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The exercise in the book goes like this: In the decimal system the fraction 1/3 is written as 0.3333... (pure periodic number). What happens in the binary system?

So I've been practising with other divisions in binary and, comparing my result with an online calculator (https://ncalculators.com/digital-computation/binary-division-calculator.htm) everything seems to go smoothly. However, when it comes to this particular division the result differs greatly:

1/11 in binary-->

  • According to me = 0,01010... This is the logic I follow: 1 divided by 11 is impossible, so I need to add a '0' to the one and a 0 to the quotient as well. Thus, 10/11. Still not possible to divide, hence I add another '0' which turns into 100/11; another 0 to the quotient. NOW I can divide.
  • According to the online calculator = 10101010,1

Why is that? Why does the comma come so further away from where I put it?

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  • $\begingroup$ You should give a link to the online calculator. Put this link in your original post, rather than as a response to this comment. $\endgroup$ – Brian Tung Jul 12 '19 at 17:55
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    $\begingroup$ Your result looks correct to me. $\endgroup$ – Martin R Jul 12 '19 at 17:55
  • $\begingroup$ Welcome to MSE! The calculator says the result is $1.01010101...$ (although 100/11 is not what you originally wanted to calculate), instead of what you wrote. Could you please edit your post and fix that? Thanks $\endgroup$ – evaristegd Jul 17 '19 at 5:36
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Yes, in binary the fraction $\frac13$ is written as $0.01010101\ldots$ That's so because what $0.01010101\ldots$ means is$$\frac1{2^2}+\frac1{2^4}+\frac1{2^6}+\cdots$$and the sum of this series is$$\frac{\frac14}{1-\frac14}=\frac13.$$

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  • $\begingroup$ Thank you, it's a different way of seeing it. $\endgroup$ – LocoVoco Jul 12 '19 at 17:59
  • $\begingroup$ Also, sorry for bothering you, but could you disclose a little bit the process that led you to that conclusion? For example, I don't know how a fraction ended up on top. (It's been very long since I went to school, please bear with me) $\endgroup$ – LocoVoco Jul 12 '19 at 18:15
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    $\begingroup$ I use the formula$$a+ar+ar^2+ar^3+\cdots=\frac a{1-r},$$with $a=r=\frac14$. $\endgroup$ – José Carlos Santos Jul 12 '19 at 18:33
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@LocoVoco Thanks for your discussion, the issue has been now fixed. binary division calculator

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    $\begingroup$ Welcome to MSE. This should be a comment, not an answer. $\endgroup$ – José Carlos Santos Jul 13 '19 at 13:16

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