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

  • #1
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...
 
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  • #2
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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.
 

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