1 Decimal (Base 10) to Binary (Base 2)

Let us start with powers of 2…

$$ \begin{array}{rrll} \frac{1}{2} &=& 2^{-1} &=& 0.100_{2} &=& 0.0111111\overline{1}_{2} \\ \frac{1}{4} &=& 2^{-2} &=& 0.010_{2} &=& 0.0011111\overline{1}_{2} \\ \frac{1}{8} &=& 2^{-3} &=& 0.001_{2} &=& 0.0001111\overline{1}_{2} \\ \end{array} $$

The third column is not immediately obvious. Let us take a few minutes to prove that:

$$ \begin{array}{rrll} \frac{1}{2} &=& 2^{-1} &=& 0.100_{2} &=& 0.0111111\overline{1}_{2} \\ \end{array} $$

is true. We will need to use the geometric series formula.

We need to prove that

\begin{array}{rl} 0.100_{2} &=& 0.0111111\overline{1}_{2} \\ 2^{-1} &=& 2^{-2} + 2^{-3} + 2^{-4} + \ldots \\ \end{array}

Notice how I changed from based to to base 10 and tweaked the right side? We now have a geometric series! We can apply the geometric series formula (after one small tweak).

\begin{array}{rl} 2^{-1} &=& 2^{-2} + 2^{-3} + 2^{-4} + \ldots \\ 2^{-1} &=& 2^{-2} \left(2^{0} + 2^{-1} + 2^{-2} + \ldots \right) \\ 2^{-1} &=& 2^{-2} \left( \frac{1}{1-2^{-1}} \right) \\ 2^{-1} &=& 2^{-2} \left( \frac{1}{2^{-1}} \right) \\ 2^{-1} &=& 2^{-2} \left( 2 \right) \\ 2^{-1} &=& 2^{-1} \\ \end{array}

2 Converting Real Numbers from Base 10 to Base 2

Example 1

Convert $\frac{1}{5} = 0.2$ to binary.

Example 2

Convert $\frac{1}{10} = 0.1$ to binary.

Example 2

Convert $\frac{1}{7} = 0.\overline{142857}$ to binary.

3 Octal (Base 8) & Hexadecimal (Base 16)

Both Octal and Hexadecimal numbers are shorthand for binary. Octal works with 3 bits and hexadecimal works with 4 bits.

Anytime I do back-of-the-envelope conversions in the middle of a problem, these tables are involved.