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

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SUMMARY

The method for converting the decimal number 0.526 from base 10 to octal (base 8) involves repeatedly multiplying the decimal part by 8. This process yields the octal representation 0.4152..., as demonstrated in K A Stroud's Engineering Mathematics 6th edition, frame 133. Each multiplication extracts the integer part, which corresponds to the digits in the octal system, while the remaining decimal part is used for the next multiplication.

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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...
 
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Ask4material said:
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.