Why 1/1010 is 0.0001100110011001

Question

Can someone please demonstrate why 1/1010 in binary is 0.0001100110011001...? I've tried doing the math and I don't get the same result.

Thanks in advance!

Answer 1

Perhaps the simplest way to check is to verify that $1010\cdot0.0001100110011\ldots=1$. You can do this quite easily working entirely in binary notation:

$$\begin{align*} 1010\cdot0.000\overline{1100}&=(1000+10)\cdot0.000\overline{1100}\\ &=1000\cdot0.000\overline{1100}+10\cdot0.000\overline{1100}\\ &=0.\overline{1100}+0.00\overline{1100}\\ &=110011001100\ldots+0.001100110011\ldots\\ &=0.\overline{1100}+0.\overline{0011}\\ &=0.\overline{1111}\\ &=0.111\ldots\\ &=1\;. \end{align*}$$

If you want, you can turn this into a statement about sums of geometric series, but this is simpler.

In case it wasn’t clear, the third step simply used the observation that multiplying by

$$1\underbrace{0\ldots0}_n=2^n$$

simply moves the binary point $n$ places to the right.

Answer 2

It's because

    0.00011001100110011...
                 × 1010 
-----------------------
    0.0011001100110011...
  + 0.11001100110011...
-----------------------
  = 0.1111111111111111...
-----------------------
  = 1.0000000000000000...

Answer 3

The proof given by @Brian M. Scott is very pedagogical but I do think that understanding the geometric series approach is as well fundamental.

It maybe difficult to obtain the adequate series, because of the interference of two "parameters": the (frequent) presence of an part of the development and its period (and, first of all, its length). In this case, both are known. Thus the only trick of the trade is to get an equivalent fraction of the form $a/\overline{0.1111}$ (we will see why), once the irregular part has been removed.

The irregular part is eliminated using multiplication by two $ =\overline{10}$.

We have now to establish that $\overline{10}/\overline{1010}=\overline{0.001100110011\cdots} \ \ \ (1)$

Let us multiply numerator and denominator of the LHS of (1) by $\overline{1.1}$.

Let us also decompose the RHS in order to put forward a geometrical series, giving :

$$\overline{11}/\overline{1111}=\overline{0.0011}+\dfrac{1}{2^4}\overline{0.0011}+\dfrac{1}{2^8}\overline{0.0011}+\cdots $$

$$\overline{0.0011}\dfrac{1}{\overline{0.1111}}=\overline{0.0011}+\dfrac{1}{2^4}\overline{0.0011}+\dfrac{1}{2^8}\overline{0.0011}+\cdots $$

Simplifying by $\overline{0.0011}$, one obtains:

$$\dfrac{1}{1-\overline{0.0001}}=1+\dfrac{1}{2^4}+\dfrac{1}{2^8}+\cdots $$

where one recognizes the formula for the sum of a convergent geometric series: $$\dfrac{1}{1-a}=1+a+a^2+a^3+\cdots \ \ \text{for} \ \ |a|<1$$

Thus the development is checked.

The main thing to remember in this approach is based on the transformation of the initial fraction into a fraction of the form $A+\dfrac{B}{\overline{0.111\cdots1}}$ where $A$ accounts for the irregular part (we didn't need such an $A$ in the previous calculations), $B$ gives the period pattern, and the number $n$ of ones in the denominator $\overline{0.111\cdots1}=1-\dfrac{1}{2^n}$ is the length of the period of this base-$2$ development.

Such an operation is always possible (here is not the place to develop any theory...), sometimes simply by elementary trial as done upwards.

A last (independant) remark: we could have worked using base 16: a well known fact (well, at least to electronicians...) is that the base 16 development is obtained by grouping four by four the base 2 digits.