# converting binary fractions with a recurring expansion to decimal (denary) system

suppose the binary numbers 0.1010…. or I mean 0.(10) ̅ a recurring binary fraction And 0.1(0101) ̅ a recurring binary fraction but with 1 digit before the period starts

can these numbers be represented in decimal (denary) form as fractions where both the numerators and the denominators are both integers??

if yes, type the answer if no, type what is the cause? is it because the digital expansion of the fractional part is finite infinite recurring infinite non-recurring

please, I'm not understanding how to convert a recurring binary fraction into decimal, I want extremely detailed answer with small gradual steps and a detailed explanation of each step

your help is very much appreciated

• You do it the same way you would convert a periodic decimal expansion to a fraction e.g. $0.1(09)=6/55$.
– dxiv
Mar 8 at 21:53
• @dxiv how do I do that? I mean I can’t divide a recurring expansion on 10^something?? Mar 8 at 21:58
• It's easier if you figure out how it works in a familiar base, like decimal, before asking about binary. But if this is the first time ever that you do such a conversion, then where did you get the problem from?
– dxiv
Mar 8 at 22:26
• @dxiv I roughly understand the decimal base recurring fraction stuff, I’m new into computer science and looking forward to learn, I got the problem from the University of London course on Coursera because I’m preparing to enroll in a CS degree next year Mar 8 at 23:14
• @dxiv I solved the first number the right way thanks to the answer of ConMan but the rule isn't going very well with the second number where the repeating pattern does not start immediately after the fraction point, do you know how to solve it? Mar 8 at 23:47

The binary number $$0.a_1 a_2 a_3 \ldots_2$$ where $$a_i \in \{0, 1\}$$ is a representation of the infinite series $$\frac{a_1}{2} + \frac{a_2}{4} + \frac{a_3}{8} + \ldots + \frac{a_n}{2^n} + \ldots$$, compared to the decimal number which would be $$\frac{a_1}{10} + \ldots + \frac{a_n}{10^n} + \ldots$$. So where you would resolve a recurring decimal by multiplying by an appropriate power of 10, for the binary you multiply by the appropriate power of 2. For example, consider $$x = 0.\overline{(101)}_2 = 0.101101101 \ldots_2$$:
$$\begin{eqnarray} x & = & 0.101101101101 \ldots_2 \\ 2^3 x & = & 101.101101101 \ldots_2 \\ & = & 101_2 + x \\ (2^3 - 1) x & = & 101_2 \\ 7 x & = & 5 \\ x & = & \frac{5}{7} \end{eqnarray}$$