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-   -   Why the method works? Change a decimal no. from denary(base 10) to octal(base 8) (http://www.physicsforums.com/showthread.php?t=559631)

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

K A Stroud Engineering Mathematics 6th ed frame 133

change $$0.526_{10}$$ to octal(base 8) form

method: every time only the decimal part multiply by 8

$$\begin{array}{r}0.526\\ \times8\\\hline\\4.208\\ \times8\\\hline\\1.664\\ \times8\\\hline\\5.312\\ \times8\\\hline\\2.496\\\mbox{... goes on}\end{array}$$

ans: 0.4152...

 HallsofIvy Dec16-11 06:55 PM

Re: Why the method works? Change a decimal no. from denary(base 10) to octal(base 8)

Quote:
 Quote by Ask4material (Post 3665620) K A Stroud Engineering Mathematics 6th ed frame 133 change $$0.526_{10}$$ to octal(base 8) form method: every time only the decimal part multiply by 8 $$\begin{array}{r}0.526\\ \times8\\\hline\\4.208\\ \times8\\\hline\\1.664\\ \times8\\\hline\\5.312\\ \times8\\\hline\\2.496\\\mbox{... goes on}\end{array}$$ ans: 0.4152...
A number, less than one, in base 8, is of the form $x= a(8^{-1})+ b(8^{-2})+ c(8^{-3})+\cdot\cdot\cdot$. Multiplying by 8 gives $8x= a+ b(8^{-1})+ c(8^{-2})+\cdot\cdot\cdot$ so the integer part is a. Removing that we get $p= b(8^{-1})+ c(8^{-2})+ \cdot\cdot\cdot$ and multiplying by 8 again gives $8p= b+ c(8^{-1})+ \cdot\cdot\cdot$ so the integer part is b.

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