Physics Forums

Physics Forums (http://www.physicsforums.com/index.php)
-   General Math (http://www.physicsforums.com/forumdisplay.php?f=73)
-   -   Why the method works? Change a decimal no. from denary(base 10) to octal(base 8) (http://www.physicsforums.com/showthread.php?t=559631)

Ask4material Dec12-11 08:09 PM

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 [tex]0.526_{10}[/tex] to octal(base 8) form

method: every time only the decimal part multiply by 8

[tex]\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}[/tex]

ans: 0.4152...

HallsofIvy Dec16-11 05:55 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 [tex]0.526_{10}[/tex] to octal(base 8) form

method: every time only the decimal part multiply by 8

[tex]\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}[/tex]

ans: 0.4152...

A number, less than one, in base 8, is of the form [itex]x= a(8^{-1})+ b(8^{-2})+ c(8^{-3})+\cdot\cdot\cdot[/itex]. Multiplying by 8 gives [itex]8x= a+ b(8^{-1})+ c(8^{-2})+\cdot\cdot\cdot[/itex] so the integer part is a. Removing that we get [itex]p= b(8^{-1})+ c(8^{-2})+ \cdot\cdot\cdot[/itex] and multiplying by 8 again gives [itex]8p= b+ c(8^{-1})+ \cdot\cdot\cdot[/itex] so the integer part is b.


All times are GMT -5. The time now is 06:20 AM.

Powered by vBulletin Copyright ©2000 - 2014, Jelsoft Enterprises Ltd.
© 2014 Physics Forums