- #1

RChristenk

- 50

- 4

- Homework Statement
- Transform ##0.7## from base ##8## to base ##5##. Transform ##0.12211## from base ##3## to base ##9##.

- Relevant Equations
- Conversion between bases

Transforming ##0.7_8## to base ##5## by multiplying ##5##:

##

\begin{array}{r}

0.7 \\

5 \\

\hline

4.3 \\

5 \\

\hline

1.7 \\

\end{array}

##

##0.7_8={0.\dot{4}\dot{1}}_5##. This is the correct answer.

Transforming ##0.12211_3## to base ##9## by multiplying ##9##:

##

\begin{array}{c}

0.&1&2&2&1&1 \\

&&&&&9 \\

\hline

12.&2&1&1&0&0 \\

\end{array}

##

Here I get two digits in the integer portion instead of one. And the correct answer is ##0.573_9##. Clearly ##12 \neq5##. And if I continued to multiply ##12.211## by ##9## to get the next digit position value in base ##9##, I would get ##1221.1## which has four integers. So the method works for me in the first problem, but not the second. I suspect there's nothing wrong with the method of multiplying by the final base, but rather I'm not understanding something or my calculations are wrong. Thanks for the help.

##

\begin{array}{r}

0.7 \\

5 \\

\hline

4.3 \\

5 \\

\hline

1.7 \\

\end{array}

##

##0.7_8={0.\dot{4}\dot{1}}_5##. This is the correct answer.

Transforming ##0.12211_3## to base ##9## by multiplying ##9##:

##

\begin{array}{c}

0.&1&2&2&1&1 \\

&&&&&9 \\

\hline

12.&2&1&1&0&0 \\

\end{array}

##

Here I get two digits in the integer portion instead of one. And the correct answer is ##0.573_9##. Clearly ##12 \neq5##. And if I continued to multiply ##12.211## by ##9## to get the next digit position value in base ##9##, I would get ##1221.1## which has four integers. So the method works for me in the first problem, but not the second. I suspect there's nothing wrong with the method of multiplying by the final base, but rather I'm not understanding something or my calculations are wrong. Thanks for the help.

Last edited: