# Percentages and decimals as rounded fractions

Related problem

Of course, every possible integer percentage $$0 \leq p \leq 1, 100p \in \mathbb{N}$$ can be expressed as a fraction $$\frac a {100}, a=100p$$. Many of these fractions (actually 61, because $$61 = 101 - \phi(100) = 101 - 40$$ with Euler's totient function $$\phi(n)$$) can be reduced to have a smaller denominator.

However, if we allow rounding, we can express most percentages with even smaller denominators, e.g. $$17\% \approx \frac 1 6$$.

Rounding

First of all, we have to define how to round. There are many ways to do this, but we have to conditions that characterize the method we'll use.

• Every number $$p$$ that isn't already an integer (in this case we wouldn't have to round) has two neighbouring integers. Unless the fractional part of $$p$$, $$p-\lfloor p \rfloor$$ is $$0.5$$ we can just round to the nearest integer.
• When working with fractions, there is a nice symmetry $$\frac a b + \frac {b-a} b = 1$$. This seems trivial, but we definitely want to keep it. So $$\frac 1 8 + \frac 7 8$$ should equal $$1$$. But when rounded, we get $$0.13 + 0.88 = 1.01$$. So, to preserve symmetry, we will always round away from $$0.5$$. $$0.125 ≈ 0.12$$, but $$0.875 ≈ 0.88$$.

These two conditions characterize our rounding method, and we can also write a formula that satisfies both:

$$p \approx \frac 1 {100} ( \text{sgn} (100p-50) \lfloor | 100p-50 | + \frac1 2 \rfloor +50 )$$

The sign-function $$\text{sgn} (x)$$ gives $$1$$ if $$x$$ is nonnegative and $$-1$$ otherwise.

If we now want to round with some other precision $$a \in \mathbb{N}$$, we can use the formula

$$p \stackrel{s}\approx \frac 1 s \left( \text{sgn} (sp-\frac s 2) \lfloor | sp-\frac s 2 | + \frac1 2 \rfloor + \frac s 2 \right)$$

To give another notation, let's define

$$r_s(p) = \frac 1 s \left( \text{sgn} (sp-\frac s 2) \lfloor | sp-\frac s 2 | + \frac1 2 \rfloor + \frac s 2 \right)$$

and in particular

$$r(p) = r_{100}(p)$$

The smallest possible denominator

With these rules in mind, we can now look at the topic of this note – expressing decimals (like percentages) as rounded fractions.

As common, we will only consider reduced fractions; and in this part mainly their denominators.

Let us define a function $$q_s(p)$$ to give the smallest denominator $$q$$ for which there exists a numerator $$a$$ so that $$\frac a q$$ equals $$p$$ when rounded with precision $$s$$.

$$q_s(p) = \min \{q \in \mathbb{N} : (\exists a \in \mathbb{N})[r_s(\frac a q) = p] \}$$

Now we can explore some properties of $$q_s(p)$$.

If we graph $$q_s(p)$$ with some fixed precision, say $$s=100$$, we see some interesting patterns .

Every $$p$$ that can be written with a small denominator $$q$$ has neighbours that need very large denominators. For example, $$p=0.5 = \frac 1 2 \Rightarrow q_{100}(0.5)=2$$, but $$q_{100}(0.49) = 35$$. This is because for every other $$q < 35$$, either the fraction could be reduced to exactly $$\frac 1 2$$ or $$\frac 1 q$$ just didn't make small enough steps to hit $$0.49$$.

We can make a table of $$q_s(p)$$ when $$s$$ goes to infinity and $$p$$ is very small, so it's influenced by $$0$$.

$$\displaystyle \lim_{s \to \infty}q_s(1) = \left\lceil \frac 2 3 s \right\rceil$$

$$\displaystyle \lim_{s \to \infty}q_s(2) = \left\lceil \frac 2 5 s \right\rceil$$

And in general,

$$\displaystyle \lim_{s \to \infty}q_s(p) = \left\lceil \frac 2 {2p+1} s \right\rceil$$

Influence

To formalize this influence of simple fractions, let's define a funtion $$i_s(p)$$ to give the number of values from $$p$$ upward for which $$q$$ is decreasing.

$$i_s(p) = max\{ p_k : sp_k \in \mathbb{N} \land q_s\left(p+\frac 1 s\right)>q_s\left(p+\frac 2 s\right)>\cdots>q_s\left(p+\frac k s\right) \} - p$$

We can look at another diagram of $$q_s(p)$$ with bigger $$s$$, in this case $$s=2000$$) to see how $$i_s(p)$$ behaves.

There are all those spikes marking the neighbours of simple fractions, and it somehow looks a little bit like a fractal.

The influence of $$0$$ is easiest to calculate. It can be done with a computer program, but there is no closed formula since it really depends on the value of $$s$$.

pseudocode:

 1 2 3 4 5 6 if (x=0) (f(1/s, s)) else if (1/floor(1/x)>x+1/(2*s) and 1/ceiling(1/x)<=x-1/(2*s)) x else f(x+1/s, s) 

What we can do, is give an approximation for $$i(0, s)$$ when $$s$$ gets very large. Since the influence counts the number of $$p$$-values whose denominators are decreasing, it will stop at the first fraction with a higher denominator, say $$p_1$$. But the one before it, $$p_0$$, had a smaller denominator and was smaller. So it necessarily also had a smaller numerator, actually $$1$$ because otherwise it could have been replaced by some other fraction with numerator $$1$$ and smaller denominator. From all this we get that the first fraction $$p_1$$ to interrup the sequence has to have a numerator greater than $$1$$. This means that there is no fraction with numerator $$1$$ that would get rounded to $$p_1$$, so it's too far away from any unit fraction.

