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I understand that converting fractional part of a decimal to binary involves this methods of repeating multiplication by 2:

For example, to convert $.375$(decimal) to binary:

  • $.375×2=0.75$ → integer part = $0$
  • $.75×2=1.5$ → integer part = $1$
  • $.5×2=1$ → integer part = $1$
  • So $.375$ = $.011_2$

My question is: Why is this specific methods the way to perform the conversion? Is there a mathematical justification or derivation behind these approaches? Or just an agreed-upon solution by convention?

PkDrew
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    Do you understand that the solution is correct, i.e. that $.375_{10} = .011_2$, i.e. that $$\frac{3}{10}+\frac{7}{100} +\frac{5}{1000} = \frac{0}{2}+\frac{1}{4}+\frac{1}{8}$$? If so, obviously this is not just a "convention", but the method leads very quickly to the right solution. – Torsten Schoeneberg Mar 12 '25 at 03:42
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    Now the task would have been here to set up $$\frac{3}{10}+\frac{7}{100} +\frac{5}{1000} = \frac{b_1}{2}+\frac{b_2}{4}+\frac{b_3}{8}+\frac{b_4}{16} + \dots $$ and now you try to find out, for each $b_i$, whether it should be $0$ or $1$. How would you find out? – Torsten Schoeneberg Mar 12 '25 at 03:44
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    Another similar question: Why do we divide or multiply by 2 when converting binary? – David K Mar 12 '25 at 04:26
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    @TorstenSchoeneberg, I see your point mate! as $b_n$ is either $0$ or $1$, each term $b_n/x^n$ is strictly less than $1$, so by each multiplication of 2 we are "extracting" the leftmost $b_i$. Many thanks for teaching ;) – PkDrew Mar 12 '25 at 04:59
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    @PkDrew: Happy to help. I was just teaching this material in a college class a few weeks ago, so I had an idea how to get the idea into students' minds, and I'm glad it worked for you too. – Torsten Schoeneberg Mar 12 '25 at 16:23

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Suppose I had some secret number, and you know that it's between 0 and 1, and has a finite length decimal expansion.

All I'm willing to do is carry out the four basic arithmetic operations, plus "take integer part", and report that result to you. How would you figure out the first digit in the decimal expansion, the one in the tenth's place?

JonathanZ
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