc}&\phantom{-\enclose{longdiv}{}}\begin{array}{ccc}\phantom{1}&\phantom{0}&\phantom{.}&\phantom{0}\end{array}&\\90&\phantom{-}\enclose{longdiv}{\begin{array}{ccc}9&0&0\end{array}}&\\&\begin{array}{lll}\end{array}&\begin{array}{c}\end{array}\end{array}$

Step 1

How many $90$'s are in $9$? The answer is $0$.

Write down the calculated result in the upper part of the table.

Now, $9-0 \cdot 90 = 9 - 0= 9$.

Bring down the next digit of the dividend.

$\require{enclose}\begin{array}{rlc}&\phantom{-\enclose{longdiv}{}}\begin{array}{ccccc}\color{Blue}{0}&\phantom{1}&\phantom{0}&\phantom{.}&\phantom{0}\end{array}&\\\color{Magenta}{90}&\phantom{-}\enclose{longdiv}{\begin{array}{ccccc}\color{Blue}{9}& 0 \downarrow&0&.&0\end{array}}&\\&\begin{array}{llll}-&\phantom{0}&\phantom{0}&\phantom{0}&\phantom{.}\\\phantom{lll}0&\phantom{.}\\\hline\phantom{lll}9&0&\phantom{.}\end{array}&\begin{array}{c}\end{array}\end{array}$

Step 2

How many $90$'s are in $90$? The answer is $1$.

Write down the calculated result in the upper part of the table.

Now, $90-1 \cdot 90 = 90 - 90= 0$.

Bring down the next digit of the dividend.

$\require{enclose}\begin{array}{rlc}&\phantom{-\enclose{longdiv}{}}\begin{array}{ccccc}0&\color{GoldenRod}{1}&\phantom{0}&\phantom{.}&\phantom{0}\end{array}&\\\color{Magenta}{90}&\phantom{-}\enclose{longdiv}{\begin{array}{ccccc}9&0& 0 \downarrow&.&0\end{array}}&\\&\begin{array}{llll}-&\phantom{0}&\phantom{0}&\phantom{0}&\phantom{.}\\\phantom{lll}0&\phantom{.}\\\hline\phantom{lll}\color{GoldenRod}{9}&\color{GoldenRod}{0}&\phantom{.}\\-&\phantom{0}&\phantom{0}&\phantom{0}&\phantom{.}\\\phantom{lll}9&0&\phantom{.}\\\hline\phantom{lll}&0&0&\phantom{.}\end{array}&\begin{array}{c}\end{array}\end{array}$

Step 3

How many $90$'s are in $0$? The answer is $0$.

Write down the calculated result in the upper part of the table.

Now, $0-0 \cdot 90 = 0 - 0= 0$.

Bring down the next digit of the dividend.

$\require{enclose}\begin{array}{rlc}&\phantom{-\enclose{longdiv}{}}\begin{array}{ccccc}0&1&\color{DarkCyan}{0}&\phantom{.}&\phantom{0}\end{array}&\\\color{Magenta}{90}&\phantom{-}\enclose{longdiv}{\begin{array}{ccccc}9&0&0&.& 0 \downarrow\end{array}}&\\&\begin{array}{llll}-&\phantom{0}&\phantom{0}&\phantom{0}&\phantom{.}\\\phantom{lll}0&\phantom{.}\\\hline\phantom{lll}9&0&\phantom{.}\\-&\phantom{0}&\phantom{0}&\phantom{0}&\phantom{.}\\\phantom{lll}9&0&\phantom{.}\\\hline\phantom{lll}&\color{DarkCyan}{0}&\color{DarkCyan}{0}&\phantom{.}\\-&\phantom{0}&\phantom{0}&\phantom{.}&\phantom{0}\\\phantom{lll}&&0&\phantom{.}\\\hline\phantom{lll}&&0&\phantom{.}&0\end{array}&\begin{array}{c}\end{array}\end{array}$

Step 4

How many $90$'s are in $0$? The answer is $0$.

Write down the calculated result in the upper part of the table.

Now, $0-0 \cdot 90 = 0 - 0= 0$.

$\require{enclose}\begin{array}{rlc}&\phantom{-\enclose{longdiv}{}}\begin{array}{ccccc}0&1&0&.&\color{DarkBlue}{0}\end{array}&\\\color{Magenta}{90}&\phantom{-}\enclose{longdiv}{\begin{array}{ccccc}9&0&0&.&0\end{array}}&\\&\begin{array}{llll}-&\phantom{0}&\phantom{0}&\phantom{0}&\phantom{.}\\\phantom{lll}0&\phantom{.}\\\hline\phantom{lll}9&0&\phantom{.}\\-&\phantom{0}&\phantom{0}&\phantom{0}&\phantom{.}\\\phantom{lll}9&0&\phantom{.}\\\hline\phantom{lll}&0&0&\phantom{.}\\-&\phantom{0}&\phantom{0}&\phantom{.}&\phantom{0}\\\phantom{lll}&&0&\phantom{.}\\\hline\phantom{lll}&&\color{DarkBlue}{0}&\phantom{.}&\color{DarkBlue}{0}\\&-&\phantom{0}&\phantom{.}&\phantom{0}\\\phantom{lll}&&&\phantom{.}&0\\\hline\phantom{lll}&&&&0\end{array}&\begin{array}{c}\end{array}\end{array}$

As can be seen, the digits are repeating with some period, therefore it is a repeating (or recurring) decimal: $\frac{900}{90}=10. \overline{0}$

Answer: $\frac{900}{90}=10.\overline{0}$