TL;DR

Rational numbers are compact and might prove better for doing a lot of stuff.

When dealing with numbers with computers, we always have to keep in mind that there is so much we can do. By default.

As an example, a fantastic number like $\pi$ cannot be represented easily in any base, neither base-10 nor base-2 or any power of 2. Considering that numbers are represented by finite strings of bits (or decimal digits on a piece of paper)â€¦ you can only get so much.

As an example, most people I know usually truncate $\pi$ to $3.14$. A few more, I think, go to $3.14159$, me included.

On the other hand, if you have to remember 6 decimal digits overall, itâ€™s probably better to remember $\frac{355}{113}$, because it gets you one digit more (with the rounding, both $\pi$ and that fraction yield $3.141593$).

The fun thing with rational numbers is that they are compact. Even if a lot of numbers are not rational, they have this interesting property that you can get as close as you finitely want to any of them with a rational representation.

Not sure about it? Letâ€™s see an example with $\pi$ itself.

We start with two fractions that are below and above $\pi$:

$B_1 = \frac{3}{1} \\ A_1 = \frac{4}{1} \\ \epsilon_1 < \frac{A_1 - B_1}{2} = \frac{1}{2}$

The first one is closer to $\pi$ so this will be our starting approximation. The error we are doing by choosing these two values is less than $\frac{1}{2}$, i.e. the distance between our starting fractions. (It cannot be equal to $\frac{1}{2}$ because $\pi$ is irrational!).

Now letâ€™s consider the mid-point between our starting fractions:

$M_1 = \frac{A_1 + B_1}{2} > \pi$

Itâ€™s still a rational number, and as it comes itâ€™s better than $A_0$ because itâ€™s closer to $\pi$. So we can build a second iteration like this:

$B_2 = \frac{3}{1} \\ A_2 = \frac{7}{2} \\ \epsilon_2 = \frac{1}{4} \\ M_2 = \frac{13}{4} > \pi$

Our initial rational approximation still wins, but now we know that the error we are doing must be less than $\epsilon_2 = 0.25$. As before, we can use $M_2$ to get a better upper rational bound and get to the second iteration:

$B_3 = \frac{3}{1} \\ A_3 = \frac{13}{4} \\ \epsilon_3 = \frac{1}{8} \\ M_3 = \frac{25}{8} < \pi$

Now we have two considerations:

• the upper bound for the error is always halving, because we are always cutting an interval in half
• this new value $M_3$ is lower than $\pi$, so we will update the lower rational bound instead of the upper one.

At this point, the algorithm is simple:

• at iteration $k$, calculate
$\epsilon_k = \frac{1}{2^k} \\ M_k = \frac{A_k + B_k}{2}$
• if $M_k < \pi$, then the boundary values for iteration $k + 1$ will be:
$A_{k+1} = M_k \\ B_{k+1} = B_k$
• otherwise:
$A_{k+1} = A_k \\ B_{k+1} = M_k$

This is it. If you fix the maximum finite error $\epsilon$ that you are willing to accept, you can then iterate until $\epsilon_k$ is lower than, or equal to this value, and youâ€™re done.

Nowâ€¦ Iâ€™m not saying that these values will be somehow optimal in terms of how compact the representation will be in terms of digits, just that you can get as close as you finitely want.

Cheers!

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