Calculate the binary equivalent of the decimal number [latex] (.3125)_{10} [/latex] using the recursive algorithm 1.2.

Let [latex] c_0=(.3125)_{10} [/latex] [latex] \begin{array}{lll} 2(.3125)_{10}=(.6250)_{10} & b_1=0 & c_1=(.6250)_{10} \\ 2(.6250)_{10}=(1.250)_{10} & b_2=1 & c_2=(.250)_{10} \\ 2(.250)_{10}=(.50)_{10} & b_3=0 & c_3=(.50)_{10} \\ 2(.50)_{10}=(1.00)_{10} & b_4=1 & c_4=(0)_{10} \end{array} [/latex] The binary equivalent of [latex] (.3125)_{10} \text { is }\left(. b_1 b_2 b_3 b_4\right)_2=(.0101)_2 [/latex] . This example has a terminating binary fraction, but not each terminating decimal fraction will give a terminating binary fraction, and this is true for other number systems also.

Question 1.10: Calculate the binary equivalent of the decimal number (.3125...

Numerical Methods: Fundamentals and Applications

Calculate the binary equivalent of the decimal number $(.3125)_{10}$ using the recursive algorithm 1.2.

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Let $c_0=(.3125)_{10}$

$\begin{array}{lll} 2(.3125)_{10}=(.6250)_{10} & b_1=0 & c_1=(.6250)_{10} \\ 2(.6250)_{10}=(1.250)_{10} & b_2=1 & c_2=(.250)_{10} \\ 2(.250)_{10}=(.50)_{10} & b_3=0 & c_3=(.50)_{10} \\ 2(.50)_{10}=(1.00)_{10} & b_4=1 & c_4=(0)_{10} \end{array}$

The binary equivalent of $(.3125)_{10} \text { is }\left(. b_1 b_2 b_3 b_4\right)_2=(.0101)_2$ . This example has a terminating binary fraction, but not each terminating decimal fraction will give a terminating binary fraction, and this is true for other number systems also.