More Issues With Floating Points

Relevant wiki: Number Base - Problem Solving

One way to represent $0.1_{10}$ in binary is by performing a long division. Another approach is to notice that $0.1 = \frac12\cdot 3\cdot \frac1{15} = \frac12\cdot 3\cdot \left(\frac1{16} + \frac1{256} + \cdots\right) = \sum_{k=0}^\infty 3\cdot 2^{-4k-5}.$ Either way, we find the repeated binary fraction

0.1 = 0.0001 1001 1001 1001 1001 1001 (1001 ...)

Because the 25th decimal is 1, rounding upward results in a smaller error than rounding downward:

0.1 ~ 0.0001 1001 1001 1001 1001 1010

This is the closest 24-decimal binary approximation of decimal 0.1. Writing this as a normal fraction, we get $\frac{110011001100110011010_2}{2^{24}} = \frac{1\:677\:722}{16\:777\:216} = \frac{838\:861}{8\:388\:608},$ so that the answer should be $\boxed{9\:227\:469}$ .

As $0.1 = \frac{1677721.6}{16777216}$ , therefore the closest integer numerator will be its rounded value, 1677722

Since the number formed of the last two digits (22, even) of the numerator is not divisible by 4 (therefore the numerator itself is neither) and the denominator ( $2^{24}$ ) has only the 2 as prime factor , therefore gcd(1677722,16777216)=2, meaning, that we can only simplify by 2 (in order to make the numerator and the denominator coprime).

Hence,

$\frac {1677722}{16777216} = \frac {838861}{8388608}$

and the solution is

$838861 + 8388608 = \boxed {9227469}$ .

Edit: See Arjen's solution below

To go from $\frac{1}{3}$ to $0.33\ldots$ , one simply divides .

Similarly, we want to divide $\frac{1}{1010_2}$ in binary. That yields $0.0001100110011\ldots_2$

In other words, we have $\Sigma^{\infty}_{i=1} \frac{3}{2^{4i + 1}} = \frac{1}{10}$

If we had 24 bits, we would set represent it as 0.000110011001100110011001 which translates to $\frac{1677721}{16777216}$