We can imagine regions in which all numbers get rounded to the same number as a grid with each cell spanning $$\frac 1 s$$ units.

This is the grid for $$s=59$$. Every point marks one unit fraction and will get rounded to the center of the interval it is in. We notice that between $$0.14$$ and $$0.16$$ there is an empty interval. This corresponds to the fact that no unit fraction will get rounded to approximately $$0.15$$ – $$\frac 1 6$$ is too big and $$\frac 1 7$$ too small. So actually, we just have to find the first interval without any unit fractions and then we can calculate $$i(0, s)$$.

Finding the empty interval

To find an empty interval, we have to compare the size of the intervals (which is constant and equals $$\frac 1 s$$) with the size of the gaps between unit fractions.

The difference between some unit fraction $$\frac 1 n$$ and the next one, $$\frac 1 {n+1}$$ is given by

$$\frac 1 n - \frac 1 {n+1} = \frac 1 {n(n+1)} = \frac 1 {n^2+n}$$

If this gap is greater than $$\frac 1 s$$, this doesn't imply that the interval is empty. $$\frac 1 n$$ could still be in the interval, $$\frac 1 {n+1}$$ in the previous and $$\frac 1 {n-1}$$ in the next.

What is sufficient, is that the gap is greater than $$\frac 2 s$$. A unit fraction could still be in this interval, but either the previous or the next have to leave an empty interval. To find an $$n$$ that satisfies this, we set up the inequality

\begin{align} \frac 1 {n^2+n} &\geq \frac 2 s \\ n^2+n &\leq \frac s 2 \\ n^2+n-\frac 1 2 s &\leq 0 \\ n &\leq \frac {-1 \pm \sqrt{1+2s}}{2} \\ n &\leq \frac {\sqrt{2s+1}-1}{2} \end{align}

At first this looks like an upper bound, but actually it is our lower bound for $$n$$ or upper bound for $$\frac 1 n$$. This is because the inequality gives us a sufficient condition, so there could be smaller poasible values for $$n$$. We have to remember that and use it later. Now, we also have to find a upper bound for $$n$$.

If the gap between two unit fractions is less than $$\frac 1 s$$, the spacing of our intervals, then there has to be a unit fraction in every interval. So we can set up a second inequality

\begin{align} \frac 1 {n^2+n} &< \frac 1 s \\ n^2+n &> s \\ n^2+n-s &> 0 \\ n &> \frac {-1 \pm \sqrt{1+4s}}{2} \\ n &> \frac {\sqrt{4s+1}-1}{2} \\ \end{align}

This is also actually an upper bound, because it is a sufficient condition that there is nos empty interval for all $$n$$ values. Ther could be a gap for $$n$$ greater than that.

If we now combine the two, rembering to swap the inequality signs, we find the following conditions for $$n$$

$$\frac {\sqrt{4s+1}-1}2 \geq n > \frac{\sqrt{2s+1}-1}2$$

Somewhere in this region there will be some $$n$$ that works. When $$s$$ gets very large, we can say

$$\frac {\sqrt{4s+1}-1}2 \sim \sqrt{s}$$

and

$$\frac {\sqrt{2s+1}-1}2 \sim \frac 1 {\sqrt{2}} \sqrt{s}$$

But, for the influence we care about $$\frac 1 n$$ and we have to multiply by $$s$$ to get the number of influenced $$p$$s, not the value. So we see, when $$s$$ is large,

$$\sqrt{s} < i(0, s) < \sqrt{2}\sqrt{s}$$

Variables and functions I used

Variables

• $$a, b, c, d \in \mathbb{N}$$: temporary variables
• $$k$$: some index
• $$0 \leq p \leq 1, p \in \mathbb{R}$$: the value we want to express
• $$s \in \mathbb{N}$$: the precision of our rounding
• $$q \in \mathbb{N}$$: the smallest possible denominator
• $$i, j$$: the influence

Functions

Note by Henry U
5 months ago

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Henry, you switched the bracket signs for the links and AWESOME NOTE!!!!

- 5 months ago

Thank you!! It's my first note of this kind and I'm glad you like it.

- 5 months ago

Actually, if you progress at this rate, YOU ARE A LEGEND!

- 5 months ago

Thank you SOO much!!

- 5 months ago

You put in so much effort and you dare say that this is your first note! Imagine you are a seasoned note writer! LEGEND!

- 5 months ago

Does anyone know how to define a function $$i(p, s)$$ in real code, maybe in Python?

 if (p=0) then (i(1/s, s)) else if (1/floor(1/p)>p+1/(2*s) and 1/ceiling(1/p)<=x-1/(2*s)) then p else i(p+1/s, s)

- 5 months ago