Station  8F

Decimals in Other Bases 

The following picture shows that $1432 \div 13 = 110 R 2$ in a $1 \leftarrow 5$ machine.

If we work with reciprocals of powers of five we can keep unexploding dots and continue the division process.

Here goes!

(This reads as $1432 \div 13 = 110.1 R 0.2$ if you like.)

(This reads as $1432 \div 13 = 110.11 R 0.02$ if you like.)

And so on.
We get, as a statement of base $5$ arithmetic,

$1432 \div 13 = 110.1111...$.

To translate this to ordinary arithmetic we have that

$1432$ in base five is $1 \times 125 + 4 \times 25 + 3 \times 5 + 1 \times 1 = 242$,

$13$ in base five is $1 \times 5 + 3 \times 1 = 8$,

$110.111...$ in base five is $30 + \dfrac{1}{5} + \dfrac{1}{25} + \dfrac{1}{125} + ...$,

so we are claiming, in ordinary arithmetic, that

$242 \div 8 = 30 + \dfrac{1}{5} + \dfrac{1}{25} + \dfrac{1}{125} + ...$.

Whoa!

What does the geometric series formula from Infinitia say about the sum $\dfrac{1}{5} + \dfrac{1}{25} + \dfrac{1}{125} + ...$. Does it equal a quarter?

To be clear, on Infinitia we showed that

$1+x+x2+x3+...=\dfrac{1}{1-x}$.

Multiplying through by $x$ then gives $x+x2+x3+...=\dfrac{x}{1-x}$.

Perhaps it is this second version of the geometric formula we need to work with now.)

Here are some (challenging) practice questions if you are up for them.

CHALLENGE