Station  8E

Irrational Numbers 

We have seen that fractions can possess finitely long decimal expansions.

For example, $\dfrac{1}{8} = 0.125$ and $\dfrac{1}{2} = 0.5$.

And we have seen that fractions can possess infinitely long decimal expansion.

For example, $\dfrac{1}{3} = 0.3333...$ and $\dfrac{6}{7} = 0.857142057142857142...$.

All the examples of fractions with infinitely long decimal expansions we’ve seen so far fall into a repeating pattern. This is curious.

We can even say that our finite examples eventually fall into a repeating pattern too, a repeating pattern of zeros after an initial start.

$\dfrac{1}{8} = 0.12500000... = 0.125\overline{0}$

$\dfrac{1}{2} = 0.50000... = 0.5\overline{0}$

$\dfrac{1}{3} = 0.\overline{3}$

$\dfrac{6}{7} = 0.\overline{857142}$

This begs the question:

Let’s go through the division process again, slowly, first with a familiar example. Let’s compute the decimal expansion of $\dfrac{1}{3}$ again in a $1 \leftarrow 10$ machine. We think of $\dfrac{1}{3}$ as the answer to the division problem $1 \div 3$, and so we need to find groups of three within a diagram of one dot.

We unexplode the single dot to make ten dots in the tenths position. There we find three groups of three leaving a remainder of $1$ in that box.

Now we can unexplode that single dot in the tenths box and write ten dots in the hundredths box.
There we find three more groups of three, again leaving a single dot behind.

And so on. We are caught in a cycle of having the same remainder of one dot from cell to cell, meaning that the same pattern repeats. Thus we conclude $\dfrac{1}{3} = 0.333...$. The key point is that the same remainder of a single dot kept appearing.

Here’s a more complicated example. Let’s compute the decimal expansion of $\dfrac{4}{7}$ in the $1 \leftarrow 10$ machine.
That is, let’s compute $4 \div 7$.

We start by unexploding the four dots to give $40$ dots in the tenths cell. There we find $5$ groups of seven, leaving five dots over.

Now unexplode those five dots to make $50$ dots in the hundredths position. There we find $7$ groups of seven, leaving one dot over.

Unexplode this single dot. This yields $1$ group of seven leaving three remaining.

Unexplode these three dots. This gives $4$ groups of seven with two remaining.

Unexplode the two dots. This gives $2$ groups of seven with six remaining.

Unexplode the six dots. This gives $8$ groups of seven with four remaining.

But this is predicament we started with: four dots in a box!

So now we are going to repeat the pattern and produce a cycle in the decimal representation. We have

$\dfrac {4}{7} = 0.571428 571428 571428...$.

Stepping back from the specifics of this problem, it is clear now that one must be forced into a repeating pattern. In dividing a quantity by seven, there are only seven possible numbers for a remainder number of dots in a cell – $0, 1, 2, 3, 4, 5,$ or $6$ – and there is no option but to eventually repeat a remainder and so enter a cycle.

In the same way, the decimal expansion of $\dfrac{18}{37}$ must also cycle. In doing the division, there are only thirty-seven possible remainders for dots in a cell ($0, 1, 2, 3, 4, 5, ..., 36$). As we complete the division computation, we must eventually repeat a remainder and again fall into a cycle.