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

[Ask4material]

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]

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.