Why does the method work? Change a decimal no. from denary(base 10) to octal(base 8)?

**BookEnquiry** · December 13th 2011, 03:57 AM

K A Stroud Engineering Mathematics 6th ed p.61 frame 133

Change $0.528_{10}$ to octal(base 8) form

Method: every time only multiply the decimal part by 8

$\begin{array}{r}0.526\\\times8\\4.208\\\times8\\1. 664\\\times8\\5.312\\\times8\\2.496\\... goes\ on\end{array}$

answer: 0.4152

**BookEnquiry** · December 13th 2011, 04:21 AM

I couldn't figure it out...

denary(base 10): $0.526_{10}=5\cdot10^{-1}+2\cdot10^{-2}+6\cdot10^{-3}$

is equal to

octal(base 8): $0.4152..._{8}=4\cdot8^{-1}+1\cdot8^{-2}+5\cdot8^{-3}+2\cdot8^{-4}+...$

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how to get the octal(base 8) answer $abcd...$ ?

denary(base 10): $0.ABC_{10}=A\cdot10^{-1}+B\cdot10^{-2}+C\cdot10^{-3}$

is equal to

octal(base 8): $0.abcd..._{8}=a\cdot8^{-1}+b\cdot8^{-2}+c\cdot8^{-3}+d\cdot8^{-4}+...$

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Why does the method work? Change a decimal no. from denary(base 10) to octal(base 8)?

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